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?

• A. 5
• B. 0.2
• C. 0.04
• D. \(\frac{1}{0.2}\)

Hint:

For the reverse reaction, the equilibrium constant is the reciprocal of the equilibrium constant for the forward reaction.

Answer 1

The Correct Option is D

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: \(\boxed{\frac{1}{0.2} \text{ or } 5}\)