Question

The value of ${K_c}$ = 4.24 at 800K for the reaction
$CO\left( g \right){H_2}O(g) \to C{O_2}(g) + {H_2}(g)$
Calculate equilibrium concentrations of $C{O_2}$ , ${H_2}$ , CO and ${H_2}O$ at 800K, if only CO and ${H_2}O$ are present initially at concentration of 0.10 M each?

Answer 1

If concentration of ${H_2}O$ and ${H_2}$ at equilibrium be taken as x, the concentration of CO and ${H_2}O$ at equilibrium becomes $\left( {0.1-x} \right)$ . and we can substitute the value in ${K_c} = \dfrac{{[C{O_2}][{H_2}]}}{{[CO][{H_2}O]}}$

Complete step by step answer:
In the reaction, $CO\left( g \right){H_2}O(g) \to C{O_2}(g) + {H_2}(g)$
Given,
Initial concentrations of CO and ${H_2}O$ = 0.1 M
Initial concentrations of $C{O_2}$ and ${H_2}$ = 0 M
So, at equilibrium, we have the following,
Concentration of $C{O_2}$ and ${H_2}$ at equilibrium = x
Concentration of CO and ${H_2}O$ at equilibrium = 0.1 – x
$\Rightarrow$ ${K_c} = \dfrac{{[C{O_2}][{H_2}]}}{{[CO][{H_2}O]}}$ = 4.24 at 800K
When we substitute the value in the expression, we get
$\Rightarrow$ 4.24 $= \dfrac{{[X][X]}}{{[0.1 - X][0.1 - X]}}$ $= \dfrac{{{X^2}}}{{{{(O.1 - X)}^2}}}$
$\Rightarrow$ 2.059 $= \dfrac{X}{{0.1 - X}}$
$\Rightarrow$ $0.2059-2.059x = x$
$\Rightarrow$ $3.059x = 0.2059$
$\Rightarrow$ x = 0.067 M
∴ Equilibrium concentration of $C{O_2}$ =(0.1 – x )= 0.033
Equilibrium concentration of ${H_2}$ = (0.1 – x) = 0.033
Equilibrium concentration of CO = x = 0.067
Equilibrium concentration of ${H_2}O$ = x = 0.067

Note:
If we know the ${K_c}$ and the initial concentrations for a reaction, then we can calculate the equilibrium concentrations. The equilibrium concentration is the sum of the initial concentration and the change which is delivered from the reaction stoichiometry. All reactant and product concentrations are constant at equilibrium.