`K_(p)` and `K_(c)` are inter related as
`K_(p)=K_(c)(RT)^(Deltan)`
Answer the following questions:
In which of the following equilibria `K_(p)` is less than `K_(c)`?

Options

A. `PCl_(5)(g) hArr PCl_(3)(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. `2H_(2)O(l) hArr 2H_(2)(g)+O_(2)(g)`

Explanation

To determine in which of the given equilibria \( K_p \) is less than \( K_c \), we need to calculate the change in the number of moles of gaseous products and reactants (\( \Delta n_g \)) for each reaction. 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 gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).

### Step-by-Step Solution:

1. **Identify the Reactions:**
- Reaction 1: \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \)
- Reaction 2: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \)
- Reaction 3: \( \text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\text{HCl}(g) \)
- Reaction 4: \( 2\text{H}_2(l) \rightleftharpoons 2\text{H}_2(g) + \text{O}_2(g) \)

2. **Calculate \( \Delta n_g \) for Each Reaction:**
- **Reaction 1:**
- Products: 1 (PCl3) + 1 (Cl2) = 2 moles
- Reactants: 1 (PCl5) = 1 mole
- \( \Delta n_g = 2 - 1 = 1 \)

- **Reaction 2:**
- Products: 2 (NH3) = 2 moles
- Reactants: 1 (N2) + 3 (H2) = 4 moles
- \( \Delta n_g = 2 - 4 = -2 \)

- **Reaction 3:**
- Products: 2 (HCl) = 2 moles
- Reactants: 1 (H2) + 1 (Cl2) = 2 moles
- \( \Delta n_g = 2 - 2 = 0 \)

- **Reaction 4:**
- Products: 2 (H2) + 1 (O2) = 3 moles
- Reactants: 2 (H2) = 0 moles (since liquid does not count)
- \( \Delta n_g = 3 - 0 = 3 \)

3. **Determine \( K_p \) and \( K_c \) Relationship:**
- **Reaction 1:** \( K_p = K_c (RT)^{1} \) → \( K_p > K_c \)
- **Reaction 2:** \( K_p = K_c (RT)^{-2} \) → \( K_p < K_c \)
- **Reaction 3:** \( K_p = K_c (RT)^{0} \) → \( K_p = K_c \)
- **Reaction 4:** \( K_p = K_c (RT)^{3} \) → \( K_p > K_c \)

4. **Conclusion:**
- The only reaction where \( K_p < K_c \) is for **Reaction 2**: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \).

### Final Answer:
The answer is **Reaction 2**.

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