Chapter 6 I Understanding Heats of Reaction - thermodynamics is the science of the relationships between heat and other forms of energy. Thermochemistry refers to the heat absorbed or evolved during a Thermochemistry chemical reaction.
LawLaw ofof ConservationConservation ofof EnergyEnergy Energy • Energy is defined as the capacity to move Energy can be converted from one form to matter. another but can neither be created nor Energy can be in many forms: destroyed. >Radiant Energy -Electromagnetic radiation. (E is constant) >Thermal Energy - Associated with random universe motion of a molecule or atom. >Chemical Energy - Energy stored within the structural limits of a molecule or atom.
1 Energy and Its Units - energy is the potential or capacity to move matter
2 1 Kinetic energy. Ek = ½ mv SI unit of energy is the joule (J) which has units kg @ m2/s2. 1 calorie (cal) is equal to the amt of heat necessary to raise 1 g of H2O by 1 oC. 1 cal = 4.184 J
Potential energy and kinetic energy. Photo Courtesy of Tennessee Valley Authority. A Problem to Consider
• Consider the kinetic energy of a person whose mass is 130 lb (59.0 kg) traveling in a car at 60 mph (26.8 m/s).
1 2 Ek = 2 (59.0 kg)×(26.8 m / s) 4 Ek = 2.12×10 J • The SI unit of energy, kg.m2/s2, is given the name Joule.
2 Energy A Problem to Consider
• Potential Energy: This energy depends on the • Consider the potential energy of 1000 lb of water “position” (such as height) in a “field of force” (453.6 kg) at the top of a 300 foot dam (91.44 m). (such as gravity). • For example, water of a given mass m at the 2 Ep = (453.6 kg)×(9.80 m / s )×(91.44 m) top of a dam is at a relatively high “position” 5 2 2 h in the “gravitational field” g of the earth. Ep = 4.06×10 kg⋅m / s 5 Ep = 4.06×10 J Ep = mgh
FirstFirst LawLaw Energy
• Internal Energy is the energy of the First Law of Thermodynamics: particles making up a substance.
•The total energy of a system is the sum of its kinetic energy, potential energy, and internal The energy of the universe is energy, U. constant. Etot = Ek + Ep + U
3 FirstFirst LawLaw TemperatureTemperature v.v. HeatHeat
∆E = q + w Temperature reflects random motions of particles, therefore related to kinetic energy ∆E = change in system’s internal energy of the system. q = heat Heat involves a transfer of energy between w = work 2 objects due to a temperature difference
Figure 6.5: A kinetic-theory WorkWork explanation of heat. work = force × distance
since pressure = force / area,
work = pressure × volume
wsystem = −P∆V
Where ∆V = Vfinal -Vinitial
4 StateState FunctionFunction First Law of Thermodynamics
·Both work & heat are ways in which energy can be Depends only on the present state of the transferred. system - not how it arrived there.
The way that energy transfer is divided between work and heat depends on the conditions of transfer - and is It is independent of pathway. called the “pathway”. However, the total amount of energy transferred (w + q) will always remain constant. Internal Energy is a State Function
Campsite to illustrate altitude. Heat of Reaction Internal Energy • Heat is defined as the energy that flows is a State into or out of a system because of a Function difference in temperature between the system and its surroundings. But work and heat • Heat flows from a region of higher are not! temperature to one of lower temperature; once the temperatures become equal, heat flow stops.
5 SystemSystem andand SurroundingsSurroundings Heat of Reaction
System: That on which we focus attention • In chemical reactions, heat is often transferred from the “system” to its Surroundings: Everything else in the universe “surroundings,” or vice versa. • The substance or mixture of substances under Universe = System + Surroundings study in which a change occurs is called the thermodynamic system (or simply system.) •Thesurroundings are everything in the vicinity of the thermodynamic system.
Illustration of a thermodynamic 1. An exothermic process is a system. chemical change in which heat Photo courtesy of American Color. is evolved (q is negative) 2. An endothermic process is a System = reaction between HCl and NaOH chemical reaction or physical HCl, NaOH, System change in which heat is NaCl, H2O The surroundings = flask, absorbed (q is positive) air, solvent (if not involved in the reaction) - “everything else”!
