DOCSLIB.ORG

Lecture Notes 7 – Fall 2013 – Gibbs & Helmholtz Free Energy

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Lecture Notes 7 – Fall 2013 – Gibbs & Helmholtz Free Energy Chem 340 - Lecture Notes 7 – Fall 2013 – Gibbs & Helmholtz Free Energy With entropy we found a state function that indicates direction of a spontaneous process, however it requires consideration of the entropy change of both system and surroundings, as chemists we prefer to focus on the system, usually not isolated, i.e. in contact with surroundings, and often we work in conditions of constant pressure and sometimes also at constant temperature. Alternatively one can imagine cases where constant volume is a characteristic. Use of these ideas helps design two new new state functions that can indicate direction of a process and one of which most useful for chemistry. The Clausius inequality will be critical in making the indication work. dS ≥ ᵭq/T T dS ≥ ᵭq recall: only equal for reversible process, dqrev recall 1st law: dU = ᵭq + ᵭw rearrange, substitute dq: TdS ≥ dU - ᵭw or TdS + ᵭw - dU ≥ 0 now there are two kinds of work, external or expansion, and non-expansion work, split up: TdS - PextdV + ᵭwnonexp - dU ≥ 0 Spontaneity now a function of state functions: U, V, S, T, and path dependent ones w Consider isothermal process: (const T) Define new state function, Helmholtz free energy: A = U - TS or equivalently: A is a state function that calculates maximum work from system on surround, i.e. A = wrev Now if add constraint: constant V and only expansion work possible then dA ≤ 0 Chemists prefer const P conditions, at const T,P: d(PV) = PdV, d(TS) = TdS, back to inequ. dU - TdS + PdV ≤ 0 d(U +PV –TS) = d(H - TS) ≤ ᵭwnonexp let new state function be Gibbs Free energy: G = H – TS, dG ≤ ᵭwnonexp or if ᵭwnonexp = 0, then dG ≤ 0 indicates spont, const T,P only PV work so now have two criterion for spontaneity, each works best under different conditions, dA for constant V, dG for constant P. Go back to Clausius dS - ᵭq/T ≥ 0 recall dSsur = - ᵭq/T so dS - dSsur ≥ 0 or the spontaneity condition 1 but for G and A - criteria for spontaneity - not depend on surroundings, also from Clausius. G = H - TS and G < 0 spontaneous isothermal, constant P process See both enthalpy and entropy contribute to spontaneity, Entropy greater impact at higher temperatures Chemical reaction spontaneous if H < 0 (exothermic) and S > 0 Chemical reaction never spontaneous if H > 0 (endothermic) and S < 0 Balance of H and S control reaction, if H < 0 & S < 0 or H > 0 & S > 0 Similar for constant V change A = U - TS and A < 0 spontaneous isothermal, constant V process Examples: -from Engel,on web site, but - sign/magnitude problems in text!! Calculating A, U, S and w for reversibly contracting collagen from l = 0.20.1 m with -1 absorption of qrev = 0.15 J at const T = 310 K, let tension = - l, = 10 Nm For this, Uw = -q, A = wrev = -0.15 J same mechanical stretch w = -∫dl = ∫ldl -4 -1 -1. 2 S = qrev/T = 0.15 J / 310 K = 4.83x10 JK 10Nm ½ (-0.03 m ) = -0.15 J If dU = TdS – PdV - dl determine (G/l)P,T and maximum work obtainable by contract dG = dH – d(TS) = d(U +PV) – d(TS) = TdS – PdV - dl + PdV + VdP – TdS – SdT = - dl + VdP – SdT sign issue! l for contraction Constant P,T: dT = dP = 0 (G/l)P,T = - l, dG =ldl Integrate and get G =∫ldl = -0.15 J = wrev and see same as A = wrev 2 Example of RNA stretching. If hook opposite ends of strand to two polystyrene beads, can pull apart and measure force required a function of distance. Need very small scale and to grab spheres (~1m) laser tweezer can hold an object and measure force exerted See (middle, black stretch, red refold) at 10 pNs-1, the path not reversible, but at 1 pNs-1 (right) black and red overlap, approximately reverses. See at ~13 pN an abrupt increase in length, this is unfolding, and length difference (~20 nm) is length of the RNA. From this work, determined: G = 174 kJ/mol, compare to melting (phase trans) at ~80 C Fundamental equations - Differential forms of U, H, A, G Start from definitions and re-express as differentials in terms of thermodynamic variables: H = U + PV method use dU = dq + dw but recognize state function A = U – TS dU = dqrev + dwrev = TdS - PdV G = H – TS assume only P,V work Take differentials: dH = dU + d(PV) = TdS – PdV + PdV + VdP = TdS + VdP dA = dU – d(TS) = TdS – PdV – TdS – SdT = -SdT – PdV dG = dH – d(TS) = TdS + VdP – TdS – SdT = -SdT + VdP Used a lot, called the fundamental equations: dU = TdS – PdV dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP - text calls this fundamental equation of chemical thermodynamics these are used to develop a set of ”natural variables”; total differential two variables, U(S,V) Coefficients of dS and dV must be same in both equations for dU 3 Same for the other equations (left): right use: (/V(U/S)V)S = (/S(U/V)S)V Left, define variances of thermodynamic right, Maxwell relationships, shows inter- state functions in terms of variables relationships of variables (T,P,V,S) for example H inc. with S,P (T,V = +) note first two have derivative at const S but G dec. w/ T and inc. w/ P (S,V = +) since dS=dqrev/T, const S dq=0 adiabatic these relations look obscure but allow express state fct. in measurable things : T,P,V, Example from text: in Notes 4 introduced internal pressure (U/V)T = T = T(P/T)V – P From complete differential: dU = (U/S)VdS + (U/V)SdV divide through by dV, at const T (U/V)T = (U/S)V(S/V)T + (U/V)S Now use (U/S)V = T and (U/V)V = - P T = (U/V)T = T(S/V)T – P and then use Maxwell relation: ( S/ V) = ( P/ T) = T( P/ T) – P (result) T V T V Dependence of Free energies on P, V, T Just described how free energy depends on “natural variables: So since S and P always positive, see A always decreases with increasing T or V Similarly for G; S and V always positive So G always decreases with T and increases with P 4 If at const T, and want change in free energy from standard state, can integrate over P Liquid and solid, V~ const, so simplify: But in gases volume strongly depends on P, so substitute in for V in integral: See Gm - ∞ as P 0, since for gas that means V∞ and S = nR ln(Vf/Vi) large +, See text, fig 3.19, 3.20 p. 125, show high T,P dependence of G for gas, more S,V Standard molar Gibbs Free Energy o Normally only use G, so absolute values not needed, but for standard state can get a G m o If we take H m = 0 in its standard state, for a pure element, then o o o o o o G m = H m - TS m = - TS m (since for elements in std. state: H mH f = 0) o o o Pure compounds are formed from elemental react – formation: G f = G m prod +iG m, react i o o o Rearrange: G m = G f + iTS m, react i - reactants in formation i - only elements o so only TS m, react i term and i all negative, since reactant o o o note for molar enthalpy, equation simpler, misses elements, no S m contribut.: H m = H f o Example. Calculate G m (a) for Ar(g) and (b) for H2O(l) at T=298 K, P=1 bar o o o o -1 -1 o (a) G m = H m - TS m = - TS m = -298Kx155 JK mol = - 4.62 kJ/mol (S m from table) o o o o o o (b) G m = G f + iTS m =G f (H2O) - TS m(H2) -1/2 TS m(O2) (- for reactants) = -237kJ/mol – 298K(131 + 205/2) J/Kmol = -307 kJ/mol o o o Can just extend this to a Hess’s law formulation as well, G f = ProdG f - ReacG f o o o using standard free energy of formation: G f = H f - TS f [- for reactant] o o o So problems solved by just looking up G f values or alternatively H f and S f in tables + For ionic solutions, need to have both cations and anions, so define H aq for reference as 0: o + G f (H aq) = 0 Text example: HCl formation in solution: + - o ½ H2 + ½ Cl2 H aq +Cl aq G f = -131 kJ/mol o o + o - o - o - But G f =G f (H aq) +G f (Cl aq) = G f (Cl aq) or G f (Cl aq ) = -131 kJ/mol + - o Now consider: Ag(s) + ½ Cl2 Ag aq +Cl aq G f = -54 kJ/mol o + This implies G f (Ag aq) = 77 kJ/mol (i.e. -54 –(-131) = 77) Remember that the process of forming an aqueous ion involves going from the element to the ion and then dissolving ion, so need to construct a cycle to model it (text p.120) 5 Gibbs-Helmholtz equations For temperature dependence, one normally looks at the variation of G/T because equilibrium constants depend on G/T. This results in the Gibbs-Helmholtz equations: This is also written, using; d(1/T)/dT = -1/T2 as: (look ahead, Keq will depend on G/T and if get variation of it with 1/T gives solution for H) For a reaction at constant P need to substitute G and H in equation above, and next “multiply” both sides by d(1/T), then if integrate both sides get: Where in the last step we assume H is constant over the T range This equation lets you determine G at some temperature, if know Go and Ho at 298K o Example: Calculate Gf for glycine, H2N(CH2)COOH, at 350K if G f = -378 kJ/mol and o o H f = -537 kJ/mol at T=298 K.
