42) ΔH° = -19.5 kJ ΔS° = 42.7 J/K T = 298 K
(a) The reaction is exothermic (the enthalpy of the products is less than the enthalpy of the reactants) because ΔH° < 0.
(b) The entropy (the disorder of the system) increases because ΔS° > 0.
(c) ΔG° = ΔH° - TΔS°
ΔG° = -19.5 kJ - 298
K x 42.7 J/ K x 1 kJ/103 J = -32.2 kJ
(d) The reaction is spontaneous (occurs without any outside intervention) because the reactants and products are present in their standard states and ΔG° < 0.