6 A test tube that contains anhydrous copper (II) sulfate and a thermometer that registers 26.1º C. Endothermic Photo courtesy of and American Color. exothermic.
Water has been added. The thermometer now registers 90.2º C. Barium hydroxide octahydrate and Photo courtesy of ammonium salt are mixed. American Color. Photo courtesy of
This dissolving American Color. process is exothermic!
H2O, -heat CuSO4 (s) º 2+ 2- Cu (aq) + SO4 (aq)
7 The flask and board are frozen ExoExo andand EndothermicEndothermic solidly together. Photo courtesy of American Heat exchange accompanies chemical reactions. Color.
. Ba(OH)2 8 H2O (s) + Exothermic: Heat flows out of the system º 2 NH4Cl (s) + heat (to the surroundings). BaCl2 (s) + 10 H2O (s) + 2 NH3 (s) Endothermic: Heat flows into the system An endothermic reaction! (from the surroundings).
Enthalpy and Enthalpy Change Enthalpy and Enthalpy Change
• The heat absorbed or evolved by a reaction • Enthalpy and Internal Energy depends on the conditions under which it – The internal energy of a system, U, is precisely defined as the heat at constant pressure plus the occurs - the “pathway”. work done by the system: Usually, a reaction takes place in an open vessel, and therefore at the constant pressure of the U = q p + w atmosphere. – In chemical systems, work is defined as a change The heat of this type of reaction is denoted qp, the in volume at a given pressure, that is: heat at constant pressure. w = −P∆V (if volume of system expands – it does work on surroundings)
8 C Enthalpy and Enthalpy Change Internal energy (U). The sum of kinetic & potential energies of the atoms making up Enthalpy is a state function. A state function a substance as well as the subatomic is a property of a system that depends only on particles of each atom. its present state, which is determined by variables such as temperature and pressure, and is independent of any previous history of
ETotal = Ek + EP + U the system. Thus, changes in enthalpy, ∆H, depend only on the initial and final states of the system not on the path (or way) in which the final state is reached.
EnthalpyEnthalpy Enthalpy and Enthalpy Change Enthalpy = H = E + PV (all state functions) • Enthalpy, denoted H, is an extensive property ∆E = ∆H −∆(PV) of a substance that can be used to obtain the ∆H = ∆E + ∆ (PV) heat absorbed or evolved in a chemical reaction. At constant pressure, –Anextensive property is one that depends on the quantity of substance. (other examples - volume, qP = ∆E + P∆V, or mass, energy). An intensive property does not )E = qp -P∆V depend on amount of substance - e.g. temperature, vapor pressure. where qP = ∆H at constant pressure, )E = Uf -Ui – Enthalpy is also a state function, a property of a system that depends only on its present state and is ∆H = energy flow as heat (at constant pressure) independent of any previous history of the system.
9 An enthalpy Enthalpy of Reaction (∆H) is the change diagram. in enthalpy for a reaction at a given temperature and pressure. 2 Na (s) + 2 H2O (l) —>
2 NaOH (aq) + H (g) ∆H = H(products) - H(reactants) 2 = qp (at constant pressure) So ∆H is essentially the heat evolved or absorbed by the system (reaction) in an open vessel where the work portion of ∆U is unmeasured.
Figure 6.9: Pressure-volume Measuring Heats of Reaction work. • To see how heats of reactions are measured, we must look at the heat required to raise the temperature of a substance, because a thermochemical measurement is based on the relationship between heat and temperature change. 2 Na (s) + 2 H2O (l) —> 2 NaOH (aq) + H2 (g) • The heat required to raise the temperature of a ) E = (-367.5 kJ - P )V) substance is its heat capacity.
