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