Q12 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Chemistry

At $500 \mathrm{~K}$, for the reaction $\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$, the $K_p$ is $0.036 \mathrm{~atm}^{-2}$. What is its $K_C$ in $\mathrm{L}^2 \mathrm{~mol}^{-1}$ ? $\left(R=0.082 \mathrm{~L}\right.$ atom $\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$.

  • A
    $2.1 \times 10^{-4}$
  • B
    $2.1 \times 10^{-5}$
  • Correct answer
    C
    $60.5$
  • D
    $605$

Solution

$K_p=K_C(R T)^{Δn}$ $\mathrm{N}_2(g)+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(g)$ $\begin{aligned} \Delta n & =2-(3+1)=2-4=-2 \\ K_p & =0.036 \quad \text { (given) }\end{aligned}$ $0.036=K_C(0.082 \times 500)^{-2}$ $\Rightarrow \quad K_C=\frac{0.036}{(0.082 \times 500)^{-2}}=60.516 \mathrm{~L}^2 \mathrm{~mol}^{-2}$

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

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