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Gibbs Free Energy & Spontaneity

In the follow...

Question

In the following equilibrium:
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
when $5$ mole of each are taken and temperature is kept at $298 K$ , the total pressure was found to be $20$ bar.
given: $Δ G_{f}^{o} {N O}_{2} = 100 k J, G_{f}^{o} N_{2} O_{4} = 50 k J$
(i) Find $Δ G$ of the reaction at $298 K$
(ii) Find the direction of the reaction.

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Solution

The reaction is:
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
Since, the number of moles of both $N_{2} O_{4}$ and ${N O}_{2}$ are same hence their partial pressure will also be same.
$P_{N_{2} O_{4}} = P_{{N O}_{2}} = \frac{20}{2} = 10 b a r$
$Q_{P} = \frac{{[P_{{N O}_{2}}]}^{2}}{P_{N_{2} O_{4}}} = 10 b a r$
$Δ G_{r e a c t i o n}^{o} = 2 Δ G_{f}^{o} {N O}_{2} - Δ G_{f}^{o} N_{2} O_{4} = 0$
We know that
$Δ G = Δ G^{o} - 2.303 R T log Q = 0 - 2.303 \times 8.314 \times 298 log 10 = - 5705 J$
Since $Δ G$ is negative, the reaction will be spontaneous and proceed in forward direction.

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