Question

Consider the reaction $A + 2B \rightleftharpoons 2C$ (all are gases). 2 moles of A, 3 moles of B and 2 moles of C were placed in sealed 2 L container. The reaction proceeds at constant temperature and reaches equilibrium. The equilibrium concentration of C is found to be 0.5 mol $L^{-1}$. The $K_c$ of the reaction is

• A. 0.025
• B. 0.075
• C. 0.05
• D. 0.15

Answer 1

Answer: (C)

0.05

Answer 2

The problem involves calculating the equilibrium constant ($K_c$) for a given reaction using initial concentrations and one equilibrium concentration.

1. Calculate Initial Concentrations: The reaction is $A(g) + 2B(g) \rightleftharpoons 2C(g)$. Initial moles: A = 2 mol, B = 3 mol, C = 2 mol. Volume of container = 2 L.

Initial concentration of A, $[A]_0 = \frac{2 \text{ mol}}{2 \text{ L}} = 1 \text{ mol/L}$ Initial concentration of B, $[B]_0 = \frac{3 \text{ mol}}{2 \text{ L}} = 1.5 \text{ mol/L}$ Initial concentration of C, $[C]_0 = \frac{2 \text{ mol}}{2 \text{ L}} = 1 \text{ mol/L}$

2. Determine the Direction of the Reaction: The equilibrium concentration of C is given as $[C]_{eq} = 0.5 \text{ mol/L}$. Comparing initial and equilibrium concentrations of C: $[C]_0 = 1 \text{ mol/L}$ $[C]_{eq} = 0.5 \text{ mol/L}$ Since $[C]_{eq} < [C]_0$, the concentration of C has decreased. This indicates that the reaction proceeded in the reverse direction (from products to reactants) to reach equilibrium.

3. Set up ICE (Initial, Change, Equilibrium) Table: Let 'x' be the change in concentration of A that occurs when the reaction proceeds in the reverse direction.

Species	Initial (I) (mol/L)	Change (C) (mol/L)	Equilibrium (E) (mol/L)
A	1	+x	1 + x
B	1.5	+2x	1.5 + 2x
C	1	-2x	1 - 2x

4. Calculate the value of 'x': We are given the equilibrium concentration of C: $[C]_{eq} = 0.5 \text{ mol/L}$. From the ICE table, $[C]_{eq} = 1 - 2x$. So, $1 - 2x = 0.5$ $2x = 1 - 0.5$ $2x = 0.5$ $x = \frac{0.5}{2} = 0.25 \text{ mol/L}$

5. Calculate Equilibrium Concentrations of all Species: Now substitute the value of x back into the equilibrium expressions: $[A]_{eq} = 1 + x = 1 + 0.25 = 1.25 \text{ mol/L}$ $[B]_{eq} = 1.5 + 2x = 1.5 + 2(0.25) = 1.5 + 0.5 = 2.0 \text{ mol/L}$ $[C]_{eq} = 0.5 \text{ mol/L}$ (given)

6. Calculate the Equilibrium Constant ($K_c$): The expression for $K_c$ for the reaction $A(g) + 2B(g) \rightleftharpoons 2C(g)$ is: $K_c = \frac{[C]_{eq}^2}{[A]_{eq}[B]_{eq}^2}$

Substitute the equilibrium concentrations: $K_c = \frac{(0.5)^2}{(1.25)(2.0)^2}$ $K_c = \frac{0.25}{(1.25)(4.0)}$ $K_c = \frac{0.25}{5.0}$ $K_c = 0.05$