Problem summary
Calculate ΔG° for the process given below at 25°C if from the given data at the same temperature.
ΔG° = ΔH° - TΔS°
ΔG° = –106 kJ mol –1 – 298 K × ΔS°
ΔG° = –106 × 103 J mol –1 – 298K × 48 J mol –1 K–1
ΔG° = –102304 J mol –1 = –102 kJ mol –1
Despite the large negative ΔG° aqueous solutions of hydrogen peroxide (H2O2) are completely stable (nonreactive) at room temperature. This is because the reaction is slow in the absence of a catalyst. However in the presence of a suitable catalyst the reaction becomes explosive.
ΔrH° = -106 kJ mol –1
ΔrS° = +48 J mol –1 K–1
ΔrS° = +48 J mol –1 K–1
H2O2(g) 
H2O(g) + ½O2(g)
H2O(g) + ½O2(g)ΔG° = ΔH° - TΔS°
ΔG° = –106 kJ mol –1 – 298 K × ΔS°
ΔG° = –106 × 103 J mol –1 – 298
ΔG° = –102304 J mol –1 = –102 kJ mol –1
Despite the large negative ΔG° aqueous solutions of hydrogen peroxide (H2O2) are completely stable (nonreactive) at room temperature. This is because the reaction is slow in the absence of a catalyst. However in the presence of a suitable catalyst the reaction becomes explosive.