Load more

Recommended publications
  • Thermodynamic Potentials and Natural Variables
    Revista Brasileira de Ensino de Física, vol. 42, e20190127 (2020) Articles www.scielo.br/rbef cb DOI: http://dx.doi.org/10.1590/1806-9126-RBEF-2019-0127 Licença Creative Commons Thermodynamic Potentials and Natural Variables M. Amaku1,2, F. A. B. Coutinho*1, L. N. Oliveira3 1Universidade de São Paulo, Faculdade de Medicina, São Paulo, SP, Brasil 2Universidade de São Paulo, Faculdade de Medicina Veterinária e Zootecnia, São Paulo, SP, Brasil 3Universidade de São Paulo, Instituto de Física de São Carlos, São Carlos, SP, Brasil Received on May 30, 2019. Revised on September 13, 2018. Accepted on October 4, 2019. Most books on Thermodynamics explain what thermodynamic potentials are and how conveniently they describe the properties of physical systems. Certain books add that, to be useful, the thermodynamic potentials must be expressed in their “natural variables”. Here we show that, given a set of physical variables, an appropriate thermodynamic potential can always be defined, which contains all the thermodynamic information about the system. We adopt various perspectives to discuss this point, which to the best of our knowledge has not been clearly presented in the literature. Keywords: Thermodynamic Potentials, natural variables, Legendre transforms. 1. Introduction same statement cannot be applied to the temperature. In real fluids, even in simple ones, the proportionality Basic concepts are most easily understood when we dis- to T is washed out, and the Internal Energy is more cuss simple systems. Consider an ideal gas in a cylinder. conveniently expressed as a function of the entropy and The cylinder is closed, its walls are conducting, and a volume: U = U(S, V ).
    [Show full text]
  • PDF Version of Helmholtz Free Energy
    Free energy Free Energy at Constant T and V Starting with the First Law dU = δw + δq At constant temperature and volume we have δw = 0 and dU = δq Free Energy at Constant T and V Starting with the First Law dU = δw + δq At constant temperature and volume we have δw = 0 and dU = δq Recall that dS ≥ δq/T so we have dU ≤ TdS Free Energy at Constant T and V Starting with the First Law dU = δw + δq At constant temperature and volume we have δw = 0 and dU = δq Recall that dS ≥ δq/T so we have dU ≤ TdS which leads to dU - TdS ≤ 0 Since T and V are constant we can write this as d(U - TS) ≤ 0 The quantity in parentheses is a measure of the spontaneity of the system that depends on known state functions. Definition of Helmholtz Free Energy We define a new state function: A = U -TS such that dA ≤ 0. We call A the Helmholtz free energy. At constant T and V the Helmholtz free energy will decrease until all possible spontaneous processes have occurred. At that point the system will be in equilibrium. The condition for equilibrium is dA = 0. A time Definition of Helmholtz Free Energy Expressing the change in the Helmholtz free energy we have ∆A = ∆U – T∆S for an isothermal change from one state to another. The condition for spontaneous change is that ∆A is less than zero and the condition for equilibrium is that ∆A = 0. We write ∆A = ∆U – T∆S ≤ 0 (at constant T and V) Definition of Helmholtz Free Energy Expressing the change in the Helmholtz free energy we have ∆A = ∆U – T∆S for an isothermal change from one state to another.
