Question

$Onemole$ of $H_{2}O$ and $onemole$ of $CO$ were heated in a $10L$ closed vessel at $1260K$. At equilibrium, $40$% of water was found to react in the equation,
$H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons H_2\left(g\right)+C{O}_2\left(g\right)$
Calculate the equilibrium constant 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 : $1$ $1$ $0$ $0$ (given)

At eqm : $(1-x)$ $(1-x)$ $x$ $x$ (let)

Given that $40$% of water is rated $\Rightarrow x=0.4$

$\Rightarrow$ At eqm, $\left[H_{2}O\right]=0.6,\left[CO\right]=0.6$

$\left[H_2\right]=0.4,\left[{CO}_2\right]=0.4$

$K_{eq}=\frac{\left[H_2\right]\left[{CO}_2\right]}{\left[H_{2}O\right]\left[CO\right]}$

$=\frac{0.4\times 0.4}{0.6\times 0.6}=0.44$