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At 298 K,Kp f...

Question

At $298 K, K_{p}$ for the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2}$ is 0.98. Predict whether the reaction is spontaneous or not.

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Solution

Spontaneity depends on the sign of change in Gibbs free energy (i.e, $△_{R} G)$ of a reaction. It's relation with $K_{p}$ ( the equillibrium constant at constant pressure) is:

$△_{r} G^{\circ} = - R T l n K_{p}$

For the given reaction

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$K_{p} = 0.98$

Then, at 298 K
$△_{r} G^{\circ} = - R T l n (0.98)$

Since, $l n (0.98) = - 0.02$ , it has a negative value. Thus, value of $△_{r} G^{\circ}$ becomes positive. (as R and T are constant with positive values)

So, the positive sign of change in gibb's free energy(i.e. $△_{r} G^{\circ})$ indicates the reaction is non-spontaneous.

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