Free Energy. Gibbs Free Energy. The spontaneity of any process. is determined by the entropy change of the universe:

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1 Free Energy Gibbs Free Energy The spontaneity of any process A B is determined by the entropy change of the universe: Note that this is S univ and not S sys. In terms of what happens to the system, we have: 1. If S sys > 0 and H sys < 0, the reaction will always be spontaneous. 2. If S sys > 0 and H sys > 0, the reaction will never be spontaneous. 3. if S sys and H sys have the same sign, spontaneity depends on specific values and temperatures. 1

2 Gibbs free energy is a state function: G = H T S and at constant temperature and pressure G = H T S A spontaneous process in now determined completely by G sys : S surr = H sys T S univ = S sys + S surr = S sys H sys T T S univ = H sys T S sys = G sys 2

3 For the vaporization of water, we write the reaction G vap is under standard conditions. H 2 O(l) H 2 O(g) and we have H vap = kj mol 1 and S vap = H vap T vap = kj mol K = J mol 1 K 1 Vaporization requires heat from the surroundings, and increases the entropy of the system. These balance out perfectly at the boiling point, when T = T vap : ( ) Hvap G vap = H vap T S vap = H vap T vap = H vap H vap = 0 T vap or G vap = H vap T S vap = J mol K J mol 1 K 1 = 0 J mol 1 If we assume that both H vap and S vap are independent of temperature, then H vap = T S vap when T = K. This makes G vap = 0, and liquid water and water vapor are in equilibrium. H vap > T S vap when T < K. This makes G vap > 0, and liquid water does not spontaneously convert into water vapor (and, indeed, water vapor will spontaneously condense to liquid.) H vap < T S vap when T > K. This makes G vap < 0, and liquid water will spontaneously vaporize. 3

4 Standard Free Energy and Free Energy of Formation The standard free energy is calculated as G = H T S Remember that standard states are 1. temperature of 298 K, 2. pressure of 1 atm, and partial pressures of 1 atm for gases, 3. 1 M concentrations for solutes, and 4. most stable forms of pure elements and compounds. 4

5 Spontaneity Once again, if we have a reaction generally written as A B examples: 2H 2 (g) + O 2 (g) 2H 2 O(g) HF(aq) + H 2 O(l) F (aq) + H 3 O + (aq) 2NaCl(s) 2Na(s) + Cl 2 (g) we can describe their spontaneity in any of the following ways: Spontaneous Equilibrium Nonspontaneous S univ > 0 S univ = 0 S univ < 0 G sys > 0 G sys = 0 G sys < 0 Q < K Q = K Q > K These are all equivalent. Free Energy and the Equilibrium Constant The relationship between G and G is easiest to write down with gases and pressures. G = nrt ln P f P i For P i = P = 1 atm and P f = P 5

6 we get which we write as G = G G = nrt ln G = G + nrt ln P P 1 atm Again, G becomes larger than G when the pressure is greater than 1 atm, and smaller when it is less. For a reaction we have aa(g) + bb(g) cc(g) + dd(g) G = G products G reactants = (cg C + dg d ) (ag A + bg B ) = [c (G C + RT ln P C ) + d (G D + RT ln P D )] [a (G A + RT ln P A ) + b (G B + RT ln P B )] = (cg C + dg D ag A bg B) + RT (c ln P C + d ln P D a ln P A b ln P B ) = G + RT (c ln P C + d ln P D a ln P A b ln P B ) Use the properties of logarithms: ln x + ln y = ln xy ln x ln y = ln x y a ln P A = ln (PA) a c ln P C + d ln P D = ln P c C ln P d D c ln P C + d ln P D a ln P A b ln P B = ln P c C ln P d D P a A ln P b B giving us which is At equilibrium, G = G + RT ln P c C ln P d D P a A ln P b B G = G + RT ln Q p Q p = K p G = 0 and so G = G + RT ln Q p 0 = G + RT ln K p G = RT ln K p So 6

7 1. A large, positive equilibrium constant ( ) means a large, negative G. 2. A tiny, fractional equilibrium constant ( ) means a large, positive G. 3. An equilibrium constant near 1 means a G near 0. We can also write G = G + RT ln Q p = RT ln K p + RT ln Q p = RT ln Q p K p so when Q p > K p, then G > 0, and so on. We will not prove that these same relations apply for K and Q as well as K p and Q p, but here they are: G = RT ln K G = G + RT ln K 7

8 Temperature Dependence of the Equilibrium Constant We have which we can break up to G = RT ln K H T S = RT ln K or ln K = H RT + S R Both H and S determine the magnitude of the equilibrium constant, but only H determines its temperature dependence: If we have values for the equilibrium constant at two temperatures, T 1 and T 2, we get ln K 1 = H RT 1 ln K 2 = H RT 2 ln K 2 ln K 1 = H R ln K 2 = H K 1 R + S R + S R ( 1 ) 1 T 2 T 1 ( ) 1 T 2 1 T 1 and we can calculate the enthalpy change of the reaction. Conversely, if we know H and K at one temperature, we can calculate K at any other temperature. Note for the reaction A(l) A(g) we have K p = P A(g) If we have a equilibrium pressure of P 1 at temperature T 1, and P 2 at T 2, we get ln K ( 2 = H 1 1 ) K 1 R T 2 T 1 8

9 ln P ( 2 = H 1 1 ) P 1 R T 2 T 1 which is exactly the Clausius-Clapeyron equation. 9