In the equilibrium constant for `N_(2)(g) + O_(2)(g)hArr2NO(g)` is K, the equilibrium constant for `(1)/(2) N_(2)(g) + (1)/(2)O_(2)(g)hArrNO(g)` will be:

Options

A. K
B. `K^(2)`
C. `K^(1//2)`
D. `(K)/(2)`

Answer

To solve the problem, we need to determine the equilibrium constant for the reaction:

\[ \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO(g) \]

given that the equilibrium constant for the reaction:

\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]

is \( K \).

### Step-by-step Solution:

1. **Identify the Original Reaction and Its Equilibrium Constant**:
The original reaction is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]
The equilibrium constant for this reaction is given as \( K \).

2. **Relate the New Reaction to the Original Reaction**:
The new reaction can be obtained by halving the original reaction:
\[ \frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO(g) \]

3. **Apply the Rule for Changing Reaction Coefficients**:
When the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor. In this case, we have halved the coefficients (multiplied by \( \frac{1}{2} \)):
\[ K' = K^{\frac{1}{2}} \]

4. **Write the Final Expression for the New Equilibrium Constant**:
Therefore, the equilibrium constant for the new reaction is:
\[ K' = K^{\frac{1}{2}} \]

### Final Answer:
The equilibrium constant for the reaction \(\frac{1}{2} N_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO(g)\) is \( K^{\frac{1}{2}} \).

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