Q14 / 180 · COMEDK 2023 (Memory Based) · Chemistry
Consider the following equilibrium,
$\begin{aligned}
& 2 \mathrm{No}(g) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2 ; K_{\mathrm{G}_1}=2.4 \times 10^{20} \\
& \mathrm{No}(g)+\frac{1}{2} \mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NoBr}(g) ; K_{C_2}=1.4
\end{aligned}$
Calculate $K_C$ for the reaction,
$\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g)$
- Correct answerA$8.96 \times 10^{-11}$
- B$9.48 \times 10^{-\theta}$
- C$8.08 \times 10^{-12}$
- D$8.96 \times 10^{11}$
Solution
For, $2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2$
$\mathrm{K}_{\mathrm{C}_1}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2}=2.4 \times 10^{20}$
For,
$\begin{aligned}
& \text { For, } \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g) \\
& K_{C_2}=\frac{[\mathrm{NOBr}]}{[\mathrm{NO}]\left[\mathrm{Br}_2\right]^{1 / 2}}=1.4 \\
& \text { and } K_C=\frac{[\mathrm{NOBr}]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]^{1 / 2}\left[\mathrm{Br}_2\right]^{1 / 2}} \\
& \text { or } K_C=\sqrt{\frac{1}{K_{\mathrm{C}_1}}} \times K_{C_2}=\sqrt{\frac{1}{2.4 \times 10^{20}}} \times 1.4 \\
& K_C=8.96 \times 10^{-11}
\end{aligned}$
Hence, the correct answer is option: 'A'.