At a given temperature, the equilibrium constant for the reactions `NO(g)+1//2O_(2)(g)hArr NO_(2)(g)` and `2NO_(2)(g)hArr 2NO(g)+O_(2)(g)` are `K_(1)` and `K_(2)` respectivel. If `K_(1)` is `4xx10^(-3)` then `K_(2)` will be

Options

A. `8xx10^(-3)`
B. `16xx10^(-3)`
C. `6.25xx10^(4)`
D. `6.25xx10^(6)`

Explanation

To find the equilibrium constant \( K_2 \) for the reaction \( 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \), we can use the relationship between the equilibrium constants of the two reactions provided.

### Step-by-Step Solution:

1. **Identify the reactions and their equilibrium constants:**
- The first reaction is:
\[ NO(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO_2(g) \]
with equilibrium constant \( K_1 = 4 \times 10^{-3} \).
- The second reaction is:
\[ 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \]
with equilibrium constant \( K_2 \).

2. **Write the expression for \( K_1 \):**
\[ K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}} \]

3. **Write the expression for \( K_2 \):**
\[ K_2 = \frac{[NO]^2[O_2]}{[NO_2]^2} \]

4. **Relate \( K_2 \) to \( K_1 \):**
- The second reaction can be derived from the first reaction by reversing it and multiplying it by 2:
- Reversing the first reaction gives:
\[ NO_2(g) \rightleftharpoons NO(g) + \frac{1}{2}O_2(g) \]
The equilibrium constant for this reversed reaction is \( \frac{1}{K_1} \).
- Multiplying this reaction by 2 gives:
\[ 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \]
The equilibrium constant for this reaction is:
\[ K_2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

5. **Substitute the value of \( K_1 \):**
\[ K_2 = \frac{1}{(4 \times 10^{-3})^2} \]

6. **Calculate \( (4 \times 10^{-3})^2 \):**
\[ (4 \times 10^{-3})^2 = 16 \times 10^{-6} = 1.6 \times 10^{-5} \]

7. **Calculate \( K_2 \):**
\[ K_2 = \frac{1}{1.6 \times 10^{-5}} = 6.25 \times 10^4 \]

### Final Answer:
Thus, the value of \( K_2 \) is:
\[ K_2 = 6.25 \times 10^4 \]