The total pressure at which $A_{(g)}$ is $50 %$ dissociated according to equation $A_{(g)} ⇌ B_{(g)} + C_{(g)}$ is numerically equal to ______ times of $K_{p}$ .

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Solution

The given reaction is :-
$A_{(g)} ⇌ B_{(b)} + C_{(g)}$
Initial : $P$ $0$ $0$ (let)
pressure
At eqm : $P (1 - α)$ $α P$ $α P$ ( $α$ is the degree of dissociation)
$= 0.5 P$ $0.5 P$ $0.5 P$ ( $α = 50$ % given)
At eqm, $P_{T o t a l} = 1.5 P \to (I)$
Now, $K_{P} = \frac{p_{C} . p_{B}}{p_{A}}$
$K_{P} = \frac{0.5 P \times 0.5 P}{0.5 P} = 0.5 P \to (I I)$
$(I)$ & $(I I)$ $\Rightarrow 3 K_{P} = P_{T o t a l}$

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