Question

For the water gas reaction,
$C\left(s\right)+H_{2}O\left(g\right)\rightleftharpoons CO\left(g\right)+H_2\left(g\right)$
the standard Gibbs free energy of reaction (at $1000K$) is $-8.1kJ/mol.$ Calculate its equilibrium constant
antilog 0.423=2.6485

Answer 1

We know that,
$K=antilog\left(\frac{-△G^{\circ }}{2.303RT}\right)...\left(1\right)$
Given, $△G^{\circ }=-8.1kJ/mol$
$R=8.314\times {10}^{-3}kJ{K}^{-1}mo{l}^{-1}$
$T=1000K$
Substituting the values in equation (1), we get
$K=antilog\left[\frac{-(-8.1)}{2.303\times 8.314\times {10}^{-3}\times 1000}\right]$
$K=antilog\left[\frac{8.1}{19.147}\right]$
$K=antilog0.423$
$K=2.65$