Question

9. Referring to the same reaction Fe2O3(s) + 3 C(graphite)→2 Fe(s) + 3 CO(g) determine the value for the equilibrium constant, Keq, for the reaction at 298 K.

Answer 1

From standard thermodynamics table:

Gof(Fe2O3(s)) = -742.2 KJ/mol

Gof(C(s,graphite)) = 0.0 KJ/mol

Gof(Fe(s)) = 0.0 KJ/mol

Gof(CO(g)) = -137.168 KJ/mol

Balanced chemical equation is:

Fe2O3(s) + 3 C(s,graphite) ---> 2 Fe(s) + 3 CO(g)

ΔGo rxn = 2*Gof(Fe(s)) + 3*Gof(CO(g)) - 1*Gof( Fe2O3(s)) - 3*Gof(C(s,graphite))

ΔGo rxn = 2*(0.0) + 3*(-137.168) - 1*(-742.2) - 3*(0.0)

ΔGo rxn = 330.696 KJ

For 2nd part we have:

T = 298 K

ΔGo = 330.696 KJ/mol

ΔGo = 330696 J/mol

use:

ΔGo = -R*T*ln Kc

330696 = - 8.314*298.0* ln(Kc)

ln Kc = -133.4758

Kc = 1.077*10^-58

Answer: 1.08*10^-58