CHAPTER THERMODYNAMICS 6 LEARNING OBJECTIVES (i) Explain the terms : system and surroundings. Discriminate between close, open and isolated systems. (ii) Explain internal energy, work and heat. Heat absorbed or evolved is q = CT. Work can be measured by w = –pexV, in case of expansion of gases. Under reversible process, we can put pex = p for infinitesimal changes in the volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT. (iii) Differentiate between extensive and intensive properties. (iv) Define spontaneous and nonspontaneous processes. (v) State first law of thermodynamics and express it mathematically. (vi) Calculate energy changes as work and heat contributions in chemical systems. (vii) Explain enthalpy. Calculate enthalpy changes for various types of reactions. Enthalpy change, H = E + ngRT, where E is change in internal energy can be found directly from the heat changes at constant pressure, H = qp. There are varieties of enthalpy changes. Changes of phase such as melting, vaporization and sublimation usually occur at constant temperature and can be characterized by enthalpy changes which are always positive. (viii) State and apply Hess’s law of constant heat summation. Enthalpy of formation, combustion and other enthalpy changes can be calculated using Hess's law. Enthalpy change for chemical reactions can be determined by
rH (a i f H products) (b i f H reactants) f i
and in gaseous state by rH bond enthalpies of the reactants – bond enthalpies of the products. (ix) Explain entropy as a thermodynamic state function and apply it for spontaneity. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive. Entropy changes can be measured by the equation q q S rev for a reversible process. rev is independent of path. T T (x) Explain Gibbs energy change (G); Establish relationship between G and spontaneity, G and equilibrium constant. Gibbs energy, G, which is related to entropy and enthalpy changes of the system by the equation: rG = rH – T rS For a spontaneous change, Gsys < 0 and at equilibrium, Gsys = 0. Standard Gibbs energy change is related to equilibrium constant by G RT ln K .
INTRODUCTION Thermodynamics (means literally flow of heat) is a physical science dealing with the quantitative relation between heat and mechanical energy. Thus, in broad sense, it deals with the relationship of heat to all other forms of energy such as electrical energy, mechanical energy, chemical energy, light, kinetic energy, etc. The entire formulation of thermodynamics is based on a few (three) fundamental laws which have been established on the basis of human experience of the experiment behaviour of macroscopic aggregates of matter collected over a long period of time. Thermodynamics helps in – (a) Determining feasibility of a particular process i.e., whether or not a particular process will occur under a given set of conditions. (b) Determining the extent to which a reaction would proceed before attainment of equilibrium. (c) Most important laws of physical chemistry such as Raoults’s law, vant’ Hoff law, distribution law, phase rule, law of equilibrium, laws of thermochemistry and expression for elevation in boiling point and depression in freezing point are in accordance with laws of thermodynamics. Thermochemistry is a branch of thermodynamics which deals with the relationships between chemical reactions and the corresponding energy changes. It is based on first law of thermodynamics. There are two laws of thermochemistry : (i) Lavoisier and Laplace law (ii) Hess’s law
Gyaan Sankalp 1 Thermodynamics SOME FUNDAMENTAL DEFINITIONS : 1. System : Thermodynamics system is defined as any portion of matter, under consideration, which is separated from the surroundings by real or imaginary boundaries. Thus, a system might be as simple as a gas contained in a flask or as complicated as a rocket shooting towards the moon. A thermodynamics system may be homogeneous or heterogeneous, but has to be macroscopic. 2. Surrounding : Surrounding is the rest of the universe around the system. A system and its surroundings are always separated by real (fixed or movable) or imaginary boundaries, through which matter and energy may be exchanged between the two. Ordinarily, surroundings means a water or air bath. 3. Boundary : The imaginary line which separates the system from the surrounding is called boundary. 4. Types of system : (i) Open System : It is the system which is capable of exchanging both matter and energy (as heat) with the surroundings. For example, plants and living beings are all open systems, since they are capable of exchanging both matter and energy with their surroundings. Similarly, an open reaction vessel (like flask, beaker, test-tube) is also an example of open system. (ii)Closed System : It is the system which is capable of exchanging only energy (as heat or work) with the surrounding, but exchange of matter is not possible. For example, a sealed flask containing a gas (or matter) is a closed system. (iii) Isolated System : It is the system which can exchange neither matter nor energy with the surroundings. For such a system, the matter and energy remain constant. We can say that in an isolated system, all matter is "trapped", i.e., no mass can escape, and no energy can leak in or out. 5. State of a System and state variable : (i) The existence of a system under a given set of conditions is called a state of systems. (ii)The properties which change with change in the state of system are called as state variables e.g., pressure, volume and temperature etc. The first and last state of a system are called initial state and final state respectively. 6. State function and Path Function : A physical quantity is said to be state function if its value depends only upon the state of the system and does not depend upon the path by which this state has been attained. For example, a person standing on the roof of a five storeyed building has a fixed potential energy, irrespective of the fact whether he reached there by stairs or by lift. Thus the potential energy of the person is a state function. On the other hand, the work done by the legs of the person to reach the same height, is not same in the two cases i.e., whether he went by lift or by stairs. Hence work is a ‘path function’. 7. Extensity and Intensive properties : An extensive property of a system is that which depends upon the amount of the substance or substances present in the system. e.g., mass, volume, energy etc. An intensive property of a system is that which is independent of the amount of the substance present in the system e.g., temperature, pressure, density, velocity etc. 8. Thermodynamic process : (a) Isothermal System : Isothermal system is a one in which reaction is carried out at constant temperature, i.e., T = 0. When a system undergoes an isothermal process, the system is usually kept in contact with a constant temperature bath (called thermo- stat) and the system maintains its temperature constant by exchange of heat with the thermostat. (b) Adiabatic System : Adiabatic system is a one in which no heat can leave or enter the system (i.e., thermally insulated). Thus, for carrying out adiabatic process, the system is carefully insulated from the surroundings. It may be pointed out that in an adiabatic process, temperature of the system may increase or decrease. (c) Isobaric Process : Isobaric Process is one in which reaction is carried out at constant pressure, i.e., P = 0. For example, a reaction taking place in an open vessel is always at atmospheric pressure and hence, such a reaction is isobaric. (d) Isochoric Process : Isochoric process is one in which the volume of the system is kept constant, i.e.,V = 0. (e) Cyclic Process : Cyclic Process is the overall process when the system in a given state goes through a number of different processes and finally returns to its initial state. (f) Reversible Process : A process which can be performed in the reverse direction, the whole series of changes constituting the process being exactly reversed, i.e., the direction of a reversible process can be reversed by an infinitesimal change in the state of the system. (g) Irreversible Process : Irreversible process is one which goes from the initial to final state in a single step and cannot be carried in reverse order. 9. Internal energy (E or U) : The total energy stored in a substance by virtue of tis chemical nature and state is called its internal energy, i.e., it is the sum of its translation, vibrational, rotational, chemical bond energy, electronic energy, nuclear energy of constituent atoms and potential energy due to interaction with neighbouring molecules. It is also called intrinsic energy. E = Et + Er + Ev + Ee + En + EPE Internal energy is a state property and its absolute value can’t be determined. However, change in internal energy (difference between the internal energies of the products and that of reactants) can be determined experimentally using a bomb calorimeter. Internal energy of a system depends upon : (a) the quantity of substance (b) its chemical nature and (c) temperature, pressure and volume.
2 Gyaan Sankalp Thermodynamics (i) For a given system, E is directly proportional to its absolute temperature. (E T) (ii) At constant volume, the quantity of heat supplied to a system (isochoric process) is equal to the increase in its internal energy, i.e., QV = E (iii) In the adiabatic expansion of a gas, it gets cooled because of decrease in internal energy. (iv) In cyclic process the change in internal energy is zero (E = 0) since E is a state function. (v) For exothermic reactions, sign of E is negative (ER > EP). (vi) For endothermic reactions, sign of E is positive (EP > ER). 10. Work : Work is expressed as the product of two factors, i.e., W = Intensity factor × capacity factor where, intensity factor is a measure of force responsible for work and capacity factor is a measure extent to which the work is done. Thus, (a) Mechanical work = Force × Displacement = F × d (b) Electrical work = Potential difference × Charge flown = V × Q = EnF (c) Expansion work = Pressure × change in volume = P × V (d) Gravitation work = Gravitational force × Height = mg × h Unit of Work : (i) Joule (J) and kilojoule (kJ) are units in SI units ; (ii) Calorie (Cal) or kilo calories (kcal) is used in SI system. (iii) while erg is the unit in CGS system. 1 J = 107 ergs = 0.2390 cal 1 cal = 4.184 J, and 1 kcal = 4.184 kJ. (i) Pressure volume work (Irreversible) : It is a kind of mechanical work. The expression for such a work may be derived as follows Consider a gas enclosed in a cylinder fitter with a weightless and frictionless piston. Suppose area of cross reaction of a cylinder = a sq. cm. Pressure on piston = P which is less than the internal pressure such that the gas expands. Let d be the distance covered by the piston when the gas expands. work done is given by dW = – Fd = – Force × distance Force = Pressure Force = Pressure × area Area dW = – P.a.d= – PdV Let the volume limits be V1 and V2. Hence integrating the equation dW = –PdV dW PdV W = – PV W = – P (V2 – V1) When expansion takes place V2 > V1 W = – ve Hence work is done by the system. When compression of gas takes place then V2 < V1, W = +ve. Hence work is done on the system. In physics convention for work is opposite. (ii) Work during isothermal Reversible Expansion of an Ideal Gas We know that, dW = – PdV
V2 dW Pdv V1 nRT For an ideal gas PV = nRT P = V
V2 V2 nRT dV v V2 W = dV W = – nRT = – nRT n V 1 = – nRT ln V V v2 V V2 V1 1
V2 W = – 2.303 nRT log V1
1 P P2 Also as V W = – 2.303 nRT log 1 W = 2.303 nRT log P P P2 1
Gyaan Sankalp 3 Thermodynamics (a) Work is not a state function because amount of work performed depends upon the path followed. (b) Positive value of work signifies that the work has been done on the system by the surroundings and it leads to an increase in the internal energy of the system. On the other hand, negative value of work indicates that work has been done by the system and it leads to decrease in the internal energy of the system. 11 Heat : It is a form of energy. The energy exchanged between a system and the surroundings when their temperatures are different is commonly known as heat It flows from higher temperature to lower temperature. When a system and surrounding are at different temperature then the heat (energy) transfer continues until the system and the surroundings come to the same tempera- ture. When the system and surrounding are at the same temperature, then these are said to be in a state of thermal equilibrium. It should be remembered that the heat is not a substance. It should also be remembered that a system does not have a definite amount of heat. It has a definite amount of energy. Unit of Heat : (i) Joule (J) and kilojoule (kJ) are units in SI units ; (ii) Calorie (Cal) or kilo calories (kcal) is used in SI system. (iii) while erg is the unit in CGS system. 1 J = 107 ergs = 0.2390 cal 1 cal = 4.184 J, and 1 kcal = 4.184 kJ. Sign conventions of heat : (i) Energy entering the system (or absorbed by the system) is positive. (ii) Any energy leaving the system (or given off by the system) is negative. 12. Enthalpy : It is defined as sum of internal energy and product of pressure volume work. It is donated by the letter H. H = E + PV Enthalpy of a system is also called heat content of system, because it is the net energy available in a system which can be converted into heat.
SPONTANEOUS AND NON-SPONTANEOUS PROCESSES : 1. Spontaneous Process : In our daily life, we come across a large number of physical and chemical processes which occur in a widely varying conditions. For example, (i) Some processes proceed on their own, e.g. Water always flows down a hill, heat flows from a body at higher temperature to a body at lower temperature. (ii) Some processes require proper initiation but once properly initiated they continue on their own e.g. Kerosene oil once ignited continues to burn till whole of it has been consumed or exhausted. (iii) Some processes proceed only so long as the external energy is available. e.g. Electrolysis of water continues so long as current is passed and stops as soon as current is cut off. “The process which can take place by itself or after proper initiation, under the given set of conditions, is called a spontaneous process.” the term “Spontaneous” simply means that given process is feasible or possible. Therefore, Spontaneous processes are also called as feasible or probable processes. Spontaneous process may or may not be instantaneous. But all instantaneous process are spontaneous. It may be pointed out that the term “Spontaneous” should not mean that the process occurs “instantaneously”. It simply implies that process has an urge to proceed or it is pratically possible. e.g. (i) Processes which occur on their own without proper initiation. (a) HCl(g) + NH3(g) NH4Cl (s) (b) H2O () H2O(g) Water keeps on evaporating from ponds and rivers etc. (c) Sugar dissolves in water and forms a solution. (d) 2NO (g) + O2 (g) 2NO2 (g) (ii) Processes which require initiation (a) In domestic oven, once coal (carbon) is ignited it keeps on burning C (s) + O2 (g) CO2 (g) (b) CH4(g) + 2O2(g) CO2(g) + 2H2O() From the above discussion we conclude that spontaneous process is the one which has natural tendency to occur. 2. Non-Spontaneous Process : A process which has no tendency to occur or which is made to occur only if energy from outside is continuously supplied e.g. (i) Decomposition of water into H2 and O2 is non-spontaneous. However, water can be decomposed by passing an electric current through it, in a process called electrolysis. Electricity H2O () 2H2 (g) + O2 (g) The process will continue as long as electric current is supplied, and as soon as the supply of electricity is cut off the decomposition stops. (ii) Water cannot be made to flow up the hill, without the help of a machine.
4 Gyaan Sankalp Thermodynamics (iii) Gold ornaments do not get tarnished in air even after a number of years. This shows that gold does not combine with oxygen in the air. Driving Force for a Spontaneous Process : the natural tendency of various processes to occur spontaneously indicated that there must be some driving force behind them.
STATE OF A SYSTEM : A thermodynamic property which depends only on the state of the system and is independent of the path chosen to bring about the change. (i) Intensive Property : A property which is independent of amount of substance present in the system, e.g., temperature, pressure, viscosity, surface tension, refractive index, density, specific heat, potential, etc. (ii) Extensive Property : A property which depends upon the quantity of matter present in the system, e.g., mass, volume, internal energy, entropy, external energy, free energy, etc. Extensive property Intensive property Volume Free energy Molar volume Freezing point Number of moles Enthalpy Pressure Density Mass Entropy Temperature Vapour pressure Energy Boiling point Refractive index.
THERMODYNAMIC EQUILIBRIUM : A system in which state function (P, V, or T) have constant values throughout the system. The criteria for equilibrium are – (a) The temperature of the system must be uniform and must be the same as the temperature of surroundings (thermal equilib- rium). (b) No mechanical work is done by one part of the system on any other part of the system. Thus, the mechanical properties must be uniform throughout the system (mechanical equilibrium). (c) The chemical composition of the system must be uniform with no net chemical change (chemical equilibrium).
ZEROTH LAW : It states that “Two system in thermal equilibrium separately with the third system are said to be in thermal equilibrium with each other” i.e., If system A and B separately are in thermal equilibrium with another system, then system A and B are also in thermal equilibrium.
FIRST LAW OF THERMODYNAMICS : The first law of thermodynamics states that “Energy can neither be created nor be destroyed although it can be converted from one form to another”. Let a system be at state I with internal energy E1, Let it be change to State II with internal energy E2 This can be achieve in the ways : (i) by heat transfer (ii) By doing work (either on system or by system) Let the heat change taking place during the change of state of system from state I to state II be ‘q’ and work done be W. E2 = E1 + q + W E2 – E1 = q + W E = q + W or E = q – PV (a) Adiabatic process (i) Adiabatic compression E = W E = + PV (ii) Adiabatic expansion : E = – W E = – PV Hence during adiabatic compression of an ideal gas internal energy of system increases and during adiabatic expansion internal energy of system decreases. (b) Isochoric process E = qv (i) Isochoric absorption of heat : E = + q internal energy of system increase (ii) Isochoric liberation of heat : E = – q internal energy of system decreases (c) Heat absorbed by system and work done by system E = +q – W (d) Heat liberation by system and work done on the system E = – q + W (e) Isobaric process (expansion) E = +qP – PV
Gyaan Sankalp 5 Thermodynamics
(i) when no gases are involved in reaction : E = +qP (ii)When gases are involved but initial and final volumes are not given E = q – ng RT where ng = nP(g) – nR(g) (f) Isothermal process : (i) E for an ideal gas E depends on temperatures. Since temperature is constant dE = 0 E = 0 (ii)According to law dE = dq + dW; since dE = 0 dq = – dW q = – W Heat absorbed is equal to work done by the system during isothermal expansion of ideal gas. Limitations of first law of thermodynamics : Though the first law of thermodynamics gives us the exact equivalence of heat and work, whenever there is a change of heat into work or vice versa, but it suffers from the following two limitations : (i) No indication is available as regards the direction in which the change will proceed. (ii) It gives no idea about the extent to which the change takes place. These limitations can be understood from the following examples : (a) This law can easily explain the heating of a bullet when it strikes a block due to the conversion of kinetic energy into heat, but it fails to explain as to why heat in the block cannot be changed into kinetic energy of bullet and make it fly back from inside of the block. (b) When a vessel of water is placed over fire, heat flows into the vessel. What prevents the heat from flowing from water into the fire, and thereby cooling the water and ultimately converting into ice. Thus direction of (flow) change is not known from first law. (c) It is practically found that whole of heat can never be converted into work. The first law has no answer to this observation. Thus, first law fails to tell extent to which the interchange of heat into work and vice versa is possible. Example 1 : One mole of an ideal gas is put through a series of changes V as shown in the graph in which A, B, C mark the three stages A of the system. At each stage the variables are shown in the graph. 24.0 (L) B (a) Calculate the pressure at three stages of the system. (b) Name the processes during the following changes: (i) A to B (ii) B to C 12.0 (L) (iii) C to A, and (iv) overall change. C Sol. (a) At stage A; V = 24.0 L; T = 300 K; n = 1 ; R = 0.0821 lit-atm K–1 mol–1 Substituting these values in the ideal gas equation, 1 0.0821 300 T P = = 1.026 atm 300 K 300 K 24.0 At stage B : Volume remains the same but temperature change from 300 K to 600 K . Thus, according to pressure law, the pressure will be double at B with respect to A. Pressure at B = 2 × 1.026 = 2.052 atm At stage C : Temperature is 300 K and volume is half that of stage A. Thus, according to Boyle’s law, the pressure at C will be double with respect to A. Pressure at C = 2 × 1.026 = 2.052 atm (b) (i) During the change from A to B, volume remains constant, the process is isochoric. (ii) During the change from B to C the pressure remains constant, the process is isobaric. (iii) During the change from C to A, the temperature remains constant, the process is isothermal. (iv) Overall, the process is cyclic as it returns to initial state. Example 2 : Calculate the work done when 1.0 mole of water at 373K vaporizes against an atmospheric pressure of 1.0 atmosphere. Assume ideal gas behaviour. Sol. The volume occupied by water is very small and thus the volume change is equal to the volume occupied by one gram mole of water vapour. nRT 1.0 0.821 373 V = = 31.0 litre P 1.0
W = –Pext × V = – (1.0) × (31.0) litre-atm = – (31.0) × 101.3 J = 3140.3 J 6 Gyaan Sankalp Thermodynamics Example 3 : Calculate the work done when 50 g of iron is dissolved in HCl at 25ºC in : (i) a closed vessel and (ii) an open beaker when the atmospheric pressure is 1 atm. Sol. (i) When the reaction is carried in a closed vessel, the change in volume is zero. Hence, the work done by the system will be zero. (ii) When iron dissolves in HCl, hydrogen is produces. Fe + 2HCl FeCl2 + H2 56 g 1 mole 50 g (1/56) × 50 mole nRT 50 0.0821 298 Volume of hydrogen produced at 25ºC = = 21.84 L P 56 1 This is equal to volume change when the reaction is carried in open beaker. Work done by the system = – PV = – 1.0 × 21.84 = – 21.84 litre–atm = – 2212.39 J Example 4 : 5 mole of an ideal gas expand isothermally and reversibly from a pressure of 10 atm to 2 atm at 300 K. What is the largest mass which can be lifted through a height of 1 metre in this expansion ? P P 1 1 10 3 Sol. Work done by the system = – nRT loge = –2.303 nRT log10 = – 2.303 × 5 × 8.314 × 300 log = – 20.075 ×10 J P2 P2 2 Let M be the mass which can be lifted through a height of 1m. Work done in lifting the mass = M g h = M × 9.8 × 1J So M × 9.8 = 20.075 × 103 = 2048.469 kg
TRY IT YOURSELF Q.1 Which is the intensive property ? (A) temperature (B) viscosity (C) density (D) all Q.2 An example of extensive property is : (A) temperature (B) internal energy (C) viscosity (D) surface tension Q.3 A gas expands isothermally and reversibly. The work done by the gas is : (A) zero (B) minimum (C) maximum (D) None Q.4 The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of 5 dm3 isothermally until the volume becomes 25 dm3 is : (A) –2.01 × 103 J (B) +2.81 × 103 J (C) –2.01 × 10–3 J (D) +2.01 ×10–6 J Q.5 If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is : (A) 20 cal (B) 50 cal (C) 40 cal (D) 30 cal Q.6 A thermodynamic process is shown in the following figure. The pressure and volumes corresponding to some point in the figure are : 4 4 –3 3 –3 3 PA = 3 × 10 Pa , PB = 8 × 10 Pa , VA = 2 × 10 m , VD = 5 × 10 m In the process AB, 600 J of heat is added to the system and in BC, 200 J of heat is added to the system. The change in internal energy of the system in the process AC would be : (A) 560 J (B) 800 J (C) 600 J (D) 640 J Q.7 In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas : (A) the temperature will decrease (B) the volume will increase (C) the pressure will remain constant (D) the temperature will increase 3 3 Q.8 20 gm of N2 at 300 K is compressed reversibly and adiabatically from 20 dm to 10 dm . Change in internal energy for the process is : (A) 284.8 J (B) 142.46 J (C) 1424.69 J (D) 3462.89 J
ANSWERS (1) (D) (2) (B) (3) (C) (4) (A) (5) (A) (6) (A) (7) (A) (8) (C)
Gyaan Sankalp 7 Thermodynamics HEAT CAPACITY OF A SYSTEM : It is the amount of heat required to raise the temperature of a system through 1ºC If ‘q’ is the amount of heat supplied to a system and as a result let the temperature rise from T1 to T2 ºC. q q Then , heat capacity = C = TTT2 1 When heat capacity varies with temperature then the value of C has to be considered over a narrow range of temperature. q Then C dT (a) Specific Heat and Molar Heat Capacity at Constant Volume Specific Heat : It is the amount of heat required to raise the temperature of 1 gm of a gas through 1º at constant volume. Molar heat capacity : It is the amount of heat required to raise the temperature of one mole of a gas through 1º at constant volume dq = dE + PdV dq dE PdV Molar heat Capacity : C dT dT At constant volume dV = 0 E dE C ; C v dT T V C v = S = Specific heat at constant volume. Molecular mass V (b) Heat Capacity at Constant Pressure : It is defined as the amount of heat required to raise the temperature of one mole of gas through 1o keeping pressure constant EV CPP ...... (a) TT PP We know that H = E + PV HEV Differentiating w.r.t. temperature P ...... (b) TTT PPP
H From (a) and (b) CP T P
CP = SP = Specific heat at constant pressure. CP is always greater than CV. M0 When gas is heated at constant volume, the pressure of gas has to increase. As the gas is not allowed to expand, therefore in case of CV heat is required for raising the temperature of one mole of a gas through 1o. When gas is heated at constant P. It expands, the gas has done some work against external pressure. More heat is therefore supplied to raise its temperature through 1o. Thus CP is heat required for the purpose of (i) Increasing temperatures of one mole of gas through 1o. (ii) For increasing the volume of the gas against external pressure. CP > CV Relationship between CP and CV : According to I law of thermodynamics. dqp = dqV + PdV CPdT = CVdT + PdV For one mole of gas PV = RT PdV + VdP = RdT Since the gas is being heated at constant pressure dP = 0 PdV = RdT CPdT = CVdT + RdT Dividing both sides by dT CP = CV + R CP – CV = R CP = CV + R 8 Gyaan Sankalp Thermodynamics
dE = nCVdT for n moles and dE = CVdT for one moles E = E2 – E1 = nCV (T2-T1) = E = n CVT For adiabatic process : q = 0 E = – W = CV (T2 – T1) [Because expansion of gas takes place] W = – CV (T2 – T1) W = – nCV (T2–T1) = – nCV T
