The Standard Gibbs Energy Change, G

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Randolph Walters
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1 The Standard Gibbs Energy Change, G S univ = S surr + S sys S univ = H sys + S sys T S univ = H sys TS sys G sys = H sys TS sys Spontaneous reaction: S univ >0 G sys < 0

2 More observations on G and Gº I. G rxn = H rxn -TS rxn H frequently wins in small molecule Rx at STP 2CH 3 OH (l ) + 3O 2 (g) 2CO 2 (g) +4H 20 (l) H = -715 kj/mol S = kj/mol K TS = 93 kj/mol G = -808 kj/mol very negative For large molecules, flexibility and weak bonds make TS competitive cytochrome C folded unfolded at T near 327 K H = 17 kj/mol S = kj/mol K TS = 17 kj/mol T= 327 K G = 0 at 327 K (54 C)

3 Gibbs Energy of Formation, G f º I. Since Gibbs energy is not on an absolute scale, we need to assume a reference state t of the elements. Convention is to choose the element reference (standard) state as the element in its most stable form at K and 1 Bar pressure. G f 0 0 II. Then for the elements in this form. G f III. For a non-elemental substance, its is the Gibbs energy for formation of 1 mole of the substance at standard T and P, made from the elements in their standard states. IV. We can then calculate G for a reaction taking place at standard conditions: G n G n G rxn i i f i j j f j products reactants V. We use this construct to determine reaction spontaneity. 0

4 Calculation of Hº, Sº and Gº I. Similar to Hess s Law H of an overall process is the sum of the H s for the individual steps S of an overall process is the sum of the S s for the individual steps G of an overall process is the sum of the G s for G of an overall process is the sum of the G s for the individual steps

5 Calculation of Hº, Sº and Gº II. Use of Δ f H, S and Δ f G (Appendix D) Δ H = Δ f H ( Products) - Δ f H (Reactants) t Δ S = S (Products) - S (Reactants) Δ G = Δ f G (Products) - Δ f G (Reactants) III. Use of equation Δ G = ΔH H - TΔ S

6 Review of Bond Formation 2 NO 2 (g) N 2 O 4 (g) H = H f H i = 55.3 kj/mol O 2 N NO 2 H i O 2 N NO 2 H f The formation of a bond releases energy The reaction is exothermic ( H)

7 Determining Reaction Spontaneity from GG Consider the formation of N 2 O at 298 K N 2 (g) + 0.5O 2 (g) N 2 O(g) 0 G f 0 H f kj/mol kj/mol S kj/mol K ConcepTest #1 Does this reaction occur spontaneously at 298 K? A. Yes B. No C. Cannot tell D. Not enough time to calculate answer

8 Sample Problem Calculation of H, S, G Consider the reaction 2NO 2 (g) N 2 O 4 (g) 1) Calculate H,, S,, and G for the reaction at 298 K H =-55.3 kj/mol, S =-176 J/mol K, G = -4.8 kj/mol 2) At what temperature would G be zero? T = 325 K

9 ConcepTest #2 For which of the following reactions is the indicated sign of HorSincorrect? S A. 2N (g) N 2 (g) H= B. 2N (g) N 2 (g) S= C. H 2 O(l) H 2 O(g) H = D. H 2 O (l) H 2 O (g) S= +

10 Temperature and Pressure dependence of G We earlier ignored the T and P dependence of G. We can learn from doing better. Pressure dependence of G Assume only P-V work is done. G = H TS = U + PV TS = q + w + PV TS dg = dq + dw + d(pv) d(ts) = dq + dw + PdV + VdP TdS SdT but dq = TdS and dw = -PdV So dg = VdP - SdT At constant T, dg = VdP

11 Temperature and Pressure dependence d of G At constant T, dg = VdP We integrate t to get G(P). Two limiting cases: Gases and non-gases Solids and liquids not compressible and V ~ constant 2 G P2 G P1 VdPVP P1 Ideal Gases P2 nrt P 2 GP2GP1 dpnrtln P1 P P1 These apply equally well to G; G thus P2 nrt P 2 GP2GP1 dpnrtln P 1 P P 1 For chemical reactions use the appropriate result for each reactant and product P

12 Temperature dependence of G Temperature dependence can be readily derived from dg = VdP SdT but I omitted that section of Chapter 3 and will only show the result. If H is constant over the temperature range of interest, G H Gibbs - Helmholtz eqn 2 T T T P frequently used in the form G G H T T 2 1 T T 2 1 We will use these results later to see how changes in G with temperature and pressure cause equilibria to shift.

13 Chemical Potential Until now, we have only considered systems of fixed composition. In order to study reaction thermodynamics, we need to know how the Gibbs energy changes with the composition of reactants and products. This is best done with a very important construct, the Chemical Potential. We generalize G to the form G = G(T, P, n 1, n 2,... ), where n i is the number of moles of substance i. Writing the differential, dg G G G dg dt dp dn 1 T P n1 Pn,, Tn,, T, P, nn, 1 1 1

14 Chemical Potential G G G dg dt dp dn 1 T P n1 Pn,, Tn,, T, P, nn, The new terms, of the form, G n i T, P, n n i, are called the Chemical Potential of species i. i depends upon P, T and the composition of a mixture. Why Composition?? (C H l ith i t f H d O Whi h h (Compare H 2 alone with a mixture of H 2 and O 2. Which has more chemical potential??) i

15 Chemical Potential (2) G G G dg dt dp dn 1 T P n1 Pn,, Tn,, T, P, nn, To see the importance of, here is an example. Consider a process in which dn i moles of component i are transferred from state A (at A ) to state B (at B ). i i At constant T and P, dg B dn A dn B A dn i i i i i i i so dg 0 if This means that reactions proceed to the lowest chemical p potential. Thus is completely analogous to the potential energy in mechanical systems or an electrical potential. B i A i

The Second Law of Thermodynamics (Chapter 4)

The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made