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Gibbs Free Energy & Spontaneity

For the water...

Question

For the water gas reaction,
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$
the standard Gibbs free energy of reaction (at $1000 K$ ) is $- 8.1 k J / m o l .$ Calculate its equilibrium constant

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Solution

We know that,
$K = a n t i l o g (\frac{- △ G^{\circ}}{2.303 R T}) . . . (1)$
Given, $△ G^{\circ} = - 8.1 k J / m o l$
$R = 8.314 \times 10^{- 3} k J K^{- 1} m o l^{- 1}$
$T = 1000 K$
Substituting the values in equation (1), we get
$K = a n t i l o g [\frac{- (- 8.1)}{2.303 \times 8.314 \times 10^{- 3} \times 1000}]$
$K = 2.65$

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