ENTHALPY : Internally energy changes are usually measured at constant volume. But in actual practice, most processes are carried out at constant pressure rather than constant volume. Hence volume changes which occurs cause changes in internal energy. To account for these changes, a new thermodynamic property is introduced called as Enthalpy. It is defined as sum of internal energy and product of pressure volume work. It is donated by the letter H. H = E + PV Enthalpy of a system is also called heat content of system, because it is the net energy available in a system which can be converted into heat. Enthalpy of Reaction : The difference between the enthalpies of the products and the reactants when molar quantities of the reactants, as indicated by the balanced chemical equation, react completely. Thus : H = Enthalpy of product – Enthalpy of reactants. = H(product) – H(reactants) Standard Enthalpy of Reaction : The enthalpy of reaction at 298 K and 1 atm. pressure when the reactants and products are in their standard states is called standard enthalpy of reaction. Characteristics : (a) It is a state functions (b) It is an extensive property (c) Change in its value can be determined by relationships H and E Let a system at state-I be transformed to state-II at constant pressure condition. State - I State-II Enthalpy H1 H2 Internal energy E1 E2 Pressure P P Volume V1 V2 H1 = E1 + PV1 H2 = E2 + PV2 H2 – H1 = H = (E2 + PV2) – (E1 + PV1) = (E2 – E1) + P (V2 – V1) H = E + PV Also for an ideal gas PV = nRT PV1 = n1 RT and PV2 = n2RT P(V2 – V1) = ng RT Then H = E + ng RT qp = qv + ng RT When ng = 0 H = E When ng > 1 H > E When ng < 1 H < E Also E = qP – PV qP = E + PV and H = qP (i.e., Enthalpy change = heat exchange at constant pressure condition) also E = qV (i.e., Internal energy change = heat exchange at constant volume condition) H = E + PV qP – qV = PV = ng RT Heat of Combustion (HC) : Heat of combustion is the enthalpy of reaction when 1 mol the substance is completely burnt in the excess of air or oxygen. Heat of combustion is always negative, because heat is always evolved. Importance of Heat of Combustion : It can be used to calculate the calorific value of a fuel (i.e., amount of heat produced when 1 g of the substance is completely burnt). It is expressed in Jg–1 or kJ g—1 or cal g–1. It may be employed to calculate the heat of reaction when heat of combustion of every reactant and product is known. Thus : Hrxn = HComb. (reactants) – HComb. (products)
Gyaan Sankalp 9 Thermodynamics Enthalpy or heat of neutralization : Enthalpy or heat of neutralization of an acid (or base) is the change in heat content (H) when one gram-equivalent of the acid (or base) in dilute aqueous solution is neutralized by one gram-equivalent of a dilute aqueous solution of a base (or acid). For examples : HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O ; H = – 57 kJ HCl(aq) + KOH (aq) KCl (aq) + H2O ; H = – 57 kJ It is obvious that heat of neutralization (H) of a strong acid (HNO3 or HCl, etc.) by a strong base (NaOH or KOH, etc.) and vice versa is – 57 kJ, no matter what acid or base is employed. This is because the heat of neutralization is in fact the heat of formation of 1 mol of water from 1 mol H+ ions (of acid) and 1 mol OH– ions (of base) ; while the other ions of the strong acid and base remain unaltered. Thus :
A H + B OH AB + H2O ; H = – 57 kJ Strong acid Strong base Salt + – or H + OH H2O ; H = – 57 kJ Na Cl e.g., H Cl + Na OH + H2O ; H = – 57 kJ Strong acid Strong acid Salt + – or H + OH H2O ; H = – 57 kJ (of acid) (of base) Heat of neutralization of weak acid = Heat of neutralization of strong acid – Heat of ionization of weak acid. In the neutralization of NH4OH with HCl, 5.6 kJ of heat is used up for the dissociation of the weak base i.e. NH4OH. Hence the heat of neutralization in this case is only 57.1 – 5.6 = 51.5 kJ. Enthalpy or heat of formation (Hf) : Enthalpy of heat of formation of a compound is the change in heat content or enthalpy (H) when one mol of the compound is formed from its elements in their commonly occurring (or standard) states. For example, the reaction : C(graphite) + 2H2(g) CH4(g) ; Hf = – 74.8 kJ indicates that when one mol of methane gas is formed from its elements, carbon and hydrogen, 74.8 kJ of heat energy is given out. Thus, enthalpy (or heat) of formation of methane is – 74.8 kJ mol–1. Heat of formation and stability : The heat of formation is an important property of a compound and is related to its stability. The stability of a compound is inversely proportional to its heat content. A compound with heat of formation (Hf) positive means that the heat content of the compound is more than heat contents of its constituent elements. Hence such a compound is unstable. A compound with heat of formation (Hf ) negative means that the heat content of the compound is less than that of its constituent elements. Consequently such a compound is stable. It is quite clear that the relative stability of various com- pounds is inversely proportional to their heats of formation. –1 For example, the heat of formation (Hf) of HI, HBr, HCl and HF are + 26.9, – 36.1 ; – 92.4 and – 161.7 kJ mol respectively. Consequently of these halides, HI is least stable ; while HF is maximum stable. In other words, the stability of these halides is in the order. HF > HCl > HBr > HI Enthalpy (or heat) of Fusion : Enthalpy of fusion of a substance is "the heat change in converting 1 mole of it from the solid to the liquid phase at its melting point." Heat of fusion is always positive, because heat is needed to overcome the intermolecular –1 forces between constituent particles of a solid. For example, heat of fusion of water, H2O(s) H2O( ), is + 0.6 kJ mol . Enthalpy (or heat) of vaporization : Enthalpy of vaporization is the heat change in converting 1 mol of the substance from liquid to the gaseous state at its boiling point. –1 For example, heat of vaporization of water, H2O( ) H2O(g), is 42.0 kJ mol . Enthalpy or heat of sublimation : Enthalpy or heat of sublimation of a solid is heat change in converting 1 mol of a solid directly into its vapour at a given temperature, below its melting point. –1 For example, heat of sublimation of iodine, I2(s) I2(g), is 62.4 kJ mol Enthalpy of transition (or heat of transition) : Enthalpy of transition is the enthalpy change (H) when 1 mol of an element changes from one allotropic form to another. –1 For Example : C(diamond) C(amorphous) ; H = + 3.3 kcal mol
Enthalpy (or heat) of solution : Enthalpy of solution is the enthalpy change (Hsol) when 1 mol of the substance is dissolved in a solvent so that further dilution does not give any change in enthalpy. –1 For example : KCl (s) + nH2O( ) KCl(aq) ; Hsol = 18.4 kJ mol In terms of lattice energy (HU) and heat of hydration (Hhyd) of ions, the heat of solution can be expressed as : Heat of solution (Hsol) = HU –Hhyd. Enthalpy of Hydration : It is the amount of heat evolved (i.e., change in enthalpy) when 1 mole of anhydrous or a partially hydrated salt combines with required number of moles of water to form a hydrate. e.g.,
10 Gyaan Sankalp Thermodynamics
CuSO4 + 5 H2O () CuSO4 . 5 H2O ; H = 18.69 Kcal CaCl2 () + 6 H2O () CaCl2 . 6 H2O ; H = 18.8 Kcal Enthalpy of ionisation : It is defined as the amount of heat absorbed when 1 mole of an electrolyte completely dissociates into + ions. CH3COOH CH3COO + H ; H = 3 Kcal + HCN H + CN ; H = 10.8 Kcal
EXOTHERMIC AND ENDOTHERMIC REACTIONS : Exothermic Reaction : These are those reactions which are accompanied by the evolution of heat. The quantity of heat produced is shown along with the products with a 'plus' sign. Some typical exothermic reactions are (i) Burning of fuels : C(s) + O2(g) CO2(g) + Heat + Light from coal in air 2C4H10 + 13O2(g) 8CO2(g) + 10H2O(g) + Heat + Light Butane : a constituent of LPG. (ii) Slackening of lime : When a small quantity of water is added to quicklime, a large amount of heat is liberated. CaO(s) + H2O() Ca (OH)2 + Large amount of heat quicklime slaked lime (iii) Reaction between an active metal and dilute acid : Zn(s) + 2HCl(dil). ZnCl2(aq) + H2(g) + Heat Exothermic reactions in terms of H H = enthalpy change. H = Heat content of products – heat content of reactants. H = HP – HR. For exothermic reaction, HR > HP. H is negative for exothermic reactions. Thus the exothermic reactions may be written in terms of H as C(s) + O (g) CO (g), H = – 393.5 kJ 2 2 REACTANTS 1
H2(g) + O2(g) H2O(), H = – 285.8 kJ ) H
2 (
Y
N2(g) + 3H2(g) 2NH3(g), H = – 93.7 kJ P L (NEGATIVE)
CH4(g) + 2O2(g) CO2(g) + 2H2O(), H = – 890.4 kJ A
H T
13 N C H (g) + O (g) 4CO (g) + 5H O(g), H = – 2658 kJ E PRODUCTS 4 10 2 2 2 2 Graphically the exothermic reaction may be represented. PROGRESS OF REACTION
Endothermic Reactions : Those reactions in which heat is absorbed are called endothermic reactions. Some typical endothermic reactions are (i) Dissolution of ammonium chloride in water NH4Cl(s) + H2O() NH4Cl(aq) – Heat (ii) Cooling effect is observed when crystals of sodium thiosulphate (Na2S2O3.5H2O) commonly called (hypo) is dissolved in water Na2S2O3.5H2O + H2O Na2S2O3(aq) – Heat (iii)Reaction between hydrated barium hydroxide and ammonium chloride. Ba (OH)2.8H2O(s) + NH4Cl(s) BaCl2.2H2O(s) + 2NH3(aq) + 8H2O(l) The heat absorbed can be written with a plus sign along with reactants, or with a minus sign if written with the products. For an endothermic reaction, HR < HP i.e. H = HP – HR = + ve H is positive for endothermic reaction. Endothermic reactions in terms of H N (g) + O (g) 2NO(g), H = + 180.7 kJ 2 2 PRODUCTS C (s) + H2O (g) CO2 (g) + H2(g), H = + 130.4 kJ
C (s) + 2S (s) CS2 (), H = + 92.0 kJ )
H (
H2 (g) + I2 (g) 2HI (g), H = + 52.5 kJ Y
2HgO (s) 2Hg () + O (g), H = + 181.6 kJ P
2 L (POSITIVE)
A H
Graphically the endothermic reaction may be represented as follows : T N Sign conventions : H E E REACTANTS Exothermic –ve –ve Endothermic +ve +ve PROGRESS OF REACTION
Gyaan Sankalp 11 Thermodynamics THERMOCHEMICAL EQUATIONS : When a balanced chemical equation not only indicates the quantities of the different reactants and products but also indicates the amount of heat evolved or absorbed, is called as thermochemical equation. For Example : 2H2(g) + O2(g) 2H2O () + 571.6 kJ 2H2(g) + O2(g) 2H2O(), H = – 571.6 kJ Some conventions about thermochemical equations. For exothermic reactions, H is negative whereas for endothermic, H is positive. Unless otherwise mentioned, H values for standard state of substances (i.e. 298 K and 1 atm. pressure). The coefficients of different substances represents the number of moles reacted and formed for the heat change represented in the equation. The physical state (s, l, g etc.) of the different substances must be mentioned as the heat evolved or absorbed depends upon the physical state. If the coefficients of the substances are multiplied or divided by some number, the value of H is multiplied or divided by the same number. If the reaction is reversed, the sign of H changes but the magnitude remains the same.
FACTORS WHICH AFFECT THE HEAT OF REACTION : (a) Physical State of Reactants and Products : The heat of reaction varies for a given reaction with the change in physical nature of reactants or products e.g.,
C O CO ; H= 94.3 kcal For reactant (diamond) 2 2 having different C O CO ; H= 97.7 kcal physical state (Amorphous) 2 2 For products having different physical state ; 1 H O g H O ( ) ; H = – 68.3 Kcal 2(g)2 2 2 1 H O g H O g ; H = – 57.0 K cal 2 g 2 2 2 (b) Reaction carried out a constant pressure or constant volume : From 1st Law of thermodynamics q = E – W at constant volume qV = E and at constant pressure qp = H Two values are related as H = E + PV where, P = Pressure at which the reaction is carried out V = Change in volume during the course of reaction. Alternatively H = E + nRT n = No. of moles of products – No. of moles of reactant (only gas phases) (c) Temperature : Heat of reaction also depends upon the temperature at which reaction is carried out. The variation in H value with temperature are due to variation in heat capacities of system with temperature. Kirchoff’s equation :
HHCTT2 1 P 2 1 and EECTT2 1 V 2 1
where CP (C p of products C P of reactants)
Cv (C V of products C V of reactants) H2, H1 are change in heat enthalpies at temperature T2 & T1 respectively E2, E1 are change in heat internal energy at temperature T2 & T1 respectively (d) Enthalpies of Solution : Enthalpies of reaction differ when in one case dry substances react and in another case when the same substance react in solution.
e.g., HSg2 I 2 g 2HIS ; H 172 Kcal
HSg2 I 2 solution 2HIsolution S; H 21.93 Kcal
CuSO4 aq CuSOHOaq 4 2 ; H 15.8 Kcal
CuSO4 .H 2 O aq CuSO 4 .5H 2 Oaq. ; H 29 Kcal
12 Gyaan Sankalp Thermodynamics BOND ENTHALPIES (BOND ENERGIES) The bond enthalpy of a diatomic molecule (H2, Cl2, O2) is equal to its dissociation energy and is defined as “the enthalpy change involved in breaking the bond between atoms of a gaseous molecule” (Bond breaking is an endothermic process). Average bond enthalpy (energy) is the average value of bond energy obtained from molecules that contain more than one bond of that type. H of molecules Av. BE = f ; BE is an additive property.. no. of bonds For bonded molecules, bond dissociation energy is directly proportional to the bond order. –1 For example, O2 molecule has a bond order of 2 and its bond dissociation energy is 494 kJ mol ; whereas N2 molecule has a bond order of 3 and its bond dissociation energy is 941 kJ mol–1. Lattice energy : Enthalpy change when one mole of gaseous ions condense to form a solid crystal lattice. eg. Na+(g) + Cl (g) NaCl (s) . Resonance energy : (observed heat of formation) (calculated heat of formation) . THERMOCHEMISTRY : Thermochemistry is the branch of physical chemistry which deals with the transfer of heat between a chemical system and its surrounding when a change of phase or a chemical reaction takes place with in the system. It is also termed as chemical energetics. Thermochemical equations : A chemical reaction which tells about the amount of heat evolved or absorbed during the reaction is called a thermochemical equation. A complete thermochemical equation supplies the following informations. (i) It tells about the physical state of the reactants and products. This is done by inserting symbols (s), () and (g) for solid, liquid and gaseous state respectively with the chemical formulae. (ii) It tells about the allotropic from (if any) of the reactant. (iii) The aqueous solution of the substance is indicated by the word aq. (iv) It tells whether a reaction proceeds with the evolution of heat or with the absorption of heat, i.e. heat change involved in the system.
LAW OF THERMOCHEMISTRY : Levoisier and Laplace Law : According to this “enthalpy of decomposition of a compound is numerically equal to enthalpy of formation of that compound with opposite sign. e.g., C (s) + O2 (g) CO2 (g) ; H = 94.3 Kcal CO2 (g) C (g) + O2 ; H = + 94.3 Kcal Hess’s law of constant heat summation : For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is equal to the sum of the enthalpy changes for the individual steps. Thus, Hess’s law enables us to break down a reaction into so many intermediate steps and passing to each step an individual enthalpy change. The sum of the individual changes must, of course, equal the overall enthalpy change provided the initial and final states are the same in each case.
H° = - x kJ mol-1 1 H2 (g) + O 2 (g) H O(l ) + O (g) Route A 22 2 -1 -1 H1 ° = - y kJ mol H2 ° = - z kJ mol
HO()2 2 l Route B For the above example, HHH 1 2 H (g) + O (g) An energy level diagram for the 2 2 above reaction cycle is shown in figure. -1 H1 ° = - 187.6 kJ mol
-1 H° = - 285.9 kJ mol HO()2 2 l
-1 H2 ° = - 98.3 kJ mol
1 H O(l ) + O (g) 22 2
Fig. Energy level diagram to illustrate Hess’s Law
Gyaan Sankalp 13 Thermodynamics Application of Hess’s law : It helps us in calculations of : (i) Heat of formation (Hf) of many substances which cannot be synthesized directly from their elements. (ii) Bond energies (iii) Enthalpy changes of slow reactions and (iv) Enthalpy of transformation, say from one allotropic from to the other.
BORN-HABER CYCLE : This thermochemical cycle was devised by Born and Haber in 1919. It relates the lattice energy of a crystalline substance to other thermochemical data. The Born-Haber cycle is the application of Hess’s law to the enthalpy of formation of an ionic solid at 298 K. Formation of cystalline sodium chloride from sodium metal and chlorine gas can be described by the reaction. 1 Na(s) Cl (g) NaCl(crystal); H H 411kJmol1 (energy evolved) 2 2 r f This overall reaction can be considered to proceed in a stepwise manner as follows : Step 1 : Conversion of solid sodium metal into isolated, gaseous sodium atoms by a process called sublimation. 1 Na(s) Na(g) ; sub H 107.3kJmol (energyabsorbed) Step 2 : Dissocation of gaseous chlorine molecules (Cl2) into gaseous Cl atoms. 1 1 1 Cl(g) Cl(g); H(Cl Cl) ( 243kJmol1 ) 121.5kJmol 1 (energy absorbed) 22 2 d 2 Step 3 : Ionisation of isolated Na atoms into Na+ ions and electrons. 1 Na(g) Na (g) e ; 1 H 495.8kJmol (energyabsorbed) Step 4 : Formation of Cl– ion from isolated Cl atoms by the addition of an electron. 1 Cl(g) e Cl (g) ; eg H 348.6 kJ mol (energy evolved) Step 5 : Formation of crystals of Na+Cl– from isolated Na+(g) and Cl–(g) ions. 1 Na (g) Cl (g) NaCl(crystal); L H 787.3kJmol (energy evolved) Then from the Hess’s law of heat summation. 1 HHHHHH f sub2 (ClCl) 1 eg L The signs of the energy involved in each step follow the rule that energy evolved is negative and energy absorbed is positive. From the Born-Haber cycle the value for any one of the steps can be calculated if data for all other other steps are known. When the Born-Haber cycle is appled to other salts, particularly those having partial covalent bonds, disrepancies arise. They indicate the mixed nature of the bonds in the crystal. For example, silver salts such as AgCl deviate as much as 5% from the calculated amount of lattice energy. This is attributed to the partial covalent bond in the AgCl crystal. STANDARD MOLAR ENTHALPY OF FORMATION : Standard Molar enthalpy of formation (or standard heat of formation), Hfº, of a compound is the standard enthalpy change, when 1 mol of that substance is formed from its constituent elements in their standard states. By convention, the standard enthalpy of formation of the stable form of any element is zero. For example, Hfº for C(graphite) is zero.
CALCULATION OF ENTHALPIES OF REACTIONS (OR HEATS OF REACTIONS) : Hrxn from standard molar enthalpies of formations (Hfº). Hrxnº = n Hfº (Products) – m Hfº (Reactants) where the symbol (sigma) means “the sum of' n and m stoichiometric coefficients of balanced equation of chemical change”. Experimental determination of heat of reaction : The apparatus used is called calorimeter. There are two types of Calorimeters : (a) Bomb Calorimeter (b) Water Calorimeter
14 Gyaan Sankalp Thermodynamics Bomb Calorimeter : The calorimeter used for determining enthalpies of Ignition Wires + _ combustion known as the bomb calorimeter is shown in figure. This apparatus was devised by Berthelot (1881) to measure the heat of combustion of organic O 2 compounds. A modified form of the apparatus shown in figure consists of a sealed combustion chamber, called a bomb, containing a weighed quantity of Insulating the substance in a dish along with oxygen under about 20 atm pressure. Container Water Sample Steel Bomb
Bomb Calorimeter JOULE - THOMSON EFFECT : The phenomenon of change of temperature produced when a gas is made to expand adiabatically from a region of high pressure to a region of extremely low pressure is called Joule Thomson effect. For ideal gas Joule Thomson effect is zero because Joule - Thomson coefficient µJT is given by H T P T µJT = = – ; µJT = 0 for an ideal gas. P H CP
INVERSION TEMPERATURE : The temperature below which a gas becomes cooler on expansion is known as inversion temperature Ti 2a T = where a & b are called Van der Waals constant. i Rb Example 5 : The standard enthalpies of formation at 298 K for CCl4(g), H2O(g), CO2(g) and HCl(g) are – 106.7, – 241.8, –393.7 and –1 –92.5 kJ mol , respectively. Calculate H298K for the reaction, CCl4(g) + 2H2O(g) CO2(g) + 4HCl(g) Sol. The enthalpy change of the given reaction will be given as 0 0 0 0 H HCOf 2, g 4H f HCl,g HCCl f 4 , g 2H f HO,g 2 = ( – 393.7 – 4 × 92.5 + 106.7 + 2 × 241.8 ) kJ mol–1 = – 173.4 kJ mol–1. Example 6 : The “heat of total cracking” of hydrocarbons HTC is defined as H at 298.15 K and 101.325 kPa for the process below : m Cn H m 2n H 2(g) nCH 4 2 g
Given that HTC is – 65.2 kJ for C2H6 and – 87.4 kJ for C3H8, calculate H for
CH(g)4 CH(g) 3 8 2CH(g) 2 6 Sol. HTC of CH4 = 0 CH4(g) + C3H8(g) 2C2H6(g) H 2 HT.C. (C 2 H 6 ) H T.C. (C 3 H 8 ) = 2 ( – 65.2 ) – ( – 87.4 ) = – 43 kJ Example 7 : Find bond enthalpy of S – S bond from the following data : 1 C2H5 — S — C2H5 Hf 147.2kJmol 1 C2H5 — S – S — C2H5 Hf 201.9kJmol 1 S(g) Hf 222.8kJmol H H H H Sol. 4C (s) + 5H2 + S H CCSCC Gyaan Sankalp 15 Thermodynamics
H H H H 4C(s) + 5H2 + 2S HS CCSCC H B.E. B.E. R P (i) – 147.2 = Heat of atomization of 4C, 10H, 1S – B.E. of 10(C – H), 2(C — S), 2(C — C) (ii)– 201.9 = Heat of atomization of 4C, 10H, 2S – B.E. of 10(C — H), 2(C — S), 2(C — C), (S — S) Subtracting (i) from (ii) – 201.9 + 147.2 = Heat of formation of 1S – B.E. of (S – S) = 222.8 kJ – B.E. of (S – S) B.E. of (S – S) = 277.5 kJ Example 8 : Compute the resonance energy of gaseous benzene from the following data. (C — H) = 416.3 kJ mol–1 (C — C) = 331.4 kJ mol–1 (C = C) = 591.1 kJ mol–1 1 sub C,graphite 718.4 kJmol
-1 Hdiss H 2 ,g 435.9kJmol
-1 Hf benzene,g 82.9kJmol ° Sol. To compute resonance energy, we compare the calculated value of ΔHf (benzene, g) with the given one. ° To calculate ΔHf (benzene, g), we add the following reactions. 6C(g) 6H(g) C6 H 6 (g) H (3 CCCCCH 3 6 ) 6C(graphite) C(g) H 6 718.4 kJ mol1 1 Add 3H2 (g) 6H(g) H 3 435.9 kJ mol 6C(graphite) + 3H2(g) C6H6(g) The corresponding enthalpy change is 1 fH (3 C C 3 C C 6 C H ) 6 718.4+3×435.9 kJ mol –1 = [ – ( 3 × 331.4 + 3 × 591.1 + 6 × 718.4 + 3 × 435.9] kJ mol ° ° –1 The given ΔHf is ΔHf (benzene, g) = 82.9 kJ mol This means benzene becomes more stable by (352.8 – 82.9) kJ mol–1, i.e., 269.7 kJ mol–1 . This is its resonance energy.
Example 9 : Calculate heat of combustion of ethene: H H C = C + 3O = O 2O=C=O + 2H –O–H H H From bond energy data : C=C C–H O=O C=O O–H K.E. KJ mol–1 619 414 499 724 460 Sol. H = sum of bond energies of reactants – Sum of bond energies of products = [H(C=C) + 4H(C–H) + 3 × H(O=O)] – [4 × H(C=O) + 4 × H(O–H)] = [619 + 4 × 414 + 3 × 499] – [ 4× 724 + 4 × 460] = – 964 kJ mol–1
16 Gyaan Sankalp Thermodynamics Example 10 : The standard enthalpy of combustion at 25°C of hydrogen, cyclohexene (C6H10) and cyclohexane (C6H12) are –241, – 3800 and – 3920 kJ/mole respectively. Calculate the heat of hydrogenation of cyclohexane. Sol. The required reaction is
C6 H 10 H 2 C 6 H 12 , H 1 ? ...... (1) Cyclohexene Cyclohexane Let us write the given facts 1 H + O 6 CO + 5 H O ; H = – 241 kJ/mole ...... (2) 2 2 2 2 2 2 17 C H O 6CO 5H O, H = – 3800 kJ/mole ...... (3) 6 102 2 2 2 3
C6H12 + 9O2 6CO2 + 6H2O, H4 = – 3920 kJ/mole ...... (4) The required reaction (1) can be obtained by adding equations (2) and (3) and subtracting (4) from the sum of (2) and (3). C6H10 + H2 C6H12 H1 = (H2 + H3) – H4 = [– 241 + (– 3800)] – (–3920) = (–241 – 3800) – ( – 3920) = – 4041 + 3920 = – 121 kJ/mole
TRY IT YOURSELF 1 Q.1 At constant P and T which statement is correct for the reaction CO(g) + O (g) CO (g) 2 2 2 (A) H = E (B) H < E (C) H > E (D) H is independent for physical state of reactant Q.2 Equal volume of C2H2 and H2 are combusted under identical condition. The ratio of their heat of combustion is : 1 H (g) + O (g) H O(g) ; H= –241.8 kJ 2 2 2 2 1 C H (g) + 2 O (g) 2CO (g) + H O(g) ; H = –1300 kJ 2 2 2 2 2 2 (A) 5.37/1 (B) 1/5.37 (C) 1 (D) none of these Q.3 Given enthalpy of formation of CO2(g) and CaO(s) are –94.0 kJ and –152 kJ respectively and then enthalpy of the reaction : CaCO3(s) CaO(s) + CO2(g) is 42 kJ. The enthalpy of formation of CaCO3(s) is : (A) –42 kJ (B) –202 kJ (C) +202 kJ (D) –288 kJ Q.4 The H values for the reaction, 1 C(s) + O (g) CO(g) ; H = – 100 kJ 2 2 1 CO(g) + O (g) CO (g) ; H = – 200 kJ 2 2 2
The heat of reaction for C(s) + O2(g) CO2(g) is : (A) –50 kJ (B) –100 kJ (C) –150 kJ (D) –300 kJ –1 Q.5 Standard heat of formation of CH4(g) , CO2(g) and water at 25º are – 17.9 , – 94.1 and – 68.3 kcal mol respectively. The heat change (in kcal) in the following reaction at 25ºC is : CH4(g) + 2O2(g) CO2(g) + 2H2O() (A) –144.5 (B) –180.3 (C) –248.6 (D) –212.8 3 1 Q.6 S + O SO + 2x kcal ; SO + O SO + y kcal 2 2 3 2 2 2 3
The heat of formation of SO2 is : (A) y – 2x (B) (2x + y) (C) x + y (D) 2x/y Q.7 Given that standard heat enthalpy of CH4, C2H4 and C3H8 are –17.9, 12.5, –24.8 kcal/mol. The H for CH4 + C2H4 C3H8 is: (A) –55.2 kcal (B) –30.2 kcal (C) 55.2 kcal (D) –19.4 kcal
Gyaan Sankalp 17 Thermodynamics -1 Q.8 The standard enthalpies of formation of NO2(g) and N2O4(g) 8 and 2 kcal mol respectively. The heat of dimerisation of NO2 in gaseous state is - (A) 10 kcal mol-1 (B) 6.0 kcal mol-1 (C) – 14 kcal mol-1 (D) – 6.0 kcal mol-1 Q.9 The bond energies of C — C, C = C, H — H and C — H linkages are 350, 600, 400 and 410 kJ per mole respectively. The heat of hydrogenation of ethylene is : (A) – 170 kJ mol–1 (B) – 260 kJ mol–1 (C) – 400 kJ mol–1 (D) – 450 kJ mol–1 + – Q.10 Hºf for the reaction, Ag (aq) + Cl (aq) AgCl(s) + –1 – –1 –1 at 25ºC. Given Hºf (Ag , aq) = 105.58 kJ mol , Hºf (Cl , aq) = –167.16 kJ mol and Hºf (AgCl, s) = –127.07 kJ mol . (A) –65.49 kJ mol–1 (B) 65.49 kJ mol–1 (C) 188.65 kJ mol–1 (D) –188.65 kJ mol–1
ANSWERS (1) (B) (2) (A) (3) (D) (4) (D) (5) (D) (6) (A) (7) (D) (8) (C) (9) (A) (10) (A)
ENTROPY : It may be defined as the measure of degree of randomness in the molecule. It is represented by the symbol S. Characteristic of Entropy : (i) It is a state of function (ii) It is an extensive property The change in entropy during a process when a system undergoes charge from one state to another is represented as S. Thus S = Sfinal - Sinitial and for chemical reaction S = Sproduct – SReactant Calculation of changes in entropy : (a) Reversible process at equilibrium q S Re v T (i) In a reversible reaction heat gained in the forward reaction is equal to heat lost in the reverse reaction. Hence in a reversible cyclic process the net charge in entropy is zero. This is called Clausius Theorem (ii)Suniverse = Ssystem +SSurrounding
qRe v(system) qRe v(surrounding) S = ; S = system T surrounding T
qRe v(system) qRe v(surrounding) qRe v(system) qRe v(system) S = + = – universe T T T T
Hence Suniverse(Rev)=0 (iii) Suniv > 0 for irreversible process (iv)Entropy change for an isothermal process E = q + w E = 0 (Isothermal process) q = – W V 2 V2 But W = –2.303 nRT log V q = 2.303 nRT log 1 V1
V2 or for a reversible process qrev = 2.303 nRT log V1 V qRev 1 V2 2 S = = (2.303 nRT log ), S = 2.303 nR log V T T V1 1
1 P1 Also V S = 2.303 nR log P P2 (v) Entropy change in an isobaric process
dqRe v(P) dS = also dq = dH T Rev(P)
18 Gyaan Sankalp Thermodynamics
But dH = CpdT dS = CpdT/T Integrating both sides
T2 S2 dT T dS T2 2 = Cp , S = Cp n T = Cp ln S1 T T1 T T1 1
T2 S = 2.303 Cp log T1 (vi) Entropy change for Isochoric process
dqRe v(v) dS = . Also dq = dE T (Rev) (v) C dT But dE = C dT ds v v T Integrating both sides one get
T T T S = C n T 2 = C ln 2 ; S = 2.303 C log 2 v T1 v v T1 T1 q (vii) Entropy changes during phase transformation S = Rev T (a) Entropy of fusion : The entropy changes taking place when 1 mole of a solid substance change into liquid form, at the melting temperature.
