What is the ΔrG for the following reaction (in kJ mol-1) at 298 K?

Question

2 Si _(s) + 3 H_{2 (g)} ⇌ Si₂H_6 (g)

The conditions for this reaction are:

P_H2 = 1.47 bar

P_Si2H6 = 0.46 bar

Answer 1

2S(s) + 3H₂(g) <==> Si₂H₆(g) ... balanced equation

Kp = (P_Si2H6) / (P_H2)³

Kp = 0.46 / (1.47)³

Kp = 0.145

∆Gº = -RT ln K = -(8.314 J/Kmol)(298K) ln 0.145

∆Gº = -2478 x -1.93

∆Gº = +4785 J/mol = 4.79 kJ / mole