The dissociation equilibrium of a gas $A B_2$ can be represented as $2 A B_2(g) \rightleftharpoons 2 A B(g)+B_2(g)$ 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

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Aicharya · Accepted Answer

$\begin{array}{cccr} 2 A B_{2(g)} & \rightleftharpoons & 2 A B_{(g)} & + & B_{2(g)} & \ 2 & & 0 & & 0 & \text { (initially) } \ 2(1-x) & & 2 x & & x & \text { (at equilibrium) } \end{array}$ $\begin{aligned} \text {Amount of moles at equilibrium } & =2(1-x)+2 x+x \ & =2+x\end{aligned}$ $\begin{aligned} & K_p= \frac{\left[p_{A B}\right]^2\left[p_{B_2}\right]}{\left[p_{A B_2}\right]^2} \ & K_p= \frac{\left(\frac{2 x}{2+x} \times P\right)^2 \times\left(\frac{x}{2+x} \times P\right)}{\left(\frac{2(1-x)}{2+x} \times P\right)^2}=\frac{\frac{4 x^3}{2+x} \times P}{4(1-x)^2} \ & K_p= \frac{4 x^3 \times P}{2} \times \frac{1}{4} \quad(\because 1-x \approx 1 \& 2+x \approx 2) \ & x=\left(\frac{8 K_p}{4 P}\right)^{1 / 3} \quad \Rightarrow x=\left(\frac{2 K_p}{P}\right)^{1 / 3}\end{aligned}$

The dissociation equilibrium of a gas $A B_2$ can be represented as $2 A B_2(g) \rightleftharpoons 2 A…

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