If the equilibrium constant for the given reaction is 0.25
`NOhArr(1)/(2)N_(2)+(1)/(2)O_(2)`, then the equilibrium constant for the reaction `(1)/(2)N_(2)+(1)/(2)O_(2)hArrNO` will be

Options

A. (a) 1
B. (b) 2
C. (c) 3
D. (d) 4

Explanation

To find the equilibrium constant for the reaction \( \frac{1}{2}N_2 + \frac{1}{2}O_2 \rightleftharpoons NO \), given that the equilibrium constant for the reaction \( NO \rightleftharpoons \frac{1}{2}N_2 + \frac{1}{2}O_2 \) is 0.25, we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:
\[ NO \rightleftharpoons \frac{1}{2}N_2 + \frac{1}{2}O_2 \]
The equilibrium constant \( K_c \) for this reaction is given as:
\[ K_c = 0.25 \]

### Step 2: Write the equilibrium constant expression for the given reaction
The equilibrium constant expression for the reaction can be written as:
\[ K_c = \frac{[\frac{1}{2}N_2]^{1/2} \cdot [\frac{1}{2}O_2]^{1/2}}{[NO]^1} \]
This simplifies to:
\[ K_c = \frac{[\sqrt{N_2} \cdot \sqrt{O_2}]}{[NO]} \]

### Step 3: Write the reverse reaction and its equilibrium constant
The reverse reaction is:
\[ \frac{1}{2}N_2 + \frac{1}{2}O_2 \rightleftharpoons NO \]
Let’s denote the equilibrium constant for this reaction as \( K_c' \).

### Step 4: Relate the equilibrium constants of the forward and reverse reactions
For a reaction and its reverse, the relationship between their equilibrium constants is:
\[ K_c' = \frac{1}{K_c} \]
This means that if we know \( K_c \) for the forward reaction, we can find \( K_c' \) for the reverse reaction by taking its reciprocal.

### Step 5: Calculate \( K_c' \)
Substituting the value of \( K_c \):
\[ K_c' = \frac{1}{0.25} = 4 \]

### Conclusion
Thus, the equilibrium constant for the reaction \( \frac{1}{2}N_2 + \frac{1}{2}O_2 \rightleftharpoons NO \) is:
\[ K_c' = 4 \]