Question

-224.2 J/K Determine the equilibrium constant for the following reaction at 549 K. CH2O(g) + 2 H2(g) → CH4(g) + H2O(g) ΔH° = -94.9 kJ; ΔS°= -224.2 J/K 1.07 x 109 481 2.08 x 10-3 9.35 x 10-10 1.94 x 10-12

Answer 1

Answer:

The equilibrium constant of the reaction is $2.08\times 10^{-3}$.

Explanation:

Formula used:

$\Delta G^o=\Delta H^o-T\Delta S^o$

$\Delta G^o=-RT\ln K_1$

where :

$\Delta G^o$ = Gibbs free energy

$\Delta H^o$ = Enthalpy of reaction

$\Delta S^o$ = Entropy of reaction

R = Gas constant = $8.314J/K mol$

T = Temperature in Kelvins

$K_1$ = equilibrium constant at T

So we have:

$\Delta G^o$ = ?

$\Delta H^o$ = -94.9 kJ/mol = -94900 J/mol

$\Delta S^o$ = -224.2 J/mol K

T = 549 K

$\Delta G^o=-94900 J/mol-549\times ( -224.2 J/mol K)=28,185.8 J/mol$

$\Delta G^o=-RT\ln K_1$

$28,185.8 J/mol=-8.314J/K mol\times 549 K\times \ln K_1$

$K_1=0.002080=2.08\times 10^{-3}$

The equilibrium constant of the reaction is $2.08\times 10^{-3}$.