Q60 / 180 · COMEDK 2025 (25 May Shift) · Chemistry
For a reaction
$2 A \rightleftharpoons 2 B+C$
2.0 moles of $A$ was placed in a one litre flask and the concentration of $B$ at equilibrium was found to be $0.4 \mathrm{~mol} \mathrm{~L}^{-1}$. The equilibrium constant at $30^{\circ} \mathrm{C}$ is $\qquad$ $\times 10^{-4}$
- Correct answerA125
- B150
- C140
- D115
Solution
The reaction is $2A \rightleftharpoons 2B + C$.
Initial moles: $[A]_0 = 2.0 \text{ mol L}^{-1}$, $[B]_0 = 0$, $[C]_0 = 0$.
At equilibrium, let the concentration of $B$ be $0.4 \text{ mol L}^{-1}$.
From the stoichiometry of the reaction, if $2x$ moles of $B$ are formed, then $x$ moles of $C$ are formed and $2x$ moles of $A$ are consumed.
Given $[B]_{eq} = 2x = 0.4$, so $x = 0.2 \text{ mol L}^{-1}$.
Equilibrium concentrations are:
$[A]_{eq} = 2.0 - 2x = 2.0 - 0.4 = 1.6 \text{ mol L}^{-1}$
$[B]_{eq} = 0.4 \text{ mol L}^{-1}$
$[C]_{eq} = x = 0.2 \text{ mol L}^{-1}$
The equilibrium constant $K_c$ is given by:
$K_c = \dfrac{[B]^2 [C]}{[A]^2}$
$K_c = \dfrac{(0.4)^2 \times (0.2)}{(1.6)^2}$
$K_c = \dfrac{0.16 \times 0.2}{2.56} = \dfrac{0.032}{2.56}$
$K_c = \dfrac{32 \times 10^{-3}}{2560 \times 10^{-3}} = \dfrac{32}{2560} = \dfrac{1}{80} = 0.0125$
$K_c = 125 \times 10^{-4}$
Answer: 125
Initial moles: $[A]_0 = 2.0 \text{ mol L}^{-1}$, $[B]_0 = 0$, $[C]_0 = 0$.
At equilibrium, let the concentration of $B$ be $0.4 \text{ mol L}^{-1}$.
From the stoichiometry of the reaction, if $2x$ moles of $B$ are formed, then $x$ moles of $C$ are formed and $2x$ moles of $A$ are consumed.
Given $[B]_{eq} = 2x = 0.4$, so $x = 0.2 \text{ mol L}^{-1}$.
Equilibrium concentrations are:
$[A]_{eq} = 2.0 - 2x = 2.0 - 0.4 = 1.6 \text{ mol L}^{-1}$
$[B]_{eq} = 0.4 \text{ mol L}^{-1}$
$[C]_{eq} = x = 0.2 \text{ mol L}^{-1}$
The equilibrium constant $K_c$ is given by:
$K_c = \dfrac{[B]^2 [C]}{[A]^2}$
$K_c = \dfrac{(0.4)^2 \times (0.2)}{(1.6)^2}$
$K_c = \dfrac{0.16 \times 0.2}{2.56} = \dfrac{0.032}{2.56}$
$K_c = \dfrac{32 \times 10^{-3}}{2560 \times 10^{-3}} = \dfrac{32}{2560} = \dfrac{1}{80} = 0.0125$
$K_c = 125 \times 10^{-4}$
Answer: 125
Hence, the correct answer is option: 'A'.