In which of the following equilibrium , the value of `K_p` is less than `K_c` ?

Options

A. `N_2O_4 hArr 2NO_2`
B. `N_2 +O_2 hArr 2NO`
C. `N_2 +3H_2 hArr 2NH_3`
D. `2SO_2 +O_2 hArr 2SO_3`

Explanation

To determine in which equilibrium the value of \( K_p \) is less than \( K_c \), we need to analyze the relationship between \( K_p \) and \( K_c \) given by the equation:

\[ K_p = K_c \cdot R^T \cdot \Delta n \]

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 \( n_p - n_r \) (the sum of the stoichiometric coefficients of the products minus that of the reactants).

### Step-by-Step Solution:

1. **Understand the Relationship**:
- From the equation \( K_p = K_c \cdot R^T \cdot \Delta n \), we can see that the value of \( K_p \) will be less than \( K_c \) if \( \Delta n < 0 \).

2. **Calculate \( \Delta n \)**:
- \( \Delta n = n_p - n_r \)
- Here, \( n_p \) is the total number of moles of gaseous products, and \( n_r \) is the total number of moles of gaseous reactants.

3. **Analyze Each Option**:
- **Option 1**:
- Products: 2 moles
- Reactants: 1 mole
- \( \Delta n = 2 - 1 = 1 \) (greater than 0, so \( K_p > K_c \))
- **Option 2**:
- Products: 2 moles
- Reactants: 2 moles
- \( \Delta n = 2 - 2 = 0 \) (so \( K_p = K_c \))
- **Option 3**:
- Products: 2 moles
- Reactants: 4 moles (1 + 3)
- \( \Delta n = 2 - 4 = -2 \) (less than 0, so \( K_p < K_c \))
- **Option 4**:
- Products: 1 mole
- Reactants: 3 moles (2 + 1)
- \( \Delta n = 1 - 3 = -2 \) (less than 0, so \( K_p < K_c \))

4. **Conclusion**:
- The equilibria where \( K_p < K_c \) are found in **Option 3** and **Option 4**.

### Final Answer:
The values of \( K_p \) are less than \( K_c \) in **Option 3** and **Option 4**.