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Derivation of Kp and Kc

In the dissoc...

Question

In the dissociation of HI, 20% of HI is dissociated at equilibrium. Calculate $K_{p}$ for.

HI(g) $⇋$ $\frac{1}{2} H_{2} (g)$ + $\frac{1}{2} I_{2} (g)$

1.25

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0.125

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12.5

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0.0125

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Solution

The correct option is B 0.125
$H I \Leftrightarrow \frac{1}{2} H_{2} + \frac{1}{2} I_{2}$
Initial mole $1$ $0$ $0$
Mole at equilibrium $(1 - α)$ $α / 2$ $α / 2$

Where $α$ is degree of dissociation and volume of container is $V$ litre.

$K_{p} = K_{c} = \frac{{(\frac{α}{2 V})}^{1 / 2} {(\frac{α}{2 V})}^{1 / 2}}{\frac{(1 - α)}{V}} = \frac{α}{2 (1 - α)}$

$K_{p} = K_{c} = \frac{0.2}{2 (1 - 0.2)}$ $(∴ α = 0.2)$

$K_{p} = K_{c} = 0.125$

Option B is correct.

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