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For the following equilibrium,
$N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$
$K_{p} is found to be equal to K_{C} . This is attained when$

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T = 12.18 K

T = 1K

T = 273 K

T = 17.15 K

answer is C.

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Detailed Solution

$K_{p} = K_{C} (RT)^{Δn}$

$Here K_{p} = K_{C}, Δn = 2 - 1 = 1$

$(RT)^{Δn} = 1$

$(0.0821 \times T) = 1$

$T = \frac{1}{0.0821} = 12.18 K$

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