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Equipartition Theorem

Δ G o for 1...

Question

$Δ G^{o}$ for $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {N H}_{3} (g)$ is $- 16.5 k J {m o l}^{- 1}$ . Find out $K_{p}$ for the reaction at $25^{o} C$ . Also report $K_{p}$ and $Δ G^{o}$ for $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ at $25^{o} C$ .

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Solution

For the reaction
$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {N H}_{3} (g)$
$log K_{p} = - \frac{Δ G^{o}}{2.303 R T}$
$log K_{p} = - \frac{(- 16.5 \times 10^{3} J / m o l)}{2.303 \times 8.314 J / m o l / K \times 298 K}$
$log K_{p} = 2.8917$
$K_{p} = 779.41$
For the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
$K_{p}$ is equal to the square of the value of $K_{p}$ for the first reaction. It is equal to ${(779.41)}^{2} = 6.07 \times 10^{5}$
$Δ G^{o} = - 2.303 \times 8.314 J / m o l / K \times 298 K log 6.07 \times 10^{5}$
$Δ G^{o} = - 32998 J {m o l}^{- 1}$
$Δ G^{o} = - 32.9 k J {m o l}^{- 1}$
$(∵ 1 k J = 1000 J)$

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