Question

In an equilibrium $A + B \rightleftharpoons C + D$ , A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reached, concentration of C was thrice the equilibrium concentration of B. Calculate ${K_C}$:
A. 0.14
B. 65
C. 3.6
D. 1.8

Answer 1

In the equilibrium constant expression, the concentrations of products appear in the numerator and the concentrations of the reactants appear in the denominator. The concentration terms are raised to appropriate stoichiometric coefficients.

Complete step by step solution:
Consider the reaction between A and B to form C and D.
$A + B \rightleftharpoons C + D$
Let $a{\text{ mol/L}}$ be the initial concentration of B.
The initial concentration of A is twice the initial concentration of B.
${\left[ A \right]_0} = 2[{\left[ B \right]_0} = 2a{\text{ mol/L}}$.
Prepare a table for the concentrations of reactants and products as shown below:

| A| B| C| D
---|---|---|---|---
Initial concentration (moles per liter) | 2a| a| 0| 0
Change in concentration (moles per liter)| -x| -x| +x| +x
Equilibrium concentration (moles per liter)| 2a-x| a-x| x| x

Write the expression for the equilibrium constant ${K_C}$
${K_C} = \dfrac{{\left[ C \right] \times \left[ D \right]}}{{\left[ A \right] \times \left[ B \right]}}$
Substitute the values of the concentrations of the reactants and products in the expression for the equilibrium constant ${K_C}$

$${K_C} = \dfrac{{\left[ C \right] \times \left[ D \right]}}{{\left[ A \right] \times \left[ B \right]}} \\\ \Rightarrow {K_C} = \dfrac{{x \times x}}{{\left( {2a - x} \right) \times \left( {a - x} \right)}} \\\ \Rightarrow {K_C} = \dfrac{{{x^2}}}{{\left( {2a - x} \right) \times \left( {a - x} \right)}}$$

At equilibrium, the concentration of C is three times the concentration of B

$$\left[ C \right] = 3\left[ B \right] \\\ \Rightarrow x{\text{ mol/L}} = 3\left( {a - x} \right){\text{ mol/L}} \\\ \Rightarrow {\text{x = }}\dfrac{{3a}}{4} \\\$$

Substitute the value of x in the equilibrium constant expression:

$$\Rightarrow {K_C} = \dfrac{{{x^2}}}{{\left( {2a - x} \right) \times \left( {a - x} \right)}} \\\ \Rightarrow {K_C} = \dfrac{{{{\left( {\dfrac{{3a}}{4}} \right)}^2}}}{{\left( {2a - \dfrac{{3a}}{4}} \right) \times \left( {a - \dfrac{{3a}}{4}} \right)}}$$ $$\Rightarrow {K_C} = \dfrac{{{{\left( {\dfrac{{3a}}{4}} \right)}^2}}}{{\left( {\dfrac{{5a}}{4}} \right) \times \left( {\dfrac{a}{4}} \right)}} \\\ \Rightarrow {K_C} = \dfrac{{9{a^2}}}{{5a \times a}} \\\ \Rightarrow {K_C} = \dfrac{{9{a^2}}}{{5{a^2}}} \\\ \Rightarrow {K_C} = 1.8$$

Hence, the value of the equilibrium constant ${K_C}$ for the given reaction is 1.8.

**Hence, the correct option is the option D.

Note:**
You can express the equilibrium constant either in terms of concentrations or in terms of partial pressures. When you express the equilibrium constant is expressed in terms of concentrations, you call it as ${K_C}$. When you express the equilibrium constant in terms of partial pressures, you call it as ${K_P}$.