Q36 / 180 · COMEDK 2022 · Chemistry
The equilibrium constant, $K_C$ for $3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})$ is $4 \mathrm{~L}^2 \mathrm{~mol}^{-2}$. If the equilil rium concentration of benzene is $0.5 \mathrm{~mol}^{-1}$ than what is the value of concentration of ethylene?
- A$0.05 \mathrm{~mol} / \mathrm{L}$
- Correct answerB$0.5 \mathrm{~mol} / \mathrm{L}$
- C$0.25 \mathrm{~mol} / \mathrm{L}$
- D$0.025 \mathrm{~mol} / \mathrm{L}$
Solution
The chemical equation for the reaction is $3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})$.
The expression for the equilibrium constant $K_{C}$ is given by:
$K_{C} = \dfrac{[\mathrm{C}_{6} \mathrm{H}_{6}]}{[\mathrm{C}_{2} \mathrm{H}_{2}]^{3}}$
Given $K_{C} = 4 \mathrm{~L}^{2} \mathrm{~mol}^{-2}$ and $[\mathrm{C}_{6} \mathrm{H}_{6}] = 0.5 \mathrm{~mol} \mathrm{~L}^{-1}$.
Substituting the values into the expression:
$4 = \dfrac{0.5}{[\mathrm{C}_{2} \mathrm{H}_{2}]^{3}}$
$[\mathrm{C}_{2} \mathrm{H}_{2}]^{3} = \dfrac{0.5}{4} = 0.125$
$[\mathrm{C}_{2} \mathrm{H}_{2}] = (0.125)^{1/3} = (0.5^{3})^{1/3} = 0.5 \mathrm{~mol} \mathrm{~L}^{-1}$.
Answer: $0.5 \mathrm{~mol} / \mathrm{L}$
The expression for the equilibrium constant $K_{C}$ is given by:
$K_{C} = \dfrac{[\mathrm{C}_{6} \mathrm{H}_{6}]}{[\mathrm{C}_{2} \mathrm{H}_{2}]^{3}}$
Given $K_{C} = 4 \mathrm{~L}^{2} \mathrm{~mol}^{-2}$ and $[\mathrm{C}_{6} \mathrm{H}_{6}] = 0.5 \mathrm{~mol} \mathrm{~L}^{-1}$.
Substituting the values into the expression:
$4 = \dfrac{0.5}{[\mathrm{C}_{2} \mathrm{H}_{2}]^{3}}$
$[\mathrm{C}_{2} \mathrm{H}_{2}]^{3} = \dfrac{0.5}{4} = 0.125$
$[\mathrm{C}_{2} \mathrm{H}_{2}] = (0.125)^{1/3} = (0.5^{3})^{1/3} = 0.5 \mathrm{~mol} \mathrm{~L}^{-1}$.
Answer: $0.5 \mathrm{~mol} / \mathrm{L}$
Hence, the correct answer is option: 'B'.