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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 $5 m o l e s$ of each is taken and the temperature is kept at $298 K$ , the total pressure was found to be $20 b a r$ .
(Given: $Δ G_{f N_{2} O_{4}} = 100 k J, Δ G_{f {N O}_{2}}^{o} = 50 k J$ )
(a) Find $Δ G$ of the reaction at $298 K$ .
(b) Find the direction of the reaction.

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Solution

Reaction Quotient $= \frac{{[p_{{N O}_{2}}]}^{2}}{p_{N_{2} O_{4}}} = \frac{100}{10} = 10$
$Δ G_{r e a c t i o n}^{o} = 2 Δ G_{f N_{2} O_{4}}^{o} - Δ G_{f {N O}_{2}}^{o} = 2 \times 50 - 100 = 0$
We know, $G = Δ G^{o} - 2.303 R T log Q_{p} = - 5705.8 J = - 5.705 k J$
Negative value shows that reaction will be in forward direction.

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