    [Show full text]
  • Chapter 3. Second and Third Law of Thermodynamics
    Chapter 3. Second and third law of thermodynamics Important Concepts Review Entropy; Gibbs Free Energy • Entropy (S) – definitions Law of Corresponding States (ch 1 notes) • Entropy changes in reversible and Reduced pressure, temperatures, volumes irreversible processes • Entropy of mixing of ideal gases • 2nd law of thermodynamics • 3rd law of thermodynamics Math • Free energy Numerical integration by computer • Maxwell relations (Trapezoidal integration • Dependence of free energy on P, V, T https://en.wikipedia.org/wiki/Trapezoidal_rule) • Thermodynamic functions of mixtures Properties of partial differential equations • Partial molar quantities and chemical Rules for inequalities potential Major Concept Review • Adiabats vs. isotherms p1V1 p2V2 • Sign convention for work and heat w done on c=C /R vm system, q supplied to system : + p1V1 p2V2 =Cp/CV w done by system, q removed from system : c c V1T1 V2T2 - • Joule-Thomson expansion (DH=0); • State variables depend on final & initial state; not Joule-Thomson coefficient, inversion path. temperature • Reversible change occurs in series of equilibrium V states T TT V P p • Adiabatic q = 0; Isothermal DT = 0 H CP • Equations of state for enthalpy, H and internal • Formation reaction; enthalpies of energy, U reaction, Hess’s Law; other changes D rxn H iD f Hi i T D rxn H Drxn Href DrxnCpdT Tref • Calorimetry Spontaneous and Nonspontaneous Changes First Law: when one form of energy is converted to another, the total energy in universe is conserved. • Does not give any other restriction on a process • But many processes have a natural direction Examples • gas expands into a vacuum; not the reverse • can burn paper; can't unburn paper • heat never flows spontaneously from cold to hot These changes are called nonspontaneous changes.
    [Show full text]
  • Thermodynamics
    ME346A Introduction to Statistical Mechanics { Wei Cai { Stanford University { Win 2011 Handout 6. Thermodynamics January 26, 2011 Contents 1 Laws of thermodynamics 2 1.1 The zeroth law . .3 1.2 The first law . .4 1.3 The second law . .5 1.3.1 Efficiency of Carnot engine . .5 1.3.2 Alternative statements of the second law . .7 1.4 The third law . .8 2 Mathematics of thermodynamics 9 2.1 Equation of state . .9 2.2 Gibbs-Duhem relation . 11 2.2.1 Homogeneous function . 11 2.2.2 Virial theorem / Euler theorem . 12 2.3 Maxwell relations . 13 2.4 Legendre transform . 15 2.5 Thermodynamic potentials . 16 3 Worked examples 21 3.1 Thermodynamic potentials and Maxwell's relation . 21 3.2 Properties of ideal gas . 24 3.3 Gas expansion . 28 4 Irreversible processes 32 4.1 Entropy and irreversibility . 32 4.2 Variational statement of second law . 32 1 In the 1st lecture, we will discuss the concepts of thermodynamics, namely its 4 laws. The most important concepts are the second law and the notion of Entropy. (reading assignment: Reif x 3.10, 3.11) In the 2nd lecture, We will discuss the mathematics of thermodynamics, i.e. the machinery to make quantitative predictions. We will deal with partial derivatives and Legendre transforms. (reading assignment: Reif x 4.1-4.7, 5.1-5.12) 1 Laws of thermodynamics Thermodynamics is a branch of science connected with the nature of heat and its conver- sion to mechanical, electrical and chemical energy. (The Webster pocket dictionary defines, Thermodynamics: physics of heat.) Historically, it grew out of efforts to construct more efficient heat engines | devices for ex- tracting useful work from expanding hot gases (http://www.answers.com/thermodynamics).
