In which one of the following gaseous equilibrium, `K_(p)` is less than `K_(c)`?

Options

A. `N_(2)O_(4)(g) rarr 2NO_(2)(g)`
B. `2SO_(2)(g) + O_(2)(g) rarr 2SO_(2)(g)`
C. `2HI(g) rarr H_(2)(g) + I_(2)(g)`
D. `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)`

Explanation

To determine in which gaseous equilibrium \( K_p \) is less than \( K_c \), we need to analyze the relationship between \( K_p \) and \( K_c \) using the formula:

\[ K_p = K_c \cdot (RT)^{\Delta n_g} \]

Where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas, calculated as the number of moles of gaseous products minus the number of moles of gaseous reactants.

### Step-by-Step Solution:

1. **Identify the Equations**: We are given four equilibrium reactions. We will analyze each one to find \( \Delta n_g \).

2. **Calculate \( \Delta n_g \)**:
- For each reaction, count the moles of gaseous products and subtract the moles of gaseous reactants.
- If \( \Delta n_g < 0 \), then \( K_p < K_c \).

3. **Evaluate Each Option**:
- **Option 1**: If the reaction has 2 moles of products and 1 mole of reactants, then:
\[ \Delta n_g = 2 - 1 = 1 \quad (\text{not valid, } K_p \text{ is not less than } K_c) \]
- **Option 2**: If the reaction has 2 moles of products and 3 moles of reactants, then:
\[ \Delta n_g = 2 - 3 = -1 \quad (\text{valid, } K_p < K_c) \]
- **Option 3**: If the reaction has 2 moles of products and 2 moles of reactants, then:
\[ \Delta n_g = 2 - 2 = 0 \quad (\text{not valid, } K_p \text{ is equal to } K_c) \]
- **Option 4**: If the reaction has 2 moles of products and 3 moles of reactants, then:
\[ \Delta n_g = 2 - 3 = -1 \quad (\text{valid, } K_p < K_c) \]

4. **Conclusion**: The reactions in options 2 and 4 both yield \( K_p < K_c \) because \( \Delta n_g \) is negative.

### Final Answer:
The correct options where \( K_p < K_c \) are **Option 2 and Option 4**.