Question

In the given below reaction, if the concentration of A, B, C are 0.2, 0.2, and 0.2
respectively and K c value is 1, then find the concentration of A, B, C at equilibrium.
${\text{2A }} \rightleftharpoons {\text{ B + C}}$

Answer 1

They have given the initial concentration of all the three components, but they have asked the concentration when the reaction reaches equilibrium. For that we should make some assumptions for the concentration terms in the equilibrium constant equation, then solve the equation and we will get the value for the unknown concentrations at equilibrium.

Complete Step by step answer: Kc is nothing but equilibrium constant, the equation for Kc can be given as
${{\text{K}}_{\text{c}}} = {\text{ }}\dfrac{{\left[ {\text{B}} \right]\left[ {\text{C}} \right]}}{{{{\left[ {\text{A}} \right]}^2}}}$
Where [A] is concentration of A
[B] is concentration of B
[C] is concentration of C
Let us take ‘x’ as initial concentration of A, and ‘y’ as the amount of A consumed in the reaction to reach equilibrium. Therefore, at equilibrium the concentration of A becomes (x-y).
Before starting the reaction the concentration of B and C will be zero as they would have not formed, and the amount of B and C formed will each be equal to ‘y’ at equilibrium.
This can be written in the table as below

| [A]| [B]| [C]
---|---|---|---
At time = 0 sec| x| 0| 0
At equilibrium| (x – y)| y| y

If we substitute these values in the equation of Kc, we get
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{\text{y}} \times {\text{y}}}}{{{{\left( {{\text{x - y}}} \right)}^2}}} \\\ \Rightarrow {{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{\text{y}}^2}}}{{{{\left( {{\text{x - y}}} \right)}^{\text{2}}}}} \\\ \Rightarrow {\left( {{\text{x - y}}} \right)^2} = \dfrac{{{{\text{y}}^2}}}{{{{\text{K}}_{\text{c}}}}} = \dfrac{{{{\text{y}}^2}}}{1} = {{\text{y}}^2} \\\$
Since, Kc = 1 (given in the question)
${{\text{x}}^2} + {{\text{y}}^2} - 2{\text{xy = }}{{\text{y}}^2} \\\ \Rightarrow {\text{ }}{{\text{x}}^2} - 2{\text{xy}} = {{\text{y}}^2} - {{\text{y}}^2} \\\ \Rightarrow {\text{ }}{{\text{x}}^2} - 2{\text{xy}} = 0 \\\ \Rightarrow {\text{ }}{{\text{x}}^2} = 2{\text{xy}} \\\$
Let substitute x = 0.2 (given in the question)

$${\text{ }}{\left( {0.2} \right)^2} = 2 \times 0.2 \times {\text{y}} \\\ \Rightarrow \dfrac{{0.2 \times 0.2}}{{2 \times 0.2}} = {\text{y}} \\\ \Rightarrow {\text{ 0}}{\text{.1 = y}} \\\$$

$\Rightarrow {\text{A = }}\left( {{\text{x - y}}} \right) = \left( {0.2 - 0.1} \right) = 0.1$
Therefore the concentration of B = C = y = 0.1 and A = 0.1

Note: In the equilibrium reactions the rate of forward reaction is equal to the rate of backward reaction and always remember that at equilibrium the concentrations of the reactants and the products formed will be constant.