In a gaseous system of the type $A B (g) ⇌ A (g) + B (g)$ at a given temperature 50% of AB is dissociated at equilibrium. What is the value of $K_{P}$ at equilibrium in terms of the equilibrium constant $K_{C}$

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Solution

Solution:-
${A B}_{(g)} ⇌ A_{(g)} + B_{(g)}$

For the above reaction,
$Δ n_{g} = n_{P} - n_{R} = (1 + 1) - 1 = 1$

As we know that,
$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

Given:-
$T = 50 ℃ = (273 + 50) K = 323 K$
$R = 0.0821 L . a t m / m o l . K$

$∴ K_{p} = K_{c} {(R T)}^{1}$
$\Rightarrow K_{p} = K_{c} R T$
$\Rightarrow K_{p} = K_{c} (0.0821 \times 323) = 26.52 K_{c}$

Hence for the given gaseous system, the value of $K_{p}$ is $26.52 K_{c}$ .

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