What is the ΔrG for the following reaction (in kJ mol-1) at 298 K?
2 Si (s) + 3 H2 (g) ⇌ Si2H6(g)
The conditions for this reaction are:
PH2 = 1.83 bar
PSi2H6 = 0.96 bar
You will also need to use Appendix II in your textbook (containing standard Gibbs energies of formation).
ΔGf(Si2H6) = 127.3 kJ/mol
ΔGf(H2) = 0 kJ/mol
2 Si (s) + 3 H2 (g) ⇌ Si2H6(g)
PH2 = 1.83 bar
PSi2H6 = 0.96 bar
Qp = PSi2H6/P^3H2
= 0.96/(1.83)^3 = 0.1566
2 Si (s) + 3 H2 (g) ⇌ Si2H6(g)
ΔG0rxn = ΔGf products - ΔGf 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
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