Q26 / 180 · COMEDK 2026 (9 May Shift 2) · Chemistry
An equilibrium mixture taken in 2litre vessel of the reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ has 4 moles of $\text{SO}_2$, 3moles of $\text{O}_2$ and 6moles of $\text{SO}_3$ then the value of equilibrium constant $(K_c)$ will be:
- A0.75 L mol$^{-1}$
- B0.15 mol L$^{-1}$
- Correct answerC1.5 L mol$^{-1}$
- D15 mol L$^{-1}$
Solution
The given equilibrium reaction is:
$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$
The volume of the vessel is $V = 2$ L.
The equilibrium concentrations of the gases are:
$[\text{SO}_2] = \dfrac{4}{2} = 2 \text{ mol L}^{-1}$
$[\text{O}_2] = \dfrac{3}{2} = 1.5 \text{ mol L}^{-1}$
$[\text{SO}_3] = \dfrac{6}{2} = 3 \text{ mol L}^{-1}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}$
Substituting the equilibrium concentrations into the expression:
$K_c = \dfrac{(3)^2}{(2)^2 \times (1.5)}$
$K_c = \dfrac{9}{4 \times 1.5} = \dfrac{9}{6} = 1.5 \text{ L mol}^{-1}$
Answer: 1.5 L mol$^{-1}$
$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$
The volume of the vessel is $V = 2$ L.
The equilibrium concentrations of the gases are:
$[\text{SO}_2] = \dfrac{4}{2} = 2 \text{ mol L}^{-1}$
$[\text{O}_2] = \dfrac{3}{2} = 1.5 \text{ mol L}^{-1}$
$[\text{SO}_3] = \dfrac{6}{2} = 3 \text{ mol L}^{-1}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}$
Substituting the equilibrium concentrations into the expression:
$K_c = \dfrac{(3)^2}{(2)^2 \times (1.5)}$
$K_c = \dfrac{9}{4 \times 1.5} = \dfrac{9}{6} = 1.5 \text{ L mol}^{-1}$
Answer: 1.5 L mol$^{-1}$
Hence, the correct answer is option: 'C'.