Question

If the equilibrium constant for ${ N2(g) + O2(g) <=> 2NO(g)}$ is K, the equilibrium constant for ${ \frac{1}{2} N_2 (g) + \frac{1}{2} O_2 (g) <=> NO(g) }$ will be :

• A. $K$
• B. $K^2$
• C. $K^{1/2}$
• D. $\frac{1}{2} K$

Answer 1

The Correct Option is C

To determine the equilibrium constant for the reaction \( \frac{1}{2} N_2 (g) + \frac{1}{2} O_2 (g) \leftrightarrow NO(g) \), given that the equilibrium constant for the reaction \( N_2(g) + O_2(g) \leftrightarrow 2NO(g) \) is K, let's analyze the situation step-by-step:

1. The original reaction is: \( N_2(g) + O_2(g) \leftrightarrow 2NO(g) \)
   The equilibrium constant for this reaction is K.
2. The new reaction given is: \( \frac{1}{2} N_2 (g) + \frac{1}{2} O_2 (g) \leftrightarrow NO(g) \)
3. Note that the new reaction is exactly half of the original reaction in terms of stoichiometry.
4. When the stoichiometry of a balanced chemical reaction is multiplied by a factor, the equilibrium constant is raised to the power of that factor. For example:
   • If the stoichiometric coefficients of all the substances in the reaction are divided by 2, the equilibrium constant is taken to the power of \(\frac{1}{2}\).
5. Therefore, the equilibrium constant for the reaction \( \frac{1}{2} N_2 (g) + \frac{1}{2} O_2 (g) \leftrightarrow NO(g) \) is \( K^{\frac{1}{2}} \).

Thus, the correct answer to the question is \(K^{1/2}\), making it the correct choice for the equilibrium constant for the new reaction.