For the equilibrium $2NO(g)+Cl_{2}(g)\rightleftharpoons2NOCl(g)$ $K_{p}$ is related to $K_{c}$ by the relation :

Question

Accepted Answer

To understand why Option 3 is the correct answer, we need to examine the relationship between the equilibrium constants $ K_p $ and $ K_c $ for the given reaction:

$$ 2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g) $$

### Explanation of the Relationship

The relationship between $ K_p $ and $ K_c $ is given by the equation:

$$ K_p = K_c (RT)^{\Delta n} $$

where:
- $ R $ is the universal gas constant.
- $ T $ is the temperature in Kelvin.
- $ \Delta n $ is the change in the number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

For the given reaction:

- Moles of gaseous products = 2 (from $ 2NOCl(g) $)
- Moles of gaseous reactants = 2 (from $ 2NO(g) + Cl_2(g) $)

Thus, $ \Delta n = 2 - 3 = -1 $.

Substituting $\Delta n = -1$ into the equation, we get:

$$ K_p = K_c (RT)^{-1} $$

This simplifies to:

$$ K_p = \frac{K_c}{RT} $$

This matches Option 3, which states:

$$ K_p = \frac{K_c}{RT} $$

### Explanation of Why Other Options are Incorrect

1. **Option 1: $ K_p = K_c (RT) $**
   - This option assumes $\Delta n = 1$, which is incorrect for this reaction. The correct $\Delta n$ is -1, not 1.

2. **Option 2: $ K_p = K_c (RT)^2 $**
   - This option assumes $\Delta n = 2$, which is incorrect. The correct $\Delta n$ is -1, not 2.

3. **Option 4: $ K_p = \frac{K_c}{(RT)^2} $**
   - This option assumes $\Delta n = -2$, which is incorrect. The correct $\Delta n$ is -1, not -2.

### Conclusion

The correct relationship between $ K_p $ and $ K_c $ for the given reaction, considering the change in moles of gas, is $ K_p = \frac{K_c}{RT} $. Therefore, Option 3 is the correct answer.