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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