Question

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

2 Si _(s) + 3 H_{2 (g)} ⇌ Si₂H_6(g)

The conditions for this reaction are:

P_H2 = 1.83 bar

P_Si2H6 = 0.96 bar

You will also need to use Appendix II in your textbook (containing standard Gibbs energies of formation).

ΔG_f(Si₂H₆) = 127.3 kJ/mol

ΔG_f(H₂) = 0 kJ/mol

Answer 1

2 Si _(s) + 3 H_{2 (g)} ⇌ Si₂H_6(g)

P_H2 = 1.83 bar

P_Si2H6 = 0.96 bar

Qp   = PSi2H6/P^3H2

        = 0.96/(1.83)^3   = 0.1566

2 Si _(s) + 3 H_{2 (g)} ⇌ Si₂H_6(g)

ΔG⁰rxn = ΔG_f products - ΔG_f reactants

           = 127.3-(2*0 + 3*0)

            = 127.3KJ/mole

             = 127300J/mole

ΔG = ΔG⁰rxn + RTlnQp

        = 127300 + 8.314*298ln(0.1566)

        = 127300 + 8.314*298*-1.854

         = 122706J/mole

         = 122.706KJ/mole >>>>answer