Sample Problem Calculation of H°, S°, G° Consider the reaction 2 NO2 (g)   N2O4 (g) 1) Calculate H°, S°, and G° for the reaction at 298 K H° = -57.2 kJ/mol, S° = -176 J/mol•K, Table 2.8 gives G° = -4.78 kJ/mol 2) At what temperature would G° be zero? T = 325 K Check it ConcepTest #1 For which of the following reactions is the indicated sign of H or S incorrect? A. 2N (g)  N2 (g) H = – B. 2N (g)  N2 (g) S = – C. H2O (l)  H2O (g) H = – D. H2O (l)  H2O (g) S = + 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. (Case where G shines) 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 Valid whenever only PV work is done – VERY useful At constant T, dT = 0, and dG = VdP Temperature and Pressure dependence of G At constant T, dG = VdP We integrate to get G(P). Two limiting cases: Gases and non-gases Solids and liquids – not compressible and V ~ constant G  P2   G  P1    VdP  V P P2 Ideal Gases P1 G  P2   G  P1    P2 P1  P2  nRT dP  nRT ln   P  P1  These apply equally well to G; thus G  P2   G  P1   P2 P 1  P2  nRT dP  nRT ln   P  P1  For chemical reactions use the appropriate result for each reactant and product Temperature dependence of G Temperature dependence can be readily derived from dG = VdP – SdT G T  H T  S If H and S are constant over the temp range of interest,    G   H       2 Gibbs - Helmholtz eqn T  T  T   P frequently integrated to obtain the form G T2  T2  G T1  T1 1 1  H    T2 T1  We will use these results later to see how changes in G with temperature and pressure cause equilibria to shift. Temperature and Pressure dependence of G dG = VdP – SdT from 3 pictures ago G  H  S T T G T2  G T1  1 1   H    T2 T1 T2 T1   P2  G  P2   G  P1   nRT ln    P1  With this background, we can now develop a key concept for reactions: The CHEMICAL POTENTIAL We employ G = G(T,P). As always, other choices are possible, but they do not lead to simplicity for constant P processes. 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. dG = VdP – SdT from 3 pictures ago Suppose we have a mixture of materials i (i=1, 2, 3, . . . ), all at temperature T and pressure P. The mixture contains ni moles of substance i, etc. We can generalize G to the form G = G(T, P, n1, n2, . . . ), where ni is the number of moles of substance i. Writing the differential, dG  G   G   G  dG   dT   dP   dn1       T P ,n1 ,  P T ,n1 ,  n1 T ,P ,n n1 , Chemical Potential  G   G   G  dG   dT   dP   dn1       T P ,n1 ,  P T ,n1 ,  n1 T ,P ,n n1 , Each of the new terms, of the form,  G   ni   i  T , P , n ni , is called the “Chemical Potential” of species i. i depends upon P, T and the composition of a mixture. Why Composition?? (Compare H2 alone with a mixture of H2 and O2. Which has more chemical potential??) Chemical Potential (2)  G  G   G  dG   dT   dP      T P ,nA ,nB  P T ,nA ,nB  nA  G  dnA    T ,P ,nB  nB  dnB  T ,P ,nA To see the importance of , here is an example. Consider a process in which dni moles of component i are transferred from state A (at  A ) to state B (at  B ). At constant T and P,   dG  B dni  Adni  B  A dni so dG  0 if B  A This means that reactions proceed to the lowest chemical potential. Thus  is completely analogous to the potential energy in mechanical systems or an electrical potential. And now for the Second Law activity!