In which of the following reactions is `K_(p)ltK_(c)`?

Options

A. `CO_(g)+CI_(2)(g)hArrCOCI_(2)(g)`
B. `CH_(4)(g)+H_(2)O(g)hArrCO(g)+3H_(2)(g)`
C. `2BrCI(g)hArrCI_(2)(g)+Br_(2)(g)`
D. `I_(2)(g)hArr2I(g)`

Answer

To determine in which of the following reactions \( K_p < K_c \), we need to analyze the relationship between \( K_p \) and \( K_c \) using the formula:

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

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.

### Step-by-Step Solution:

1. **Identify the Reactions**: We need to analyze each reaction provided in the question.

2. **Calculate \( \Delta n \)** for each reaction:
- **Reaction A**: \( \text{CO (g)} + \text{Cl}_2 (g) \rightarrow \text{COCl}_2 (g) \)
- Moles of products = 1 (COCl2)
- Moles of reactants = 1 (CO) + 1 (Cl2) = 2
- \( \Delta n = 1 - 2 = -1 \)

- **Reaction B**: \( \text{CH}_4 (g) + \text{H}_2\text{O} (g) \rightleftharpoons \text{CO} (g) + 3\text{H}_2 (g) \)
- Moles of products = 1 (CO) + 3 (H2) = 4
- Moles of reactants = 1 (CH4) + 1 (H2O) = 2
- \( \Delta n = 4 - 2 = 2 \)

- **Reaction C**: \( 2\text{BrCl} (g) \rightleftharpoons \text{Br}_2 (g) + \text{Cl}_2 (g) \)
- Moles of products = 1 (Br2) + 1 (Cl2) = 2
- Moles of reactants = 2 (BrCl)
- \( \Delta n = 2 - 2 = 0 \)

- **Reaction D**: \( \text{I}_2 (g) \rightleftharpoons 2\text{I} (g) \)
- Moles of products = 2 (I)
- Moles of reactants = 1 (I2)
- \( \Delta n = 2 - 1 = 1 \)

3. **Substitute \( \Delta n \) into the \( K_p \) and \( K_c \) relationship**:
- For **Reaction A**:
\[ K_p = K_c \cdot (RT)^{-1} \implies K_p < K_c \]
- For **Reaction B**:
\[ K_p = K_c \cdot (RT)^{2} \implies K_p > K_c \]
- For **Reaction C**:
\[ K_p = K_c \cdot (RT)^{0} \implies K_p = K_c \]
- For **Reaction D**:
\[ K_p = K_c \cdot (RT)^{1} \implies K_p > K_c \]

4. **Conclusion**: The only reaction where \( K_p < K_c \) is **Reaction A**.

### Final Answer:
**The correct answer is Reaction A: \( \text{CO (g)} + \text{Cl}_2 (g) \rightarrow \text{COCl}_2 (g) \)**.