Aim # 18: What is the Gibbs Free Energy? H.W. # 18 Study pp. 798-805 (sec. 17.4 – 17.5) Ans. ques. p. 824 # 36,37,39,41,42,45, 46c,47 Do Now: Zumdahl (8th ed.) p. 812 # 84

I Gibbs Free Energy - We define a new state function G = H – TS where G is the free energy H is the enthalpy T is the Kelvin temp. S is the entropy (of the system) For a process occurring at constant temperature, ΔG = ΔH - TΔS

For a spontaneous process, If ∆Suniv > 0 ∆Ssys + ∆Ssurr > 0 ∆S - ∆H > 0 T and T∆S - ∆H > 0 ∆H -T∆S< 0 since ∆G = ∆H - T∆S, ∆G < 0 Therefore, if ∆G is neg., the process is spontaneous in the forward direction

If ∆Suniv < 0, ∆S - ∆H < 0 T T∆S - ∆H < 0 ∆H - T∆S > 0 and ∆G > 0 If ∆G is positive, the process is not spontaneous. The reverse reaction is spontaneous.

If ∆Suniv = 0 ∆S - ∆H = 0 T T∆S - ∆H = 0 ∆H - T∆S = 0 and ∆G = 0 If ∆G is zero, neither direction is favored, and the system is in equilibrium.

II The effect of temperature on ΔG and reaction spontaneity Situation Signs of ΔH and ΔS ΔG Characteristics 1 ΔH = (-) ΔS = (+) Always spontaneous at favorable favorable neg. all temps. 2 ΔH = (+) ΔS = (-) Always nonspontaneous unfavorable unfavorable pos. (cannot occur) at any temp. 3 ΔH = (-) ΔS = (-) Neg. at Reaction favorable unfavorable low temps. occurs only Pos. at at sufficiently high temps. low temps. 4 ΔH = (+) ΔS = (+) Pos. at low Reaction unfavorable unfavorable temps. occurs only Neg. at high at sufficiently temps. high temps.

Problem: For 2H2S(g) + SO2(g) ↔ 3S(s) + 2H2O(g) At 298 K, the standard enthalpy change, ΔH0, for the reaction is -145 kJ. a) Predict the sign of the standard entropy change, ΔS0, for the reaction. Explain the basis for your prediction. Ans: ΔS0 is negative. 3 moles of gas 2 moles of gas

b) At 298 K, the forward reaction is spontaneous b) At 298 K, the forward reaction is spontaneous. What change, if any, would occur in the value of ΔG0 for this reaction as the temperature is increased? Explain your reasoning using thermodynamic principles. Ans: ΔG0 = ΔH0 – TΔS0 As T↑, -TΔS0 becomes larger, and ΔG0 -(+)(-) becomes less neg. c) The absolute temperature at which the forward reaction becomes nonspontaneous can be predicted. Write the equation that is used to make the prediction. Ans: 0 = ΔH0 – TΔS0 and T = ΔH0 ΔS0

III Entropy is expected to increase for processes in which 1. Liquids or solutions are formed from solids 2. Gases are formed from solids and liquids 3. The number of molecules of gas increases during a chemical reaction 4. the temperature of a substance is increased IV The third law of thermodynamics The entropy of a pure crystalline substance at absolute zero is zero.

V Calculation of entropy changes Standard molar entropy(S0) – the entropy value for 1 mole of a substance in its standard state. The standard state of a substance is its state at 1 atm and 298 K (250C). For a chemical reaction, ΔS0 = ∑npS0(products) - ∑nrS0(reactants) where np and nr are the coefficients in the chemical equation.

substance S0(J/mol•K) SO2(g) 248 NiO(s) 38 Problem: Calculate ΔS0 for the reaction 2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s) given the following standard entropy values. substance S0(J/mol•K) SO2(g) 248 NiO(s) 38 O2(g) 205 NiS(s) 53 Ans: ΔS0 = ∑npS0(products) - ∑nrS0(reactants) = (2S0SO2(g) + 2S0NiO(s)) – (2S0NiS(s) + 3S0O2(g)) = 2 mol(248 J/mol•K) + 2 mol(38 J/mol•K) -2 mol(53 J/mol•K) – 3 mol(205 J/mol•K) = 496 J/K + 76 J/K – 106 J/K – 615 J/K ΔS0 = -149 J/K We would expect ΔS0 to be negative because the number of gaseous molecules decreases in this reaction.

Practice Problems Zumdahl (8th ed.) p.809 # 47,44,81