Question

At 100 K, then value of $ {K_c} $ ​ for the reaction $ {C_{(s)}} + {H_2}{O_{(g)}} \rightleftharpoons C{O_{(g)}} + {H_{2(g)}} $ is $ 3.0 \times {10^{ - 2}} $ . Calculate equilibrium concentrations of $ {H_2}O,C{O_2}\& {H_2} $ in the reaction mixture obtained by heating 6.0 mole of steam and an excess of solid carbon in a 5.0 L container. What is the molar composition of the equilibrium mixture?
A) $ [CO] = [{H_2}O] = 0.18M;[{H_2}O] = 1.02M $
B) $ [CO] = [{H_2}O] = 1.02M;[{H_2}O] = 0.18M $
C) $ [CO] = [{H_2}O] = 0.36M;[{H_2}O] = 2.04M $
D) None of These

Answer 1

Hint: We are given the no. of moles and volume of the reactant $ {H_2}{O_{(g)}} $ i.e., steam, from here we can find the concentration of the reactant (M) which can be given by the formula: $ [\operatorname{Re} ac\tan t] = \dfrac{n}{V} $ where n is the no. of moles and V is the volume (in litres).

Complete Step By Step Answer:
In the reaction given to us $ {C_{(s)}} + {H_2}{O_{(g)}} \rightleftharpoons C{O_{(g)}} + {H_{2(g)}} $ , the active components are only three; they are $ {H_2}O,C{O_2}\& {H_2} $ . Always remember that if there are two or more states present, the one that is more scattered is considered as the active component. Here we are given a mixture of gas and solid. Hence the concentration of C will be neglected (also because it is given in access) . Let us first find the concentration of $ {H_2}{O_{(g)}} $
Concentration of $ {H_2}{O_{(g)}} $ can be given as: $ [{H_2}O] = \dfrac{{6.0mol}}{{5.0L}} = 1.2mol/L $
1.2 mol/L is the initial concentration of the reactant. Let us assume that after time ‘t’, when equilibrium is achieved, the concentration of the products $ C{O_{(g)}}\& {H_{2(g)}} $ can be assumed as ‘x’ mol/L. The equilibrium concentrations can be shown as:
 $ {C_{(s)}}{\text{ }} + {\text{ }}{H_2}{O_{(g)}}{\text{ }} \rightleftharpoons {\text{ }}C{O_{(g)}}{\text{ }} + {\text{ }}{H_{2(g)}} $

T=0	-	$ 1.2 $	-	-
T=equilibrium	-	$ 1.2 - x $	$ x $	$ x $

We are given the value of equilibrium constant $ {K_c} $ which can be given as the ratio of the concentrations of the active products to the concentration of the active reactants. The value of $ {K_c} $ for the given reaction hence can be given as: $ {K_c} = \dfrac{{[CO][{H_2}]}}{{[{H_2}O]}} $
 $ {K_c} = \dfrac{{[CO][{H_2}]}}{{[{H_2}O]}} = \dfrac{{(x)(x)}}{{(1.2 - x)}} = 3.0 \times {10^{ - 2}} $
 $ {K_c} = \dfrac{{{{(x)}^2}}}{{(1.2 - x)}} = 3.0 \times {10^{ - 2}} $
 $ {x^2} = 3.0 \times {10^{ - 2}}(1.2 - x) $
 $ {x^2} - 3.6 \times {10^{ - 2}} - 3.0 \times {10^{ - 2}}x = 0 $
On solving the quadratic equation, we get the value of x as: $ x = 0.18M $
We know that $ [CO] = [{H_2}] = x = 0.18M $ and the concentration of $ {H_2}{O_{(g)}} $ is $ = 1.2 - x = 1.2 - 0.18 = 1.02M $
Hence the correct answer is Option (A).

Note:
In case we are given many reactants and products, for finding the value of $ {K_c} $ , only the concentration of active reactants and products are taken into consideration. In some cases instead of $ {K_c} $ we can also be given $ {K_p} $ , which is the ratio of partial pressures of the reactants to the partial pressures of the product.