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Arrhenius Acid

A reaction, A...

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A reaction, $A (g) + 2 B (g) ⇌ 2 C (g) + D (g)$ was studied using an initial cojncentration of B which was 1.5 times that of A. But the equilibrium concentrations of A and B were found to be equal. The value of $K_{p}$ for the equilibrium is

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Solution

$A (g) + 2 B (g) ⇌ 2 C (g) + D (g)$
$\begin{matrix} Initial conc. & x & 1.5 x & 0 & 0 At equilibrium & x - y & 1.5 x - 2 y & 2 y & y \end{matrix}$
Given: $x - y = 1.5 x - 2 y; y = 0.5 x .$ Thus,
$[A] = 0.5 x [B] = 0.5 x [C] = x [D] = 0.5 x K_{C} = K_{p} = \frac{(x^{2}) (0.5 x)}{(0.5 x) (0.5 x)^{2}} = 4$

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