Q19 / 180 · COMEDK 2023 (Morning Shift) · Chemistry
Consider the following equilibrium,
$2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g); K_{C_1} = 2.4 \times 10^{20}$
$\text{NO}(g) + \frac{1}{2}\text{Br}_2(g) \rightleftharpoons \text{NOBr}(g); K_{C_2} = 1.4$
Calculate $K_C$ for the reaction,
$\frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) + \frac{1}{2}\text{Br}_2(g) \rightleftharpoons \text{NOBr}(g)$
- A$8.96 \times 10^{11}$
- B$8.08 \times 10^{-12}$
- Correct answerC$8.96 \times 10^{-11}$
- D$9.48 \times 10^{-9}$
Solution
Target reaction: $\dfrac{1}{2}N_2(g) + \dfrac{1}{2}O_2(g) + \dfrac{1}{2}Br_2(g) \rightleftharpoons NOBr(g)$
Reverse reaction 1 and multiply by $\dfrac{1}{2}$:
$\dfrac{1}{2}N_2(g) + \dfrac{1}{2}O_2(g) \rightleftharpoons NO(g)$, $K'' = \dfrac{1}{\sqrt{K_1}} = \dfrac{1}{\sqrt{2.4 \times 10^{20}}}$
Add reaction 2 as it is:
$NO(g) + \dfrac{1}{2}Br_2(g) \rightleftharpoons NOBr(g)$, $K_2 = 1.4$
$K_C = K'' \times K_2 = \dfrac{1.4}{\sqrt{2.4 \times 10^{20}}} = \dfrac{1.4}{\sqrt{2.4} \times 10^{10}}$
$= \dfrac{1.4 \times 0.64}{10^{10}} \approx \dfrac{0.896}{10^{10}} = 8.96 \times 10^{-11}$
Reverse reaction 1 and multiply by $\dfrac{1}{2}$:
$\dfrac{1}{2}N_2(g) + \dfrac{1}{2}O_2(g) \rightleftharpoons NO(g)$, $K'' = \dfrac{1}{\sqrt{K_1}} = \dfrac{1}{\sqrt{2.4 \times 10^{20}}}$
Add reaction 2 as it is:
$NO(g) + \dfrac{1}{2}Br_2(g) \rightleftharpoons NOBr(g)$, $K_2 = 1.4$
$K_C = K'' \times K_2 = \dfrac{1.4}{\sqrt{2.4 \times 10^{20}}} = \dfrac{1.4}{\sqrt{2.4} \times 10^{10}}$
$= \dfrac{1.4 \times 0.64}{10^{10}} \approx \dfrac{0.896}{10^{10}} = 8.96 \times 10^{-11}$
Hence, the correct answer is option: 'C'.