Question

For the water gas reaction
$\text{C}\left(S\right)+{\text{H}}_{2}O\left(g\right)\rightleftharpoons \text{CO}\left(g\right)+{\text{H}}_2\left(g\right)$
At $1000\text{K}$ the standard Gibbs free energy change of the reaction is$-8.314\text{kJ/mol}$. Therefore at $1000K$ the equilibrium constant of the above water gas reaction is:
$(\text{Where}e=2.718=\text{Euler's number})$

• A. $1$
• B. $10$
• C. $\frac{1}{e}$
• D. $e$

Answer 1

The correct option is D $e$

We know the equation,
$-\Delta G^{\circ }=RT\ln K_{eq}$
Where, $\Delta G^{\circ }=\text{standard Gibbs free energy change}$
$K_{eq}=\text{equilibrium constant}$
${\Delta }_r{G}^{\circ }=-8.314\times {10}^3$
$\therefore -{\Delta }_r{G}^{\circ }=8.314\times {10}^3=RT\ln K_{eq}\Rightarrow 8.314\times {10}^3=8.314\times {10}^3\ln K_{eq}\Rightarrow \ln K_{eq}=1=\ln e\Rightarrow K_{eq}=e$