Ch. 19 Entropy and Free Energy: Spontaneous Change

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1 Ch. 19 Entropy and Free Energy: Spontaneous Change 19-1 Spontaneity: The Meaning of Spontaneous Change 19-2 The Concept of Entropy 19-3 Evaluating Entropy and Entropy Changes 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics 19-5 Standard Free Energy Change, ΔG 19-6 Free Energy Change and Equilibrium 19-7 ΔG and K eq as Functions of Temperature

2 19-1 Spontaneity: The Meaning of Spontaneous Change There is a natural direction for change in chemical processes. The natural direction is the one in which the system will proceed if left to its own devices i.e. not subject to external influence. We call these Spontaneous; changes that proceed without apparent external influence, force, cause or treatment.

3 The direction of spontaneous change generally depends on temperature N 2 (g) + O 2 (g) 2 NO(g) At Equilibrium: Reactants are favoured at low temperatures, but products are favoured at high temperatures 2NO(g) + O 2 (g) 2 NO 2 (g) At Equilibrium: Products are favoured at low temperatures, but Reactants are favoured at high temperatures

4 A Plausible Criteria for Spontaneous Processes Potential energy decreases spontaneous process are always downhill (maybe?) For chemical systems the internal energy, U, is analogous to mechanical potential energy. Berthelot and Thomsen 1870 s. Spontaneous change occurs in the direction in which the enthalpy of a system decreases. Mainly true but there are many exceptions.

5 A spectacular exception to this rule

6 19-2 The Concept of Entropy; S ΔU = ΔH = 0 Entropy, S. A statistical description: The greater the number of ways there are to arrange the microscopic particles among the energy levels in a particular system, the greater the entropy of the system. ΔS > 0; spontaneous?

7 The Boltzmann Equation for Entropy States. The microscopic energy levels available in a system. Multiplicity, W. S = k lnw The number of ways in which the microscopic particles can be distributed amongst the available states. The total number of microstates The Boltzmann constant, k. Effectively the gas constant per molecule = R/N A.

8 Boltzmann Distribution

9 But, which configuration is observed? Why don t all the gas molecules just end up on one side of the vessel (one possible configuration) rather than evenly distributed (another possible configuration)? We observe the configuration that is most likely; i.e. many different microstates will give us the same macrostate, or observed configuration. In general one configuration is statistically dominant; i.e. produced by the overwhelming majority of possible microstates. This is the configuration we observe at equilibrium.

10 Coin toss

11 Several Thousand Tosses

12 Entropy Change ΔS = q rev T For changes occurring at constant T

13 Processes in which Entropy increases Pure liquids or liquid solutions are formed from solids Gases are formed from solids or liquids The number of molecules in a gas increases due to a chemical reaction The temperature of a substance increases

14 19-3 Evaluating Entropy and Entropy Changes Phase transitions. Exchange of heat can be carried out reversibly. ΔS tr = ΔH tr T tr H 2 O(s, 1 atm) H 2 O(l, 1 atm) ΔH fus = 6.02 kj at K ΔS fus = ΔH fus T tr = 6.02 kj mol K = kj mol -1 K -1

15 Trouton s Rule ΔS = ΔH vap T bp 87 kj mol -1 K -1

16 Absolute Entropies Third law of thermodynamics: The entropy of a pure perfect crystal at 0 K is zero. Standard molar entropy. Tabulated in Appendix D of your text. ΔS = [ Σν p S (products) - Σν r S (reactants)]

17 Entropy as a Function of Temperature

18 Vibrational Energy and Entropy NO molecule NO 2 molecule In general, the larger the molecule (i.e. number of atoms) the greater the molar entropy.

19 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics. ΔS total = ΔS universe = ΔS system + ΔS surroundings The Second Law of Thermodynamics: ΔS universe = ΔS system + ΔS surroundings > 0 All spontaneous processes produce an increase in the entropy of the universe.

20 Free Energy and Free Energy Change Second law is very difficult to apply in form given on previous slide Consider hypothetical process: only pressure-volume work, at constant T and P. q surroundings = -q p = -ΔH sys Make the enthalpy change reversible. infinite surroundings, infinitesimal change in temperature. Under such conditions we can calculate entropy.

