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

Options

A. `PCl_(3(g))+Cl_(2(g))hArrPCl_(4(g))`
B. `H_(2(g))+Cl_(2(g))hArr2HCl_((g))`
C. `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`
D. `CaCO_(3(s))hArrCaO_((s))+CO_(2(g))`

Explanation

To determine for which of the given reactions \( K_p = K_c \), we need to analyze the relationship between \( K_p \) and \( K_c \). The relationship is given by the equation:

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

where:
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas, calculated as:

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

### Step-by-Step Solution:

1. **Identify the Reactions**: We will analyze each reaction provided to find the value of \( \Delta n \).

2. **Calculate \( \Delta n \) for Each Reaction**:
- **Reaction 1**: \( \text{PCl}_3(g) + \text{Cl}_2(g) \rightarrow \text{PCl}_4(g) \)
- Moles of gaseous products: 1 (PCl4)
- Moles of gaseous reactants: 2 (PCl3 + Cl2)
- \( \Delta n = 1 - 2 = -1 \)

- **Reaction 2**: \( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \)
- Moles of gaseous products: 2 (2 HCl)
- Moles of gaseous reactants: 2 (H2 + Cl2)
- \( \Delta n = 2 - 2 = 0 \)

- **Reaction 3**: \( 3 \text{H}_2(g) + \text{N}_2(g) \rightarrow 2 \text{NH}_3(g) \)
- Moles of gaseous products: 2 (2 NH3)
- Moles of gaseous reactants: 4 (3 H2 + 1 N2)
- \( \Delta n = 2 - 4 = -2 \)

- **Reaction 4**: \( \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \)
- Moles of gaseous products: 1 (CO2)
- Moles of gaseous reactants: 0 (CaCO3 is solid)
- \( \Delta n = 1 - 0 = 1 \)

3. **Determine \( K_p = K_c \)**:
- From the calculations:
- Reaction 1: \( \Delta n = -1 \) (not equal)
- Reaction 2: \( \Delta n = 0 \) (this is the case where \( K_p = K_c \))
- Reaction 3: \( \Delta n = -2 \) (not equal)
- Reaction 4: \( \Delta n = 1 \) (not equal)

### Conclusion:
The reaction for which \( K_p = K_c \) is **Reaction 2**: \( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \).