Question

Gibbs free energy change AG is defined as AG= AH-TAS. For a spontaneous process, AG<0. For a. reversible reaction at equilibrium, AG = 0. The equilibrium constant K of the reversible reaction is relate- AG"=-RTIn(K). Symbol 40" refers to thermodynamics standard condition: 298 K and 1 atm. Exercise:consider the following reaction: bo:pfoiex 2 HNO:(aq)+NO(g)- 3 NO2(g) + H2O(I) AH=+136.5 kJ; AS = +287.5 J/K a. Below what temperature does the following reaction becomes nonspontaneous? niliod sis anotsole s ob io 9199r b. Is this reaction spontaneous at 298 K and 1 atm? (RO c.Calculate the equilibrium constant of the reaction at 298 K CIS)M 9219 79 boHAlohuia A1 ror

Answer 1

(a) For a reaction to be non-spontaneous, $\Delta$G > 0

i.e.$\Delta$H - T$\Delta$S >0

$\Delta$H > T$\Delta$S

136.5 kJ > T * 287.5 J/K

136500 J / (287.5 J/K) > T

T < 475 K

Thus, For a temperature less than 475 K, reaction will be non- spontaneous

(b) As done earlier

For a temperature less than 475 K, reaction will be non- spontaneous

Thus, Non - spontaneous

Alternate method

$\Delta$G = .$\Delta$H - T$\Delta$S = 136.5 kJ - (298 K* (287.5 J/K)) = 136500 J - 85675 J = 50825 = 50.825 kJ

Thus,$\Delta$G > 0. Hence Non - spontaneous

(c) $\Delta$G^o = -RT lnK

K = exp (-$\Delta$G^o / RT) = exp( - 50825 J / (8.314 J/mol.K * 298 K) = 1.23 *10^-9