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Derivation of Kp and Kc

For the equil...

Question

For the equilibrium in a closed vessel $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g),$
$K_{p}$ is found to be double of $K_{c} .$ This is attained when:

A

T = 2 K

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B

T = 12.18 K

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C

T = 24.36 K

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D

T = 27.3 K

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Solution

The correct option is C $T = 24.36 K$
We know the relation
$K_{p} = K_{c} (R T)^{△ n}$
Given $K_{p} = 2 K_{c}$
$2 K_{c} = K_{c} (R T)^{△ n}$
$2 = (R T)^{1}$
$T = \frac{2}{0.0821} = 24.36 K$

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Derivation of Kp and Kc

Standard XIII Chemistry

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