Hfusion Sfusion = Tfusion (b) Entropy of vaporization is the entropy change when one mole of a liquid changes into vapours at boiling point.
Hvapourisation Svap = Svap – Sliquid = Tboiling point (c) Entropy of sublimation is the entropy change when one mole of a solid changes into vapours at sublimation temperature.
Hsub Ssub = Tsub (d) Hence for any physical transformation
Htransition Stransition = Ttransition
(viii) Let a given mass of a liquid be heated from temperature T1 to T2. Assuming specific heat of liquid to be constant between T1 and T2 and that no change occurs the amount of heat required to raise the temperature by dT is given by dq = mCdT T S = mC ln 2 T1 Trouton’s rule : Entropy of vaporization of non-associated or non-dissociated liquid is constant and may be taken as about 87.3 Jk–1 mol–1.
SECOND LAW OF THERMODYNAMICS : According to it all forms of energy can be converted into heat, but no process is available for converting the heat so-obtained completely to other forms of energy. An other statement is : In a spontaneous process, the entropy of universe increases and remains unchanged in an equilibrium process. Mathematically, we can express the second law of thermodynamics as follows : (i) A spontaneous process : Suniv = Ssys + Ssurr > 0 (ii) An equilibrium process : Suniv = Ssys + Ssurr = 0 (iii) A non-spontaneous process : Suniv < 0.
Gyaan Sankalp 19 Thermodynamics CARNOT CYCLE : A Isothermal Carnot devised an ideal engine which is based on a reversible cycle of four q expansion
n 1
operations in succession : isothermal expansion, adiabatic expansion, isother- o c i i n c t s T
1 i mal compression and adiabatic compression. o t a s i a e b B s r b a
There are four thermodynamical processes isothermal expansion, n i p a a i d p m
adiabatic expansion, isothermal compression and adiabatic compression. d
A D
P x o A
T e Main parts of Carnot's engine are : c 2 Source of Heat : A hot body of high temperature T from which the heat can be drawn. 1 Isothermal q C 2 It is a hot body of very large heat capacity kept at a constant high temperature T1 K. I compression ts upper surface is perfectly conducting so that working substance can take heat V1 V V2 from heat. Mechanical Arrangements and working substance : A cylinder whose walls are perfectly non-conducting but base perfectly conducting fitted with non-conducting piston which can move without any frictional losses. Ideal gas enclosed in these system as a working substance. Heat sink : A cold body at low temperature T2 K to which the heat can be rejected. It is a body of large heat capacity. Its upper surface is highly conducting so that working substance can reject heat to it. Stand : It is made by perfectly insulating material so that when cylinder placed on it working substance can expended or compressed adiabatically . Efficiency of Carnot Engine, Net work Heat added TT 1 2 (T1 Source of temp. , T2 Sink temp.) T1 TT 1 2 100% T1 q q 1 2 100% = (q1 Heat absorbed, q2 Heat rejected) q1 It can be shown that the efficiency for the Carnot engine is the best that can be obtained for any heat engine and eqn. gives an upper limit to the efficiency of any heat engine operating between temperatures T1 and T2. The efficiency depends upon the temperatures T1 and T2 which approaches unity only when the temperature of the cold reservoir approaches absolute zero. A steam engine using steam at 373 K and with the cold reservoir at 273 K has a best possible efficiency of 1 – (273/373) 27%. The efficiency of a Carnot engine is never 100% because it is 100% only if temperature of sink T2 = 0 which is impossible.
Example 11 : A Carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. What is the amount of heat energy supplied to the engine from source per cycle?
Sol. W = 800 J, T1 = 600 K T2 = 300 K T W 300 800 800 = 1 – 1 ; 1 or 0.5 = T2 q 600 q q 800 Heat energy supplied by source, q = 1600 Joule per cycle . 0.5
GIBB’S FREE ENERGY : It is defined as, the energy available in the system for conversion into useful work. It is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the useful work done by the system. G = H – TS where H is the heat content, T is the absolute temperature and S is the entropy of the system. As before, for the isothermal processes, we have G = H – TS (Gibb’s Helmotz equation) H = H2 – H1 is the enthalpy change of the system 20 Gyaan Sankalp Thermodynamics Deriving the criteria for spontaneity from Gibbs-Helmholtz equation. According to Gibbs-Helmholtz equation G = H – TS The equation combines in itself both the factor which decide the spontaneity of a process, namely (i) the energy factor, H (ii) the entropy factor, TS Depending upon the signs of H and TS and their relative magnitudes, the following different possibilities arise. 1. When both H and TS are negative i.e., energy factor favours the process but randomness factor opposes it, Then (i) If H > TS , the process is spontaneous and G is negative. (ii) If H < TS , the process is non-spontaneous and G is positive. (iii) If H = TS , the process is in equilibrium and G is zero. 2. When both H and TS are positive i.e., energy factor opposes the process but randomness factor favours it. Then (i) If H > TS , the process is non-spontaneous and G is positive. (ii) If H < TS, the process is spontaneous and G is negative. (iii) If H = TS, the process is in equilibrium and G is zero. 3. When H is negative but TS is positive i.e., energy factor as well as the randomness factor favour the process. The process will be highly non-spontaneous and G will be highly positive. An important advantage of free energy criteria over the entropy criteria lies in the fact that the former requires free energy change of the system only whereas the latter requires the total entropy change for the system and the surroundings. Standard free energy change (G°) : The standard free energy change G° is defined as the free energy change for a process at a specified temperature in which the reactants in their standard state are converted to the products in their standard state. It is denoted by G°. Like the standard enthalpy of formation of an element “the standard free energy of formation of an element in its standard state 0 0 0 is zero.” And so, Gr G f (products) G f (reactants) . Physical significance of Gibb’s free energy : E = q + w E = q + wexpansion + wnonexpansion E = q – PV + wnonexpansion (because wexpansion = –PV) E + PV = q + wnonexpansion H = qRev + wnonexpanison q (For a reversible process taking place at constant temperature. S = Re v q = TS) T Rev
As qRev = TS and E + PV = H we get H = TS + wuseful [wnonexpansion = wuseful] H – TS = wuseful G = wuseful (i) When useful work is done by the system wuseful = –ve value G = –ve value and G = wuseful Gproduct – Greactant = –ve Capacity to do useful work by product is less than the capacity to do useful work by reactant Product more stable than reactant. Hence according to II law of thermodynamics the process is a spontaneous process as every substance wants to be in the state of maximum stability. (ii) When work is done on the system wuseful = +ve G = +ve GP > GR Capacity to do useful work by product is more than the capacity to do useful work by reactant Reactant more stable than product Process non-spontaneous according to II law of thermodynamics Calculation of the free energy changes : Consider the following reaction occurring under standard stable conditions : aA + bB cC + dD The standard free energy change is given by : Grxnº = [c Gº(C) + d Gº(D)] – [a Gº(A) + b Gº(B)]
Gyaan Sankalp 21 Thermodynamics THE STANDARD FREE ENERGY OF FORMATION OF A COMPOUND (Gº) : The standard free energy of formation of a compound is the free energy change of the reaction when 1 mol of the compound is formed from its constituent elements in their standard states.
PHASE TRANSITIONS : At the transition temperature, e.g., at the melting or the boiling point, the system is at equilibrium, (G = 0), therefore, the equation G = H – T S H becomes : 0 = H – T S or S = T
For the ice water transition, H is enthalpy of fusion (Hfusion), T is the melting point, and Hfusion is the entropy of fusion.
Hfusion Sfusion = Tm
For the water () water (g), transition H is the enthalpy of vaporization ( Hvap), T is the boiling point, and S is entropy of vaporization ( Svap.)
Hvap Svap = Tb FREE ENERGY CHANGE (G) AND EQUILIBRIUM CONSTANT (K) : For a reaction : G = Gº + 2.303 RT log Q where G = free energy change of the reaction at TK ; Gº = standard free energy change at 298 K ; R = gas constant (= 8.314 J mol–1 K–1), and Q is the quotient of reaction. For reaction : aA + bB cC + dD, [C]c[D]d Q = [A]a[B]b At equilibrium, G = 0, so Q must be equal to equilibrium constant, K. 0 = Gº + 2.303 RT log K or Gº = – 2.303 RT log K Thus, if Gº is (i) negative, K > 1, (ii) zero, K = 0, and (iii) positive, K < 1 Also Gº = Hº – TSº for standard states – 2.303 RT log K = Hº – TSº Hº T = Sº 2.303 R log K EFFECT OF TEMPERATURE (T) ON EQUILIBRIUM CONSTANT (K) : Gº = – 2.303 RT log K and Gº = Hº – TSº – 2.303 RT log K = Hº – TSº Hº Sº or log K = 2.303RT 2.303R
If K1 and K2 are the values of equilibrium constants at T1K and T2K respectively and Hº and Sº are constant in temperature range T1K to T2K, then :
K2 Hº L 1 1 O Hº LT2 T1 O log = M P = M P K1 2.303R NT2 T1 Q 2.303R N T1T2 Q This equation is known as Von't Hoff equation.
FREE ENERGY AND ELECTROMOTIVE FORCE OF A CELL : Gº and G are related to Eºcell and Ecell by equation. Gº = – nF Eºcell G = – nF Ecell where n stands for number of electrons exchanged, F the Faraday, Eºcell the emf of a cell in the standard state and Ecell the emf of a cell in the given state.
22 Gyaan Sankalp Thermodynamics
2.303 RT Eº = log K cell nF 0.0591 Eº = log K at 298 K cell n Gibbs-Helmholtz equations : At constant pressure, it relates G with H d( G) GH d( G) d( G) G = H + T or = or – S = T P T T P T P
d Ecell Since, G = – nF Ecell or S = nF dT P
VARIATION OF EQUILIBRIUM VAPOUR PRESSURE OF A LIQUID WITH TEMPERATURE :
K2 Hvap L 1 1 O H vap LT2 T1 O log = M P = M P K1 2.303R NT2 T1 Q 2.303R N T1T2 Q The equation is called Clausius – Clapeyron equation.
THIRD LAW OF THERMODYNAMICS : It states that, “Entropy of all perfectly crystalline solids may be taken as zero as the absolute zero of temperature.” Third law of thermodynamics may also be defined as (i) The entropy of a solid is zero at the absolute zero of temperature. (ii) It is impossible to reduce the temperature of any system to absolute zero by any process. (iii) At any pressure, the entropy of every crystalline solid in thermodynamic equilibrium at absolute zero is zero. (iv)At the absolute zero, increment in entropy for isothermal process in crystalline approaches zero as the limit. (v) At absolute zero, every crystal becomes ideal crystal. Example 12 : Show that the reaction, CO(g) + 1/2 O2(g) CO2(g) at 300 K is spontaneous and exothermic, when the standard entropy change is – 0.094 kJ mol–1 K–1. The standard Gibb’s free –1 energies of formation of CO2 and CO are – 394.4 and –137.2 kJ mol respectively. Sol. The given reaction is, CO(g) + (1/2) O2(g) CO2(g)
1 G° (for reaction) = GGGCO CO O = – 394.4 – ( – 137.2 ) – 0 = – 257.2 kJ mol–1 22 2 Gº = Hº – TS° – 257.2 = H° – 298 × (0.094) or Hº = –288.2 kJ Gº is –ve, hence the process is spontaneous, and Hº is also –ve, hence the process is also exothermic. Example 13 : Calculate G for the expansion of one mole of hydrogen in going from 1 atm to 0.2 atm at 30ºC. Sol. For the isothermal change, if we assume an ideal gas, G = H – TS or dG = dH – TdS or dG = dE + PdV + VdP – TdS considering work involved is mechanical one, dqrev = TdS = dE + PdV nRT dG = VdP = dP P
P2 On integrating, G = nRT ln P1
P2 0.2 or G = nRT × 2.303 log or G = 1 × 1.987 × 303 × 2.303 log or G = – 969.2 cals. P1 1
Gyaan Sankalp 23 Thermodynamics Example 14 : When 10gm of hydrogen gas at 1 atm pressure and 30ºC is brought to a reduced pressure of 1/2 atm also at 30ºC, calculate entropy change involved. Sol. For an isothermal process : E = 0, w = q, q = wmax
P1 q P1 10 1 q = 2.303 nRT log ; S = rev = 2.303 × n × R × log or S = 2.303 × × 1.987 × log or S = 6.9 e.u. P2 T P2 2 0.5 TRY IT YOURSELF Q.1 The enthalpy and entropy change for a chemical reaction are –2.5 × 103 cal and 7.4 cal deg–1 respectively. Nature of reaction at 298 K is – (A) spontaneous (B) reversible (C) irreversible (D) non–spontaneous Q.2 For the reaction : 2NO(g) + O2(g) 2NO2(g) the enthalpy and entropy changes are -113.0 kJ mol-1 and -145 J K-1 mol-1 respectively. Find the temperature above which the reaction is spontaneous (A) 432.3 K (B) 570.5 K (C) 1035.7 K (D) 779.3 K Q.3 The enthalpy change for a given reaction at 298 K is -x cal/mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature - (A) can be negative but numerically larger than x/298 cal K-1 (B)can be negative, but numerically smaller than x/298 cal K-1 (C)cannot be negative (D)cannot be positive Q.4 The favorable condition for a process to be spontaneous is : (A) TS > H, H = +ive, S = -ive (B) TS > H, H = +ive, S = +ive (C) TS = H, H = -ive, S = -ive (D) TS = H, H = +ive, S = +ive Q.5 For a process at H and TS both are positive in what condition, the process is spontaneous (A) H > TS (B) H < TS (C) H = TS (D) not predicted Q.6 The entropy change for the conversion of one gram of ice to water at 273 K and one atmospheric pressure is : [ Hfusion = 6.025 kJ mol-1 ] (A) 7.30 J K-1 mol-1 (B) 1.226 J K-1 g-1 (C) 1.226 J K-1 mol-1 (D) 7.30 J K-1 g-1 -1 Q.7 H and S for Br2() + Cl2(g) 2BrCl(g) and 29.37 kJ and 104.0 J K respectively. Above what temperature will this reaction become spontaneous ? (A) T > 177.8 K (B) T > 354.1 K (C) T > 282.4 K (D) T > 141.2 K Q.8 The equilibrium constant Kc for the following reaction at 400 K -1 2NOCl(g) 2NO(g) + Cl2(g) is : [ Given Hº = 77.2 kJ and Sº = 122 J K at 400 K ] (A) 2.577 × 10-4 (B) 1.958 × 10-4 (C) 28.4 × 10-3 (D) 1.466 × 10-2 ANSWERS (1) (A) (2) (B) (3) (B) (4) (B) (5) (B) (6) (B) (7) (C) (8) (B)
USEFUL TIPS 1. Pressure is an intensive property. 2. A rerversible process is always quasi-static. 3. The work done by a gas during free expansion is equal to zero. 4. First law of T.D. is applicable to all processes irrespective to whether they are reversible or irreversible. 5. All spontaneous processes proceed in one direction only. 6. A real crystal has higher entropy than the ideal crystal. 7. According to IUPAC conventions work done on the surroundings is –ve. 8. A carnot cycle uses only two thermal reservoir. 9. A carnot cycle consists of only reversible processes. 10. The efficiency of a carnot engine can be increased by decreasing sink temperature when the source temperature is held constant. 11. For a reversible adiabatic process, S = constant and hence it is called as an isentropic. 12. Entropy change of a system is determined by the initial and final states only, irrespective of how the system has changed its states. 13. Solidification of liquid shows decreases in entropy. 14. When Fe(s) is dissolved in a aqueous HCl in a closed rigid vessel, the work done is zero. 15. For non-spontaneous process at constant T and P, G is positive. 24 Gyaan Sankalp Thermodynamics MISCELLANEOUS SOLVED EXAMPLES Example 1 : In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process ? Sol. Heat absorbed by the system, q = + 701 J Work done by the system, w = – 394 J Then, change in the internal energy, E = + 701J – 394J or E = + 307 J Example 2 : –1 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, –393, 81 and 9.7 kJ mol respectively. Find the value of rH for the reaction : N2O4(g) + 3CO(g) N2O(g) + 3CO2(g) Sol. Using the given data, one can write rH = f H(products) – f H(reactants) = [81 kJ mol–1 + 3 × (–393 kJ mol–1)] – [9.7 kJ mol–1 + 3 × (–110 kJ mol–1)] = (– 1098 kJ mol–1) – (– 320.3 kJ mol–1) = – 777.7 kJ mol–1 Example 3 : The heat of combustion of ethylene at 18ºC and at constant volume is –335.8 kcal when water is obtained in liquid state. Calculate the heat of combustion at constant pressure and at 18ºC. Sol. The chemical equation for the combustion of C2H4 is C2H4(g) + 3O2(g) = 2CO2(g) + 2H2O (); E = –335.8 kcal 1mole 3 mole 2 mole No. of moles of gaseous reactants = (1 + 3 ) = 4 No. of moles of gaseous products = 2 So,n = (2 – 4) = –2 Given E = –335.8 kcal, n = –2, R = 2 × 10–3 kcal and T = (18 + 273) = 291 K H = E + nRT = –335.8 + (–2) (2×10–3) (291) = –336.964 kcal Example 4 : The enthalpy of formation of ethane, ethylene and benzene from the gaseous atoms are – 2839.2, – 2275.2 and –5536 KJ mol– respectively. Calculate the resonance energy of benzene, compared with Kekule structure. The bond enthalpy of C – H bond is given as equal to 410.87 KJ mol–1. Sol. Bond enthalpy of C – C bond = Enthalpy required to break C2H6 into gaseous atoms – 6 × bond enthalpy of C – H bond = 2839.2 KJ mol–1 – 6 × 410.87 kJ mol–1 = 373.98 kJ mol–1 Bond enthalpy of C = C bond = Enthalpy required to break C2H4 into gaseous atoms – 4 × bond enthalpy of C – H bond = 2275.2 kJ mol–1 – 4 × 410.87 kJ mol–1= 631.72 kJ mol–1 For the formation of benzene having Kekule structure, we have to form 3C – C bonds, 3C = C bonds and 6C – H bonds for which enthalpy released is [3 (–373.98) + 3 (–631.72) + 6(–410.87)] = – 5482.32 kJ mol–1 –1 But the given value of fH is fH (actual) = – 5536 kJ mol Hence resonance energy compared to Kekule structure = f H (actual) – fH (Kekule structure) = (–5536 + 5482.32) = – 53.68 kJ mol–1 Example 5 : A gas expands by 0.5 litre against a constant pressure one atmosphere. Calculate the work done in joule and calorie. Sol. Work = – Pext × volume change = – 1 × 0.5 = – 0.5 litre-atm = – 0.5 × 101.328 J = – 50.644 J 0.5 lit-atm = – 0.5 × 24.20 cal = – 12.10 cal
Example 6 : Using bond enthalpy data given below, estimate enthalpy of formation of acetic acid Bond Bond enthalpies C – H 413.38 kJ mol–1 C – C 347.69 kJ mol–1 C = O 728.02 kJ mol–1 C – O 351.46 kJ mol–1 O – H 462.75 kJ mol–1
Gyaan Sankalp 25 Thermodynamics Enthalpy of atomization C 718.39 kJ atom–1 H 217.94 kJ atom–1 O 247.52 kJatom–1 –1 The observed fHº for acetic acid is –438.15 kJ mol . Compute the resonance energy of acetic acid. Sol. The desired reaction for the formation of CH3COOH is 2C(s) + 2H2(g) + O2(g) CH3COOH() –1 fHº = 3 × 413.38 – 1 × 347.69 – 728.02 – 351.46 – 462.75 + 2 × 718.39 + 4 × 217.94 + 2 × 247.52 = – 326.48 kJ mol –1 Resonance energy = Hºobs – Hºcal = – 438.15 + 326.48 = – 111.67 kJ mol Example 7 : –1 –1 The bond dissociation enthalpies of H2(g) and N2(g) are + 435.95 k mol and + 941.8 kJ mol and enthalpy of formation of –1 NH3(g) is – 46.024 kJ mol . (a) What is the enthalpy of atomization of NH3(g) ? (b) What is the bond enthalpy of N – H bond ? Sol. (a) The desired equation is NH3(g) N(g) + 3H(g) We can consider the reaction as 1 3 NH (g) H1 N(g) + 3H(g) H2 N (g) + H (g) 3 2 2 2 2 –1 H1 + H2 = + 46.024 KJ mol 3 1 Now, H = – (435.95) – (941.8) 2 2 2
or H1 = 46.024 + 653.925 + 470.9 –1 or H1 = 1170.849 KJ mol –1 or H1 = 1.170849 MJ mol 1170.849 (b) The average bond enthalpy of N – H bond = = 390.283 KJ mol–1. 3 Example 8 : The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. 3 NH CN(s) + O (g) N (g) + CO (g) + H O() 2 2 2 2 2 2 3 1 Sol. From the chemical equation, n = n – n = 2 – = Products reactants 2 2 Given, E = – 742.7 kJ mol–1 We know, H = E + nRT = 742.7 kJ mol–1 + (1/2 × 8.3 J K–1 mol–1 × 298 K) = – 742.7 kJ mol–1 + 1.2 kJ mol–1= – 741.5 kJ mol–1 Example 9 : Sulphur exists in more than one solid form. The stable form. The stable form at room temperature is rhombic sulphur. But above room temperature the following reaction occurs : s (rhombic) s(mono clinic) –1 –1 Thermodynamic measurements reveal that at 101.325 KPa and 298 K, rH = 276.144 J mol and rG = 75.312 J mol (a) compute rSat 298 K (b) Assume that rH and rS do not vary significantly with temperature, compute Teq, the temperature at which rhombic and monoclinic sulphur exist in equilibrium with each other. Sol. (a) Since rG = rH – TrS, 1 1 HG 276.144J mol 75.312J mol Therefore, S = r r = = 0.674 JK–1 mol–1 r T 298K (b) When the rhombic sulphur is in equilibrium with monoclinic sulphur, we would have rG = 0 = rH – Teq rS 1 rH 276.144 J mol Thus Teq = = 1 = 409.7 K rS 0.674 J mol 26 Gyaan Sankalp Thermodynamics Example 10 : Calculate the number of kJ necessary to raise the temperature of 60.0 g of aluminum from 35 to 55ºC. Molar heat capacity of Al is 24 J mol–1 K–1. Sol. Mass of aluminium, m = 60.0 g Rise in temperature, t = (55ºC – 35ºC) = 20ºC Molar heat capacity, C = 24 J K–1 mol–1 60.0g 60 24 20 Heat required = × 24 J K–1 mol–1 × 20ºC = J = 1066.7 J = 1.07 kJ 27g / mol 27
Example 11 : –1 Enthalpy of combustion of carbon to CO2 is – 393.5 kJ mol . Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. Sol. C(s) + O2(g) CO2(g) 1mol (= 44g) 1mol Amount of CO in 35.2 g = × 35.2g = 0.8 mol 2 44g –1 So, heat released during the formation of 35.2 g CO2 = 393.5 kJ mol × 0.8 mol = 314.8 kJ
Example 12 : Will S be positive or negative in the following processes ? Discuss qualitatively (a) H2O(s) H2O() (b) H2O() H2O(g) (c) H2(g) + Cl2(g) 2HCl(g) 1 3 (d) N (g) + H (g) NH (g) (e) 2H (g) + N (g) N H () (f) Cl (g) 2Cl(g) 2 2 2 2 3 2 2 2 4 2
Sol. As we have discussed that the entropy of reaction is more if there is a change in value of vg (the change in the stoichiometric number of gaseous species), since the entropy of gases is much larger than the entropy of the condensed phases. for process (a) S is +ve for process (b) S is +ve for process (c) S is zero for process (d) S is negative for process (e) S is negative for process (f) S is positive Example 13 : Calculate the standard enthalpy of formation of CH3OH() from the following data : 3 (i) CH OH() + O (g) CO (g) + 2H O() H = – 762 kJ mol–1 3 2 2 2 2 C –1 (ii) C(g) + O2(g) CO2(g) CH = – 393 kJ mol 1 (iii) H (g) + O (g) H O() H = – 286 kJ mol–1 2 2 2 2 f
Sol. The standard enthalpy of formation of CH3OH() is equal to the enthalpy change for the reaction C(s) + 2H2(g) + 1/2 O2(g) CH3OH() To get this chemical equation, rewrite the given equations as follows : –1 Eq. (ii) C(s) + O2(g) CO2(g) H1 = – 393 KJ mol –1 –1 2 × Eq. (iii) 2H2(g) + O2(g) 2H2O() H2 = 2 × (–286 kJ mol ) = – 572 kJ mol 3 Eq. (i) CO (g) + 2H O(l) CH OH() + O (g) H = – (– 726 kJ mol–1) = + 726 kJ mol–1 2 2 3 2 2 3 Adding these rewritten equations, one gets C(s) + 2H2O() + 1/2 O2(g) CH3OH() 0 0 0 –1 –1 fHº = H1 + H2 + H3 = (– 393 – 572 + 726) kJ mol = – 239 kJ mol –1 As per definition, fHº (CH3OH()) = rHº = – 239 kJ mol
Gyaan Sankalp 27 Thermodynamics Example 14 : 12 A B The diagram shows a P-V graph of a thermodynamic )
behaviour of an ideal gas. Find out form this graph : 2 10 m (i) work done in the process A B, B C, C D and D A / n 8 o (ii) work done in the complete cycle A B C D A. t w
Sol. Work done in the process AB (the process is expansion, e 6 N hence work is done by the gas) 5 0
5 –3 1 4 = – P × V = –12 × 10 × 5 × 10 = – 6000 J ( P Work done in the process B C is zero as volume remains constant. 2 D C Work done in the process C D (The process is contraction, hence work is one on the gas) 1 2 3 4 5 6 = P × dV = 2 × 105 × 5 × 10–3 = 1000 J V (litre) Work done in the process D A is zero a volume remains constant. Net work one in the whole cycle = – 6000 + 1000 = – 5000 J i.e. net work is done by the gas. Example 15 : Calculate the enthalpy change for the process CCl4(g) C(g) + 4Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g) –1 vap H (CCl)4 = 30.5 kJ mol –1 fH (CCl4) = – 135.5 kJ mol –1 aH (C) = 715.0 kJ mol –1 aH (Cl2) = 242 kJ mol Where aH is enthalpy of atomisation 0 0 0 0 Sol. rH = [fH (C(g)) + 4 × fH (Cl(g))] – [fH (CCl4(g))] 0 0 0 –1 –1 –1 But, fH (CCl4(g) = fH (CCl4()) + vap H (CCl4) = – 135.5 kJ mol + 30.5 kJ mol = – 105 kJ mol Substituting the values in first equation, 0 –1 –1 –1 –1 fH = (715 kJ mol + 2 × 242 kJ mol ) – (– 105 kJ mol ) = 1304 kJ mol 1304kJ mol1 Then, bond enthalpy of C – Cl bond in CCl (g) = = 326 kJ mol–1 4 4 Example 16 : For the reaction at 298 K, 2A + B C H = 400 kJ mol–1and S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering H and S to be constant over the temperature range. Sol. For a reaction to be spontaneous, G must be negative. Using Gibb’s-Helmholtz equation, for a spontaneous reaction, one can write. 1 H 400kJmol T or T or T 2000 K S 0.2kJ K1 mol 1 Example 17 : For the reaction, 2Cl(g) Cl2(g), what are the signs of H and S ? Sol. For the given reaction, H is negative S is negative. Example 18 : For the reaction, 2A(g) + B(g) 2D(g) E = – 10.5 kJ mol–1 and S = – 44.1 J K–1 mol–1 Calculate G for the reaction, and predict whether the reaction may occur spontaneously. Sol. For a reaction, E0 and H0 are related by the relationship 0 0 H = E + ng (RT) [ng = 2 – (2 + 1) = – 1] = – 10.5 kJ mol–1 + (–1) × 8.3 J mol–1 K–1 × 298 K) = – 10.5 kJ mol–1 – 2.47 kJ mol–1 = – 12.97 kJ mol–1 Then, G0 = H0 – T S0 = – 12.97 kJ mol–1 – 298 K × (–44.1 J K–1 mol–1) = – 12.97 kJ mol–1 + 1.31 kJ mol–1 So, G0 = 11.67 kJ mol–1 Since, for this reaction, G0 is positive, it is not a spontaneous reaction.
