Question

The dissociation equilibrium of a gas $\mathrm{AB}_{2}$ can be represented as:
$$2 \mathrm{AB}_{2(g)} \leftrightharpoons 2 \mathrm{AB}_{(g)}+\mathrm{B}_{2(g)}$$
The degree of dissociation is $x$ and is small compared to 1 . The expression relating the degree of dissociation $(x)$ with equilibrium constant $K_{p}$ and total pressure $P$ is
(a) $\left(2 K_{p} / P\right)^{1 / 2}$
(b) $\left(K_{p} / P\right)$
(c) $\left(2 K_{p} / P\right)$
(d) $\left(2 K_{p} / P\right)^{1 / 3}$

Answer 1

The correct option is: (d) $\left(2 K_{P} / P\right)^{1 / 3}$
Explanation:
$$\begin{aligned} &\begin{array}{c} 2 A B_{2(g)} \rightleftharpoons 2 A B_{(g)}+B_{2(g)} \\ 0 \end{array} \quad \begin{aligned} & \text { (initially) } \\ 2(1-x) & 2 x \quad x \quad & \text { (at equilibrium) } \end{aligned} \\ &\text { Amount of moles at equilibrium }=2(1-x)+2 x+x \\ & & =2+x \\ &K_{p}= & \frac{\left[p_{A B}\right]^{2}\left[p_{B_{2}}\right]}{\left[p_{A B_{2}}\right]^{2}} \\ &K_{p}= & \frac{\left(\frac{2 x}{2+x} \times P\right)^{2} \times\left(\frac{x}{2+x} \times P\right)}{\left(\frac{2(1-x)}{2+x} \times P\right)^{2}}=\frac{\frac{4 x^{3}}{2+x} \times P}{4(1-x)^{2}} \\ &K_{p}= & \frac{4 x^{3} \times P}{2} \times \frac{1}{4} \quad(\because \quad 1-x \approx 1 \& 2+x \approx 2) \\ & & x=\left(\frac{8 K_{p}}{4 P}\right)^{1 / 3} \quad \Rightarrow x=\left(\frac{2 K_{p}}{P}\right)^{1 / 3} \end{aligned}$$

Answer 2

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Answer 3

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