Question:
At T(K) the $K_c$ for the reactions (I) and (II) are $2.4 \times 10^{30}$ and 1.4 respectively. What is the $K_c$ value for: $\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) + \frac{1}{2}Br_2(g) \rightleftharpoons NOBr(g)$?
At T(K) the $K_c$ for the reactions (I) and (II) are $2.4 \times 10^{30}$ and 1.4 respectively. What is the $K_c$ value for: $\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) + \frac{1}{2}Br_2(g) \rightleftharpoons NOBr(g)$?
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When reversing a reaction, $K$ becomes $1/K$; when multiplying coefficients by $n$, $K$ becomes $K^n$.
Updated On: Jun 10, 2026
- $\frac{1.4}{\sqrt{2.4}} \times 10^{14}$
- $\frac{1.4}{\sqrt{2.4}} \times 10^{-15}$
- $\frac{\sqrt{1.4}}{2.4} \times 10^{-15}$
- $\frac{\sqrt{1.4}}{2.4} \times 10^{-16}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Equilibrium constant manipulation.
Step 2: Analysis
Reaction I: $2NO \rightleftharpoons N_2 + O_2$, $K_1 = 2.4 \times 10^{30}$. Target: $1/2 N_2 + 1/2 O_2 + 1/2 Br_2 \rightleftharpoons NOBr$. This is reverse of $1/2$ Reaction I + Reaction II: $(N_2+O_2 \rightleftharpoons 2NO)^{1/2} + (NO + 1/2 Br_2 \rightleftharpoons NOBr)$. $K = (1/K_1)^{1/2} \times K_2 = 1/\sqrt{2.4 \times 10^{30}} \times 1.4 = 1.4 / (\sqrt{2.4} \times 10^{15}) = \frac{1.4}{\sqrt{2.4}} \times 10^{-15}$.
Step 3: Conclusion
The value is $\frac{1.4}{\sqrt{2.4}} \times 10^{-15}$.
Final Answer: (B)
Equilibrium constant manipulation.
Step 2: Analysis
Reaction I: $2NO \rightleftharpoons N_2 + O_2$, $K_1 = 2.4 \times 10^{30}$. Target: $1/2 N_2 + 1/2 O_2 + 1/2 Br_2 \rightleftharpoons NOBr$. This is reverse of $1/2$ Reaction I + Reaction II: $(N_2+O_2 \rightleftharpoons 2NO)^{1/2} + (NO + 1/2 Br_2 \rightleftharpoons NOBr)$. $K = (1/K_1)^{1/2} \times K_2 = 1/\sqrt{2.4 \times 10^{30}} \times 1.4 = 1.4 / (\sqrt{2.4} \times 10^{15}) = \frac{1.4}{\sqrt{2.4}} \times 10^{-15}$.
Step 3: Conclusion
The value is $\frac{1.4}{\sqrt{2.4}} \times 10^{-15}$.
Final Answer: (B)
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