At temperature $T$ , the compound $A B_{2} (g)$ dissociates according to the reaction,
$2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)$
with a degree of dissociation, $x$ . which is small compared with unity. Deduce the expression for $x$ in terms of the equilibrium constant, $K_{p}$ and the total pressure, $P$ .

{(\frac{K_{P}}{P})}^{1 / 3}

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{(\frac{2 K_{P}}{P})}^{1 / 3}

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{(\frac{2 K_{P}}{P})}^{1 / 2}

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{(\frac{K_{P}}{P})}^{1 / 2}

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Solution

The correct option is B ${(\frac{2 K_{P}}{P})}^{1 / 3}$
$\begin{matrix} 2 A B_{2} (g) & ⇌ & 2 A B (g) & + & B_{2} (g) At equilibrium & (1 - x) & x & x / 2 Total moles at equilibrium & = 1 - x + x + x / 2 = \frac{2 + x}{2} \end{matrix}$

$P_{A B_{2}} = \frac{2 (1 - x)}{(2 + x)} \cdot P; P_{A B} = \frac{2 x}{(2 + x)} \cdot P; p_{B_{2}} = \frac{x}{(2 + x)} \cdot P;$
$K_{p} = \frac{(p_{A B})^{2} (p_{B_{2}})}{(p_{A B_{2}})^{2}} = \frac{{[\frac{2 x}{(2 + x)} P]}^{2} (\frac{x}{2 + x}) P}{{(\frac{2 (1 - x)}{(2 + x)} P)}^{2}}$
$= \frac{x^{3} P}{(2 + x) (1 - x)^{2}}$
As $x$ is very small, $(2 + x) \to 2$ and $(1 - x) \to 1$
So, $K_{P} = \frac{x^{3} P}{2}$
$\Rightarrow x^{3} = \frac{2 K_{P}}{P}$
$\Rightarrow x = {(\frac{2 K_{P}}{P})}^{1 / 3}$

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AT temperature T, a compound $A B_{2} (g)$ dissociates according to the reaction,
$2 A B_{2 (g)} ⇌ 2 A B_{(g)} + B_{2 (g)}$ with a degree of dissociation, x (where x << 1). If KP and P are equilibrium constant and equilibrium pressure respectively, the value of x is