Q54 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry

For the reaction $\mathrm{P}+\mathrm{Q} \leftrightharpoons \mathrm{R}+\mathrm{S}$, carried out at 298 K , the equilibrium constant was found to be 169 and the initial concentrations of all reactants and products was 1.0 M . What is the equilibrium concentration of the reactants?

  • A
    0.698 M
  • B
    0.857 M
  • Correct answer
    C
    0.143 M
  • D
    0.462 M

Solution

The reaction is $\mathrm{P} + \mathrm{Q} \leftrightharpoons \mathrm{R} + \mathrm{S}$. The equilibrium constant $K_c = 169$.

Initial concentrations are $[P]_0 = 1.0 \text{ M}$, $[Q]_0 = 1.0 \text{ M}$, $[R]_0 = 1.0 \text{ M}$, $[S]_0 = 1.0 \text{ M}$.

Let $x$ be the change in concentration at equilibrium. Since the reaction quotient $Q_c = \dfrac{[R][S]}{[P][Q]} = \dfrac{1 \times 1}{1 \times 1} = 1 < K_c = 169$, the reaction proceeds in the forward direction.

At equilibrium:
$[P] = 1 - x$
$[Q] = 1 - x$
$[R] = 1 + x$
$[S] = 1 + x$

The equilibrium constant expression is $K_c = \dfrac{[R][S]}{[P][Q]} = \dfrac{(1+x)^2}{(1-x)^2} = 169$.

Taking the square root of both sides:
$\dfrac{1+x}{1-x} = \sqrt{169} = 13$.

$1 + x = 13(1 - x)$
$1 + x = 13 - 13x$
$14x = 12$
$x = \dfrac{12}{14} = \dfrac{6}{7} \approx 0.857$.

The equilibrium concentration of the reactants $[P] = [Q] = 1 - x = 1 - 0.857 = 0.143 \text{ M}$.

Answer: 0.143 M

Hence, the correct answer is option: 'C'.

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