28 Gyaan Sankalp Thermodynamics Example 19 : The equilibrium constant for a reaction is 10. What will be the value of G ? R = 8.314 J K–1 mol–1, T = 300 K. Sol. We know, G0 = – 2.303 RT log K = – 2.303 × 8.314 J K–1 K–1 mol–1 × 300 K log 10 or G0 = – 5744.1 J mol–1 = – 5.7 kJ mol–1 Example 20 : Calculate the entropy change in surroundings when 1.00 mol of H2O() is formed under standard conditions. –1 fH = – 286 kJ mol . 1 q H0 286 kJ mol Sol. S = surr = f = = 0.9597 kJ mol–1 = 959.7 J mol–1. T T 298 K Example 21 : Show that the volume of a fixed amount of an ideal gas is a state function. nRT Sol. For an ideal gas V = , therefore P
V nR V nRT = and = – T P P P T P2 V = (P, T)
V V dV = dP + dT P T T P
nRT nR we have, dV = dP + dT P2 P If V is to be a state function, dV must be an exact differential, for which the Euler reciprocity condition states that
V V P = T T T P P P T
V nRT nR P = 2 = 2 T T P T P P P
V nR nR T = = 2 T P T P P T P Volume of an ideal gas is a state function. Example 22 : –1 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0ºC to ice at –10.0ºC. fus H = 6.03 kJ mol at 0ºC. –1 –1 CP [H2O(l)] = 75.3 J mol K –1 –1 CP [H2O(s)] = 36.8 J mol K Sol. The whole process can be described by the flowsheet diagram given below : H1 H2O (at 10ºC) Water (at 0ºC)
H2 Ice (– 10ºC) H3 Ice (at 0ºC) –1 –1 –1 –1 H1 = CP (H2O (l)) × T = 75.3 J mol K × (– 10ºC) = – 753 J mol = – 0.753 kJ mol –1 H2 = fus H = – 6.03 kJ mol –1 –1 –1 –1 H3 = CP (H2O(s)) × T = 36.8 J mol K × (–10ºC) = – 368 J mol = – 0.368 kJ mol Total enthalpy change = – 0.753 kJ mol–1 – 6.03 kJ mol–1 – 0.368 kJ mol–1 = – 7.15 kJ mol–1
Gyaan Sankalp 29 Thermodynamics Example 23 : From the following thermodynamic relations G = H – TS, H = E + PV, dqrev = TdS, dE = dqrev – PdV G G V S Show that = – S; = V and = – T P P T T P P T Sol. G = H – TS Since H = E + PV Therefore G = E + PV – TS ...... (1) Differentiating the equation (1) on both sides, dG = dE + PdV + VdP – TdS – SdT or dG = dE + PdV + VdP – dqrev – SdT or dG = dqrev + VdP – dqrev – SdT or dG = VdP – SdT ...... (2) From equation (2) it is obvious that G = (P, T)
G G dG = dP + dT ...... (3) P T T P By comparing equation (2) and equation (3) we get, G G = V and = – S P T T P G is a state function of P & T
G V G S P = and T = – T T P T P P P T P T
V S = – T P P T Example 24 : Coefficients of thermal expansion and compressibility factor k are defined as 1 V 1 V = k = – V T P V P T Taking V as a state function, derive the equation of state for which K (V a) 3(V a) (i) V = 1 keeping T constant and V = K T, keeping P constant. (ii) = and k = , where a is constant. P 2 TV 4PV K Sol. (i) Since V = 1 for constant temperature, from where we can get P
V K1 PV V = – 2 = – 2 = – P T P P P
V V Also V = K2T for constant pressure, from where = K2 = T P T
V V V V since V = f(P, T), therefore dV = dP + dT or dV = dP + dT P T T P P T Dividing by V and rearranging, we get dV dP dT dV dP dT = – + or + = ...... (1) V P T V P T on integrating (1) we get, ln V + ln P = In T + In C where ln C in an integrating constant, ln PV = ln TC or PV = TC ...... (2)
30 Gyaan Sankalp Thermodynamics If we replace C by R, the equation (2) become PV = RT (ii) Taking V = (P, T)
V V dV = dP + dT or dV = – Vk dP + V dT P T T P 3(V a) V (V a) 3(V a) (V a) or dV = – V dP + dT or dV = – dP + dT 4PV TV 4P T
dV 3 dP dT = + ...... (3) (V a) 4 P T Integrating both sides of equation (3), we get, 3 ln (V – a) = ln P + ln T + ln C' where ln C’ is integrating constant 4 or ln (V – a) = ln P–3/4 + ln T + ln C' or ln (V – a) = ln [(P–3/4) TC’] or P3/4 (V – a) = TC’ P3/4 (V – a) = TC’ is the required equation of state. Example 25 : An ideal gas undergoes a single-stage expansion against a constant opposing pressure P2 from P1, V1, T to P2, V2, T. What is the largest mass m which can be lifted through a height h in this expansion ? Sol. The maximum work done by the system= P2(V2 – V1) nRT nRT P2(V2 – V1) = mgh or P2 = mgh PP2 1
P2 nRT P2 or mgh = nRT 1 or m = 1 P1 gh P1 Example 26 : An ideal gas undergoes a single-stage expansion against a constant opposing pressure P2 from P1, V1, T to P2, V2, T. Then the system is restored to its initial stage by a single stage compression. What is the net mass lowered through height h in the cyclic transformation ?
nRT P2 Sol. The maximum mass which can be lifted due to expansion, m = 1 gh P1 The smallest mass which must fall through the height h = m' nRT nRT P 1 1 wcomp = P1(V2 – V1) = P1 = nRT PP2 1 P2
P1 nRT P1 nRT 1 = m'gh or m' = 1 P2 gh P2 net mass lowered through a height h, m”.
nRT P1 nRT P2 = m' – m = 1 – 1 gh P2 gh P1
nRT (PP)1 2 nRT (PP)1 2 = – gh P2 gh P1
nRT 1 1 nRT(P P )2 1 2 = (P1 – P2) = . gh PP2 1 P2 P 1 gh
Gyaan Sankalp 31 Thermodynamics Example 27 : The enthalpy of formation of methane at constant pressure and 300 K is –75.83 kJ. What will be the heat of formation at constant volume ? [ R = 8.3 JK–1 mol–1 ] Sol. The equation for the formation of methane is C(s) + 2H2(g) = CH4(g) ; H = –75.83 kJ 2 mole 1 mole n = (1 – 2) = –1 Given H = –75.83 kJ, R = 8.3 × 10–3 kJ K–1 mol–1 T = 300 K Applying H = E + nRT –75.83 = E + (–1) (8.3 × 10–3) (300) E = –75.83 + 2.49 = –73.34 kJ Example 28: Show mathematically that the magnitude of work involved in a reversible expansion of an ideal gas from volume V1 and V2 is larger than the corresponding work involved in an irreversible expansion against a constant pressure of P2. Sol. We have the relation
V V2 2 1 1 |wrev| = nRT n = nRT n V1 V1 n(1 + x) x (by neglecting the other terms) {by using n (1 + x) expansion x2 x3 n (1 + x) = x – + – ...... ; neglecting the higher terms 2 3 thus n (1 + x) = x} V nRT 2 1 |wrev| nRT = (V2 – V1) = P1 (V2 – V1) and |wirr| = P2 (V2 – V1) V1 V1
|wrev| – |wirr| = P1 (V2 – V1) – P2 (V2 – V1) nRT nRT = (V2 – V1) (P1 – P2) = (V2 – V1) VV1 2
2 (VV)2 1 nRT(V2 V 1 ) = nRT (V2 – V1) = VV1 2 VV1 2 2 V2 V1, therefore (V2 – V1) is always positive quantity |wrev| – |wirr| is always positive |wrev| > |wirr| The magnitude of work involved in a reversible expansion is large than the corresponding work involved in an irreversible expansion. Example 29 : Calculate rH for the process at 25ºC of dissolving 1.00 mol of KCl in a large excess of water. Does this process represent an ionisation reaction ? Explain. 0 + –1 f H [K (aq)] = – 251.2 kJ mol 0 – –1 f H [Cl (aq)] = – 167.08 kJ mol 0 –1 f H [KCl] = – 437.6 kJ mol + – 0 Sol. KCl(s) + H2O() K (aq) + Cl (aq) rH = ? 1 mol excess 0 0 + 0 – 0 r H = [f H (K (aq) + f H (Cl (aq))] – [f H (KCl)] = [–251.2 + (–167.08)] kJ mol–1 – [–437.6 kJ mol–1] = – 418.28 kJ mol–1 + 437.6 kJ mol–1 0 –1 rH = + 19.3 kJ mol This process does not represent an ionisation reaction. The process here involves dissolution of KCl(s) in water and formation of K+(aq) and Cl–(aq) by dissociation.
32 Gyaan Sankalp Thermodynamics Example 30 : For 1 mole of helium which is expanding isothermally at 30ºC and reversibly from 1 atm to 0.1 atm. Calculate the following : w, q, E, H, S and G. Sol. (i) For the reversible isothermal change,
P1 w = 2.303 × nRT × log P2 1 w = 2.303 × 1 × 1.987 × 303 × log = 1390 cals 0.1 (ii) For the isothermal change of ideal gas E = 0 E = q – w q = E + w = 0 + 1390 = 1390 cals (iii) H = E + (PV) As per Boyle’s law, PV = constant, Therefore, PV = 0 H = E = 0 q 1390 (iv) S = rev = = 4.58 eu T 303 (v) G = H – TS or G = 0 – 303 (4.58) = 1390 cals the process is spontaneous. Example 31 : State the conditions at which any of the following may be zero : q, w, E, H, S and G. Sol. (i) For a reversible adiabatic expansion of an ideal gas q = 0 E = –w (ii) For a system capable of performing only expansion work, if dV = 0, then w = 0 w = PV Under constant volume (Isochoric conditions) V = 0, and hence w = 0 First law of thermodynamic says, E = qv Heat flow under isochoric condition is a direct measure of E. (iii) For a reversible isothermal expansion of an ideal gas E = 0 (iv) Because the internal energy content of an ideal gas is a function only of temperature, under isothermal conditions, E = 0 H = E + (PV) or H = 0 + (nRT) = 0 Thus for isothermal and isobaric (constant pressure) expansion of an ideal gas. E = H = 0 (v) The entropy S, of a perfect crystalline substance at absolute zero (– 273 K) is zero. This is the statement of third law of thermodynamics. For a reversible process, S(system) = – S(surroundings) and S(universe) = S(system) + S(surroundings) S (universe) = 0 for a reversible process. (vi) For a reaction in equilibrium G = 0
Example 32 : A cylindrical container of volume 44.8 litres is containing equal no. of moles of an ideal monoatomic gas in two sections A and B separated by an adiabatic frictionless piston as shown in figure. The initial temperature and pressure of gas in both section is 273. and 1 atm. Now gas in section ‘A’ is slowly heated till the volume of section B becomes (1/8)th of initial volume. Find change in internal energy (E) for section A in cal. 3 Given : R = 2 cal/mol-K, C of monoatomic gas = R , AT STP ideal gas occupy 22.4 litre. v,m 2
Gyaan Sankalp 33 Thermodynamics
Frictionless adiabatic piston Frictionless adiabatic piston
The gas q in section Adiabatic Adiabatic 'A' is heated A B wall A B wall reversibly 22.4L 22.4L
Initial state Final state Sol. Initial volume in both section = 22.4 litre The gas in sec B is compressed reversibility & adiabatically
1 V 1 1 T T1 27.3 (8)2/3 4T TVTV11 2 2 2 1 1 V2
VT Final pressure 1 2 P2 P 1 1 8 4 32 atm VT2 1 Final temperature in section A PV 32 T2 2 27.3 (1.875) 1638K 1 P1 V 1 1
3 E nC T 0.1 R (1638 27.3) 483.21cal A v,m 2 Example 33 : 0 Calculate fH for chloride ion from the following data : 1 1 H (g) + Cl (g) HCl(g) H = – 92.8 kJ mol–1 2 2 2 2 + – –1 HCl(g) + H2O() H3O (aq) + Cl (aq) diss H = – 75.2 kJ mol Sol. From the second equation, 0 + 0 – 0 diss H = [f H (H3O ) + f H (Cl )] – [f H (HCl(g))] –1 0 – –1 –75.2 kJ mol = [0 + f H (Cl (aq)] + 92.8 kJ mol 0 – –1 –1 So, fH (Cl (aq)) = – 75.2 kJ mol – 92.8 kJ mol 0 – –1 or f H (Cl (aq)) = 168.0 kJ mol Example 34 : The external pressure 2 atm is applied on frictionless movable piston, fitted in a vessel containing 100g of X () at 450 K. Now heat is supplied keeping pressure constant till 40g of X is evaporated to form X (g) at 500 K (boiling point). Calculate change in internal (E) energy in kJ for overall process. Assume internal energy of liquid is depend only any temperature.
Given : Molar heat capacity of X () = 60 J/mol K, Hvaporisation = 30 kJ/mol, R = 8.3 J/mol-K, At. weight of X = 20g/mol.
2 atm
2 atm
X(g):40g
T = 450K T = 500K Initial state Final state 34 Gyaan Sankalp Thermodynamics
Sol. Etotal nC T n E m where H E ng RT
Etotal 5 60 50 2 (30,000 8.3 500) = 15000 + 51700 J = 66700 J 66.7 kJ Example 35 : One mole of an ideal gas at 300 K is heated at constant volume (V1) until it’s temperature is doubled, then it is expanded isothermally till it reaches the original pressure. Finally the gas is cooled at the constant pressure till system reached to the half of original volume (V1/2). Determine total work done ( |w| ) in cal. [Use ln 2 = 0.70, R = 2 Cal/K mol] P V T C(P,2V,2T),D(2, ,1 ) B Sol. A (P1 , V 1 ,T 1 ), B (2P 1 , V 1 , 2T 1 ) , 1 1 1 1 21 2
wnet w AB w BC w CD wAB = 0 2V D A C w 1R 2Tln 223000.7 840cal BC 1 V1 V
T1 wCD P 1 V R T R 2T 1 2 3 3 RT 2 300 900cal 21 2
WTotal 840 900 60 cal Example 36 : How much heat is required to change 10 g ice 0ºC to steam at 100ºC? Latent heat of fusion and vaporization for H2O are 80cal/g and 540cal/g respectively. Specific heat of water is 1 cal/g. Sol. Total heat absorbed = Hfusion + Htemp.rise + Hvap. = 10 × 80 + 10 × 1 ×100 + 10 + 540 = 7200 cal. Example 37 : 10 g of argon gas is compressed isothermally and reversibly at a temperature of 27oC from 10L to 5 L. Calculate q, w and E for this process. (At wt. of Ar =40) 10 Sol. No. of moles of argon = = 0.25 mole and V = 10 litre, V = 5 litres, T = 300 K 40 1 2
V2 For isothermal reversible compression w = – 2.303 nRT log V1 10 = 2.302 2.5 2 300 log = 103.6 cal 5 Amount of heat absorbed =103.6 cal. Now we know that during isothermal reversible process, internal energy remains constant throughout the process, hence the change in energy (E) will be zero. Example 38 : Estimate the maximum possible temperature of a Bunsen burner flame. Assume that the gas is pure methane and that it is premixed with sufficient air to permit complete combustion. Given that –1 CH4(g) + 2O2(g) 2H2O() + CO2(g) rHº298 K = – 890.36 kJ mol –1 H2O() H2O(g) rHº298 K = 44.02 kJ mol –1 –1 CP of CO2(g) = 37.11 JK mol –1 –1 CP of H2O(g) = 33.58 JK mol Sol. To calculate the temperature of the flame we have to consider adiabatic conditions where there is no loss of heat from the system. The rise in temperature can be calculated by considering the reaction to take place in the following two steps, both at constant pressure.
Gyaan Sankalp 35 Thermodynamics Step I : Reactants (T , P) Products (T , P) H 0 0 r T0 Step II : Products (T0, P) Products (Tf, P) rH2 Tf
rH2 = CP (products) dt T0 Since the overall reaction is the sum of the two above steps, i.e., Reactants (T0, P) products (Tf, P) The net heat change is given by H = H + H r r T0 r 2 Since the reaction is considered to be adiabatic, Therefore, rH = 0 H + H = 0 r T0 r 2 or H = – H r 2 r T0 Tf or – H = C dT r T0 P T0
If CP is independent of temperature, then
rH T – H = C (T – T ) or T = – 0 + T r T0 P f 0 f 0 Cp For our problems, –1 rHº298K = – 890.36 + 44.02 × 2 = – 802.32 kJ mol For the reaction, CH4(g) + 2O2(g) 2H2O(g) + CO2(g) –1 –1 CP of products = 37.11 + 2 × 33.58 = 104.27 JK mol 802.32 T = × 103 + 298 = 7995.59 K 8000 K. f 104.23 Example 39 : Calculate the heat of formation of anhydrous aluminium chloride, Al2Cl6, from the following data. (i) 2Al(s) + 6HCl(aq.) Al2Cl6(g); 3H2(g) ; H =-240 kcal (ii) H2(g) + Cl2(g) 2HCl(g) ; H=-44.0 kcal (iii)HCl(g) + aq . HCl(aq.) ; H=-17.5 kcal (iv) Al2Cl6(s) + aq. Al2Cl6 (aq) ; H=-153.7 kcal Sol. The required equation is 2Al (s) + 3Cl2(g) Al2Cl6(s) ; H = ? For obtaining this , Multiply (ii) by 3, (iii) by 6 and add the resulting equations to (i) (i) 2Al (g) + 6HCl(aq.) Al2Cl6(aq.) + 3H2(g) ; H = 240.0 kcal Subtract (iv) from (v) and rearrange the product 2Al (s) + 3Cl2(g) + aq. A2Cl6(aq.) ; H = -477.0 kcal –Al2Cl6(s) - aq. –Al2Cl6(aq) ; H = +153.7 kcal 2Al (s) + 3Cl2(g) -– Al2Cl6(s) ; H = -323.3 kcal or 2Al(s) + 3Cl2(g) Al2Cl6 ; H=-323.3 kcal Hence the heat of formation of anhydrous aluminium chloride = – 323.3 kcal. Example 40 : 3 10 dm of O2 at 101.325 KPa and 298 K is heated to 348 K. Calculate the heat absorbed, H and E of this process at (a) at constant pressure and (b) at constant volume –1 –1 –6 2 Given : CP / JK mol = 25.72 + 0.013 (T/K) – 3.86 × 10 (T/K) Assume ideal behaviour. PV (101.325) (10) Sol. Amount of the gas, n = or n = = 0.409 mol. RT (8.314) (298)
36 Gyaan Sankalp Thermodynamics
T2 (a) Constant pressure, q = H = n dT p Cp T1
Here T2 = 348 K, T1 = 298 K TTTT2 2 3 3 (25.72)(T T) 0.0132 1 (3.86 106 ) 2 1 or q = n 2 1 p 2 2 3 3
or qp = 0.409 × 1475.775 = 603.59 J. E = H – (PV) = H – PV – VP at constant pressure E = H – PV = H – nRT = 603.59 – 0.409 × 8.314 × 50 = 433.57 J. (b) At constant volume
T2 T2 T2 n CP nR qV = E = n CV dT = dT – dT = 603.59 – 170.02 = 433.57 J T1 T1 T1 H = E + (PV) = u + nRT = 603.59 J. Example 41 : 2 mol of an ideal diatomic gas at 300 K and 0.507 MPa are expanded adiabatically to a final pressure of 0.203 MPa against a constant pressure of 0.101 MPa. Calculate the final temperature. Sol. For an adiabatic process q = 0 nRT nRT 2 1 w = – Pext (V2 – V1) = – Pext PP2 1 E = w TT 1 2 i.e., nCV (T2 – T1) = Pext nR PP1 2
3 R 5 Assuming no contribution from vibration C for a diatomic molecule = R + 2 = R V 2 2 2
Substituting the expression of Cv in the previous expression, we get
5 TT1 2 5 TT1 2 R n (T2 – T1) = Pext nR or (T2 – T1) = Pext 2 PP1 2 2 PP1 2 Substituting the values, we get
5 300 T2 (T – T ) = (0.101) – 2 2 1 0.507 0.203
Solving for T2 we get T2 = 270 K. Example 42 : (a) Nitric oxide (NO) reacts with molecular oxygen as follows : 2NO (g) + O2 (g) –––––2 NO2 (g) Initially NO and O2 are separated as shown below. When the valve is opened, the reaction quickly goes to completion. Determine what gases remains at the end of reaction and calculate their partial pressures. Assume that the temperature remains constant at 27°C. [Given : R = 0.08 atm L/mole-K] (b) Two mole of an ideal gas is expanded irreversibly and isothermally at 27°C until its volume doubled and 3.3 kJ heat is absorbed from
surrounding. Determine Ssystem and Ssurrounding . NO O2
0.5 4 2 1 Sol. (a) Initially n , n 2L at 1 atm NO 0.080 300 O2 0.080 300 4L at 0.5 atm & 300 K & 300 K
Gyaan Sankalp 37 Thermodynamics
2NO (g) + O2 (g) –––––2 NO2 (g) 2 2 Initial 0 24.0 24.0 1 2 1 at the end of reaction – 24 24 12
NO2 and O2 gases remains at the end of reactions. Vtotal = 6L 1 0.080 300 1 P.Pr of O atm or 0.166 atm 2 24 6 6 1 0.080 300 1 P.Pr of NO 0.333 atm or atm 2 12 6 3
V2 (b) Ssystem nRln 2 R ln2 11.52J/K V1 3.3 1000 S 11 J / K surrounding 300 Example 43 : Calculate the heat of formation of acetic acid from the following data:
(i) CH3COOH() + 2O2(g) 2CO2(g) +2H2O() ; H = –207.9 kcal
(ii) C(s) + O2(g) CO2(g) H = –94.48 kcal (iii)H (g) + 1 O (g) H O() H = –68.4 kcal 2 2 2 2 Sol. First method : The required equation is 2C(s) + 2H2(g) + O2(g) = CH3COOH(); H = ? This equation can be obtained by multiplying Eq. (ii) by 2 and also Eq. (iii) by 2 and adding both and finally substracting Eq.(i) [2C + 2O2 + 2H2 + O2 – CH3COOH() –2O2 = 2CO2 + 2H2O –2CO2 – 2H2O] H CH3 COOH( ) = 2 × (–94.48) + 2 (–68.4)– (–207.9) = –188.96 – 136.8 + 207.9 = –325.76 + 207.9 = –117.86 kcal Second method : From eq. (ii) and (iii) Enthalpy of CO2 = – 94.48 kcal, Enthalpy of H2O = – 68.4 kcal, Enthalpy of O2 = 0 (by convention) H of Eq. (i) = Enthalphies of products – Enthalpies of reactants H –207.9 = 2 × (–94.48) + 2(–68.4) – CH3 COOH( ) H CH3 COOH( ) = –188.96 – 136.8 + 207.9 = –235.76 + 207.9 = –117.86 kcal Example 44 : (a) 2 mole of an ideal gas undergoes is isothermal compression along three different paths (i) reversible compression from Pi = 2 bar and Vi = 4L to Pf = 20 bar (ii) a single stage compression against a constant external pressure of 20 bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until Pgas = Pext, followed by compression against a constant pressure of 20 bar until Pgas = Pext. Calculate the work (in bar L) for each of these processes and for which of the irreversible processes is the magnitude of the work greater ? [Given : R = 0.08 bar L/mole K] (b) A heating coil is immersed in a 100 g sample of H2O liquid at 1 atm and 100°C in a closed vessel as shown in figure. In this process, 30% of the liquid is converted to the gaseous form at constant pressure of 1 atm. A current of 2A flows through the heater from a 12V battery for 1000s to effect the transforma- tion. The densities of liquid & gaseous water under these conditions are coil 1000 kg/m³ 0.60 kg/m3 respectively. Calculate q and w (in kJ) for the process. [Use 1 atm = 105 Pa]
38 Gyaan Sankalp Thermodynamics Sol. (a) PV = nRT 2 × 4 = 2 × 0.080 × T ; T = 50 K
P1 1 (i) wrev 2.303 n R Tlog ; wrev 2.303 2 0.08 Tlog 18.424 bal L P2 10
nRT nRT 1 1 w(irr) P(V V) 20 208 72barL (ii) ext 2 1 P2 P 1 20 2
nRT nRT nRT nRT (iii) wirr(total) w 1 w 2 10 20 = nRT (5 – 1) + nRT (2 – 1) = 5 × nRT = 5 × 8 = 40 bar.L 10 2 20 10 Magnitude of work is maximum in single stage compression. (b) q = V × I × t = 12 × 2 × 1000 = 24000 J or 24 kJ
w Pext (V f V i )
3 3 3 5 30 10 70 10 100 10 w 10 = – 105 (50 × 10–3 + 0.07 × 10–3 – 0.1 × 10–3) 0.60 1000 1000 = – 105 (49.97 × 10–3) = – 4997 J = – 4.997 kJ Example 45 : A gas expands from a volume of 3.0 dm3 to 5.0 dm3 against a constant pressure of 3.0 atm. The work done during expansion is used to heat 10.0 mole of water of temperature 290.0 K. Calculate the final temperature of water . [ specific heat of water = 4.184 JK–1 g–1 ] Sol. Work done = P ×dV = 3.0 × (5.0 – 3.0) = 6.0 litre–atm = 6.0 × 101.3 J = 607.8 J Let T be the change in temperature. Heat absorbed = m × S × T = 10.0 × 18 × 4.184 × T Given P × dV = m × S × T P dV 607.8 or T = = 0.807 m S 10.0 18.0 4.184 Final temperature = 290 + 0.807 = 290.807 K
Example 46 : From the following data at 25ºC –1 Reaction r Hº/kJ mol 1 1 H (g) + O (g) OH (g) 42.09 2 2 2 2 1 H (g) + O (g) H O(g) –241.84 2 2 2 2 H2(g) 2H(g) 435.88 O2(g) 2O(g) 495.04 Calculate r Hº for the following reactions (a) OH(g) H(g) + O(g) (b) H2O(g) 2H(g) + O(g) (c) H2O(g) H(g) + OH(g) Sol. (a) The desired equation is OH(g) H(g) + O(g) We have to develop the desired equation by using the four given equations 1 435.88 (i) H (g) H(g) Hº = 2 2 r 2 1 1 (ii) OH(g) H (g) + O (g) Hº = – 42.09 2 2 2 2 r 1 495.04 (iii) O (g) O(g) Hº = 2 2 r 2
