In which one of the following equilibrium ,` K_p ne K_c ? `

Options

A. ` 2NO(g) hArr N_2(g) +O_2(g)`
B. ` 2C(s) + O_2(g) hArr 2CO( g)`
C. ` 2Hl (g) hArr H_2 (g) +I_2(g)`
D. ` NO_2(g) + SO_2(g) hArr NO(g) +SO_3(g).`

Answer

To determine in which equilibrium \( K_p \neq K_c \), we need to analyze the given reactions and calculate the change in the number of gaseous moles, denoted as \( \Delta N_g \). The relationship between \( K_p \) and \( K_c \) is given by the equation:

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

where:
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta N_g \) is the difference in the number of moles of gaseous products and gaseous reactants.

### Step 1: Analyze each option for \( K_p \) and \( K_c \)

1. **Option 1**:
\[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \]
- Moles of reactants: 2 (1 from A + 1 from B)
- Moles of products: 2 (1 from C + 1 from D)
- \( \Delta N_g = 2 - 2 = 0 \)
- Here, \( K_p = K_c \).

2. **Option 2**:
\[ A(g) + B(s) \rightleftharpoons C(g) + D(g) \]
- Moles of reactants: 1 (only A is gaseous)
- Moles of products: 2 (1 from C + 1 from D)
- \( \Delta N_g = 2 - 1 = 1 \)
- Here, \( K_p \neq K_c \) because \( \Delta N_g \) is non-zero.

3. **Option 3**:
\[ 2A(g) \rightleftharpoons 2B(g) \]
- Moles of reactants: 2 (2 from A)
- Moles of products: 2 (2 from B)
- \( \Delta N_g = 2 - 2 = 0 \)
- Here, \( K_p = K_c \).

4. **Option 4**:
\[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \]
- Moles of reactants: 2 (1 from A + 1 from B)
- Moles of products: 2 (1 from C + 1 from D)
- \( \Delta N_g = 2 - 2 = 0 \)
- Here, \( K_p = K_c \).

### Conclusion:
From the analysis, we find that only in **Option 2** does \( K_p \) not equal \( K_c \) because \( \Delta N_g \) is non-zero.

### Final Answer:
The correct option is **2**.

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