Question

In a reaction,
 $ A\left( g \right) + B\left( g \right) \rightleftharpoons C\left( g \right) + D\left( g \right) $
If the initial concentration of A is twice the initial concentration of B. After the equilibrium is reached, concentration of C is thrice the concentration of B. Calculate $ {{\text{K}}_{\text{c}}} $

Answer 1

Hint: To answer this question, you must recall the formula for finding the equilibrium constant of a reaction in terms of the concentrations of the constituents of the reaction mixture. The equilibrium constant of a reaction is given as the ratio between the concentrations of the products to the concentrations of the reactants and each term raised to the power of their stoichiometric coefficients in the balanced chemical equation.

Formula used: For a reaction, $ aA + bB \rightleftharpoons cC + dD $
We can write, $ {K_c} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}} $
Where, $ {K_c} $ denotes the equilibrium constant of the given reaction.
 $ \left[ A \right] $ denotes the concentration of reactant A in the reaction mixture at equilibrium.
 $ \left[ B \right] $ denotes the concentration of reactant B in the reaction mixture at equilibrium.
 $ \left[ C \right] $ denotes the concentration of product C in the reaction mixture at equilibrium.
 $ \left[ D \right] $ denotes the concentration of product C in the reaction mixture at equilibrium.
And, $ a,b,c,d $ are the stoichiometric coefficients of $ A,B,C,D $ respectively in the balanced chemical equation of the reaction.

Complete step by step solution:
In the question, we are given the equation of the reaction as,

t =0	2c	c	0	0
t = equilibrium	$ 2c - x $	$ c - x $	x	x

 $ A\left( g \right) + B\left( g \right) \rightleftharpoons C\left( g \right) + D\left( g \right) $
Considering the initial concentration of B as c, then the initial concentration of A will be 2c. Let the amount of reactants consumed and products formed be x each.
At equilibrium, we are given that $ \left[ C \right] = 3\left[ B \right] $ . So, we have
 $ \Rightarrow x = 3\left( {c - x} \right) $
 $ \therefore x = \dfrac{{3c}}{4} $
For the given equation, we can write the equilibrium constant as $ {K_c} = \dfrac{{\left[ C \right]\left[ D \right]}}{{\left[ A \right]\left[ B \right]}} $
 $ \Rightarrow {K_c} = \dfrac{{\left( x \right)\left( x \right)}}{{\left( {2c - x} \right)\left( {c - x} \right)}} = \dfrac{{\left( {3c/4} \right)\left( {3c/4} \right)}}{{\left( {2c - 3c/4} \right)\left( {c - 3c/4} \right)}} $
 $ \therefore {K_c} = 1.8 $

Note:
Equilibrium is a thermodynamic property of a reaction. It tells us the extent to which the reaction is completed. Equilibrium does not tell us anything about the rate of the reaction.