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…

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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 $$ \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) ightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) $$ Its $\mathrm{K}_{\mathrm{c}}$ value is

Aicharya · Accepted Answer

$\Rightarrow$ 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; $$ \begin{aligned} & {\left[\mathrm{H}_2 \mathrm{O}ight]=\frac{0.6 \mathrm{~mol}}{1 \mathrm{~L}}=0.6 \mathrm{M}} \\ & {[\mathrm{CO}]=\frac{0.6 \mathrm{~mol}}{1 \mathrm{~L}}=0.6 \mathrm{M}} \\ & {\left[\mathrm{H}_2ight]=\left[\mathrm{CO}_2ight]=\mathrm{c} \alpha=1 \times 0.4=0.4 \mathrm{M}} \\ & \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{H}_2ight]\left[\mathrm{CO}_2ight]}{\left[\mathrm{H}_2 \mathrm{O}ight]}=\frac{(0.4)(0.4)}{(0.6)(0.6)} \\ & =0.444 . \end{aligned} $$

One mole $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ and one mole $\mathrm{CO}(\mathrm{g})$ are taken in $1…

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