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Charles' Law

Cl2g ⇌ 2ClgFo...

Question

$C l_{2} (g) ⇌ 2 C l (g)$

For the given reaction at equilibrium, moles of $C l_{2} (g)$ is equal to the moles of Cl(g) and the equilibrium pressure is 1 atm. If $K_{p}$ of this reaction is x $\times 10^{- 1}$ , find x.

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Solution

According to the Question : $C l_{2} ⇌ 2 C l [P_{t o t a l} = 1 a t m] .$
moles of $C l_{2} =$ Moles of Cl (at equilibrium).
Given : $K_{p} = X \times 10^{- 1}$
$n_{C l_{2}} = n_{C l}$
Therefore, $P_{C l_{2}} = P_{C l}$
Hence, $P_{C l_{2}} = P_{C l} = 0.5$ atm
$K_{p} = \frac{(P_{C l})^{2}}{P_{C l_{2}}} = 0.5 = 5 \times 10^{- 1}$
So, X=5

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