Gyaan Sankalp 39 Thermodynamics By adding equation (i), (ii) & (iii), we get OH(g) H(g) + O(g) 435.88 495.04 Hº = – 42.09 + or Hº = 423.37 KJ/mol. r 2 2 r
(b) The desired equation is H2O(g) 2H(g) + O(g) 1 495.04 (i) O (g) O(g) Hº = 2 2 r 2
(ii) H2(g) 2H(g) rHº = 435.88 1 (iii) H O(g) H (g) + O (g) Hº = 241.88 2 2 2 2 r The net equation is, H2O(g) 2H(g) + O(g) rH = 925.28 KJ/mol. (c) The desired equation is H2O(g) H(g) + OH(g) (i) H2O(g) 2H(g) + O(g) rHº = 925.28 (ii) H(g) + O(g) OH(g) rHº = 423.37 By adding equation (i) and (ii) H2O(g) H(g) + OH(g) rHº = 501.91
Example 47 : From the given data prove that graphite is more state allotrope of carbon than diamond from the given data Cdia + O2 CO2 Hº1 = – 94500 cal Cgra + O2 CO2 Hº2 = – 94050 cal Sol. C + O (g) CO (g) Hº = – 94500 cal dia 2 2 1 Diamond Cgra + O2(g) CO2(g) Hº2 = – 94050 cal on substraction H H = – 450 Cdia Cgra rHº = – 450 cal since heat content of diamond > graphite. Graphite Therefore diamond is less stable than graphite. Example 48 : You are given Avogadro’s number of “X” atoms. If half of the atoms of X transfer one electron to other half of ‘X’ atoms, 409 KJ must be added. If these X– ions are subsequently converted to X+, an addition 733 KJ must be added. Calculate IP and EA of X –19 23 in eV. Use (1 eV = 1.602 × 10 J and NA = 6.023 × 10 ) + Sol. X X + e H = IP1 = a eV – X + e X H = EA1 = – b eV
N N 409 103 2 409 103 a × A – b A = eV or a – b = × or a – b = 8.477 eV 2 2 1.602 1019 6.023 1023 1.602 1019 N Now A of X– lose two electrons to give X+ 2 – X X + e H = + EA1 = + b + X X + e H = + IP1 = + a
N N 733 103 2 733 103 a × A + b × A = eV or a + b = × 2 2 1.602 1019 6.023 1023 1.602 1019 or a + b = 15.194 eV on solving a = 11.835 eV b = 3.358 eV IP of X 11.835eV and EAof X 3.358eV
40 Gyaan Sankalp Thermodynamics Example 49 : An athlete takes 20 breaths per minute at room temperature. The air inhaled in each breath is 200 ml which contains 20% oxygen by volume, while exhaled air contains 10% oxygen by volume. Assuming that all the oxygen consumed if used for converting glucose into CO2 and H2O(), how much glucose will be burnt in the body in one hour and what is the heat produced (Room temperature = 27ºC and enthalpy of combustion of glucose is – 2822.5 kJ mol–1 at 0ºC) 200 20 200 10 Sol. Inhaled O in one breath = ml, Exhaled O in one breath = ml 2 100 2 100 200 20 200 10 used O in one breath = – = 20 ml 2 100 100 Now 20 breaths are taken in one minute 24,000 Breaths taken in 1 hour = 20 × 1200 = 24000 ml at 27ºC = × 273 ml at 0ºC = 21.84 lit. 300
C6H12O6 + 6O2 6CO2 + 6H2O H = – 2822.5 kJ 6 × 22.4 litre O2 is used for 180 gm glucose 180 21.84 21.84 litre O is used for = = 29.25 gm 2 6 22.4
Also 6 × 22.4 litre O2 with glucose gives = 2822.5 kJ 2822.5 21.84 21.84 litre O with glucose gives = = 458.66 kJ 2 6 22.4 Example 50 : How much heat will be required to make 2 kg of calcium carbide (CaC2) according to the following reaction ? CaO (s) + 3C(s) CaC2(s) + CO(g) The heats of formation of CaO (s), CaC2(s) and CO(g) are –151.6, –14.2 and –26.4 kcal respectively. Sol. H = H°f(products) – H°f(reactants) = [H°f(CaC2) + H°f(CO)] – [H°f(CaO) + 3H°f(C)] = [–14.2 –26.4] – [–151.6 + 3 × 0] = – 40.6 + 151.6 = 111.0 kcal For formation of 64 g of CaC2 111.0 ckal of heat is required. So, heat required for making 2000 g of 111.0 CaC = × 2000 = 3468.75 kcal 2 64 Example 51 : –1 The heat evolved on combustion of 1gm starch (C6H10O5)n into CO2 and H2O() is 17.49 kJ gm . Compute the enthalpy of –1 –1 formation of 1gm starch. Given Hf of H2O() = – 285.85 kJ mol , Hf of CO2 = – 293.7 kJ mole . 1 Sol. Given, H + O H O() H = – 285.85 kJ 2 2 2 2
C + O2 CO2() H = – 293.7 kJ Since 1 gm starch on combustion given 17.49 kJ 162 n gm starch on combustion gives 17.49 × 162 n = 2833.38 n kJ Since molecular weight of starch (C6H10O5)n = 162 n Therefore,
(C6H10O5)n + 6n O2 6n CO2 + 5n H2O H = – 2833.38 n kJ 5n 5n H + O + 6n C + 6n O 5n H O + 6n CO H = – 3791.45 n kJ 2 2 2 2 2 2
(C6H10O5)n + 6n O2 6n O2 + 5n H2O H = – 2833.38 n kJ 5n 6n + 5nH + O + (C H O ) H = – 958.07 n kJ 2 2 2 6 10 5 n 958.07 n Heat of formation of 1gm of starch = – = – 5.91 kJ 162n
Gyaan Sankalp 41 Thermodynamics Example 52 : The enthalpy of atomization of carbon is 171.4 k cal mol–1. Bond energy of H – H bond is 104.2 k cal mol–1. Bond energies of C –1 –1 –1 –1 – H : 98.8 k cal mol , C – C : 83.1 k cal mol , C = C : 147 k cal mol , Hf of benzene is 19.8 k cal mol . Calculate the resonance energy of benzene.
Sol. Hf of benzene, 6C(s) + 3H2 (g) C6H6(l) Enthalpy of formation of benzene = (Sum of bond energy of H2 + enthalpy of atomisation of C) – (total bond energy of benzene) (3 × 104.2 + 171.4 × 6) – (6 × 98.8 + 3 × 83.1 + 3 × 147) = 1341.0 – 1283.1 = 57.9 k.cal. Resonance energy = Hf, calculated – Hf, experimental = 57.9 – 19.8 = 38.1 k.cal. Example 53 : Find the bond enthalpy of S – S bond from the following data : –1 C2H5 – S – C2H5(g) fHº = – 147.23 kJ mol –1 C2H5 – S – S – C2H5(g) fHº = – 201.92 kJ mol –1 S(s) S(g) fHº = 222.80 kJ mol Sol. Given that –1 (i) 4C(s) + 5H2(g) + S(s) C2H5 – S – C2H5 fHº = – 147.23 kJ mol –1 (ii) 4C(s) + 5H2(g) + 2S(s) C2H5 – S – S – C2H5 fHº = – 201.92 kJ mol Substracting equation (i) from equation (ii), we get –1 C2H5 – S – C2H5(g) + S(s) C2H5 – S – S C2H5(g) fHº = – 54.69 kJ mol –1 S(g) S(s) fHº = – 222.80 kJ mol Adding the above two equations we get, –1 C2H5 – S – C2H5(g) + S(g) C2H5 – S – C2H5(g) fHº = – 277.49 kJ mol In the last equation 277.49 kJ of heat is evolved because of the S – S bond formation. Hence, bond enthalpy of S – S is 277.49 kJ mol–1. Example 54 : One litre sample of a mixture of methane gas and O2 measured at 25ºC and 740 torr. was allowed to reacts at constant pressure in a calorimetre which together with its content had a heat capacity of 1260 cal/degree. The complete combustion of methane to CO2 and H2O caused a temperature rise in calorimeter of 0.667 K. What was the percentage of CH4 in original mixture ? Given H –1 combustion of CH4 is – 210.8 kcal mole . 740 740 1 Sol. V = 1 lit. P = atm ; T = 298 K n = × = 3.985 × 10–2 moles 760 t 760 0.082 298 n –3 CH4 × 210.8 = 1200 × 0.667 × 10 n –3 CH4 = 3.796 × 10 moles
3.796 103 mole % of CH = × 100 = 9.525 4 3.985 102
percentage of CH4 in original mixture = 9.525%. Example 55: 0.138 gm sample of solid magnesium (molar mass = 24.3 gm mol–1) is burned in a constant volume bomb calorimeter that has a heat capacity of 1.77 kJ/ºC. The calorimeter contains 300 ml of water (density, 1gm/ml) and its temperature is raised by 1.126ºC. Calculate enthalpy of combustion of magnesium at 289 K. Sol. We have, qwater = mwater Cwater T or qwater = 300 × 1 × 4.184 × 1.126 = 1413 J = 1.413 kJ qbomb = (mbomb . Cbomb) T = 1.77 × 1.126 = 1.992 kJ qcomb = – (qwater + qbomb) = – (1.413 + 1.992) = – 3.405 kJ 3.405KJ qcomb –1 CE = = 0.138 = – 599.6 kJ mol msubs / M subs 24.3 The combustion reaction is Mg(s) + 1/2 O2(g) MgO(s) ng = – 1/2 –3 –1 CH = CE + (ng) RT = – 599.6 + (–1/2) (8.314 × 10 ) (298) = – 5.99.6 – 1.24 = – 600.84 kJ mol
42 Gyaan Sankalp Thermodynamics Example 56 : Calculate the lattice energy for the reaction : Li+ (g) + Cl–(g) LiCl (s) From the following data : H = 160.67 kJ mol–1; 1 (Cl ) = 122.17 kJ mol–1 sub(Li) 2 2 –1 –1 –1 I.P. (Li) = 520.07 kJ mol ; E.A. (Cl) = –365.26 kJ mol and Hºf (LiCl) = – 401.66 kJ mol Sol. Applying the equation, – Q = H + 1 D + I.P. – E.A. + U 2 And substituting the respectrive values, – 401.66 = 160.67 + 122.17 + 520.07 – 365.26 + U ; U = – 839.31 kJ mol–1 Example 57 : Calculate G0 of the reaction: 0 CH4(g) + 2O2(g) CO2(g) + 2H2O( ) at 25 C Using the following data : CH4(g) O2(g) CO2(g) H2O() S0 : 44.5 49.0 51.06 16.75 cal. deg–1 mol–1 H0 ; –17.9 0 –94 –68.3 k. cal. mol–1 Sol. CH4(g) + 2O2(g) CO2(g) + 2H2O() S0 = [51.06 + (2 x 16.75 )] – [44.5 + (2 x 49)] = – 57 .9 cal deg–1 mol–1 H0 = [–94.0 + (–2 x 68.3)] – [– 17.9 + 0] = – 212.7 k.cal At 250C, TS0 = – 298 x 57.9 = – 17254 cal G0 = H0 – TS0 = – 212.7 + 17.25 = – 195.45 k.cal. Example 58 : –1 –1 Bond dissociation enthalpies of H2(g) & N2(g) are 436.0 kJ mol and 941.8 kJ mol and enthalpy of formation of NH3(g) is –1 – 46 kJ mol . What is enthalpy of atomization of NH3(g)? What is the average bond enthalpy of N–H bond? Sol. N2(g) + 3H2(g) 2NH3(g) ; H = –2 × 46 kJ/mol H = S(B.E.)R – S(B.E.)P = (941.8 + 3 × 436) – (6x) = –2 × 46 kJ/mol (here x = B.E. of N–H bonds) –1 x = 380.3 kJ mol NH3 N + 3(H) Heat of automization = 3 × 390.3 = 1170.9 kJ mol–1 Example 59 : The mean molar heat capacities at constant pressure of H2(g), O2(g) and H2O(g) in the temperature range 25°C to 100°C are 6.92, 7.04 and 8.03 cal deg–1 mol–1. Calculate the standard heat of formation of water vapour at 1000C. Give standard heat of formation at 25°C is –57.8 k.cal. [– 57.98 k.cal mol–1] 1 Sol. H (g) + O (g) H O (g) at 1000C 2 2 2 2 For this reaction 1 Cp = Cp (H2O,g) – Cp (H 2 ,g) C p O 2 ,g 2 = 8.03 – (6.92 + 3.52) = – 2.41 cal deg–1 T2 – T1 = 100 – 25 = 75 For the above reaction at 25°C, H1 = – 57.8 k.cal
H2 – H1 = Cp (T2– T1 ) –3 H2 + 57.8 = – 2.41 × 10 × 75 –1 H2 = – 57.98 k. cal . mol Example 60 : One hundred and fifty grams of liquid benzene are vapourised at its boiling point 80ºC and 1 atm pressure. If the heat of vapourisation of the liquid at this temperature is 95 cal gm–1, calculate (a) E, (b) H, (c) Q, (d) W. Sol. The process is carried under thermodynamically reversible conditions. (b) H = QP = 95 × 150 = 14.25 kcal; (c) Q = 14.25 kcal 150 (d) W = P(V – V ) = PV = nRT = × 2 × 353 = 1.36 kcal V V 78 (a) E = Q – W = 14.25 – 1.36 = 12.89 kcal.
Gyaan Sankalp 43 Thermodynamics Example 61 : The enthalpy of neutralization of HNO3 by NaOH = – 13680 cal/eqt. When one equivalent of NaOH is added to a dilute solution containing one equivalent of HNO3 and one equivalent of a certain monoprotic weak acid, 13960 cals are evolved. Assume that the base is distributed between HNO3 and the weak acid in the ratio 3 : 1 and the weak acid is practically nonionized. Calculate the enthalpy of ionization of the weak acid. (A) – 1120 cals (B) – 2110 cals (C) – 1210 cals (D) + 1210 cals Sol. (A). Suppose the enthalpy of neutralization of the weak acid is – Q cal/eqt. 3 1 Then × (– 13680) + × (– Q) = – 13960 cal 4 4 41040 + Q = 55840 cals Q = 14800 cals Hneutralization = – 14800 cal – 14800 = DHionization – 13680 Hionization = (– 14800 + 13680) cal Hionization = – 1120 cals Example 62 : The enthalpies of formation of SO2(g), H2O(), HCl(g) and H2SO4() are – 70.97, – 68.32, – 22.1 and – 188.84 (all in kcal/mol). The enthalpies of solution of SO2, H2SO4 and HCl in dilute solution are –8.56, –22.05 and –17.63 (all in kcal/mol) respectively. Calculate the enthalpy change for the reaction. SO2 (aq) + Cl2 (g) + 2 H2O () = H2SO4 (aq) + 2 HCl (aq) aq Sol. SO2(g) SO2 (aq) H = – 8.56 kcal/mol – 70.97 x x + 70.97 = – 8.56 x = – 70.97 – 8.56 = – 79.53 kcal/mol aq H2SO4 (liq) H2SO4 (aq) H = – 22.05 – 188.84 y y + 188.84 = – 22.05 y = – 188.84 – 22.05 = – 210.89 kcal/mol HCl (g) aq HCl (aq) H = – 17.63 – 22.1 z z + 22.1 = – 17.63 z = – 22.1 – 17.63 = – 39.73 kcal SO2 (aq) + Cl2 (g) + 2H2O () = H2SO4 (aq) + 2HCl (aq) – 79.53 0 2 × (– 68.32) – 210.89 2 × (– 39.73) H = (– 79.46 – 210.89) – (– 2 × 68.32 – 79.53) = – 74.18 kcal Example 63 :
(a) Given that C2H6 (g) + H2(g) 2 CH4 (g), H = – 15.6 kcal and C3H8 (g) + 2 H2(g) 3 CH4 (g), H = – 20.9 kcal Calculate H for CH4 (g) + C3H8 (g) 2 C2 H6 (g). (The values given are called HTC i.e. heat of total cracking.) (b) The enthalpies of neutralization of an organic acid (HX) and HCl are –13.350 kcal/eqt and –13.700 kcal/eqt. When one mole of NaX in solution is treated with one mole of HCl there is an evolution of 280 calories. What conclusion may be drawn regarding the relative strengths of HX and HCl ?
Sol. (a) C3H8 (g) + 2H2 (g) 3CH4 (g); H = – 20.9 kcal 4CH4 (g) 2H2 (g) + 2C2H6 (g) ; H = 2 × 15.6 kcal ––––––––––––––––––––––––––––––––––––
C3 H 8 (g) CH 4 (g) 2C 2 H 6 (g) H = 10.3 kcal
(b) HX + NaOH NaX + H2O H = – 13.350 k cal/eqt
HCl + NaOH NaCl + H2O H = – 13.700 k cal/eqt Reverse the first equation and add to the second NaX + HCl NaCl + HX; H = – 350 cal
44 Gyaan Sankalp Thermodynamics But this is equivalent to one mole of NaX getting converted into HX or one mole of X– in forming HX. But since only 280 calories 280 are evolved, 0.8 mole of X– was converted into HX. 350 This implies that X– + H+ HX proceeds partially (upto 80%) because of the ionization of HX. Example 64 : The enthalpies of formation of PCl5, HCl(aq) and H2O are –105 kcal/mol, –39.3 kcal/mol and –68.4 kcal/mol respectively. Given aq that H for the reaction PCl5 + 4 H2O H3PO4(aq) + 5 HCl (aq) is –123.4 kcal/mol. Check up whether the given data are self-consistent. Enthalpy of formation of H3PO4 (aq) = –305 kcal/mol. ad Sol. PCl5 + 4H2O H3PO4 (aq) + 5HCl (aq) H = – 123.4 – 105 4 × (– 68.4) – 305 5 × (– 39.3) H = [– 196.5 – 305 + 105 + 273.6] kcal = – 122.9 kcal. This agrees closely with the given value – 123.4 kcal. the data given are self-consistent. Example 65 : Five moles of oxygen gas at 25ºC expand adiabatically and reversibly from 100 atm to 10 atm. Calculate (a) W, (b) E, (c) H, (d) –1 –1 Final temperature of the gas (Cv for O2 gas = 5.0 cal deg mol ). 1 T2 P CP 7 Sol. = 2 where = = T C 1 P1 V 5
2 / 7 T2 10 = (d) T = 154.3 K = – 118.7ºC 298 100 2
(b) E = nCv (T2 – T1) = – 5 × 5 × 143.7 = – 35.925 kcal (a) E = – W; W = – E = 35.925 kcal (c) H = nCp (T2 – T1) = – 5 × 7 × 143.7 = – 50.295 kcal Example 66 : Calculate the standard internal energy change for the following reaction at 25ºC. 2H2O2() 2H2O () + O2(g) –1 –1 Hfº at 25ºC for H2O2() = –188 kJ mol , H2O () = –286 kJ mol Sol. Hº = Hº (product) – Hº(reactants) = 2(–286) + 0 – 2 (–188) = – 572 + 376 = –196 kJ n(g) = 1 – 0 = 1 Hº = Eº + n(g)RT 8.314 298 E 196 kJ 198.47kJ 1000 Example 67 : –1 The heat of formation, Hf of N2O is 82 kJ mol . The bond energies of N N, O = O and N= N, N = 0 are 946, 498, 418, 607 kJ respectively. Calculate the resonance energy of N2O. Sol. The heat of formation may be calculated from bond energy data 1 N (g) + O (g) N O (g) 2 2 2 2
Hf = B.E of reactants – B.E of products 1 = 946 x498 – (418 + 607) = 1195 – 1025 = 170 KJ 2 –1 Resonance energy = Hf (Calculated) – Hf (expected) = 170 – 82 = 88 KJ. mol
Gyaan Sankalp 45 Thermodynamics Example 68 : 1.0 mole of monatomic perfect gas, initially at 1.0 atm. pressure and at 300 K is heated reversibly to 400 K at constant volume. Calculate the change in internal energy and the work done. 3 Sol. For a monoatomic perfect gas C = R v 2 The change in internal energy , E = Cv T 3 E = x 8.314 x 100 = 1247.1 J = 1.247 KJ 2 Since the volume remains constant, w = 0 Example 69 : 0 –1 –1 –1 For sublimation of I2(s)at 25 C and 1 atm, pressure H = 9.41 k. cal mol and the change in entropy S = 20.6 cal deg . mol . At what temperature, solid l2 will be in equilibrium with iodine vapour ? Sol. G0 = H0 – TS0 At equilibrium, G0 = 0 H0 = TS0 H0 9.41x103 T = = = 456.8 K = 183.80C S0 20.6 Example 70 : –1 –1 Enthalpy of neutralization of HCl by NaOH is – 57.32 kJ mol and by NH4OH is –51.34 kJ mol . Calculate the enthalpy of dissociation of NH4OH. Sol. Given that + + –1 H (aq) + NH4OH (aq) NH4 (aq) + H2O() rH = – 51.34 kJ mol We may consider neutralization in two steps : + – (i) Ionization : NH4OH(aq) NH4 (aq) + OH (aq) fH1 ? + – –1 (ii) Neutralization : H (aq) + OH (aq) H2O(l) rH2 = – 57.32 kJ mol Thus, rH = rH1 + rH2 –1 Therefore, rH1 = rH – rH2 = – 51.34 + 57.32 = 5.98 KJ/mol Example 71 : C(s) + O2(g) CO2(g) + 394 kJ C(s) + 1/2 O2(g) CO + 111 kJ (a) In an oven using coal (assume the coal is 80% carbon in weight), insufficient oxygen is supplied such that 60% of carbon is converted to O2 and 40% carbon is converted to CO. Find out the heat generated when 10 kg of coal is burnt in this fashion. (b) Calculate the heat generated if a more efficient oven is used such that only CO2 is formed. (c) Calculate the percentage loss in heating value for the inefficient oven. Sol. (a) C(s) + O2(g) CO2(g) + 394 kJ ...... (i) C(s) + 1/2 O2(g) CO + 111 kJ ...... (ii) Weight of C in 10 kg coal = 10000 × 0.8 = 8000 g Weight of C converted into CO2 = 8000 × 0.6 = 4800 g weight of C converted into CO = 8000 × 0.4 = 3200 g 12 g (1 mole) C on conversion into CO2 liberates = 394 kJ 4800 g of c on conversion into CO2 liberates = 153,600 kJ 12 g(1 mole) C on conversion into CO liberates = 111 kJ 111 3200 3200 g of c on conversion into CO liberates = = 29600 kJ 12 Total heat liberated = 153600 + 29600 kJ = 183200 kJ (b) C(s) + O2(g) CO2(g) + 294 kJ 12 g carbon liberates heat = 394 kJ 294 8000 8000 g of carbon liberates heat = = 262666.67 kJ 12
46 Gyaan Sankalp Thermodynamics (c) Heat lost by oven = 26266.67 – 183200 = 79466.67 kJ 79466.67 100 % lost of heat = = 32.25% 262666.67 Example 72 : At 25ºC, 1 mole MgSO4 was dissolved in water. The heat evolved was found to be 91.211 kJ. One mole of MgSO4. 7H2O on dissolution gives a solution of the same composition accompanied by an absorption of 13.807 kJ. Find the enthalpy of hydration, i.e., H for the reaction : MgSO4(s) + 7H2O() MgSO4.7H2O(s) –1 Given that (i) MgSO4(s) + aq MgSO4(aq) rH1 = – 91.211 kJ mol –1 (ii) MgSO4.7H2O(s) + aq MgSO4(aq) rH2 = + 13.807 kJ mol Sol. Equation (i) can be written as follows : –1 (iii) MgSO4(s) + 7H2O() + aq MgSO4(aq) rH = – 91.211 kJ mol equ (iii) – (ii) will give –1 MgSO4(s) + 7H2O() MgSO4.7H2O(s) rH = – 91.211 – 13.807 = – 105.018 kJ mol Example 73 : (a) Calculate Gº and k for the reaction, NO + O3 NO2 + O2 Given Gºf (NO2) = 12.39 kcal, Gºf(NO) = 20.7 kcal, Gºf(O3) = 39.0 kcal. (b) Is the size of the equilibrium constant principally a consequence of the Hº or of the Sº for this reaction ? Given Hºf(NO2) = 8.09 kcal, Hºf (NO) = 21.6 kcal, Hºf(O3) = 34.0 kcal. Sol. G = Gºf (NO2) + Gºf(O2) – Gºf(NO) – Gºf (O3) = 12.39 + 0 – 20.7 – 39.0 = – 47.3 kcal k = 10Gº/2.3RT = 10(47,300)/1360 = 5 × 1034 (using Gº = – RT ln k)
S0 /R Hº/RT k = e Hº = Hºf(NO2) + Hºf(O2) – Hºf(NO) – Hºf(O3) = 8.09 + 0 – 21.6 – 34.0 = – 47.5 kcal Sº = Sº (NO2) + Sº (O2) – Sº (NO) – Sº (O3) = 57.5 + 49.0 – 50.3 – 56.8 = – 0.6 cal/deg k = 10Sº/2.3R × 10– Hº/2.3RT k = 10–0.13 × 10+34.8 k = 4.67 × 1034 The entropy change for the reaction is very small, since geometrical molecular structures of products and reactants are similar. The real ‘driving force’ is that products are energetically more stable than the reactants. Example 74 : Calculate the electron affinity of the hydrogen atom using the following enthalpy data –1 H2(g) 2H(g) ; H = +436 kJ mol –1 H2(g) + 2K(s) 2KH(s) ; H = –118 kJ mol K(s) K(g) ; H = +83 kJ mol–1 K(g) K+(g) + e– ; H = +413 kJ mol–1 H–(g) + K+(g) KH(s) ; H = –742 kJ mol–1 Sol. It can be solved by forming the Born Haber’s cycle
–118 kJ = (+83 ×2) + (413 ×2) + (436) + (–2x) + (–2 ×742) 2x = 62 kJ mol–1 x = 31 kJ mol–1
Gyaan Sankalp 47 Thermodynamics QUESTION BANK
EXERCISE - 1 ONLY ONE OPTION IS CORRECT Q.11 Using only the following data:
