In a reversible reaction $A \underset{K_{2}}{\overset{K_{1}}{⇌}} B$ the initial concentration of A and B are b in moles per litre and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively; Express x in terms of $K_{1}$ , $K_{2}$ , $α$ and $b$ .

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$\frac{K_{1} a - K_{2} b}{K_{1} + K_{2}}$

$\frac{K_{1} a - K_{2} b}{K_{1} - K_{2}}$

$\frac{K_{1} a - K_{2} b}{K_{1} K_{2}}$

$\frac{K_{1} a + K_{2} b}{K_{1} + K_{2}}$

answer is A.

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Detailed Solution

Here;

$A \underset{K_{1}}{\overset{K_{2}}{⇌}} B$

$t = 0 a b$

$t = t_{eq} a - x b + x$

At equilibrium, $r_{f} = r_{b}$

$K_{1} (a - x) = K_{2} (b + x)$

$\Rightarrow x = \frac{K_{1} a - K_{2} b}{K_{1} + K_{2}}$

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