Question

In a reaction $ {\text{A + 2B}} \rightleftarrows {\text{2C}} $ , 2 moles of A, 3 moles of B, and 1 mole of C are placed in a 2L flask and the equilibrium concentration of C is 1 $ {\text{M}} $ the equilibrium constant of the reaction is:
(A) $ \dfrac{1}{2} $
(B) $ \dfrac{4}{3} $
(C) $ \dfrac{5}{3} $
(D) $ \dfrac{2}{3} $

Answer 1

Hint: A reaction is said to be in chemical equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. It is equal to the ratio of concentration of products to the reactants raised to powers of their stoichiometric coefficients.

Complete step by step solution:
The equilibrium constant of any dynamic chemical reaction can be defined as equal to the ratio of the concentration of the reactants raised to their stoichiometric coefficient in the reaction to the products raised to their stoichiometric coefficient.
The reaction can be written as follows:
 $ {\text{A + 2B }} \rightleftharpoons {\text{ 2C}} $
2 moles 3 moles 1 mole (The initial concentration of the reactants and products)
1mole $ \dfrac{3}{2} $ moles $ \dfrac{1}{2} $ mole (initial concentration, for 1 litre volume) in $ {\text{M}} $ .
 $ \left( {1 - x} \right) $ $ \left( {\dfrac{3}{2} - 2x} \right) $ $ \left( {\dfrac{1}{2} + 2x} \right) $ (Equilibrium concentrations of the reactants and the products)
Now, as per the question, $ \left( {\dfrac{1}{2} + 2x} \right) $ = 1 $ {\text{M}} $
Therefore, x = $ \dfrac{1}{4} $ .
Therefore, the concentration of reactant A at equilibrium = $ \left( {1 - \dfrac{1}{4}} \right) $ = $ \dfrac{3}{4} $ $ {\text{M}} $
And that of reactant B is equal to $ \left( {\dfrac{3}{2} - 2x} \right) $ = $ \left( {\dfrac{3}{2} - \dfrac{1}{2}} \right) $ = 1 $ {\text{M}} $
Hence the equilibrium constant of the reaction,
K = $ \dfrac{{{{\left[ {\text{C}} \right]}^{\text{2}}}}}{{\left[ {\text{A}} \right]{{\left[ {\text{B}} \right]}^{\text{2}}}}} $ = $ \dfrac{{{{\left[ 1 \right]}^2}}}{{\left[ {\dfrac{3}{4}} \right]{{\left[ 1 \right]}^2}}} $ = $ \dfrac{4}{3} $
Hence, the correct answer is option B.

Note:
The molarity of a solution is defined as the number of moles of a solute dissolved in one litre of a solvent and is designated by the symbol $ {\text{mol }}{{\text{L}}^{{\text{ - 1}}}} $ . The equilibrium constant is a unit less quantity as it is a ratio of two quantities.