At equilibrium for the reaction $\mathrm{A}_2(g)+\mathrm{B}_2(g) \rightleftharpoons 2 \mathrm{AB}(g)$ the…

At equilibrium for the reaction $\mathrm{A}_2(g)+\mathrm{B}_2(g) \rightleftharpoons 2 \mathrm{AB}(g)$ the concentrations of $\mathrm{A}_2, \mathrm{~B}_2$ and AB respectively are $1.5 \times 10^{-3} \mathrm{M}, 2.1 \times 10^{-3} \mathrm{M}$ and $1.4 \times 10^{-3} \mathrm{M}$ in a sealed vessel at 800 K . What will be $\mathrm{K}_{\mathrm{p}}$ for the decomposition of AB at same temperature?
  1. 0.62
  2. 1.6
  3. 0.44
  4. 2.27

Solution

$\begin{aligned} & \text { } \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g}) \\ & \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{AB}]^2}{\left[\mathrm{~A}_2\right]\left[\mathrm{B}_2\right]} \\ & \mathrm{K}_{\mathrm{c}}=\frac{1.4 \times 10^{-3} \times 1.4 \times 10^{-3}}{1.5 \times 10^{-3} \times 2.1 \times 10^{-3}} \\ & \mathrm{~K}_{\mathrm{c}}=0.622 \\ & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \mathrm{x}(\mathrm{RT})^{\Delta \mathrm{n}} \\ & \Delta \mathrm{n}=0 \\ & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}=0.622 \end{aligned}$ $\begin{aligned} \mathrm{K}_{\mathrm{p}} \text { for the decomposition of } \mathrm{AB}\left(\mathrm{~K}^{\prime}\right) & =\frac{1}{\mathrm{~K}_{\mathrm{p}}} \\ & =\frac{1}{0.622} \\ \mathrm{~K}^{\prime} & =1.60 \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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