Question

The value of ${K_c}$ for the reaction: $2A \rightleftarrows B + c$ is $2X{10^{ - 3}}$. At the given time the composition of reaction mixture is $[A] = [B] = [C] = 3X{10^{ - 4}}M$ . In which direction the reaction will proceed?
A. Forward
B. Backward
C. Equilibrium
D. Forward or Backward

Answer 1

Hint: The reaction quotient Q is used to measure the amount of products and reactant present in the reaction at a particular time. Comparing the ${K_c}$ and ${Q_c}$ value of a reaction helps us to determine in which direction the reaction will proceed to reach the equilibrium state.


Step by step answer: ${K_c}$ is the equilibrium constant which determines the reactant and the finally derived products of a reaction whereas ${Q_c}$is the reaction quotient.
For the reversible reaction $2A \rightleftarrows B + c$${K_c}$=$2X{10^{ - 3}}$ the ${Q_c}$can be measured by using the below mentioned equation:

The reaction Quotient ${Q_c}$ \[
  \dfrac{{ = [B][C]}}{{{{[A]}^2}}} \\
    \\
\]

As explained in Le Chatelier’s principle, when the stress is applied to a particular reaction it will move away from the equilibrium and the reaction mixture will try to adjust them to bring it back into the equilibrium state. By comparing the value of ${K_c}$and ${Q_c}$ we can easily know that is our reaction consuming products to convert them into reactants or vice versa to attain equilibrium. The three possible scenario are:
${Q_c} \geqslant {K_c}$ : it suggests that we have more product as required to reach at the equilibrium state. So our reaction will proceed in the backward reaction forming reactions to reach equilibrium.
${Q_c} \leqslant {K_c}$ : It suggests that we have more reactants and product formation will take place to reach equilibrium state moving reaction into forward direction.
${Q_c} = {K_c}$ It means that our reaction is already at equilibrium making it move neither in forward nor in backward direction.


 In our case :
${Q_{c }}$ = \[\dfrac{{(3X{{10}^{ - 4}})(3X{{10}^{ - 4}})}}{{(3X{{10}^{ - 4)}}}}\] $ = 1$

Since ${Q_c} \geqslant {K_c}$ which means that our products are more than required to attain equilibrium state, the reaction will proceed in the backward reaction.

Hence the correct option is (B), backward direction.

Note: When the forward and backward reaction occurs at the same time a dynamic equilibrium state is achieved. At a given temperature the equilibrium constant is denoted with the ${K_c}$which is the ratio of product concentration with the ratio of reactants concentration whereas the ${Q_c}$is the reaction coefficient which means the ratio of instantaneous concentration of products with the ratio of instantaneous concentration of reactants mixture in the reaction.