Question:
At 700 K, the equilibrium constant $K_e$ for the reaction $ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $ is 0.2 mol L$^{-2}$. What is the value of $K$ for the reverse reaction?
At 700 K, the equilibrium constant $K_e$ for the reaction $ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) $ is 0.2 mol L$^{-2}$. What is the value of $K$ for the reverse reaction?
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For the reverse reaction, the equilibrium constant is the reciprocal of the equilibrium constant for the forward reaction.
Updated On: Mar 18, 2026
- 5
- 0.2
- 0.04
- \(\frac{1}{0.2}\)
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The Correct Option is D
Solution and Explanation
The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.
Given that the equilibrium constant for the forward reaction is \(K_e = 0.2 \, \text{mol} \, \text{L}^{-2}\), the equilibrium constant for the reverse reaction is:
\[
K_{\text{reverse}} = \frac{1}{K_e} = \frac{1}{0.2} = 5
\]
Final answer
Answer: \(\boxed{\frac{1}{0.2} \text{ or } 5}\)
Final answer
Answer: \(\boxed{\frac{1}{0.2} \text{ or } 5}\)
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