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Question: The value of \( {K_c} = 4.24\; \) at \( 800K \) for the reaction \( CO\left( g \right) + {H_2}O\l...

The value of Kc=4.24  {K_c} = 4.24\; at 800K800K for the reaction
CO(g)+H2O(g)⇌CO2(g)+H2(g)CO\left( g \right) + {H_2}O\left( g \right) \rightleftharpoons C{O_2}\left( g \right) + {H_2}\left( g \right)
Calculate equilibrium concentrations of CO2C{O_2} ​, H2{H_2} , COCO and H2O  {H_2}O\; at 800K800K , if only COCO and H2O  {H_2}O\; are present initially at concentration of 0.10M  0.10M\; each?

Explanation

Solution

Equilibrium refers to a condition when the rate of forward reaction is equal to the rate of reverse reaction. The equilibrium constant, denoted by KK , expresses the relationship between reactants and products of a reaction at an equilibrium condition with respect to a specific unit.

Complete step by step solution:
For a generalised chemical reaction taking place in a solution:
aA+bB⇌cC+dD  aA + bB \rightleftharpoons cC + dD\;
The equilibrium constant can be expressed as follows:
K=[C]c[D]d[A]a[B]bK = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}
where [A], [B], [C] and [D] refer to the molar concentration of species A, B, C, D respectively at equilibrium. The coefficients like a, b, c, and d in the generalised chemical equation become exponents as seen in the above expression. The given reaction in the question is:
CO(g)+H2O(g)⇌CO2(g)+H2(g)CO\left( g \right) + {H_2}O\left( g \right) \rightleftharpoons C{O_2}\left( g \right) + {H_2}\left( g \right)
0.1 0 0 (Initial concentration) (Given)
0.1-x 0.1-x x x (At equilibrium) (Assumption)
In the question we are given Kc=4.24  {K_c} = 4.24\; .
Now, substituting all the values in the formula of equilibrium constant:
Kc=[CO2][H2][CO][H2O]{K_c} = \dfrac{{[C{O_2}][{H_2}]}}{{[CO][{H_2}O]}}
4.24=x2(0.1−x)24.24 = \dfrac{{{x^2}}}{{{{\left( {0.1 - x} \right)}^2}}}
x2=4.24(0.01+x2−0.2x){x^2} = 4.24(0.01 + {x^2} - 0.2x)
x2=0.0424+4.24x2−0.848x{x^2} = 0.0424 + 4.24{x^2} - 0.848x
∴3.24x2−0.848x+0.0424=0\therefore 3.24{x^2} - 0.848x + 0.0424 = 0
As we can see we have got the equation in the form of quadratic equation i.e. ax2+bx+c=0a{x^2} + bx + c = 0
So here, a=3.24a = 3.24
b=−0.848b = - 0.848
c=0.0424c = 0.0424
After solving this quadratic equation from the formula: x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} , we get two values i.e.
x1=0.067{x_1} = 0.067
x2=0.194{x_2} = 0.194
We will neglect the second value because the value of ‘x’ cannot be greater than the initial concentration value.
Hence the equilibrium concentration values will be as follows:
[CO]=[H2O]=0.1−x=0.1−0.067=0.033M\left[ {CO} \right] = \left[ {{H_2}O} \right] = 0.1 - x = 0.1 - 0.067 = 0.033M
[CO2]=[H2]=x=0.067M\left[ {C{O_2}} \right] = \left[ {{H_2}} \right] = x = 0.067M

Note:
While calculating the value of KK , always remember a few points such as (i) KK is constant for a particular reaction at a particular temperature. KK changes on changing the temperature, (ii) Pure solids or pure liquids are not generally included in the expression of equilibrium and (iii) The chemical reaction must always be balanced including the coefficients (lowest possible integer values) to attain the correct value for KK .