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Butoxors [25]
4 years ago
9

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 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
Chemistry
1 answer:
Dominik [7]4 years ago
8 0

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}.

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