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For the equilibrium,

equilibrium constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of these two gases be taken as represented by the point P in the figure then

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Above reaction will go in the forward direction till it attains equilibrium.

$\underset{r}{Δ G}$ increases during the process.

Above reaction will go in the backward direction till it attains equilibrium.

Above equilibrium must be attained with in micro seconds.

answer is C.

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Detailed Solution

$[\frac{P_{i s o b u \tan e}}{P_{n - b u \tan e}}] = 1.732 - K_{C}$
at equilibrium
from the graph
$\frac{[i s o - b u \tan e]}{[n - b u \tan e]} = T a n 75 = Q_{C}$
= 3.73
> 1. 732
As $Q_{C} < K_{C}$ reaction occurs in backward direction

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