The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol)

Question

Substance:	H2O2(l)	H2O2(g)
ΔH°f (kJ/mol):	–187.7	–136.3
ΔG°f (kJ/mol):	–120.4	–105.6
S° (J/K • mol):	109.6	232.7

The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J/K • mol)

H2O2(l)  H2O2(g)

Use the following thermodynamic information at 298 K to estimate this temperature.

Answer 1

∆Gº = ∆Hº - T∆Sº

At equilibrium ∆Gº = 0

0 = ∆Hº - T∆Sº

T = ∆Hº / ∆Sº

∆Hrxn = -136.3 - (-187.7) = 51.4 kJ/mol

∆Srxn = -.2327 - 0.1096 = 0.1231 kJ/Kmol

T = 51.4 kJ/mol / 0.1231 kJ/Kmol

T = 477.55K

T = 144ºC