The dissociation equilibrium of a gas \(A B_2\) can be represented as \[ 2 A B_{2(g)} \rightleftharpoons 2 A…

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
  1. \(\left(2 K_p / p\right)^{1 / 2}\)
  2. \(K_p / p\)
  3. \(2 K_p / p\)
  4. \(\left(2 K_p / p\right)^{1 / 3}\).

Solution

$2 A B_{2(g)} \rightleftharpoons 2 A B_{(g)}+B_{2(g)}$ $\begin{array}{lccc}\text { Initial } & 2 & 0 & 0 \\ \text { Equilibrium } & 2(1-x) & 2 x & x\end{array}$ $\begin{aligned} & \text { Moles at equilibrium }=2(1-x)+2 x+x \\ & =2-2 x+2 x+x=x+2 \\ & \begin{aligned} K_p & =\frac{\left[P_{A B}\right]^2\left[P_{B_2}\right]}{\left[P_{A B_2}\right]}=\frac{\left(\frac{2 x}{x+2} \times p\right)^2\left(\frac{x}{2+x} \times p\right)}{\left[\frac{2(1-x)}{x+2} \times p\right]^2} \\ & =\frac{\frac{4 x^3}{x+2} \times p}{4(1-x)^2}=\frac{4 x^2 \times p}{2} \times \frac{1}{4}\end{aligned}\end{aligned}$ $x=\left(\frac{2 K_p}{p}\right)^{1 / 3}$ (as $1-x \approx 1,2+x \approx 2$ )

Asked in: NEET 2006

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