For the reaction taking place at certain temperature $N H_{2} C O O N H_{4} (s) ⇌ 2 N H_{3} (g)$ if equilibrium pressure is $3 X$ bar then $Δ G^{\circ}$ would be:

- R T l n 9 - 3 R T l n X

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R T l n 4 - 3 R T l n X

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- 3 R T l n 4 X

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none of these

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Solution

The correct option is D $- 3 R T l n 4 X$
$N H_{2} C O O N H_{4} \to 2 N H_{3} + C O_{2}$
$2 X$ + $X$ = $3 X$
$K = {2 X}^{2} \times X = {4 X}^{3}$
Standard gibbs free energy $= - R T \times l n K$
$Δ G^{o} = - R T \times l n {4 X}^{3}$
$Δ G^{o} = - 3 R T l n 4 X$

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