    [Show full text]
  • Thermodynamic Analysis Based on the Second-Order Variations of Thermodynamic Potentials
    Theoret. Appl. Mech., Vol.35, No.1-3, pp. 215{234, Belgrade 2008 Thermodynamic analysis based on the second-order variations of thermodynamic potentials Vlado A. Lubarda ¤ Abstract An analysis of the Gibbs conditions of stable thermodynamic equi- librium, based on the constrained minimization of the four fundamen- tal thermodynamic potentials, is presented with a particular attention given to the previously unexplored connections between the second- order variations of thermodynamic potentials. These connections are used to establish the convexity properties of all potentials in relation to each other, which systematically deliver thermodynamic relationships between the speci¯c heats, and the isentropic and isothermal bulk mod- uli and compressibilities. The comparison with the classical derivation is then given. Keywords: Gibbs conditions, internal energy, second-order variations, speci¯c heats, thermodynamic potentials 1 Introduction The Gibbs conditions of thermodynamic equilibrium are of great importance in the analysis of the equilibrium and stability of homogeneous and heterogeneous thermodynamic systems [1]. The system is in a thermodynamic equilibrium if its state variables do not spontaneously change with time. As a consequence of the second law of thermodynamics, the equilibrium state of an isolated sys- tem at constant volume and internal energy is the state with the maximum value of the total entropy. Alternatively, among all neighboring states with the ¤Department of Mechanical and Aerospace Engineering, University of California, San Diego; La Jolla, CA 92093-0411, USA and Montenegrin Academy of Sciences and Arts, Rista Stijovi¶ca5, 81000 Podgorica, Montenegro, e-mail: [email protected]; [email protected] 215 216 Vlado A. Lubarda same volume and total entropy, the equilibrium state is one with the lowest total internal energy.
    [Show full text]
  • Standard Thermodynamic Values
    Standard Thermodynamic Values Enthalpy Entropy (J Gibbs Free Energy Formula State of Matter (kJ/mol) mol/K) (kJ/mol) (NH4)2O (l) -430.70096 267.52496 -267.10656 (NH4)2SiF6 (s hexagonal) -2681.69296 280.24432 -2365.54992 (NH4)2SO4 (s) -1180.85032 220.0784 -901.90304 Ag (s) 0 42.55128 0 Ag (g) 284.55384 172.887064 245.68448 Ag+1 (aq) 105.579056 72.67608 77.123672 Ag2 (g) 409.99016 257.02312 358.778 Ag2C2O4 (s) -673.2056 209.2 -584.0864 Ag2CO3 (s) -505.8456 167.36 -436.8096 Ag2CrO4 (s) -731.73976 217.568 -641.8256 Ag2MoO4 (s) -840.5656 213.384 -748.0992 Ag2O (s) -31.04528 121.336 -11.21312 Ag2O2 (s) -24.2672 117.152 27.6144 Ag2O3 (s) 33.8904 100.416 121.336 Ag2S (s beta) -29.41352 150.624 -39.45512 Ag2S (s alpha orthorhombic) -32.59336 144.01328 -40.66848 Ag2Se (s) -37.656 150.70768 -44.3504 Ag2SeO3 (s) -365.2632 230.12 -304.1768 Ag2SeO4 (s) -420.492 248.5296 -334.3016 Ag2SO3 (s) -490.7832 158.1552 -411.2872 Ag2SO4 (s) -715.8824 200.4136 -618.47888 Ag2Te (s) -37.2376 154.808 43.0952 AgBr (s) -100.37416 107.1104 -96.90144 AgBrO3 (s) -27.196 152.716 54.392 AgCl (s) -127.06808 96.232 -109.804896 AgClO2 (s) 8.7864 134.55744 75.7304 AgCN (s) 146.0216 107.19408 156.9 AgF•2H2O (s) -800.8176 174.8912 -671.1136 AgI (s) -61.83952 115.4784 -66.19088 AgIO3 (s) -171.1256 149.3688 -93.7216 AgN3 (s) 308.7792 104.1816 376.1416 AgNO2 (s) -45.06168 128.19776 19.07904 AgNO3 (s) -124.39032 140.91712 -33.472 AgO (s) -11.42232 57.78104 14.2256 AgOCN (s) -95.3952 121.336 -58.1576 AgReO4 (s) -736.384 153.1344 -635.5496 AgSCN (s) 87.864 130.9592 101.37832 Al (s)
    [Show full text]
  • Chapter 19 Chemical Thermodynamics
    Chapter 19 Chemical Thermodynamics Entropy and free energy Learning goals and key skills: Explain and apply the terms spontaneous process, reversible process, irreversible process, and isothermal process. Define entropy and state the second law of thermodynamics. Calculate DS for a phase change. Explain how the entropy of a system is related to the number of possible microstates. Describe the kinds of molecular motion that a molecule can possess. Predict the sign of DS for physical and chemical processes. State the third law of thermodynamics. Compare the values of standard molar entropies. Calculate standard entropy changes for a system from standard molar entropies. Calculate the Gibbs free energy from the enthalpy change and entropy change at a given temperature. Use free energy changes to predict whether reactions are spontaneous. Calculate standard free energy changes using standard free energies of formation. Predict the effect of temperature on spontaneity given DH and DS. Calculate DG under nonstandard conditions. Relate DG°and equilibrium constant (K). Review Chapter 5: energy, enthalpy, 1st law of thermo Thermodynamics: the science of heat and work Thermochemistry: the relationship between chemical reactions and energy changes Energy (E) The capacity to do work or to transfer heat. Work (w) The energy expended to move an object against an opposing force. w = F d Heat (q) Derived from the movements of atoms and molecules (including vibrations and rotations). Enthalpy (H) Enthalpy is the heat absorbed (or released) by a system during a constant-pressure process. 1 0th Law of Thermodynamics If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A.
