Question

When $1mole{H}_2{O}_{\left(g\right)}$ and $1moleC{O}_{\left(g\right)}$ are heated at ${1000}^{\circ }C$ in a closed vessel of $5litre$, it was found that $40\%$ of $H_{2}O$ react at equilibrium according to $H_2{O}_{\left(g\right)}+C{O}_{\left(g\right)}\rightleftharpoons H_{2\left(g\right)}+C{O}_{2\left(g\right)}$. Find $K_C$ of the reaction.

Answer 1

The given reaction is :-

$H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons H_2\left(g\right)+{CO}_2\left(g\right)$

Initial moles : $1$ $1$ $0$ $0$

At eqm : $(1-\alpha )$ $(1-\alpha )$ $\alpha$ $\alpha$ (let)

Now, given that, at equilibrium,

$40$% of $H_{2}O$ has reacted $\Rightarrow \alpha =0.4$

$\therefore$ At eqm, no. of moles of $H_{2}O$ $=0.6$

No. of moles of $H_2=4$

No. of moles of ${CO}_2=0.4$

So, $K_C=\frac{\left[H_2\right]\left[{CO}_2\right]}{\left[H_{2}O\right]\left[CO\right]}=\frac{0.4\times 0.4}{0.6\times 0.6}=\frac{4}{9}=0.44$