10 HeatHeat CapacityCapacity Measuring Heats of Reaction
• Heat Capacity and Specific Heat heat absorbed –The heat capacity, C, of a sample of substance C = is the quantity of heat required to raise the mole substance increase in temperature temperature of one mole of the sample of substance one degree Celsius. J J – Changing the temperature of the sample = or requires heat equal to: mole° C mole K q = nC∆T
SomeSome HeatHeat ExchangeExchange TermsTerms
specific heat capacity, s Measurement of Heat of Reaction may be heat capacity per gram = J/°C g or J/K g done using a calorimeter. A calorimeter is a q = s m )T , where m is in g, )T = T -T f i device used to measure the heat absorbed or evolved during a physical or chemical change molar heat capacity, C heat capacity per mole = J/°C mol or J/K mol
q = nC )T, where n = no. moles, )T = Tf -Ti
11 Heats of Reaction: Calorimetry Coffee-cup calorimeter. •Acalorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. (see Figure Reaction is done at 6.11) constant pressure, so • The heat absorbed by the calorimeter and its we are measuring qp or )H contents is the negative of the heat of reaction.
qcalorimeter = −qrxn
A Problem to Consider A Problem to Consider
• Suppose one mole of iron requires 6.70 J of • Calculate the heat absorbed by a heat to raise its temperature by one degree calorimeter when the temperature of 15.0 Celsius. The quantity of heat required to grams of water is raised from 20.0 oC to raise the temperature of the piece of iron 50.0 oC. (The specific heat of water is from 25.0 oC to 35.0 oC is: since C x n = 4.184 J/g.oC.) (6.70 J/mole C) x 1 mole q = s×m× ∆T q = nC∆T = (6.70 J / o C)×(35.0 oC − 25.0 oC) J o q = (4.184 o )× (15.0g)× (50.0 − 20.0 C) q = 67.0 J g⋅ C q = 1.88×103 J
12 A Problem to Consider Heats of Reaction: Calorimetry
– First, let us calculate the heat absorbed • When 23.6 grams of calcium chloride, CaCl2, was dissolved in water in a calorimeter, the by the calorimeter. temperature rose from 25.0 oC to 38.7 oC. o o o qcal = (1258 J/ C )(38.7 C - 25.0 C) If the heat capacity of the solution and the calorimeter o 4 is 1258 J/ C, what is the enthalpy change per mole qcal = 1.72×10 J of calcium chloride? • Now we must calculate the heat per mole of calcium chloride.
Heats of Reaction: Calorimetry • Calcium chloride has a molecular mass of 111.1 g, so A bomb (1 mol CaCl ) calorimeter (23.6 g CaCl )× 2 = 0.212 mol CaCl 2 111.1 g 2 qv = )E • Now we can calculate the heat per mole of since w = 0 calcium chloride, using qcal = -qrxn. q −17.2 kJ ∆H = rxn = = −81.1 kJ / mol mol CaCl2 0.212 mol
13 Applying Stoichiometry and Thermochemical Equations - the chemical equation for a reaction (including phase Heats of Reactions • Consider the reaction of methane, CH4, burning labels) in which the equation is given a in the presence of oxygen at constant pressure. molar interpretation, and the enthalpy of Given the following equation, how much heat reaction for these molar amounts is written could be obtained by the combustion of 10.0 directly after the equation grams CH4?
o CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l); ∆H = -890.3 kJ 2 Na (s) + 2 H2O (l)± 2 NaOH (aq) 1mol CH4 −890.3 kJ + H2 (g); ∆H = - 367.5 kJ 10.0 g CH × × = −556 kJ 4 16.0g 1 mol CH4
Hess’sHess’s LawLaw Hess’s Law
• Hess’s law of heat summation states that Reactants → Products for a chemical equation that can be written as the sum of two or more steps, the The change in enthalpy is the same whether enthalpy change for the overall equation is the reaction takes place in one step or a the sum of the enthalpy changes for the series of steps. individual steps.
14 CalculationsCalculations viavia Hess’sHess’s LawLaw
1. If a reaction is reversed, ∆H is also reversed. What is the heat of formation for the reaction? 2 C (graphite) + O2 (g) —> 2 CO (g) Given: N2(g) + O2(g) → 2NO(g) ∆H = 180 kJ C (graphite) + O (g) Æ CO (g) ) H = -393.5 kJ 2NO(g) → N (g) + O (g) ∆H = −180 kJ 2 2 2 2 2 CO (g) + O2 (g) Æ 2 CO2 (g) ) H = -566 kJ 2. If the coefficients of a reaction are multiplied So: 2 C (graphite) + 2 O (g) —> 2 CO (g) by an integer, ∆H is multiplied by that same 2 2 ) H = (-393.5 kJ) x 2 integer. 2CO2 (g) —> 2 CO (g) + O2 (g) ) H = (-566 kJ) x -1 2 C (graphite) + O2 (g) —> 2 CO (g) ) H = -221.0 kJ 6NO(g) → 3N2(g) + 3O2(g) ∆H = −540 kJ
Enthalpy diagram illustrating Hess’s Law • For example, suppose you are given the Hess’s law. following data:
Net equation: o 2 C (graphite) + O (g) Æ 2 CO (g) 2 S(s) + O2 (g) → SO2 (g); ∆H = -297 kJ o 2SO3 (g) → 2SO2 (g) + O2 (g); ∆H = 198 kJ
• Could you use these data to obtain the enthalpy change for the following reaction?
