For which of the following reaction, `K_(p) ne K_(c)`?

Options

A. `2NOCl(g) hArr 2NO(g)+Cl_(2)(g)`
B. `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
C. `H_(2)(g)+Cl_(2)(g) hArr 2HCl(g)`
D. `2N_(2)O_(4)(g) hArr 2NO_(2)(g)`

Explanation

To determine for which of the following reactions \( K_p \neq K_c \), we will use the relationship between \( K_p \) and \( K_c \):

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

where \( \Delta n \) is defined as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants:

\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

### Step-by-Step Solution

1. **Identify the Reactions**: We need to analyze each reaction given in the options to find \( \Delta n \).

2. **Calculate \( \Delta n \) for Each Reaction**:

- **Reaction 1**: \( 2 \text{NOCl} (g) \rightarrow 2 \text{NO} (g) + \text{Cl}_2 (g) \)
- Moles of products: \( 2 + 1 = 3 \)
- Moles of reactants: \( 2 \)
- \( \Delta n = 3 - 2 = 1 \)
- Since \( \Delta n \) is positive, \( K_p > K_c \).

- **Reaction 2**: \( \text{N}_2 (g) + 3 \text{H}_2 (g) \rightarrow 2 \text{NH}_3 (g) \)
- Moles of products: \( 2 \)
- Moles of reactants: \( 1 + 3 = 4 \)
- \( \Delta n = 2 - 4 = -2 \)
- Since \( \Delta n \) is negative, \( K_c > K_p \).

- **Reaction 3**: \( \text{H}_2 (g) + \text{Cl}_2 (g) \rightarrow 2 \text{HCl} (g) \)
- Moles of products: \( 2 \)
- Moles of reactants: \( 1 + 1 = 2 \)
- \( \Delta n = 2 - 2 = 0 \)
- Since \( \Delta n \) is zero, \( K_p = K_c \).

- **Reaction 4**: \( 2 \text{N}_2\text{O}_4 (g) \rightarrow 2 \text{N}_2 (g) + 2 \text{O}_2 (g) \)
- Moles of products: \( 2 + 2 = 4 \)
- Moles of reactants: \( 2 \)
- \( \Delta n = 4 - 2 = 2 \)
- Since \( \Delta n \) is positive, \( K_p > K_c \).

3. **Conclusion**: From the calculations:
- For Reaction 1 and Reaction 4, \( K_p > K_c \).
- For Reaction 2, \( K_c > K_p \).
- For Reaction 3, \( K_p = K_c \).

Thus, the reactions for which \( K_p \neq K_c \) are Reaction 1 and Reaction 2.