For the following gases equilibrium, $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g), K_{p}$ is found to be equal to $K_{c}$ . This is attained when :

0^{o} C

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273 K

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1 K

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12.18 K

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Solution

The correct option is D $12.18 K$
By using the relation,
$K_{p} = K_{c} (R T)^{Δ n}$
Here, $Δ n = 1$
$K_{p} = K_{c} (R T)$
So,
$T = \frac{K_{p}}{K_{c}} . \frac{1}{R} = \frac{1}{0.0821} = 12.18 K$

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