Q.1 The heat of formation of HCl(g) from the reaction (I) Fe2O3(s) + 3CO (g) 2Fe(s) + 3CO2 (g) ; H° = – 26.8 kJ H2(g) + Cl2(g) ; H = – 44 kcal is (A) + 44 kcal (B) – 44kcal (II) Fe(s) + CO(g) FeO(s) + CO(g) ; (C) + 22 kcal (D) – 22 kcal Hº = + 16.5 kJ Q.2 Among mass, volume, density and specific volume of a the H° value, in kilojoules, for the reaction gas, the intensive properties are Fe2O3(s) + CO(g) 2FeO(s) + CO2(g) is calculated (A) Density and specific volume to be: (B) Volume and density (A) –43.3 (B) –10.3 (C) Specific volume and mass (C) +6.2 (D) +10.3 (D) Density only Q.12 82 litres of carbon dioxide are produced at a pressure of 1 Q.3 If a chemical change is brought about by one or more atm by the action of acid on a metal carbonate. The work methods in one or more steps, then the amount of heat done by the gas (in calories) in pushing back the absorbed or evolved during the complete course of reaction atmosphere is (R = 0.082 litre-atm deg–1 mol–1) is same, which ever method was followed. This law is known (A) 1000 (B) 820 as – (C) 1640 (D) 2000 (A) Le Chatelier’s principle (B) Hess’s law Q.13 Among the following, the reaction for which H = E is- (C) Joule Thomson effect (D) Trouton’s law (A) PCl5(g) PCl3(g) + Cl2(g) Q.4 If 1.00 kcal of heat is added to 1.2 L of oxygen in a cylinder (B) H2(g) + Cl2(g) 2HCl(g) at constant pressure of 1.000 atm, the volume increases to (C) C2H5OH( ) + 3O2(g) 2CO2(g) + 3H2O( ) 1.5 L, Hence E for this process is: (D) C2H4(g) + H2(g) C2H6(g) (A) 0.993 kcal (B) 1.0073 kcal Q.14 For which change HE (C) 0.0993 kcal (D) 1.00073 kcal (A) H + I 2HI Q.5 Heats of combustion of CH , C H , C H are – 890, – 1411 2 2 4 2 4 2 6 and – 1560 kJ/mole respectively. Which has the lowest fuel (B) HCl + NaOH NaCl + H2O value in kJ/gm ? (C) C(s) + O2(g) CO2(g) (A) CH (B) C H 4 2 4 (D) N + 3H 2NH (C) C H (D) All same 2 2 3 2 6 Q.15 Enthalpy change when 1.00 g water is frozen at 0ºC, is : Q.6 The standard heat of combustion of a hydrocarbon com- –1 pound is an/a- (Hfus = 1.435 kcal mol ) (A) Extensive property (B) Colligative property (A) 0.0797 kcal (B) –0.0797 kcal (C) Intensive property (D) Constitutive property (C) 1.435 kcal (D) – 1.435 kcal Q.7 The Kirchhoff’s equation gives the effect of ..... on heat of Q.16 One mole of hydrogen gas at 25ºC and 1 atm pressure is heated at constant pressure until its volume has doubled. reaction. –1 –1 (A) Pressure (B) Temperature Given that Cv for hydrogen is 3.0 cal deg mol , the H (C) Volume (D) Molecularity and E for this process are- Q.8 The temperature of 5 ml of a strong acid increases by 5º (A) H = 1490 cal and E = 894 cal when 5 ml of a strong base is added to it. If 10 ml of each is (B) H = E = 1490 cal mixed, temperature should increase by- (C) H = 894 cal and E = 1490 cal (A) 5º (B) 10º (D) H = E = 894 cal (C) 15º (D) cannot be known Q.17 The enthalpy change of the reaction Q.9 The heat of neutralisation of HCl by NaOH is – 55.9 kJ/ Al2Cl6(s) + 6Na(s) 2Al(s) + 6NaCl(s) is – 257 kcal. mole. If the heat of neutralization of HCN by NaOH is –12.1 Given the heat of formation of NaCl(s) is –98.0 kcal, the kJ/mole, then energy of dissociation of HCN is- heat of formation of Al2Cl6(s) is- (A) – 43.8 kJ (B) 43.8 kJ (A) 158 kcal (B) –331 kcal (C) 68 kJ (D) – 68 kJ (D) –166 kcal (D) 316 kcal Q.10 For a reaction at equilibrium- Q.18 Some of the thermodynamic parameters are state variables (A) G = Gº = 0 (B) G = 0 but not Gº while some are process variables. Some grouping of the (C) Gº = 0 but not G (D) G = Gº 0 parameters are given. Choose the correct one – (A) State variables : Temperature, No. of moles Process variables : Internal energy, work done by the gas
48 Gyaan Sankalp Thermodynamics (B) State variables : Volume, Temperature (C) H > H Process variables : Internal energy, work done by the gas Pr oducts Reac tan ts (C) State variables : work done by the gas, heat rejected by (D) H = 0 but H is positive. the gas. Pr oducts Reac tan ts Process variables : Temperature, volume Q.26 All the natural process in this universe produce (D) State variables : Internal energy, volume (A) A decrease in entropy of the universe Process variables : Work done by the gas, heat absorbed (B) An increase in entropy of the universe by the gas (C) No change in entropy Q.19 One mole of a real gas is subjected to heating at constant (D) Sometimes increase or sometimes decrease in entropy volume from (P1, V1, T1) state to (P2, V1, T2) state. Then it Q.27 For the reaction, 2N2(g) + O2(g) 2N2O, at 298K H is is subjected to irreversible adiabatic compression against 164 KJ mol–1. The E of the reaction is –1 –1 constant external pressure of P3 atm till system reaches (A) 166.5 kJ mol (B) 161.5 kJmol –1 –1 the final state (P3, V3, T3). If the constant volume molar (C) 164.0 kJmol (D) 169 kJ mol –1 –1 –1 heat capacity of real gas is CV. Find out correct expression Q.28 For A B , H = 4 kcal mol ,S = 10 cal mol K . for H from state 1 to state 3 – Reaction is spontaneous when temperature can be : (A) 400K (B) 300K (A) C(TT)(PVPV)v 3 1 3 1 1 1 (C) 500K (D) none is correct (B) Cv (T 2 T 1 ) (P 3 V 2 P 1 V 1 ) Q.29 Which of the following statement is true in the case of Hess’s law ? (C) C(TT)(PVPV)v 2 1 3 1 1 1 (A) It is not relavant to first law of thermodynamics (D) C(Tp 2 T) 1 (PV 3 1 PV) 1 1 (B) Enthalpy change at constant pressure is dependent only on the thermodynamic state and not on the path. 1 (C) It is not related to enthalpy changes associated with H (g) O(g) HO(g); H 241.8kJ 22 2 2 chemical reactions. Q.20 1 (D) Heat of a reaction is dependent on the path. CO(g) O2 (g) CO(g); 2 H 283kJ 2 Q.30 44.0 kJ of heat is required to evaporate one mole of water at –1 0 The heat evolved in the combustion of 112 litres of water 298 K. If Hf of H2O() is –286 kJ mol , H f of H2O(g) gas (mix of equal volumes of H and CO) is 2 –1 –1 (A) 241.8 kJ (B) 283 kJ (A) – 330 kJ mol (B) + 242 kJ mol –1 –1 (C) 1312 kJ (D) 1586 kJ (C) –242 kJ mol (D) –198 kJ mol Q.21 In which case of mixing of a strong acid and a base, each of Q.31 The molar heat capacity, Cv of an ideal gas whose energy 1(N) concentration, temperature-increase is the highest ? is that of translational motion only is –1 –1 –1 –1 (A) 20 ml acid and 30 ml alkali (A) 2.98 J deg mol (B) 12.47J deg mol –1 –1 –1 –1 (B) 10 ml acid and 40 ml alkali (C) 6.43 J deg mol (D) 9.41 J deg mol –1 (C) 25 ml acid and 25 ml alkali Q.32 The lattic energy of NaCl is –780 kJ mol The enthalpies + – –1 (D) 35 ml acid and 15 ml alkali of hydration of Na (g) and Cl (g) ions are – 406 kJ mol –1 Q.22 The heats of neutralisation of four acids A, B, C , D are and –364 kJ mol . The enthalpy of solution of NaCl(s) is –1 –1 – 13.7, – 9.4, – 11.2 and – 12.4 kcal respectively when they (A) 738 kJ mol (B) 10 kJ mol –1 –1 are neutralised by a common base. The acidic character (C) – 10 kJ mol (D) – 822 kJ mol –1 obeys the order : Q.33 Enthalpy of fusion of a liquid is 1.435 kcal mol and –1 –1 (A) A > B > C > D (B) A > D > C > B molar entropy change is 5.26 cal mol K . Hence melting (C) D > C > B > A (D) D > B > C > A point of liquid is : (A) 100ºC (B) 0ºC Q.23 The Hf for CO2(g), CO(g), and H2O(g) are – 393.5, (C) 373 K (D) –273ºC –110.5 & – 241.8 kJ mol–1 respectively. The standard Q.34 Following reaction occurs at 25ºC : –5 –2 enthalpy change (in kJ) for the reaction , 2NO(g, 1 × 10 atm) + Cl2(g, 1 × 10 atm) –2 CO2(g) + H2(g) H2O(g) is : 2NOCl(g, 1 × 10 atm) Gº is- (A) 524.21 (B) 41.2 (A) –45.65 kJ (B) – 28.53 kJ (C) – 262.5 (D) – 41.2 (C) – 22.82 kJ (D) – 57.06 kJ Q.24 The dissociation energy of CH4 and C2H6 are respectively Q.35 When 1 mole of an ideal gas to 20 atm pressure and 15 L 360 and 620 Kcal/mole. The bond energy of C – C is- volume expands such that the final pressure becomes 10 (A) 260 Kcal/mole (B) 180 Kcal/mole atm and the final volume become 60 L. Calculate entropy (C) 130 Kcal/mole (D) 80 Kcal/mole change for the reaction (Cp.m = 30.96) Q.25 For an endothermic reaction - (A) 80.2 J.k–1 mol–1 (B) 62.42 kJ. k–1 mol–1 2 –1 –1 –1 –1 H H H H (C) 120 × 10 Jk mol (D) 27.22 J.k mol (A) Pr oducts = Reac tan ts (B) Pr oducts < Reac tan ts
Gyaan Sankalp 49 Thermodynamics
Q.36 If a process is both endothermic and spontaneous , then : d ( G) (A) S > 0 (B) S < 0 Q.46 If G = H –TS and G = H + T dT then (C) H < 0 (D) G > 0 P Q.37 The bond energies of C = C and C – C at 298 K are 590 and variation of EMF of a cell E, with temperature T, is given by –1 331 kJ.mol respectively. The enthalpy of polymerization H G (A) (B) per mole of ethylene is nF nF (A) – 70 kJ (B) – 72 kJ (C) 72 kJ (D) – 68 kJ S S (C) (D) – Q.38 1 mole of NH3 gas at 27ºC is expanded adiabatic condition nF nF to make volume 8 times (= 1.33). Final temperature and Q.47 The standard heat of combustion of Al is –837.8 kJ mol–1 work done respectively are- at 25ºC which of the following releases 250 kcal of heat ? (A) 150 K, 900 cal (B) 150 K, 400 cal (A) The reaction of 0.624 mol of Al (C) 250 K, 1000 cal (D) 200 K, 800 cal (B) The formation of 0.624 mol of Al2O3 Q.39 Temperature of 1 mol of a gas is increased by 1º at constant (C) The reaction of 0.312 mol of Al pressure. Work done is- (D) The formation of 0.150 mol of Al2O3 (A) R (B) 2R Q.48 CP – Cv = R . This R is : (C) R/2 (D) 3R (A) Change in K.E. Q.40 When 0.16 g of glucose was burnt in a bomb calorimeter, (B) Change in rotational energy the temperature rose by 4 deg. Calculate the calorimeter (C) work done which system can do on expanding the constant (water equivalent of the calorimeter) given that gas per mol per degree increase in temperature Hº = – 2.8 × 106 J mol–1. [molar enthalpy of combustion]. (D) All correct Molar mass of glucose = 180 mol–1. Q.49 Following reaction occurs at 25ºC : (A) 5.73 × 102 J/deg (B) 7.53 × 102 J/deg –5 –2 2NO(g, 1 × 10 atm) + Cl2(g, l × 10 atm) (C) 6.22 × 102 J/deg (D) 3.57 × 102 J/deg 2NOCl(g, l × 10–2 atm) Q.41 The C–Cl bond energy can be calculated from: ºG is: (A) H° (CCl , ) only (B) º (CCl , ) and (Cl ) f 4 f 4 2 (A) –45.65 kJ (B) –28.53 k J (C) H° (CCl , ) (Cl ) f 4 2 (C) –22.82 kJ (D) –57.06 kJ (D) H° (CCl , ) (Cl ), H° (C, g) and H° (CCl ) f 4 2 f vap 4 Q.50 Heat of neutralisation of oxalic acid is – 25.4 Kcal mol–1 –1 Q.42 Given Hf of DyCl3 (s) = – 994.30 kJ mol using strong base, NaOH. Hence enthalpy change of the process is H C O 2H+ + C H 2– is- 1 1 aq 2 2 4 2 4 H (g) + Cl (g) HCl (aq. 4 M); (A) 2.0 kcal (B) – 11.8 kcal 2 2 2 2 Hº = – 158.31 kJ mol–1 (C) 1.0 kcal (D) – 1.0 kcal aq Q.51 5 moles of nitrogen gas are enclosed in an adiabatic cylin- DyCl (s) DyCl (aq. in 4 M HCl); 3 HCl 3 drical vessel. The piston itself is a rigid light cylindrical Hº = – 180.06 kJ mol–1 container containing 3 moles of Helium gas. There is a 3 HCl 3 heater which gives out a power 100 cal to the nitrogen gas. Dy (s) DyCl (aq. 4 M HCl) + H (g); A power of 30 cal is transferred to Helium through the 3 2 2 aq.4 M bottom surface of the piston. Hº = x, calculate x. The rate of increment of (A) – 966.5 kJ/mol (B) – 699.43 kJ/mol temperature of the nitrogen gas assuming He (C) – 596.6 kJ/mol (D) – 569.6 kJ/mol that the piston moves slowly : Q.43 1 g H gas at S.T.P. is expanded so that volume is doubled. N2 2 (A) 2K/sec (B) 4K/sec Hence work done is: (C) 6K/sec (D) 8K/sec (A) 22.4 L atm (B) 5.6 L atm (C) 11.2 L atm (D) 44.8 L atm Q.52 During the isothermal mixing of ideal gases at pressure, p, Q.44 H for the reaction 2C (s) + 3H (g) C H (g) is – 20.24 2 2 6 the entropy change per mole for the mixing process is – R kcal/mol. To what value of the enthalpy of sublimation of x n x where x , x , ...... , x are the mole fractions of the C(s) does this point given that the bond energies of C – C, i i 1 2 i components, 1, 2, ...... , i of the mixture. Assuming ideal C – H and H – H are 63 kcal/mol, 85.6 kcal/mol and 102.6 gas behavior, calculate S for the mixing of 0.8 mole of N kcal/mol. 2 and 0.2 mole of O . (at 25ºC and 0.9 atm) [1 eu = cal/deg] (A) 124.3 kcal/mol (B) 185.3 kcal/mol 2 (A) 0.9943 eu (B) 0.7533 eu (C) 158.3 kcal/mol (D) 211.5 kcal/mol (C) 0.6798 eu (D) 0.7112 eu Q.45 The gas absorbs 100 J heat and is simultaneously com- Q.53 The standard heat of combustion of solid boron is equal pressed by a constant external pressure of 1.50 atm from 8 to: lit. to 2 lit. in volume. Hence E will be- (A) – 812 J (B) 812 J 1 (A) Hº (B O ) (B) Hº (B O ) (C) 1011 J (D) 911 J f 2 3 2 f 2 3 50 Gyaan Sankalp Thermodynamics (A) – 150.6 kJ.mol–1 (B) – 109.8 kJ mol–1 1 –1 –1 (C) 2H°f (B2O3) (D) – Hºf (B2OP3) (C) – 130.2 kJ mol (D) – 141.8 kJ mol 2 Q.63 In an adiabatic expansion the product of pressure and vol- Q.54 An ideal gas can be expanded from an initial state to a ume – certain volume through two different processes, (A) decreases (B) increases 2 2 (A) PV = K and (B) P = KV , where K is a positive (C) remains constant constant. Then, choose the correct option from the (D) first increase then decreases following. Q.64 The molar entropy of vaporisation of acetic acid is (A) Final temperature in (A) will be greater than in (B) 14.4 cal K–1 mol–1 at its boiling point 118ºC. The latent (B) Final temperature in (B) will be greater than in (A) heat of vaporisation of acetic acid is (C) Work done by the gas in both the processes would be (A) 49 cal g–1 (B) 64 cal g–1 equal (C) 94 cal g–1 (D) 84 cal g–1 (D) Total heat given to the gas in (A) is greater than in (B) Q.65 100 ml of 0.3 M HCl solution is mixed with 100 ml of 0.35 M Q.55 For the reaction : X2O4 () 2XO2(g) NaOH solution. The amount of heat liberated is E = 2.1 kcal. S = 20 cal/K at 300 K. Hence G is : (A) 7.3 kJ (B) 5.71 kJ (A) 2.7 kcal (B) –2.7 kcal (C) 10.42 kJ (D) 1.713 kJ (C) 9.3 kcal (D) –9.3 kcal Q.66 Five moles of ideal gas expand isothermally and reversibly Q.56 Heat of hydrogenation of ethene is x1 and that of benzene from an initial pressure of 100 atm to a final pressure of is x2. Hence resonance energy is- 1 atm at 27°C. The wrok done by the gas is (A) x1 – x2 (B) x1 + x2 (A) 3455 cal (B) 6909 cal (C) 3x1 – x2 (D) x1 – 3x2 (C) 0 (D) 69010 cal Q.57 Which of the following statement is true if the reaction Q.67 The quantity q i.e. heat absorbed an infinitesimal process quotient, Q is equal to 1 ? is (A) G = 0 (B) Gº = 0 (A) Dependent on the path of transformation (C) Sº = 0 (D) G = Gº (B) Dependent on the thermodynamic state of the system Q.58 C2H6(g) + 3.5 O2(g) 2CO2(g) + 3H2O(g) (C) Independent of both (D) An exact differential –1 Svap (H2O, ) = x1 cal K (b.p. + T1) Q.68 An ideal gas absorbs 2000 cals of heat from a heat reser- Hf (H2O, ) = x2 ; Hf (CO2) = x3, Hf(C2H6) = x4 voir and does mechanical workequivalent 4200J. The change Hence H for the reaction is- in internal energy of the gas is (A) 2x3 + 3x2 – x4 (B) 2x3 + 3x2 – x4 + 3x1T1 (A) 3000 cals (B) 2000 cals (C) 2x3 + 3x2 – x4 – 3x1 T1 (D) x1 T1 + x2 + x3 – x4 (C) 1500 cals (D) 1000 cals Q.59 The bond energies of C C, C – H, H – H and C = C are 198, Q.69 For an ideal gas four 98, 103, 145 kcal respectively. The enthapy change of the processes are marked as reaction HC CH + H2 C2H4 is 1, 2, 3 and 4 on P-V P (A) 48 kcal (B) 96 kcal diagram as shown in (C) –40 kcal (D) –152 kcal figure. The amount of heat 1 supplied to the gas in the Q.60 H2(g) + O2(g) H2O() process 1, 2, 3 and 4 are 2 A B Q1, Q2, Q3 and Q4 B.E. (H – H) = x1; B.E. (O = O) = x2 B.E. (O – H) = x3 respectively, then correct E D 700K Latent heat of vaporization of water liquid into water vapour 600K = x , then H (heat of formation of liquid water) is- order of heat supplied to T 4 f the gas is – x2 x2 [AB is process-1, AC is V (A) x + – x + x (B) 2x – x – – x 1 2 3 4 3 1 2 4 process-2, AD is adiabatic process-3 and AE is process-4] (A) Q > Q > Q > Q (B) Q > Q > Q > Q x x 1 2 3 4 1 2 4 3 2 2 (C) Q > Q > Q > Q (D) Q < Q < Q < Q (C) x1 + – 2x3 – x4 (D) x1 + – 2x3 + x4 1 4 2 3 1 2 3 4 2 2 Q.70 The latent heat of vaporiztion of water at 250 C is 10.5 kcal Q.61 In a process the pressure of a gas is inversely proportional mol–1 and the standard heat of formation of liquid water is to the square of the volume. If temperature of the gas is –68.3 kcal. The enthalpy change of the reaction increases, then work done by the gas – H2(g) + (1/2) O2(g) H2O(g) is therefore, (A) is positive (B) is negative (A) –57.8 kcal (B) –78.8 kcal (C) is zero (D) may be positive (C) 78.8 kcal (D) –47.3 kcal Q.62 The enthalpies of formation of CO2(g) and CO(g) at 298 K Q.71 Which of the following thermodynamic quantities is an are in the ratio 2.57 : 1. For the reaction, outcome of the second low of thermodynmics ? (A) Work (B) Enthalpy CO2(g) + C(s) CO (g), H = 172.5 kJ, (C) Internal energy (D) Entropy Hf of CO(g) is
Gyaan Sankalp 51 Thermodynamics
Q.72 From the following data of H, of the following reactions, Sº(NO) = 50.3 cal/deg, Sº (O3) = 56.8 cal/deg. (A) – 10.12 kJ (B) – 28.32 kJ 1 C(s) + O2(g) CO(g) ; H = – 110 kJ (C) – 18.72 kJ (D) – 8.21 kJ 2 Q.78 The heat of combustion of ethylene at 17ºC and at con- C(s) + H2O(g) CO(g) + H2(g) ; H = 132 kJ stant volume is –332.19 kcals. What is the value at con- What is the mole composition of the mixture of steam and stant pressure, given that water is in liquid state ? oxygen on being passed over coke at 1273 K, keeping tem- (A) – 131.25 k cals (B) – 412.23 k cals perature constant. (C) – 534.12 k cals (D) – 333.35 k cals (A) 0.5 : 1 (B) 0.6 : 1 Q.79 The enthalpies of the following reactions are shown (C) 0.8 : 1 (D) 1 : 1 alongwith. Q.73 State which of the following statements is true ? 1 1 (A) First law of thermodynamics is not adequate in pre- H (g) + O (g) OH(g) ; H = 42.09 kJ mol–1 dicting the direction of the process . 2 2 2 2 –1 (B) In an exothermic reaction ,the total enthalpy of the H2(g) 2H(g) ; H = 435.89 kJ mol –1 products is greater than that of reactants . O2(g) 2O(g) ; H = 495.05 kJ mol (C) The standard enthalpy of diamond is zero at 298K and Calculate the O –– H bond energy for the hydroxyl radical. 1 atm pressure . (A) 223.18 kJ mol–1 (B) 423.38 kJ mol–1 (D) It is possible to calculate the value of H for the (C) 513.28 kJ mol–1 (D) 113.38 kJ mol–1 Q.80 The bond dissociation enthalpy of gaseous H , Cl and reaction H (g) + Br () 2HBr(g) from the bond 2 2 2 2 HCl are 435, 243 and 431 kJ mol–1, respectively. Calculate enthaly data. –1 the enthalpy of formation of HCl gas. Q.74 If H of a reaction is 100 kJ mol , then the activation (A) – 92 kJ mol–1 (B) – 82 kJ mol–1 energy must be- –1 –1 –1 –1 (C) – 21 kJ mol (D) – 55 kJ mol (A) Greater than 100 kJ mol (B) Less than 100 kJ mol Q.81 The standard enthalpy of combustion at 25 ºC of hydro- (C) Equal to 100 kJ mol–1 (D) None is correct gen, cyclohexene (C6H10) and cyclohexane (C6H12) are Q.75 The intermediate SiH2 is formed in the thermal decomposi- –241, –3800 and –3920 kJ mol–1, respectively. Calculate the standard enthalpy of hydrogenation of cyclohexene. tion of silicon hydrides. Calculate Hf of SiH2 given the (A) – 131 kJ mol–1 (B) – 155 kJ mol–1 following reactions (C) – 167 kJ mol–1 (D) – 121 kJ mol–1 Si H (g) + H (g) 2SiH (g); Hº = – 11.7 kJ/mol 2 6 2 4 Q.82 A gas mixture consisting of 3.67 litres of ethylene and meth- SiH (g) SiH (g) + H (g); Hº = + 239.7 kJ/mol 4 2 2 ane on complete combustion at 25ºC produces 6.11 litres H , Si H (g) = + 80.3 kJ mol–1 f 2 6 of CO . Find out the amount of heat evolved on burning (A) 353 kJ/mol (B) 321 kJ/mol 2 one litre of the gas mixture. The heats of combustion of (C) 198 kJ/mol (D) 274 kJ/mol ethylene and methane are – 1423 and – 891 kJ mol–1, re- Q.76 A certain vessel X has water and nitrogen gas at a total spectively, at 25 ºC. pressure of 2 atm. and 300 K. All the contents of the vessel (A) 30.88 kJ (B) 20.28 kJ are transferred to another vessel Y having half the capacity (C) 50.88 kJ (D) 60.18 kJ. of the vessel X. The pressure of N in this vessel was 3.8 2 Q.83 Iodine molecule dissociates into atoms after absorbing light atm. at 300 K. The vessel Y is heated to 320 K and the total of 4500 Å. If one quantum of radiation is absorbed by each pressure observed was 4.32 atm. Calculate the enthalpy of molecule, calculate the kinetic energy of iodine atoms. Bond vapourization of water assuming it to be independent of energy of I is 240 kJ mol–1. temperature. Also assume the volume occupied by the gases 2 (A) 2.16 × 10–20 J (B) 0.16 × 10–20 J in a vessel is equal to the volume of the vessel. (C) 5.03 × 10–20 J (D) 8.32 × 10–20 J (A) 39.637 kJ mol–1 (B) 19.531 kJ mol–1 Q.84 From the following data, calculate the enthalpy change for (C) 396.37 kJ mol–1 (D) 3.9127 kJ mol–1 the combustion of cyclopropane at 298 K. The enthalpy of Q.77 For a reaction, A + B AB, C is given by the equation P formation of CO (g), H O() and propene (g) are – 393.5, 40 + 5 × 10–3 T JK4 in the temperature range 300 – 600 K. 2 2 – 285.8 and 20.42 kJ mol–1 respectively. The enthalpy of The enthalpy of the reaction at 300 K is –25.0 kJ. Calculate isomerisation of cyclopropane to propene is – 33.0 kJ mol–1. the enthalpy of the reaction at 450 K. (A) – 1021.32 kJ mol–1 (B) – 2091.32 kJ mol–1 Also Sº (NO ) = 57.5 cal/deg, Sº(O ) = 49.0 cal/deg, 2 2 (C) – 5021.32 kJ mol–1 (D) – 3141.32 kJ mol–1
