The equilibrium constant, $K_{p}$ for the reaction $A \Leftrightarrow 2 B$ is released to degree of dissociation $α$ of A and total pressure P as

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$\frac{4 α^{2} P}{1 - α}$

$\frac{4 α^{2} P}{1 - α^{2}}$

$\frac{4 α^{2} P^{2}}{1 - α^{2}}$

answer is A.

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Detailed Solution

$A \Leftrightarrow 2 B 1 0 (1 - α) 2 α Total moles = 1 - α + 2 α = 1 + α K_{p} = \frac{{P_{B}}^{2}}{P_{A}} \frac{{(\frac{2 α}{1 + α} P)}^{2}}{(\frac{1 - α}{1 + α} P)} = \frac{4 α^{2}}{1 - α^{2}} P$

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