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For the water gas reaction
$C (s) + H_{2} O (g) \overset{}{⇆} CO (g) + H_{2} (g)$
At 1000 K, the standard Gibbs free energy change of the reaction is $- 8.314 kJ / mol .$ Therefore, at 1000 K the equilibrium constant of the above water gas reaction is

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$\frac{1}{e}$

2.718

answer is D.

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Detailed Solution

$∆_{r} G^{0} = - 8.314 \times 10^{3} ∴ - ∆_{r} G^{0} = 8.314 \times 10^{3} = RT \ln K_{eq} or 8.314 \times 10^{3} = 8.314 \times 10^{3} \ln K_{eq} or \ln K_{eq} = 1 = \ln e or K_{eq} = e = 2.718$

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