At $\mathrm{T}(\mathrm{K})$, the $\mathrm{K}_{\mathrm{p}}$ for the reaction $\mathrm{A}_2…

At $\mathrm{T}(\mathrm{K})$, the $\mathrm{K}_{\mathrm{p}}$ for the reaction $\mathrm{A}_2 \mathrm{~B}_6(\mathrm{~g}) \rightleftharpoons \mathrm{A}_2 \mathrm{~B}_4(\mathrm{~g})+$ $\mathrm{B}_2(\mathrm{~g})$ is $0.04 \mathrm{~atm}$. The equilibrium pressure (in atm) of $\left.\mathrm{A}_2 \mathrm{~B}_6\right) \mathrm{g}$ ) when it is placed in a flask at $4 \mathrm{~atm}$ pressure and allowed to come to above equilibrium is
  1. $0.362$
  2. $0.380$
  3. $3.62$
  4. $2.62$

Solution

$\mathrm{A}_2 \mathrm{~B}_6(\mathrm{~g}) \rightleftharpoons \mathrm{A}_2 \mathrm{~B}_4(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g})$ $\begin{aligned} & \text {Initial } \begin{array}{l} \mathrm{p}=4 \mathrm{~atm} \quad 0.0 \\ \begin{array}{l} \text {Equilibrium } \\ =4 \alpha \end{array} \quad \mathrm{p} \alpha=4 \alpha \end{array} \quad \mathrm{p} \alpha=4 \alpha \\ & \qquad \begin{array}{l} \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{A}_2 \mathrm{~B}_4\right) \times \mathrm{P}\left(\mathrm{B}_2\right)}{\mathrm{P}\left(\mathrm{A}_2 \mathrm{~B}_6\right)}=0.04 \mathrm{~atm} \\ =\frac{(4 \alpha)(4 \alpha)}{4(1-\alpha)}=\frac{4 \alpha^2}{1-\alpha} \approx 4 \alpha^2 \\ \Rightarrow \alpha=0.1 \end{array} \end{aligned}$ Therefore, $\mathrm{P}\left(\mathrm{A}_2 \mathrm{~B}_6\right)$ at equilibrium $=4(1-\alpha)$ $\begin{aligned} & =4(1-0.1) \\ & =3.62 \end{aligned}$

Asked in: AP EAMCET 2023 (19 May Shift 1)

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