Question

One mole of $H_{2}O$ and one mole of $CO$ are taken in a $10$ litre vessel and heated to $725K$. At equilibrium, $40$ percent of water (by mass) reacts with carbon monoxide according to the equation:
$H_2{O}_{\left(g\right)}+C{O}_{\left(g\right)}\rightleftharpoons H_{2\left(g\right)}+C{O}_{2\left(g\right)}$
Calculate the equilibrium constant for the reaction.

Answer 1

$H_2{O}_{\left(g\right)}+C{O}_{\left(g\right)}\rightleftharpoons H_{2\left(g\right)}+C{O}_{2\left(g\right)}$
$1$ $1$ $0$ $0$ --- Initial no. of moles
$(1-0.4)$ $(1-0.4)$ $0.4$ $0.4$ --- Moles at equilibrium

$(40\%$ of $H_{2}O$ react at equilibrium)

$K_c=\frac{\left[C{O}_2\right]\left[H_2\right]}{\left[H_{2}O\right]\left[CO\right]}=\frac{0.4\times 0.4}{0.6\times 0.6}=0.44$.