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Question: At 100 K, then value of \( {K_c} \) ​ for the reaction \( {C_{(s)}} + {H_2}{O_{(g)}} \rightleftharpo...

At 100 K, then value of Kc{K_c} ​ for the reaction C(s)+H2O(g)⇌CO(g)+H2(g){C_{(s)}} + {H_2}{O_{(g)}} \rightleftharpoons C{O_{(g)}} + {H_{2(g)}} is 3.0×10−23.0 \times {10^{ - 2}} . Calculate equilibrium concentrations of H2O,CO2&H2{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]=[H2O]=0.18M;[H2O]=1.02M[CO] = [{H_2}O] = 0.18M;[{H_2}O] = 1.02M
B) [CO]=[H2O]=1.02M;[H2O]=0.18M[CO] = [{H_2}O] = 1.02M;[{H_2}O] = 0.18M
C) [CO]=[H2O]=0.36M;[H2O]=2.04M[CO] = [{H_2}O] = 0.36M;[{H_2}O] = 2.04M
D) None of These

Explanation

Solution

We are given the no. of moles and volume of the reactant H2O(g){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: [Re⁡actan⁡t]=nV[\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)+H2O(g)⇌CO(g)+H2(g){C_{(s)}} + {H_2}{O_{(g)}} \rightleftharpoons C{O_{(g)}} + {H_{2(g)}} , the active components are only three; they are H2O,CO2&H2{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 H2O(g){H_2}{O_{(g)}}
Concentration of H2O(g){H_2}{O_{(g)}} can be given as: [H2O]=6.0mol5.0L=1.2mol/L[{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 CO(g)&H2(g)C{O_{(g)}}\& {H_{2(g)}} can be assumed as ‘x’ mol/L. The equilibrium concentrations can be shown as:
C(s) + H2O(g) ⇌ CO(g) + H2(g){C_{(s)}}{\text{ }} + {\text{ }}{H_2}{O_{(g)}}{\text{ }} \rightleftharpoons {\text{ }}C{O_{(g)}}{\text{ }} + {\text{ }}{H_{2(g)}}

T=0-1.21.2--
T=equilibrium-1.2−x1.2 - xxxxx

We are given the value of equilibrium constant Kc{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 Kc{K_c} for the given reaction hence can be given as: Kc=[CO][H2][H2O]{K_c} = \dfrac{{[CO][{H_2}]}}{{[{H_2}O]}}
Kc=[CO][H2][H2O]=(x)(x)(1.2−x)=3.0×10−2{K_c} = \dfrac{{[CO][{H_2}]}}{{[{H_2}O]}} = \dfrac{{(x)(x)}}{{(1.2 - x)}} = 3.0 \times {10^{ - 2}}
Kc=(x)2(1.2−x)=3.0×10−2{K_c} = \dfrac{{{{(x)}^2}}}{{(1.2 - x)}} = 3.0 \times {10^{ - 2}}
x2=3.0×10−2(1.2−x){x^2} = 3.0 \times {10^{ - 2}}(1.2 - x)
x2−3.6×10−2−3.0×10−2x=0{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=0.18Mx = 0.18M
We know that [CO]=[H2]=x=0.18M[CO] = [{H_2}] = x = 0.18M and the concentration of H2O(g){H_2}{O_{(g)}} is =1.2−x=1.2−0.18=1.02M= 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 Kc{K_c} , only the concentration of active reactants and products are taken into consideration. In some cases instead of Kc{K_c} we can also be given Kp{K_p} , which is the ratio of partial pressures of the reactants to the partial pressures of the product.