52 Gyaan Sankalp Thermodynamics EXERCISE - 2 ONE OR MORE THAN ONE CHOICE MAY Q.8 Select incorrect statement(s) – 0 BE CORRECT (A) S values for all elements at 298 K and 1 bar are posi- tive Q.1 For an ideal gas- (B) For the reaction 2H (g) ––––H2 (g), H and S are E E positive (A) = 0 (B) = 0 (C) S is negative for reversible adiabatic compres- T P T V system sion of an ideal gas E E (D) |w|adiabatic < |w|isothermal in compression process for an (C) = 0 (D) = 0 ideal gas. P T V T Q.2 Which of the following statement(s) is/are correct ? Q.9 C(s) 1 O CO(g), H° = –26 kcal mol–1 2 2 (A) the system of constant entropy and constant volume will attain the equilibrium in a state of minimum energy 1 –1 CO(g) + O2(g) CO2(g), H° = –68 kcal mol (B) the entropy of the universe is on the increase 2 Which is/are correct statement(s) ? (C) the process would be spontaneous when (S) < 0, E,V (A) heat of formation of CO is –68 kcal mol–1 (E) > 0 2 S, V (B) heat of combustion of C(s) is –26 kcal mol–1 (D) the process would be spontaneous when (S) > 0, E,V (C) heat of combustion of CO(g) –68 kcal mol–1 (E) < 0 S, V (D) heat of formation of CO(g) is –26 kcal mol–1 Q.3 Which of the following processes have positive value for Q.10 When a real gas expends adiabatically against a finite pres- H ? sure- (A) H (g) 2H(g) 2 (A) Its internal energy increases (B) H+(aq) + OH–(aq) H O() 2 (B) Its internal energy decreases (C) H(g) H+(g) + e (C) Its temperature always decreases (D) H O() H O(s) 2 2 (D) Its entropy always remains constant Q.4 When a solid melts there will be : Q.11 An adiabatic process is one in which (A) an increase in enthalpy (B) a decrease in free energy (A)all energy is transferred as heat (C) no change in enthalpy (B) no energy is transferred as heat (D) a decrease in internal energy (C) the temperature of a gas decreases in a reversible Q.5 One mole of an ideal diatomic gas (C = 5 cal) was trans- v adiabatic expansion formed from initial 25ºC and 1 L to the state when tempera- (D) dU dW ture is 100ºC and volume 10 L. Then for this process (R = 2 Q.12 The factors that influence the heat of reaction are- calories/mol/K) (take calories as unit of energy and kelvin (A) The physical state of reactants and products for temp) (B) The temperature 373 (C) The pressure of volume (A) H = 525 (B) S = 5 ln + 2 ln 10 298 (D) The method by which the final products as obtained (C) E = 525 H is perfectly differential, therefore, it does not depend (D) G of the process can not be calculated using given on the method by which the final products are obtained. information. Q.13 Which of the following statements are correct ? Q.6 Variation of heat of reaction with temperature is given by (A) when G = 0, the system is at equilibrium Kirchoff’s equation which is : (B) when G < 0, the process will be spontaneous (C) when G is negative, the process is said to be exergonic HH (A) H = H + C (T – T ) (B) 2 1 = C (D) when G is positive, the process is said to be 2 1 P 2 1 T P endergonic d( H) (C) = C (D) None of these T P ASSERTION AND REASON QUESTIONS Q.7 Which of the following statements are correct ? (Q.14-Q.21) (A) the entropy of an isolated system increases in an Note : Each question contains STATEMENT-1 (Assertion) irreversible process and STATEMENT-2 (Reason). Each question has 5 choices (B) the entropy of an isolated system remains unchanged (A), (B), (C), (D) and (E) out of which ONLY ONE is cor- in a reversible process rect. (C) Entropy can never decrease (A) Statement-1 is True, Statement-2 is True; Statement-2 (D) S(system) as well as (surroundings) are negative is a correct explanation for Statement-1. quantities (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement -1 is True, Statement-2 is False. Gyaan Sankalp 53 Thermodynamics (D) Statement -1 is False, Statement-2 is True. Q.23 Match the column correctly – (E) Statement -1 is False, Statement-2 is False. Column I Column II Q.14 Statement 1 : Internal energy of a system is an extensive (A) Extensive properties (p) density (at constant property. P, V & T pressure) Statement 2 : The internal energy of a system depends (B) closed system (q) bursting of tyre upon the the amount and physical state of the substance (C) path function (r) additive in nature (D) adiabatic (s) E constant Q.15 Statement - 1: As solid changes to liquid and then to vapour state, entropy increases. (E) intensive properties (t) heat Statement-2: As going from solid to liquid and then to Q.24 Match the column correctly – vapour state, disorder increases. Column I Column II (i) Enthalpy change at constant (a) E = q + w Q.16 Statement - 1:The enthalpy of formation of gaseous oxygen molecules at 298 K and under a pressure of one pressure atm. is zero. (ii) First law of thermodynamics (b) qP – ngRT (iii) Enthalpy change at constant (c) E + PV Statement- 2 : The entropy of formation of gaseous oxygen molecules under the same condition is zero. volume Q.17 Statement - 1 : Heat energy is completely transformed into qrev work during reversible isothermal expansion of an ideal (iv) Change in degree of randomness (d) T gas. initial Statement - 2 : During a reversible isothermal process of Q.25 Match the column an ideal gas, the changes in internal energy due to decrease Column I Column II in pressure are nullified by the changes due to increase in (A) For spontaneous (p) (B.E.)R – (B.E.)p volume. reaction Q.18 Statement - 1 : The standard free energy of all spontaneity (B) For endothermic (q) H = E occurring reactions are positive reaction because (C) Bond dissociation (r) G = –ve Statement - 2 : The standard states at 760 mm Hg and 298 energy K are taken as zero. (D) For solids and liquids (s) Hp > HR Q.19 Statement-1: For an adiabatic process PV = constant, in a thermochemical reaction where is the ratio of CP and CV) Statement-2: The slope of isothermal process curve = x slope of adiabatic curve PASSAGE BASED QUESTIONS Q.20 Statement-1: If a system goes from state A to state B by Passage 1- (Q.26-Q.28) following reversible and irreversible process respectively J.W. Gibbs and H.Von Helmoltz had given two equation then change in internal energy in both the process should and are known as Gibbs-Helmholtz equation. One equa- be same. tion can be expressed in terms of change in free energy Statement–2 : Change in internal energy is a state function. (G) and enthalpy (H) while other can be expressed in Q.21 Statement-1: The enthalpy of formation of H2O(l) is greater terms of change in internal energy (E) and work function than that of H2O(g) (W) Statement-2: Enthalpy change is negative for the G condensation reaction : H2O(g) H2O(l) G = H + T . d ...... (1) dT P
MATCH THE COLUMN TYPE QUESTIONS W (Q.22-Q.25) W = E + T . d ...... (2) dt V Each question contains statements given in two columns Where T is temperature equation (1) is obtained at con- which have to be matched. Statements stant pressure while equation (B) is obtained at constant (A, B, C, D) in column I have to be matched with state- volume system. It is observed that for the reaction. ments (p, q, r, s) in column II. Q.22 Match the column correctly – N2(g) + 3H2(g) 2NH3(g) Column I Column II Free energy change at 25ºC is –33 kJ (A) Isochoric process (p) E = 0 while at 35ºC is –28 kJ which are at a constant pressure. (B) Isobaric process (q) Q = 0 Q.26 What would be the difference between enthalpy change at (C) Isothermal process (r) W = 0 25ºC and 35ºC for a given reaction : (D) Adiabatic process (s) W0 (A) 4 kJ (B) 5 kJ (C) 3 kJ (D) zero Q.27 What would be the free energy change at 30ºC – (A) 30.5 kJ (B) 33 kJ
54 Gyaan Sankalp Thermodynamics (C) –28 kJ (D) –30.5 kJ Q.33 In the following reaction, involving neutralisation of Q.28 Internal energy change at 25ºC is E1 while at 35ºC is E2 HF (a weak acid) with NaOH (a strong base). then – HF (aq) + NaOH (aq) NaF(aq) + H2O () –1 (A) E1 = E2 (B) E2 > E1 Hº = –68.6 kJ mol (C) E1 > E2 (D) None of these This values is much higher than the heat of neutralisation of strong acid with strong base ( = – 57.3 kJmol–1). This is Passage 2- (Q.29-Q.31) due to : Entropy is the degree of disorder or randomness of the (A) in aqueous solution, HF behaves as a strong acid molecule of a system. (B) there is hydration of F– ion in aqueous solution, and Characteristic property of a spontaneous process is al- process is exothermic ways accompanied by an increase in the randomness of (C) both (A) and (B) (D) none of the above –1 the molecules of the system. Q.34 Heat of dissociation of CH3COOH is 0.005 kcal g , hence The entropy change depends only on the initial and final enthalpy change when 1 mole of Ca(OH)2 is completely states and not on the path taken by the system. neutralised by CH3COOH is: Q.29 Entropy is ____ function (A) – 27.4 kcal (B) – 13.6 kcal (A) state (B) path (C) – 26.8 kcal (D) – 27.1 kcal (C) state and path (D) None Q.30 The entropy and entropy change for pure crystalline solid Passage 4 (Q.35-Q.37) substance at 0°K is Enthalpy of reaction : It is used for the heat change (A) increases (B) decreases accompaining any reaction. depending upon the nature of (C) zero (D) Maximum the reaction the enthalpy of the reaction is named Q.31 The increase in entropy of the system is _____ in any accordingly reversible cyclic process For example : In the reaction (A) Maximum (B) Minimum CH4(g) + O2(g) CO2(g) + H2O(); (C) zero (D) 100 H= –890.4 kJ/mole This reaction shown that 890.4 kJ of heat is produced Passage 3- (Q.32-Q.34) when methane is completely burnt it is known as heat of Enthalpy of Neutralisation is defined as the enthalpy combustion. For example: In the reaction change when one mole of H+ in dilute solution combines C(s) + O2(g) CO2(g) H = –393.5 kJ/mole with one mole of OH– to give rise to undissociated water when CO2(g) is formed from its elements i.e. from C(s) and + – O2(g) 393.5 kJ of heat is produced and the process in heat i.e. H (aq) + OH (aq) H2O() H = – 57.54 kJ/mole or –13.7 Kcal/mole. of formation. When ever one mole of strong monoprotic acid is mixed Q.35 The enthalpy of formation of methane from the following with one mole of strong base, the above neutralisation data reaction takes place. (i) C(s) + O2(g) CO2(g) When ever a weak acid (or) base reacts with a strong base H = –393.7 kJ/mole (or) acid, the release of heat is less than 57.54 kJ mol–1. It 1 (ii) H2(g) + O2(g) H2O() is because of the fact that these acids (or) bases are not 2 completely ionized in the solution. Some heat is consumed H = –285.8 kJ/mole in ionising these acids and bases. (iii) CH4(g) + 2O2(g) CO2(g) + 2H2O() Q.32 When 1 equivalent reacts with 0.33 equivalent of HCl, H = – 890.4 kJ/mol enthalpy change is – 4.6 cal. When 0.66 equivalent of HCl (A) –74.9 kJ/mole (B) 54.6 kJ/mole is further added, enthalpy change would be: (C) 99.7 kJ/mole (D) –89.6 kJ/mole (A) –13.7 (B) – 9.2 cal (C) 13.7 cal (D) 9.2 cal EXERCISE - 3 SUBJECTIVE QUESTIONS Q.3 Calculate the work done when 1 mol of zinc dissolves in hydrochloric acid at 273 K in (a) an open beaker (b) a closed Q.1 Calculate the inversion temperature of CO2 gas. Given : a = 3.64 dm6 bar mol–2 and b = 0.04267 dm3 mol–1 beaker at 300 K. Q.4 The first IP of lithium is 5.41 eV and electron affinity of Cl is Q.2 Calculate rHº for the reaction –1 + – –3.61 eV. Calculate H in kJ mol for the reaction. Ag (aq) + Cl (aq) AgCl(s) at 25ºC. + – + –1 Li(g) + Cl(g) Li (g) + Cl (g) Given : fHº (Ag , aq) = 105.58 kJ mol , – –1 Q.5 The equilibrium constant Kp for the reaction fHº (Cl , aq) = – 167.16 kJ mol and Hº (AgCl, s) = – 127.07 kJ mol–1 f C(s) + H2O(g) CO(g) + H2(g)
Gyaan Sankalp 55 Thermodynamics at 4270 is 2.8 × 10–2. Calculate G for the reactant of steam 2NO(g) + O2(g) 2NO2(g) is exotermic or endother- at 200 atm with carbon to form CO and H2 each at a pres- sure 50 atm. Will the reaction occur spontaneously under mic. What is the H of the reaction ? the given conditions ? Q.14 Calculate the amount of work done by 2 mole of an ideal Q.6 Calculate the maximum work done when pressure on 10 g gas at 298 K in reversible isothermal expansion from 10 litre of hydrogen is reduced from 20 to 1 atm at a constant 20 litre. temperature of 273 K. The gas behaves ideally. Will there Q.15 The heat of combustion of ethylene at 250 C and under be any change in internal energy. Also calculate Q. constant volume conditions is 332.0 kcal. What will be the Q.7 The heat of combustion of glycogen is about 476 kJ/mol heat of combustion if the reaction is carried out in con- of carbon. Assume that average heat loss by an adult male stant pressure ? is 150 watt. If we were to assume that all the heat comes Q.16 At 1 atm and 27ºC, will the vaporisation of liquid water be from oxidation of glycogen, how many units of glycogen spontaneous ? Given H = 9710 cal and s = 26 eu. (1 mole carbon per unit) must be oxidised per day to provide Q.17 Calculate resonance energy for CO2 from the following : for this heat loss ? C = 0 = 339 KJ Q.8 Three moles of an ideal monoatomic gas initially at 300 K O = 0 = 498 KJ and 20 atm pressure exands isothermally and reversible to C(s) C(g) = 718 KJ a final pressure of 2 atm. Calculate E,Q, W, and S in the Heat of combustion of carbon = – 393 kJ. process Q.18 Gases Gºf (Cal/mole) Q.9 Calculate w and E for the conversion of 0.5 mole of water CO –32.80 at 100ºC to steam at 1 atm pressure. Heat of vaporisation of H2O – 54.69 –1 water at 100ºC is 40670 J mol . CO2 – 94.26 0 Q.10 Two moles of liquid water at 100 C and 1 atm are converted H2 0 into vapour at the same temperature and pressure. Calcu- Estimate the standard free energy change in the chemical late W, Q, E, H and S in this transformation. (Assume reaction : CO + H2O = CO2 + H2 water vapour to behave ideally. Latent heat of vaporiza- Q.19 For the reaction : X (g) Y(g) 2XY(g) tion of water is 9720 cal mol–1) 2 2 –1 Q.11 The heat of atomisation of Bi is 207 kJ mol . It forms Bi2 G0 and S0 are –20 kJ/mol and 20 JK–1 mol–1 respec- molecules in the gas phase and for the reaction, r r –1 –1 –1 tively at 200 K. If rC p is 20 JK mol then calculate 2Bi(s) Bi2(g), H = 125 kJ mol of Bi(s). Calculate Bi – Bi bond energy in Bi (g). 2 | H0 | in kJ/mol at 220 K. Q.12 The specific heat of a monatomic gas at constant pressure r –1 is 248.2 J kg–1 K–1 and at constant volume it is Q.20 Two moles of an ideal monoatomic gas (CV = 12.55 JK –1 149.0 J kg–1 K–1. Calculate the mean molar mass of the gas. mol ) expands irreversibly and adiabatically from an ini- tial pressure of 1.013 MPa against a constant external pres- Q.13 The molar heats of formation of NO and NO2 are 2.15 and 7.5 kcal respectively. Use these data to find out whether sure 0.1013 MPa, until the temperature drop from its initial the reaction, value of 325 K to a final value of 275 K. How much work is done and what is its final volume ? EXERCISE - 4 PREVIOUS YEAR IIT-JEE QUESTIONS Q.3 A sample of argon gas at 1 atm pressure and 27 ºC expands 3 3 Q.1 Diborane is a potential rocket fuel which undergoes reversibly and adiabatically from 1.25 dm to 2.50 dm . combustion according to the reaction, Calculate the enthalpy change in this process. –1 –1 Cv.m. for argon is 12.48 jK mol . [2000] B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) From the following data, calculate the enthalpy change for Q.4 Show that the reaction CO(g) + 1/2 O2 (g) CO2(g) at the combustion of diborane : [2000] 300 K is spontaneous and exothermic, when the standard entropy change is –0.094 kJ mol–1 K–1. the standard Gibbs 3 2B(s) + O (g) B O (s) ; H = –1273 kJ free energies of formation for CO2 and CO are –394.4 and 2 2 2 3 –137.2 kJ mol–1, respectively. [2000] Q.5 Which one of the following statements is false ? [2000] 1 (A) work is a state function H2(g) + O2(g) H2O(l) ; H = –286 kJ 2 (B) temperature is a state function H2O(l) H2O(g) ; H = 44 kJ (C) work appears at the boundary of the system 2B(s) + 3H2(g) B2H6(g) ;H = 36 kJ (D) change in the state is completely defined when the Q.2 Calculate q , W , E and H for the isothermal reversible initial and final states are specified. expansion of one mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K. [2000] 56 Gyaan Sankalp Thermodynamics Q.6 In thermodynamics, a process is called reversible when : (A) 4.98 kJ (B) 11.47 kJ [2000] (C) –11.47 kJ (D) 0 kJ (A) surroundings and system change into each other Q.13 100 ml of liquid contained in an isolated container at a (B) there is no boundary between system and surroundings pressure of 1 bar. The pressure is steeply increased to 100 (C) the surroundings are always in equilibrium with the bar. The volume of the liquid is decreased by 1 ml at this system constant pressure. Find the H and U. [2004] (D) the system changes into the surroundings Q.14 Which of the following isomers of phosphorus are spontaneously thermodynamically least and most stable? [2005] Q.7 1 mole of an ideal gas undergoes reversible isothermal (A) White (least), Red (most) (B) Yellow (least), Red (most) expansion from an initial volume V1 to a final volume, 10 7 (C) Red (least), White (most) V1, and does 10 kJ of work, The initial pressure was 1 × 10 Pa. (D) White (least), Black (most) Q.15 One mole of a monoatomic ideal gas expands adiabatically (a) Calculate V1, (b) If there were 2 mol of gas, what must its temperature have been ? [2001] at initial temperature T against a constant external pressure Q.8 One mole of a non-ideal gas undergoes a change of state of 1 atm from one litre to three litres. The final temperature (2.0 atm, 3.0 L, 95 K) (4.0 atm, 5.0 L, 245 K) with a of the gas is: [2005] (5R/2) change in internal energy, U = 30.0 L atm. The change in (A) T (B) T/3 enthalpy (H) of the process in L atm is : [2002] (C) T – 2K / (1.5 × 0.082) (D) T + 2K / (1.5 ×0.082) (A) 40.0 (B) 42.3 (C) 44.0 Q.16 A process A B is difficult to occur directly instead it (D) not defined, because pressure is not constant. takes place in three successive steps. [2006] C Q.9 Two moles of a perfect gas undergo the following D S(A C) = 50 processes : [2002] S(C D) = 30 (a) a reversible isobaric expansion from (1.0 atm, 20.0 L) to A B S(B D) = 20 (1.0 atm, 40.0 L) (b) a reversible isochoric change of state from (1.0 atm, The entropy change for the process A B is 40.0 L) to (0.5 atm, 40.0 L) (A) 100 (B) 60 (c) a reversible isothermal compression from (0.5 atm, 40.0 (C) 100 (D) +60 L) to (1.0 atm, 20.0 L) Q.17 Find the molar heat capacity of an ideal monoatomic gas (i) Sketch with labels each of the processes on the same P for a process for which p/v is constant and is equal to I – V diagram. (A) 3/2 R (B) 4/2 R [2006] (ii) Calculate the total work (w) and the total heat change (C) 5/2 R (D) 0 (q) involved in the above processes. Q.18 For the process H2O () (1 bar, 373 K) H2O(g) (g) (1 bar, (iii) What will be the values of U, H and S for the 373 K), the correct set of thermodynamic parameters is – overall process ? [2007] Q.10 Which of the reaction defines Hfº ? [2003] (A) G = 0, S = +ve (B) G = 0, S = –ve (C) G = +ve, S = 0 (D) G = –ve, S = +ve 1 1 Q.19 STATEMENT-1: There is a natural asymmetry between (A) H2(g) + F2(g) HF(g) 2 2 converting work to heat and converting heat to work. 1 and (B) CO(g) + O (g) CO (g) 2 2 2 STATEMENT-2: No process is possible in which the sole result is the absorption of heat from a reservoir and its (C) C + O (g) CO (g) (diamond) 2 2 complete conversion into work. [2008] (D) N (g) + 3H (g) 2NH (g) 2 2 3 (A) STATEMENT-1 is True, STATEMENT-2 is True; Q.11 H = 30 kJ/mole and S = 75J mol–1 K–1. vap vap STATEMENT-2 is correct explanation for STATEMENT-1 Find temperature of vapour, at one atmosphere. [2004] (B) STATEMENT-1 is True, STATEMENT-2 is True; (A) 400 K (B) 350 K STATEMENT-2 is NOT a correct explanation for (C) 298 K (D) 250 K STATEMENT-1 Q.12 2 mol of an ideal gas expanded isothermally & reversible (C) STATEMENT-1 is True, STATEMENT-2 is False from 1 litre to 10 litres at 300 K. What is the enthalpy (D) STATEMENT-1 is False, STATEMENT-2 is True change? [2004]
Gyaan Sankalp 57 Thermodynamics HINTS & SOLUTIONS
