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One mole $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ and one mole $\mathrm{CO}(\mathrm{g})$ are taken in $1 \mathrm{~L}$ flask and heated to $725 \mathrm{~K}$. At equilibrium, $40 \%$ (by mass) of water reacted with $\mathrm{CO}(\mathrm{g})$ as follows
$\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})$
Its $\mathrm{K}_{\mathrm{c}}$ value is
Options:
$\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})$
Its $\mathrm{K}_{\mathrm{c}}$ value is
Solution:
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Verified Answer
The correct answer is:
0.444
$\Rightarrow$ Number of moles of $\mathrm{H}_2 \mathrm{O}$ reacting
$=40 \%$ of 1 mole $=0.4$ moles $($ degree of dissociation $=\alpha)$
$\Rightarrow 0.4$ moles of $\mathrm{CO}$ and 0.4 moles of $\mathrm{H}_2 \mathrm{O}$ react.
Let $\mathrm{C}=$ initial concentrations of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{CO}$.
$\Rightarrow$ At equilibrium;
$\begin{aligned}
& {\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{0.6 \mathrm{~mol}}{1 \mathrm{~L}}=0.6 \mathrm{M}} \\
& {[\mathrm{CO}]=\frac{0.6 \mathrm{~mol}}{1 \mathrm{~L}}=0.6 \mathrm{M}} \\
& {\left[\mathrm{H}_2\right]=\left[\mathrm{CO}_2\right]=\mathrm{c} \alpha=1 \times 0.4=0.4 \mathrm{M}} \\
& \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{CO}_2\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]}=\frac{(0.4)(0.4)}{(0.6)(0.6)} \\
& =0.444 .
\end{aligned}$
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