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At $\mathrm{T}(\mathrm{K})$ when one $\mathrm{mol}$ of $\mathrm{X}$ and one mol of $\mathrm{Y}$ are heated in a $1 \mathrm{~L}$ flask, 0.5 moles of $\mathrm{Z}$ is formed at the equilibrium. The $K_c$ value for the reaction is
$\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})+\mathrm{A}(\mathrm{g})$
Options:
$\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})+\mathrm{A}(\mathrm{g})$
Solution:
1420 Upvotes
Verified Answer
The correct answer is:
$1.0$
$[\mathrm{X}]=\frac{1 \mathrm{~mol}}{1 \mathrm{~L}}=1 \mathrm{M} ;[\mathrm{Y}]=\frac{1 \mathrm{~mol}}{1 \mathrm{~L}}=1 \mathrm{M}$
$\begin{aligned}
& {[\mathrm{Z}]=\frac{0.5 \mathrm{~mol}}{1 \mathrm{~L}}=0.5 \mathrm{M}} \\
& \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})+\mathrm{A}(\mathrm{g}) \\
& \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{Z}][\mathrm{A}]}{[\mathrm{X}][\mathrm{Y}]}
\end{aligned}$
$\begin{array}{|c|c|c|c|c|}
\hline & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{A} \\
\hline \text{Before} & 1 & 1 & 0 & 0 \\
\hline \text{At equili.} & 1-\alpha & 1-\alpha & 0.5 & \alpha \\
\hline
\end{array}$
$\begin{aligned}
\Rightarrow \alpha & =0.5 \text { and }[\mathrm{X}]=[\mathrm{Y}]=1-\alpha=1-0.5 \\
& =0.5 \mathrm{M} . \text { Also, }[\mathrm{A}]=0.5 \mathrm{M} .
\end{aligned}$
$\begin{aligned}
& {[\mathrm{Z}]=\frac{0.5 \mathrm{~mol}}{1 \mathrm{~L}}=0.5 \mathrm{M}} \\
& \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})+\mathrm{A}(\mathrm{g}) \\
& \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{Z}][\mathrm{A}]}{[\mathrm{X}][\mathrm{Y}]}
\end{aligned}$
$\begin{array}{|c|c|c|c|c|}
\hline & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{A} \\
\hline \text{Before} & 1 & 1 & 0 & 0 \\
\hline \text{At equili.} & 1-\alpha & 1-\alpha & 0.5 & \alpha \\
\hline
\end{array}$
$\begin{aligned}
\Rightarrow \alpha & =0.5 \text { and }[\mathrm{X}]=[\mathrm{Y}]=1-\alpha=1-0.5 \\
& =0.5 \mathrm{M} . \text { Also, }[\mathrm{A}]=0.5 \mathrm{M} .
\end{aligned}$
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