EXERCISE - 1 Q 1 2 3 4 5 6 7 8 9 10 11 A D A B A B C B A B B C Q 12 13 14 15 16 17 18 19 20 21 22 A D B D B A B D C C C B Q 23 24 25 26 27 28 29 30 31 32 33 A B D C B A C B C B B B Q 34 35 36 37 38 39 40 41 42 43 44 A A D A B A A C D B C A Q 45 46 47 48 49 50 51 52 53 54 55 A C C B C A A A A B B B Q 56 57 58 59 60 61 62 63 64 65 66 A C D B C C B B A C D B Q 67 68 69 70 71 72 73 74 75 76 77 A A D B A D B A A D A C Q 78 79 80 81 82 83 84 A D B A D C A B (1) (D). The heat of formation of HCl is to be calculated on per base net, heat evolved = H mole of HCl produced H temperature increase = 44 ms heat of formation of HCl = – = – 22 kcal. 2 when 10 ml of each are reaction, the heat evolved = 2H; (2) (A) Density is mass per unit volume and specific volume is but mass of the solution = 2m volume per unit mass. 2 H H (3) (B). The statement is definition of Hess’s law temp. increase = = (4) (A). H = E + PV 2ms ms (9) (B). The heat of neutralisation = – 55.9 kJ 1(1.5 1.2)L atm 1.00 = E + × 2 × 10–3 kcal The heat of neutralisation of HCN by NaOH = the heat 0.082 L atm of neutralisation + the energy of dissociation of HCN E = 0.993 kcal = – 12.1 kJ (5) (B). The heat of combustion of per gm of CH4 or the energy of dissociation of HCN 890 = + 55.9 – 12.1 = 43.8 kJ = – = – 55.625 kJ (10) (B). For a reaction at equilibrium G = 0 and not Gº. 16 (11) (C). from equation (I) + (2 × II) ; Hº = 6.2 kJ The heat of combustion of per gm of C2H4 (12) (D). The work done by a gas is given by 1411 W = P V = = – 50.393 kJ 28 (13) (B). When n = 0, H = E because H = E + RT n. (14) (D). H = E + nRT The heat of combustion of per gm of C2H6 1.435 1560 (15) (B). H (per g) = kcal = 0.0797 kcal g = – = – 52 kJ 18 30 (16) (A). H = C T and E = C T. (6) (C). The standard heat of combustion is defined as the p V C – C = R using these equations after calculating the heat evolved on burning one mole of the compound in p v final temperature, the desired result is obtained. presence of O2 to give CO2 and water. Since the amount of H the substance is specified, it becomes an intensive prop- (17) (B). H of reaction = HNaCl × 6 – Al2 Cl 6 where HNaCl erty. H (7) (B). Kirchhoff’s equation is : and Al2 Cl 6 are the heats of formation of NaCl and Al2Cl6. H2 H1 = CP (T2 T1) (8) (A). When 5 ml of strong acid reacting with 5 ml of strong (18) (D). Internal energy and volume depend upon states. 58 Gyaan Sankalp Thermodynamics (19) (C). 3 H = C (T – T ) + (P V – P V ) (31) (B) E = RT for 1 mole I v 2 1 2 1 1 1 tr 2 HII = U + P3V2 – P2V1 = P (V – V ) + P V – P V 3 2 1 3 2 2 1 3 = P V – P V RT 3 1 2 1 E 2 H = C (T – T ) + (P V – P V ) 3 –1 –1 overall v 2 1 3 1 1 1 Cv = = T = R = 12.47 J deg mol T v 2 112 (20) (C). 112 litre (at NTP) contains no. of moles = = 5 22.4 (32) (B). NaCl(s) Na+(g) + Cl–(g) 2.5 moles of H2 and 2.5 moles of CO present. Hence total H = + 780 kJ amount of heat evolved Na+(g) + aq Na+(aq) = 2.5 × 241.8 + 2.5 × (283) = 1312 kJ H = – 406 kJ (21) (C). When 25 ml acid is reacting with 25 ml of alkali, maxi- Cl–(g)+ aq Cl– (aq) mum equivalent of acid is reacting with alkali since the H = – 364 kJ volume of the resulting solution is the same in every cases, The net reaction is dissolution of NaCl(s). therefore, increase in temperature is the highest where –1 Hsolution = 780 – (406 + 364) = 10 kJ mol maximum equivalents of acids and alkalies are reacting. H H 1435cal (22) (B). Lower is heat of neutralisation , more is dissociation (33) (B). S = T = = 273 K 0ºC energy, weaker is acid T S 5.26 (23) (B). Given 2 PNOCl C + O2 CO2 ; Hº = – 393.5 kJ ...... (i) 8 (34) (A). K = 2 = 10 eq PP C + (1/2)O2 CO ; Hº = – 110.5 kJ ...... (ii) NO Cl2 H + (1/2)O H O ; Hº = – 241.8 kJ ...... (iii) 2 2 2 Gº = – RT ln K = – 2.303 × 8.314 × 298 × 8 = – 45.65 kJ By (ii) + (iii) – (i), eq CO2 + H2 CO + H2O ; Hº = + 41.2 PV1 1 PV2 2 T2 6 (24) (D). (35) (D). = ; = T1 T2 T1 3 CH4 C(s) + 2H2(g) H = 360 Kcal C H 2C(s) + 3H (g) H = 620 2 6 2 TP2 1 S = 2.303 × n Cp log 10 R log 10 360 TP1 2 The avg C – H bond energy = = 90 Kcal 4 6 20 C – C bond energy = 620 – 90 × 6 = 620 – 540 = 80 Kcal S = 2.303 × 1 30.96log10 Rlog 10 (25) (C). For an endothermic reaction, H is positive 3 10 = H(products) – H(reactants). S = 27.22 J.k–1 mol–1 (26) (B). For any natural process there is always an increase in (36) (A). As G = H – TS entropy. For spontaneous process, G = negative (27) (A). H = E + n RT. For endothermic process, H = positive n = – 1 Therefore S > 0 164 x 103 = E – 1 x 08.314 x 298 (37) (B). Polymerization reaction 3 = E – 2.478 x 10 n CH = CH E = 164 x 103 + 2.478 x 103 2 2 –1 one mole of C = C bond is broken and two moles of C – C = 166.5 KJ mol bonds are formed per mole of ethylene. (28) (C).G = –ve for spontaneous charge H = 590 – 2 × 331= 590 – 662 G = H – TS = – 72 kJ per mole of ethylene. H 4000 (38) (A). Temperature and volume are related for adiabatic TS > H, T > T > T > 400 K – 1 –1 S 10 process as : T2V2 = T1V1 (30) (C). H O () H O(g) H = 44 KJ 1 1 2 2 TV 1/3 1 1 V1 1 1 T2 = 1 = T1 = 300 H (g) + O (g) H O() ; H0 = – 286 KJ V1 V2 8 2 2 2 2 f 1 1 = 300 × = 150 K Adding H2(g) + O2 (g) H2O(g) 2 2 CP 4 0 –1 = = CV = 3R & CP = 4R H f = – 286 + 44 = – 242 KJ mol CV 3 Adiabatic work done w = – CV T = – 3 × 2 × (150 – 300) = 900 cal
Gyaan Sankalp 59 Thermodynamics (39) (A). Temperature and volume are related for adiabatic pro- d( G) cess is as (46) (C). On comparison :S = W = P (V) dT PV = RT d( nFE) dE dE S P(V + V) = R(T + 1) Þ S = = nF = PV = R(T + 1) – RT = R dT dT dT nF 6 (40) (C). 180 gms of glucose 2.8 × 10 J of heat evolved 3 1 (47) (B). Al + O = Al O ; H = – 837.8 kJ 2.8 106 4 2 2 2 3 0.16 gms would yield × 0.16 J 180 250 kcal = 250 × 4.2 kJ = 1050 kJ 1050 kJ heat indicated the formation of If the calorimeter constant = W, then 1 1050 2.8 0.16 106 2.8 1.6 103 × = 0.624 moles of Al O . W × 4 = = J 2 837.8 2 3 180 1.8 Thus, the liberation of 250 kcal energy indicating the 3 2.8 1.6 10 formation of 0.624 moles of Al2O3 W = Jdeg–1 = 6.22 × 102 Jdeg–1 (48) (C). PV = RT at temp T for one mol 1.8 4 P(V + V) = R(T + 1) at temp. (T + 1) for one mol (41) (D). C(s) + 2Cl2(g) CCl4() PV = R Hf (CCl4, ) = Hº[C(s) C(g)] + 2(BE)Cl–Cl 2 – [H° (CCl ) + 4 (BE) ] PNOCl vap 4 Cl–Cl (49) (A). G° = –2.303 RT log Keq Keq = = 108 PP2 3 NO Cl2 (42) (B).Dy(s) + Cl (g) DyCl (s) Hence Gº = – 45.65 kJ 2 2 3 (50) (A). The heat of neutralisation of strong acid and strong H = – 994.30 kJmol–1 base = – 13.7 kcal/equiv. aq. HCl –25.4 = – 2 × 13.7 × Hdisso Dy Cl3 (s) DyCl3 (aq. in 4.0 M HCl) or H = 2 × 13.7 – 25.4 = 2 kcal H = – 180.06 kJmol–1 disso (51) (A). Net power given to N2 gas = 100 – 30 = 70 cal/sec 3 3 The nitrogen gas expands isobarically. 3 HCl(aq.4M) H (g) + Cl (g) 2 2 2 2 dT 7 dT Q nCP or 70 5 R H = 3 × 158.31 KJmol–1 dt 2 dt dT 2K / sec 3 dt Dy (s) + 3 HCl DyCl + H (g) 3 2 2 (52) (A). Mole fractions are 0.8 and 0.2 H = – 699.43 kJmol–1 = x entropy of mixing per mole (aq. 4 M) (aq. 4 M HCl) = – 1.987[0.8 n (0.8) + 0.2 n (0.2)] = – 1.987[–0.1785 – 0.3219] = 0.9943 eu (43) (C). V1 (volume of 1 g H2) = 11.2 L at NTP V (volume of 1 g H ) = 22.4 L nRT 1 2 2 (54) (B). Process (A). VK2 ; TV = constant ; T W = PV = 11.2 L atm V V (44) (A). nRT 2 T 3 2 C (s) + 3 H2 (g) C2H6 (g) ; Process (B) KV ; constant ; TV H = – 20.24 kcal/mol V V3 (55) (B). H = E + ng RT = 2.1 + 2 × 0.002 × 300 = 3.03 kcal G = H – TS = 3.3 – 300 × (0.02) = –2.7 kcal (56) (C). C6H6 + 3H2 = C6H12 H = x2 3C2H4 + 3H2 = 3C2H6 H = x1 Resonance energy of benzene (contains three double bond) = 3x – x [2 × Hsublim ] + 307.8 – 63 – 513.6 = – 20.24 1 2 2 × H = [–20.24 + 63 + 513.6 – 307.8] kcal (57) (D). G = Gº + RT n Q sublim If reaction quotient = 1 Hsublim = 124.3 kcal/mole of carbon. (45) (C). H = E + PV G = Gº (58) (B). H = 3 × Hf (H2O, g) + 2Hf (CO2, g) – Hf (C2H6, g) 8.314 H O() H O(g) H 100 = E + 1.5 (2 – 8) × 2 2 f 0.0821 Hf = Hf (H2O, g) – Hf (H2O, ) E = 1011.4 J Hf = (H2O, g) = T1x1 + Hf (H2O, ) = T1x1 + x2 H = 3T1x1 + 3x2 + 2x3 – x4 60 Gyaan Sankalp Thermodynamics
(59) (C). The enthalpy change for the given reaction is calcu- (67) (A). q is a path dependent property and is also an inexact lated by considering the number of the various bonds bro- differential. ken in the reactants and the number of new bonds formed (68) (D). The change in internal energy E is related to heat (Q) in the products. and work terms (W) according to E = Q – W 1 (60) (C). H (g) + O (g) H O(g) H (69) (B). In process-1 heat supplied 2 2 2 2 = area under AB curve + n × cv × 100 (isobaric process) H = – 2x bond formation energy of (O – H) + bond break- In process-2 heat supplied = area under AC curve 1 (isothermal process) ing energy of H + bond breaking energy of O 2 2 2 In process-3 heat supplied = 0 (adiabatic process) In process-4 heat supplied = n × cv (T – 600) x2 (isobaric process) H = x + – 2x 1 2 3 (70) (A). The result can be arrived from the equations H O() = H O(g) H = 10.5 kcal mol–1 H2O(g) H2O() H1 = – x4 2 2 1 1 H (g) + O (g) H O() H H (g) + O (g) = H O() ; H = 68.3 kcal mol–1 2 2 2 2 2 2 2 2 2 by suitable algebraic manipulation. x H = x + 2 – 2x – x (71) (D). The entropy concept is direct outcome of the second 2 1 2 3 4 law of thermodynamics through the Carnot cycle approach. 1 k (72) (B). The first reaction is exothermic and the second one is P P PV2 k (61) (B). 2 2 endothermic. If a mixture of steam and O2 is passed over VV coke and temperature is constant, the conversion of each to CO should not show any heat change, i.e., total heat PV.V k nRTV k TV k1 Since temperature increases therefore volume decreases. evolved in I = total heat absorbed in II. n1 × 2 × 100 = n2 × 132 H CO2 n1 132 0.6 (62) (B). = 2.57; HCO = 2.57 HCO HCO 2 = = n2 220 1 172.5 = H – H = H – 2.57 H CO CO2 CO CO (73) (A). The statements given under (B), (C) are not correct because statement (B) refers to an endothermic reaction 172.5 and the standard state for carbon is graphite under (C). For H = – = – 109.8 kJ mol–1 CO 1.57 the calculation of H of the reaction under (D), additional (63) (A). In an adiabatic expansion, internal energy decreases data in the heat of vapourisation of Br2 (l) is necessary. and hence temperature decreases. (74) (A). The reaction is endothermic therefore, potential en- from equation of state of ideal gas, PV = nRT ergy diagram of the given reaction is The product of P and V decreases. H H (64) (C). s = ; 14.4 = T(K) 391 Eact G Product 5630 –1 H = 391 × 14.4 = 5630 cal. mol–1 = = 94 cal. g–1 H = 100 KJ mol 60 Reactant 100 0.3 (65) (D). Amount of HCl added = = 0.03 mol 1000 Progress of reaction –1 100 0.35 Eact > 100 kJ mol Amount of NaOH added = = 0.035 mol 1000 (75) (D). Si2H6 (g) + H2 (g) 2 SiH4 (g) Limiting reagent = 0.03 mol of HCl H = – 11.7 kJ/mol amount of heat evolved = 57.1 × 0.03 = 1.713 kJ +80.3 kJ/mol 0 x (66) (B). The work done in the isothermal reversible expansion 2x – 80.3 = – 11.7, 2x = 80.3 – 11.7 = 68.6 kJ/mol of n moles of an ideal gas is given by the equation. x = 34.3 kJ/mol SiH (g) SiH (g) + H (g) ; H = + 239.7 kJ/mol V2 P1 4 2 2 +34.3 y 0 W = nRT n V = nRT n P 1 2 y – 34.3 = 239.7 –1 where V1, V2 are the initial and final volumes of the gas, y = (239.7 + 34.3) kJmol = 274 kJ/mol the corresponding pressure P1 and P2. Gyaan Sankalp 61 Thermodynamics (76) (A). Pressure of nitrogen in Y = 3.8 atm. 1 1 Pressure of nitrogen in X = 1.9 atm. = 435 243 431 kJ mol–1 = – 92 kJ mol–1. 2 2 Pressure of H2O(g) in X at 300 K = 2 – 1.9 = 0.1 atm (81) (D). The given data are : 3.8 Pressure of N at 320 K : × 320 = 4.05 atm. 2 300 1 –1 (i) H2 (g) + O2(g) H2O () ; Hº = – 241 kJ mol Total pressure at 320 K : 4.32 atm. 2 Pressure of water vapour at 320 K = 4.32 – 4.05 = 0.27 atm. 17 (ii) C H (g) + O (g) 6CO (g) + 5H O () 0.27 H 1 1 6 10 2 2 2 2 ln = , –1 0.1 R 300 320 Hº = – 3800 kJ mol –1 (iii) C6H12(g) + 9O2(g) 6CO2(g) + 6H2O () H = 39.637 kJ mol –1 (77) (C). According to Kirchhoff’s equation, Hº = – 3920 kJ mol We have to calculate the enthalpy change for the reaction T2 C6H10(g) + H2(g) C6H12(g) CP This equation can be obtained by the following manipula- H2 = H1 + dT T1 tions. Eq. (ii) + Eq. (i) – Eq. (iii) 450 = – 25000 + (40 5 103 T) dT = – 18.72 kJ Carrying out the corresponding manipulations on H s, 300 we get (78) (D). The equation for combustion of C2H4 is º º º C2H4(g) + 3O2(g) 2CO2(g) + 2H2O() Hº = H(ii) + H(i) – H(iii) 1 mole 3 mol 2 moles = (– 3800 – 241 + 3920) kJ mol–1 = – 121 kJ mol–1. H = E + 2nT = –332190 + 2 × (–2) × (273 + 17) (82) (C). The combustion reactions are = –333350 cals = – 333.35 k cals C H (g) + 3O (g) 2CO (g) + 2H O() (79) (B). We have to calculate the enthalpy of the reaction 2 4 2 2 2 CH (g) + 2O (g) CO (g) + 2H O() OH (g) O(g) + H(g) 4 2 2 2 Let V be the volume of C H (g) in the gaseous mixture of From the given reactions, this can be obtained as follows. 2 4 3.67 L. From the chemical equations, we find that 1 1 Volume of CO2 (g) produced due to the combustion of – H2 (g) O 2 (g) OH(g) ; H=– 42.09 kJ mol–1 2 2 C2H4 (g) = 2V Volume of CO2 (g) produced due to the combustion of 1 1 CH (g) = 3.67 L – V + [H (g) 2H(g)] ; H = × 435.89 kJ mol–1 4 2 2 2 Equating the latter with 6.11 L – 2 V, we get 3.67 L – V = 6.11 L – 2V or V = 2.44 L 1 1 + [O (g) 2O(g)] ; H = × 495.05 kJ mol–1 Hence, in the original mixture, we have 2 2 2 Volume of C2H4(g) per litre of the mixture Add 2.44 L OH(g) H(g) O(g) = (1 L) = 0.665 L 3.67 L 1 H = 423.38 kJ mol Volume of CH4 (g) per litre of the mixture = 1.0 L – 0.665 L = 0.335 L (80) (A). The given data are –1 Now, Volume of 1 mol of any gas at 25 ºC (i) H2(g) 2H(g) H = 435 kJ mol –1 (ii) Cl2(g) 2Cl(g) H = 243 kJ mol 298 K (iii) HCl(g) H(g) + Cl(g) H = 431 kJ mol–1 = (22.414 L) = 24.467 L 273 K We have to find H for the reaction Hence, Heat released due to the combustion of 1 1 H (g) + Cl (g) HCl(g) 0.665 L 2 2 2 2 C2H4(g) = (1423 kJ) = 38.68 kJ This equation can be obtained by the following manipula- 24.467 L tion. Heat released due to the combustion of 1 1 0.335 L Eq. (i) + Eq. (ii) – Eq. (iii) CH (g) = (891 kJ) = 12.20 kJ 2 2 4 24.467 L Hence, carrying out the corresponding manipulation on Total heat released = (38.68 + 12.20) kJ = 50.88 kJ. Hs, we get 1 1 H = + H + H – H 2 (i) 2 (ii) (iii) 62 Gyaan Sankalp Thermodynamics
(83) (A). The dissociation reaction is I2 2I (84) (B). For the reaction Bond energy per single bond is 9 (i) CH CH = CH (g) + O (g) 3CO (g) + 3H O() 240 kJ mol1 3 2 2 2 2 2 –22 –19 = 23 1 = 3.985 × 10 kJ = 3.985 × 10 J 6.023 10 mol rH = 3fH(CO2, g) + 3fH(H2O, ) – fH(CH3CH = CH2) = [3(– 393.5) + 3(– 285.8) – (20.42)] kJ mol–1 Energy of 1 photon of wavelength 4500 Å is = – 2058.32 kJ mol–1 34 8 1 h c (6.626 10 Js)(3 10 ms ) To the reaction (i), if we add the reaction E = = (4500 1010 m) CH3CH = CH2(g) = 4.417 × 10–19 J Of the energy 4.417 × 10–19 J, the energy consumed to –1 rH = – 33.0 kJ mol break I molecules is 3.985 × 10–19 J. The remaining energy 2 9 is used in imparting kinetic energy to the two iodine atoms. we get (g) + O (g) 3CO (g) + 3H O() Hence, Kinetic energy per iodine atom 2 2 2 2 –1 (4.417 3.985) 1019 J Hence, cH(cyclopropane) = (– 2058.32 – 33.0) kJ mol = = 2.16 × 10–20 J. = – 2091.32 kJ mol–1 2 EXERCISE - 2 (1) (CD). G E E G2 = H2 + T2 d As per the properties of ideal gases = = 0 dT P P T V T (2) (ABD) G G2 – G1 = (H2 – H1) + (T2 – T1) (3) (AC). If the stability of products is lesser than that of reac- dT P tants, then reactions will be endothermic and H will be positive. GG2 1 (H2 – H1) = (G2 – G1) – (T1 – T2) = 0 (4) (AB) TT2 1
Tf Vf G GG2 1 (5) (AB). S = nCv ln + nR ln Ti Vi (27) (D). S = – d = – dT P TT2 1 H = nCpT E = nCVT 28 ( 33) = – = – 0.5 kJ/K (6) (ABC) (7) (ABC) (8) (BCD) 308 298 (9) (CD) (10) (BCD). dq = du + dw G G1 = H1 + T1 d If dq = 0, dw = – du dT P Internal energy decreases –33 = H + 298 × 0.5 Its temperature always decreases. The reversible adiabatic 1 H = –33 – 149 = –182 kJ/mole process is only known as isoentropic process. 1 (11) (BC) G Now G = H + T.d (12) (ABC). H is perfectly differential, therefore, it does not dt P depend on the method by which the final products are G = –182 + 303 × 0.5 = –30.5 kJ obtained. This is average of G at 25ºC and 35ºC (13) (ABCD) (28) (C). H1 = E1 + ngRT H = E + n RT (14) (A) (15) (A) (16) (C) 2 2 g 2 H – H = 0 So H = H (17) (C) (18) (D) (19) (C) 2 1 2 1 So E + n RT = E + n RT (20) (A) (21) (A) 1 g 1 2 g 2 (E – E ) = n R(T – T ) (22) (A) r, (B) s, (C) p, s, (D) q, s 1 2 g 2 1 But (E – E ) > 0 (23) (A) r, (B) t, (C) s (D) q, (E) p 1 2 So (E > E ) (24) (A) r, (B) p, (C) q (D) s 1 2 (29) (A) (30) (C) (31) (C) (25) (A) - r, (B) - s, (C) - p, (D) - q (32) (B) (33) (B) (34) (C) G (35) (A) (36) (C) (37) (D) (26) (D). G1 = H1 + T1 d dT P
Gyaan Sankalp 63 Thermodynamics EXERCISE - 3 Q = W = 4145.4 cal 2a (1) Ti = Q 4145.4 Rb S = rev = = 13.82 cal deg–1 T 300 2 3.64 T = ; T = 2053.42 K (9) Volume of 0.5 mole of steam at 1 atm pressure i 0.082 1.01325 0.04267 i nRT 0.5 0.0821 373 (2) For the reaction = = 15.3 L Ag+(aq) + Cl–(aq) AgCl(s) P 1.0 We have Change in volume = Vol. of steam – vol. of water + – = 15.3 –negligible = 15.3 L fHº = (AgCl, s) – fHº (Ag , aq) – fHº (Cl , aq) = [– 127.07 – 105.58 – (–167.16)] = – 65.49 kJ mol–1 Work done by the system, (3) (a) From one mole of zinc, w = Pext × volume change = 1 × 15.3 = 15.3 litre-atm the no. of moles of H2 gas evolved = 1 Hence volume of hydrogen gas evolved = 22.4 lit. = 15.3 × 101.3 J = 1549.89 J (when P = 1 atm & T = 273 K) ‘w’ should be negative as the work has been done by the w = – PV = – 1 × 22.4 lit. atm. system on the surroundings. w = –1549.89 J 8.314 = – 22.4 × J = – 2271.14 J Heat required to convert 0.5 mole of water in 100ºC to steam 0.082 = 0.5 × 40670 J = 20335J (b) For a closed system Pext = 0, therefore, w = 0 According to first law of thermodynamics, (4) H/ion pair of Li+ and Cl– E = q + w = IP1 + EA = 5.41 – 3.61 = 1.8 eV = 20335 – 1549.89 = 18785.11 J H/mol of ion pair = 1.8 × 6.023 × 1023 eV 0 = 1.8 × 6.023 × 1023 × 1.602 × 10–19 J= 173.7 kJ (10) 2H2O() 2H2O(r) at 100 C and 1 atm pressure. 0 –2 (5) G = –RT ln Kp = –2.303 × 1.987 × 700 × log 2.8 × 10 Since it is a constant pressure process, = (–3203.2) × (–1.555) = + 4981 cal Qp = H = 2 × 9720 = 19440 cal W = PV = P (V – V ) P .V = nRT Q ext V L ext V p [CO][H2 ] [50][50] (This expression assumes that the volume of the liquid is G0 = RT ln Q = = = 12.5 Kp p [HO]2 [200] negligible in comparison to that of vapour and also water vapour is ideal). 12.5 G = 2 × 700 × 2.303 log = 3224.24 × 2.65 W = 2 × 2 × 373 = 1492 cal 2.8 102 E = Q – W = 19440 – 1492 = 17948 cal = 8544.13 cal Q 19440 Since G is positive, the reaction is not spontaneous un- S = rev = = 52.12 cal deg–1 T 373 der the conditions. (11) Bi (s) Bi (g) H = 207 kJ P1 (6) W = 2.303 nRT log P 2Bi (s) Bi2(g) H = 125 kJ per 2 mole of Bi(s) 10 20 Reversing the first reaction and multiplying by 2, = 2.303 2 273 log = 8180 calories 2 1 2Bi (g) 2Bi(s) H = – 2 × 207 kJ Since the change is taking place constant temperature, 2Bi (s) Bi (g) H = 2 × 125 kJ internal energy will not change, i.e. 2 Adding, 2Bi(g) Bi2(g) H = – 414 + 250 E = 0 Q = E + W = 0 + 8180 = 8180 calories = – 164 kJ per mole (7) Total energy required in the day of Bi2(g) The bond energy of Bi – Bi bond = 164 kJ mol–1 ×24×60×60 –1 kJ (1 watt = J sec ) = 12960 kJ (12) Molar heat capacity at constant pressure, 1000 –1 Cp = 248.2 × M J.kg 12960 where M is the molar mass of the gas. Units of glycogen required = 27.22 units. –1 476 Similarly, Cv = 149 × M J. kg Cp – Cv = R V2 P1 20 248.2 × M – 149M = 8.314 (8) W = nRT ln = nRT ln = 3 × 2 × 300 × 2.303log V1 P2 2 8.314 M = = 0.0838 kg/mol = 4145.4 cal 248.2 149 For an ideal gas E (at constant t) = 0 64 Gyaan Sankalp Thermodynamics
Molar mass of the gas = 83.8 g/mol (17) The heat of combustion of carbon to CO2(g) can be calcu- lated as, (13) 2NO(g) + O2(g) 2NO2(g) C(s) + O2(g) CO2(g) H = (2 × 7.5) – (2 × 21.5) = (2 × –14) = –28 kcal Hº (Calculated) = + 718 – 2 × 339 + 498 = + 538 KJ Since H is –ve, it is an exothermic reaction. f R.E. or CO2 = Hºf, (exp) – Hºf, (cal) (14) Amount of work done in reversible isothermal expansion = – 393 – 538 = – 931 KJ
V2 (18) Using the necessary data from the table. w = –2.303nRT log CO H O CO H V1 2 2 2 –1 –1 G – –32.8 –54.69 –94.26 0 Kcal Given n = 2, R = 8.314 JK mol , Gº = – 94.26 + 0 – (–32.8) – (– 54.69) T = 298 K, V2 = 20 L and V1 = 10L. = – 6.8 Kcal / mol Substituting the values in above equation (19) G0 H 0 T. S 0 20 r r r w = –2.303 × 2 × 8.314 × 298 log 10 0 200 20 20 r H = –2.303 × 2 × 8.314 × 298 ×0.3010 1000 = –3434.9 J H at 200K 16 kJ / mole i.e., work is done by the system. r rHHC(TT) 2 r 1 r p,m 2 1 (15) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O() E = –332.0 kcal ; H = E + RT n 20 (220 200) H ( 16) where n = no. of gaseous products – no. of gaseous r 220 1000 reactants H 16 0.4 H 15.6 kJ / mole = 2 – (1 + 3) = – 2 r 220 r 220 –3 H = –332.0 – 2 × 2 × 10 × 298 |r H 220 | 15.6 kJ / mole = –332.0 –1.2 = –333.2 kcal (20) We have (16) H O() = H O(g) (P = 1 atm) 2 2 u = nCV (T2 – T1) G = H – TS = 9710 – 26 × 300 = + 1910 cal = 2 × 12.55 × (275 – 325)= – 1255 J since G is positive, at 1 atm, vaporisation is not possible. W=u = – Pext (V2 – V1)= – 0.1013 MPa (V2–V1)=– 1255 J Rather the reverse process of condensation will occur. nRT 2.8.314 325 The temperature at which the liquid and vapour will be in 1 3 Now V1 = = = 5334.75 cm equiv. can be obtained, by putting G = 0, i.e., P 1013 G = 9710 – 26T = 0 1255 T = 373.4ºC Therefore V = + V = 12388.94 cm3 + 5334.75 cm3 2 0.1013 1 This indeed is the boiling point of water at 1 atm. = 17723.69 cm3 = 17.72 dm3 EXERCISE - 4 (1) H = – 2035 kJ/mol (10) (A) (11) (A) (2) H = 0 ; W = – 5227.2 J ; E = 0, q = 5227.2 J (12) (D) (13) H = 9.9 lit atm ; U= 1 lit atm (3) – 117.62 J (5) (A) (14) (D) (15) (C) (6) (C) (7) (a) 4.34 × 10–4 m3, (b) 261.13 K (16) (D) (17) (B) (8) (B) (18) (A) (19) (B) (9) (i) W = – 620.77 J , q = + 620.77 J (ii) U = 0 , H = 0 , S = 0
Gyaan Sankalp 65