One mole of \mathrm{H}_{2} \mathrm{O} and one mole of \mathrm{CO} are taken in 10 \mathrm{~L} vessel and heated to 725 \mathrm{~K}. At equilibrium, 60 \% of water (by mass) reacts with \mathrm{CO} according to the equation, \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{co}_{2}(\mathrm{~g}) Calculate the equilibrium constant for the reaction.
One mole of \mathrm{H}_{2} \mathrm{O} and one mole of \mathrm{CO} are taken in 10 \mathrm{~L} vessel and heated to 725 \mathrm{~K}. At equilibrium, 60 \% of water (by mass) reacts with \mathrm{CO} according to the equation, \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{co}_{2}(\mathrm{~g}) Calculate the equilibrium constant for the reaction.

Answer:

As we know that the equilibrium constant, Kc, is defined as the product of the equilibrium concentrations of products over the equilibrium concentrations of reactants, each raised to the power of the stoichiometric coefficients of the reactants and products.

The given reaction in the question is,

\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})

Therefore, the equilibrium constant for the reaction,

\mathrm{K}_{\mathrm{C}}=\left(\left[\mathrm{H}_{2}\right]\left[\mathrm{CO}_{2}] \mathrm{~J}\right) /\left(\left[\mathrm{H}_{2} \mathrm{O}\right][\mathrm{CO}]\right)=\left(0.4 \times 0.4\right) /(0.6 \times 0.6)=0.444\right.