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Derivation of Kp and Kc

For the react...

Question

For the reaction,
$S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g), K_{p} > K_{x}$ is obtained at:

0.5 a t m

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0.8 a t m

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1 a t m

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2 a t m

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Solution

The correct option is D $2 a t m$
$K_{p} = \frac{P_{C}^{c} \cdot P_{D}^{d}}{P_{A}^{a} \cdot P_{B}^{b}} = \frac{(X_{C} \times P)^{c} (X_{D} \times P)^{d}}{(X_{A} \times P)^{a} (X_{B} \times P)^{b}}$

$= \frac{X_{C}^{c} \cdot X_{D}^{d}}{X_{A}^{c} \cdot X_{B}^{b}} \times P^{(c + d) - (a + b)}$

$\Rightarrow K_{p} = K_{x} \cdot P^{(c + d) - (a + b)}$

For reaction,
$S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)$

$(c + d) - (a + b) = (1 + 1) - 1 = 1$

$K_{p} = K_{x} \cdot P$

Thus, $\frac{K_{p}}{K_{x}} = P$

$\Rightarrow K_{p} > K_{x}$

$\Rightarrow P > 1$

$∴ P = 2 a t m$

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