In which of the following reaction `K_(p) gt K_(c)`

Options

A. `PCl_(3) + Cl_(2) to PCl_(5)`
B. `H_(2) + I_(2) to 2HI`
C. `2SO_(3) to O_(2) + 2SO_(2)`
D. `N_(2) + 3H_(2) to 2NH_(3)`

Explanation

To determine in which of the given reactions \( K_p \) is greater than \( K_c \), we can use the relationship between these equilibrium constants:

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

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

The key points to remember are:
- If \( \Delta N_g > 0 \), then \( K_p > K_c \).
- If \( \Delta N_g < 0 \), then \( K_p < K_c \).
- If \( \Delta N_g = 0 \), then \( K_p = K_c \).

Now, let's analyze each reaction one by one to find \( \Delta N_g \).

### Reaction 1: \( PCl_3(g) + Cl_2(g) \rightarrow PCl_5(g) \)

1. **Identify moles of reactants and products:**
- Reactants: \( PCl_3 \) (1 mole) + \( Cl_2 \) (1 mole) = 2 moles
- Products: \( PCl_5 \) (1 mole) = 1 mole

2. **Calculate \( \Delta N_g \):**
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 1 - 2 = -1 \]

3. **Conclusion:**
Since \( \Delta N_g < 0 \), \( K_p < K_c \).

### Reaction 2: \( H_2(g) + I_2(g) \rightarrow 2HI(g) \)

1. **Identify moles of reactants and products:**
- Reactants: \( H_2 \) (1 mole) + \( I_2 \) (1 mole) = 2 moles
- Products: \( 2HI \) (2 moles) = 2 moles

2. **Calculate \( \Delta N_g \):**
\[ \Delta N_g = 2 - 2 = 0 \]

3. **Conclusion:**
Since \( \Delta N_g = 0 \), \( K_p = K_c \).

### Reaction 3: \( 2SO_3(g) \rightarrow O_2(g) + 2SO_2(g) \)

1. **Identify moles of reactants and products:**
- Reactants: \( 2SO_3 \) (2 moles) = 2 moles
- Products: \( O_2 \) (1 mole) + \( 2SO_2 \) (2 moles) = 3 moles

2. **Calculate \( \Delta N_g \):**
\[ \Delta N_g = 3 - 2 = 1 \]

3. **Conclusion:**
Since \( \Delta N_g > 0 \), \( K_p > K_c \).

### Reaction 4: \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \)

1. **Identify moles of reactants and products:**
- Reactants: \( N_2 \) (1 mole) + \( 3H_2 \) (3 moles) = 4 moles
- Products: \( 2NH_3 \) (2 moles) = 2 moles

2. **Calculate \( \Delta N_g \):**
\[ \Delta N_g = 2 - 4 = -2 \]

3. **Conclusion:**
Since \( \Delta N_g < 0 \), \( K_p < K_c \).

### Final Conclusion:
The only reaction where \( K_p > K_c \) is the third reaction:

**Answer: Reaction 3: \( 2SO_3(g) \rightarrow O_2(g) + 2SO_2(g) \)**

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