o 2S(s) + 3O2 (g) → 2SO3 (g); ∆H = ?
15 2 S (s) + 2 O (g) —> 2 SO (g) )H = (-297 kJ) x (2) 2 2 Standard Enthalpies of 2 SO (g) + O (g) —> 2 SO (g) )H = (198 kJ) x (-1) 2 2 3 Formation 2 S (s) + 3 O (g) —> 2 SO (g) )H = - 792 kJ 2 3 • The standard enthalpy of formation of a We have multiplied the equation for the formation of sulfur o dioxide by 2 and we have multiplied the heat of formation for substance, denoted ∆Hf , is the enthalpy sulfur dioxide by 2. change for the formation of one mole of a substance in its standard state from its We have reversed the reaction for the formation of sulfur trioxide from sulfur dioxide and oxygen (and changed the sign component elements in their standard for the corresponding heat of reaction) state.
We have added the reactions and the heats of reaction to –Note that the standard enthalpy of formation obtain the values for the heat of formation of sulfur trioxide. for a pure element in its standard state is zero.
StandardStandard StatesStates
Compound For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure liquid or solid. Element
The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.
16 Allotropes of sulfur.
o Photo courtesy of James Scherer. What is )H f for
HCl (aq)?
Remember that HCl is a strong acid and HCl (aq) º H+(aq) + Cl- (aq)
o So to use the )H f for Cl- (aq) and An allotrope is one of two or more distinct forms of an element + H (aq) (-167.5 kJ + in the same physical state. Shown here two allotropes of S8, 0 kJ = - 167.5 kJ) rhombic (left) and monoclinic (right). The rhombic form is more stable at room temperature and is the reference form.
Standard Enthalpies of Formation •The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants. o o o ∆H = ∑n∆Hf (products) − ∑m∆Hf (reactants) Σ is the mathematical symbol meaning “the sum of”, and m and n are the coefficients of the substances in the chemical equation. 68
17 A Problem to Consider A Problem to Consider • You record the values of ∆H o under the • Large quantities of ammonia are used to f formulas in the equation, multiplying them prepare nitric acid according to the by the coefficients in the equation. following equation:
4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2O(g) 4NH3 (g) + 5O2 (g) → 4NO(g) + 6H2O(g) - What is the standard enthalpy change for this 4(−45.9) 5)(0) 4)(90.3 6(−241.8 reaction? Use Table 6.2 for data. - You can calculate ∆Ho by subtracting the values for the reactants from the values for the products.
Figure 6.10: A pathway for the A Problem to Consider combustion of ammonia. • Using the summation law: o o o ∆H = ∑n∆Hf (products) − ∑m∆Hf (reactants) ∆Ho = [4(90.3) + 6(−241.8)] kJ −[4(−45.9) + 5(0)] kJ
∆Ho = −906 kJ
- Be careful of arithmetic signs as they are a likely source of mistakes.
18 Practice Problem 6.45 Practice Problem 6.46
CS2 is used in the m anufacture of p lastics and cellophane. Keep in mind the following C alculate ) H o fo r th e reactio n o b elow using the ) H f for the concepts. r eactants and products shown b elow :
C S 2 (g ) + 3 O 2 (g) [1 m ol (87.9 kJ/m ol) + 3 mol (0) ] C O 2 (g ) + 2 S O 2 (g) [1 mol (-393.5 kJ/mol)+2 mol(-296.8 kJ/mol)
o o o ) H = [E m ) H f (products)]-[n) H f (reactant
= [(-393.5 kJ)+(-593.6 kJ)]-[87.9 kJ] = - 1075.0 kJ
19