Q9 / 160 · TS EAMCET 2023 (13 May Shift 2) · Chemistry

At $T(K), K_C$ value for the reaction $\frac{1}{3} \mathrm{~N}_2(\mathrm{~g}) \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \frac{2}{3} \mathrm{NH}_3(\mathrm{~g})$ is 50 . The $\mathrm{K}_{\mathrm{C}}$ value for the reaction $2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$ at the same temperature is

  • A
    $4 \times 10^{-6}$
  • Correct answer
    B
    $8 \times 10^{-6}$
  • C
    $6 \times 10^{-6}$
  • D
    $8 \times 10^{-3}$

Solution

$\frac{1}{3} \mathrm{~N}_2+\mathrm{H}_2 \rightleftharpoons \frac{2}{3} \mathrm{NH}_3 \mathrm{~K}_{\mathrm{C}}=50$ given. Multiply by 3 in eq ${ }^{\text {n }}$ $\therefore \quad \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \text {. }$ Reverse eq (2) $\therefore \mathrm{K}_{\mathrm{C}}^{\prime}=(50)^3$ $\begin{aligned} & \therefore \quad 2 \mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2+3 \mathrm{H}_2 \mathrm{~K}_{\mathrm{C}^{\prime}}=\left(\frac{1}{50}\right)^3 \\ & \therefore \quad \mathrm{K}_{\mathrm{C}^{\prime}}=\frac{1}{50} \times \frac{1}{50} \times \frac{1}{50}=8 \times 10^{-6} . \end{aligned}$

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

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