21 Free Energy (G) and Free Energy Change (ΔG) For the universe: TΔS univ. = TΔS sys ΔH sys -TΔS univ. = ΔH sys TΔS sys Define the system Free Energy, G: G = H - TS ΔG = ΔH - TΔS ΔG sys = - TΔS universe

22 Criteria for Spontaneous Change ΔG sys < 0 the process is spontaneous. ΔG sys = 0 the process is at equilibrium. ΔG sys > 0 the process is non-spontaneous. J. Willard Gibbs

24 19-5 Standard Free Energy Change, ΔG The standard free energy of formation, ΔG f The change in free energy for a reaction in which a substance in its standard state is formed from its elements in reference forms in their standard states. ΔG = [ Σν p ΔG f (products) - Σν r ΔG f (reactants)] The standard free energy of reaction, ΔG

25 19-6 Free Energy Change and Equilibrium

26 Free Energy Change and Equilibrium Condensation Equilibrium Vaporization

27 Relationship of ΔG to ΔG for Non-standard Conditions 2 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ΔG = ΔH - TΔS ΔG = ΔH - TΔS For ideal gases ΔH = ΔH ΔG = ΔH - TΔS

28 Relationship Between S and S q rev = -w = RT ln V f V i ΔS = q rev T = R ln V f V i ΔS = S f S i = R ln V f V i = R ln P i P f = -R ln P f P i S = S - R ln P P = S - R ln P 1 = S - R ln P

29 N 2 (g) + 3 H 2 (g) S N2 = S H2 = S N2 Rln P N2 S H2 Rln P H2 2 NH 3 (g) S NH3 = S NH3 Rln P NH3 ΔS rxn = 2 (S NH 3 Rln P NH3 ) (S N 2 Rln P N2 ) 3 (S H 2 Rln P H2 ) ΔS rxn = 2 S NH 3 S N2 3S H2 + Rln P N2 P H2 ΔS rxn = ΔS rxn + Rln P N2 P 3 H2 P 2 NH P NH 3

30 ΔG Under Non-Standard Conditions ΔG = ΔH - TΔS ΔS rxn = ΔS rxn + Rln P 2 N2 P 3 H2 2 ΔG = ΔH - TΔS rxn TR ln P 2 N2 P 3 H2 2 P 2 ΔG = ΔG + RT ln NH 3 PN 2 P 3 2 H2 P NH 3 P NH 3 ΔG = ΔG + RT ln Q

31 ΔG and the Equilibrium Constant K eq ΔG = ΔG + RT ln Q If the reaction is at equilibrium then: ΔG = ΔG + RT ln K eq = 0 ΔG = -RT ln K eq

32 Criteria for Spontaneous Change: A graphical perspective The direction of spontaneous change in a chemical system is the one that decreases the free energy

33 Significance of the Magnitude of ΔGo

34 The Thermodynamic Equilibrium Constant: Activities For ideal gases at 1.0 bar: S = S - R ln P P = S - R ln P 1 PV=nRT or P=(n/V)RT, pressure is an effective concentration Therefore, in solution: S = S - R ln c c = S - R ln a The effective concentration in the standard state for an ideal solution is c = 1 M.

35 Activities For pure solids and liquids: a = 1 For ideal gases: a = P (in bars, 1 bar = atm) For ideal solutes in aqueous solution: a = c (in mol L -1 )

36 The Thermodynamic Equilibrium Constant, K eq A dimensionless equilibrium constant expressed in terms of activities. Often K eq = K c or K P, but not always. Can be used to determine ΔG. ΔG = ΔG + RT ln a ga h a a a b

37 19-7 ΔG and K eq as Functions of Temperature ΔG = ΔH -TΔS ΔG = -RT ln K eq ln K eq = -ΔG RT = -ΔH RT + TΔS RT ln K eq = -ΔH RT + ΔS R

38 Temperature Dependence of K eq Assume ΔH and ΔS do not vary significantly with T ln K eq = -ΔH + RT ΔS R slope = -ΔH R -ΔH = R slope = J mol -1 K K = kj mol -1

39 Van t Hoff Equation Write equation on previous slide at two temps. and subtract ({K eq1, T 1 } and {K eq2, T 2 }): K eq2 ln = K eq1 -ΔH RT 2 ΔS + - R -ΔH RT 1 + ΔS R ln K eq2 K eq1 = -ΔH R T 2 T 1

Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry

Chemistry 123: Physical and Organic Chemistry Topic 2: Thermochemistry Recall the equation. w = -PΔV = -(1.20 atm)(1.02 L)( = -1.24 10 2 J -101 J 1 L atm Where did the conversion factor come from? Compare two versions of the gas constant and calculate. 8.3145 J/mol K 0.082057