    [Show full text]
  • Energy and Enthalpy Thermodynamics
    Energy and Energy and Enthalpy Thermodynamics The internal energy (E) of a system consists of The energy change of a reaction the kinetic energy of all the particles (related to is measured at constant temperature) plus the potential energy of volume (in a bomb interaction between the particles and within the calorimeter). particles (eg bonding). We can only measure the change in energy of the system (units = J or Nm). More conveniently reactions are performed at constant Energy pressure which measures enthalpy change, ∆H. initial state final state ∆H ~ ∆E for most reactions we study. final state initial state ∆H < 0 exothermic reaction Energy "lost" to surroundings Energy "gained" from surroundings ∆H > 0 endothermic reaction < 0 > 0 2 o Enthalpy of formation, fH Hess’s Law o Hess's Law: The heat change in any reaction is the The standard enthalpy of formation, fH , is the change in enthalpy when one mole of a substance is formed from same whether the reaction takes place in one step or its elements under a standard pressure of 1 atm. several steps, i.e. the overall energy change of a reaction is independent of the route taken. The heat of formation of any element in its standard state is defined as zero. o The standard enthalpy of reaction, H , is the sum of the enthalpy of the products minus the sum of the enthalpy of the reactants. Start End o o o H = prod nfH - react nfH 3 4 Example Application – energy foods! Calculate Ho for CH (g) + 2O (g) CO (g) + 2H O(l) Do you get more energy from the metabolism of 1.0 g of sugar or
    [Show full text]
  • Thermodynamic Potential of Free Energy for Thermo-Elastic-Plastic Body
    Continuum Mech. Thermodyn. (2018) 30:221–232 https://doi.org/10.1007/s00161-017-0597-3 ORIGINAL ARTICLE Z. Sloderbach´ · J. Paja˛k Thermodynamic potential of free energy for thermo-elastic-plastic body Received: 22 January 2017 / Accepted: 11 September 2017 / Published online: 22 September 2017 © The Author(s) 2017. This article is an open access publication Abstract The procedure of derivation of thermodynamic potential of free energy (Helmholtz free energy) for a thermo-elastic-plastic body is presented. This procedure concerns a special thermodynamic model of a thermo-elastic-plastic body with isotropic hardening characteristics. The classical thermodynamics of irre- versible processes for material characterized by macroscopic internal parameters is used in the derivation. Thermodynamic potential of free energy may be used for practical determination of the level of stored energy accumulated in material during plastic processing applied, e.g., for industry components and other machinery parts received by plastic deformation processing. In this paper the stored energy for the simple stretching of austenitic steel will be presented. Keywords Free energy · Thermo-elastic-plastic body · Enthalpy · Exergy · Stored energy of plastic deformations · Mechanical energy dissipation · Thermodynamic reference state 1 Introduction The derivation of thermodynamic potential of free energy for a model of the thermo-elastic-plastic body with the isotropic hardening is presented. The additive form of the thermodynamic potential is assumed, see, e.g., [1–4]. Such assumption causes that the influence of plastic strains on thermo-elastic properties of the bodies is not taken into account, meaning the lack of “elastic–plastic” coupling effects, see Sloderbach´ [4–10].
