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Derivation of Kp and Kc

For the follo...

Question

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 at

A

0^{\circ}

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B

273 K

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C

1 K

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D

12.18 K

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Solution

The correct option is D 12.18 K
We know,
$K_{p} = K_{c} (R T)^{Δ n_{g}} = K_{c} (R T) [Δ n_{g} = 2 - 1 = 1]$
So, $T = \frac{K_{p}}{K_{c}} \frac{1}{R} F o r, K_{p} = K_{c} T = \frac{1}{0.0821} = 12.18 K$

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Derivation of Kp and Kc

Standard XII Chemistry

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