    [Show full text]
  • Entropy and Free Energy
    Spontaneous Change: Entropy and Free Energy 2nd and 3rd Laws of Thermodynamics Problem Set: Chapter 20 questions 29, 33, 39, 41, 43, 45, 49, 51, 60, 63, 68, 75 The second law of thermodynamics looks mathematically simple but it has so many subtle and complex implications that it makes most chemistry majors sweat a lot before (and after) they graduate. Fortunately its practical, down-to-earth applications are easy and crystal clear. We can build on those to get to very sophisticated conclusions about the behavior of material substances and objects in our lives. Frank L. Lambert Experimental Observations that Led to the Formulation of the 2nd Law 1) It is impossible by a cycle process to take heat from the hot system and convert it into work without at the same time transferring some heat to cold surroundings. In the other words, the efficiency of an engine cannot be 100%. (Lord Kelvin) 2) It is impossible to transfer heat from a cold system to a hot surroundings without converting a certain amount of work into additional heat of surroundings. In the other words, a refrigerator releases more heat to surroundings than it takes from the system. (Clausius) Note 1: Even though the need to describe an engine and a refrigerator resulted in formulating the 2nd Law of Thermodynamics, this law is universal (similarly to the 1st Law) and applicable to all processes. Note 2: To use the Laws of Thermodynamics we need to understand what the system and surroundings are. Universe = System + Surroundings universe Matter Surroundings (Huge) Work System Matter, Heat, Work Heat 1.
    [Show full text]
  • Chemistry 130 Gibbs Free Energy
    Chemistry 130 Gibbs Free Energy Dr. John F. C. Turner 409 Buehler Hall [email protected] Chemistry 130 Equilibrium and energy So far in chemistry 130, and in Chemistry 120, we have described chemical reactions thermodynamically by using U - the change in internal energy, U, which involves heat transferring in or out of the system only or H - the change in enthalpy, H, which involves heat transfers in and out of the system as well as changes in work. U applies at constant volume, where as H applies at constant pressure. Chemistry 130 Equilibrium and energy When chemical systems change, either physically through melting, evaporation, freezing or some other physical process variables (V, P, T) or chemically by reaction variables (ni) they move to a point of equilibrium by either exothermic or endothermic processes. Characterizing the change as exothermic or endothermic does not tell us whether the change is spontaneous or not. Both endothermic and exothermic processes are seen to occur spontaneously. Chemistry 130 Equilibrium and energy Our descriptions of reactions and other chemical changes are on the basis of exothermicity or endothermicity ± whether H is negative or positive H is negative ± exothermic H is positive ± endothermic As a description of changes in heat content and work, these are adequate but they do not describe whether a process is spontaneous or not. There are endothermic processes that are spontaneous ± evaporation of water, the dissolution of ammonium chloride in water, the melting of ice and so on. We need a thermodynamic description of spontaneous processes in order to fully describe a chemical system Chemistry 130 Equilibrium and energy A spontaneous process is one that takes place without any influence external to the system.
    [Show full text]
  • Lecture 13. Thermodynamic Potentials (Ch
    Lecture 13. Thermodynamic Potentials (Ch. 5) So far, we have been using the total internal energy U and, sometimes, the enthalpy H to characterize various macroscopic systems. These functions are called the thermodynamic potentials: all the thermodynamic properties of the system can be found by taking partial derivatives of the TP. For each TP, a set of so-called “natural variables” exists: −= + μ NddVPSdTUd = + + μ NddPVSdTHd Today we’ll introduce the other two thermodynamic potentials: theHelmhotzfree energy F and Gibbs free energy G. Depending on the type of a process, one of these four thermodynamic potentials provides the most convenient description (and is tabulated). All four functions have units of energy. When considering different types of Potential Variables processes, we will be interested in two main U (S,V,N) S, V, N issues: H (S,P,N) S, P, N what determines the stability of a system and how the system evolves towards an F (T,V,N) V, T, N equilibrium; G (T,P,N) P, T, N how much work can be extracted from a system. Diffusive Equilibrium and Chemical Potential For completeness, let’s rcall what we’vee learned about the chemical potential. 1 P μ d U T d= S− P+ μ dVd d= S Nd U +dV d − N T T T The meaning of the partial derivative (∂S/∂N)U,V : let’s fix VA and VB (the membrane’s position is fixed), but U , V , S U , V , S A A A B B B assume that the membrane becomes permeable for gas molecules (exchange of both U and N between the sub- ns ilesystems, the molecuA and B are the same ).
    [Show full text]