In which one of the following gaseous equilibrium, `K_(p)` is less than `K_(c)`?

Options

A. `N_(2)O_(4)(g) rarr 2NO_(2)(g)`
B. `2SO_(2)(g) + O_(2)(g) rarr 2SO_(2)(g)`
C. `2HI(g) rarr H_(2)(g) + I_(2)(g)`
D. `N_(2)(g) + O_(2)(g) rarr 2NO(g)`

Explanation

To determine in which gaseous equilibrium \( K_p \) is less than \( K_c \), we can follow these steps:

### Step 1: Understand the relationship between \( K_p \) and \( K_c \)
The relationship between the equilibrium constants \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c (RT)^{\Delta N_g} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta N_g \) is the change in the number of moles of gas, calculated as:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

### Step 2: Identify when \( K_p < K_c \)
From the equation, we can deduce that:
- \( K_p < K_c \) when \( \Delta N_g < 0 \)
This means that the number of moles of gaseous products is less than the number of moles of gaseous reactants.

### Step 3: Analyze the given reactions
We need to evaluate the provided reactions to calculate \( \Delta N_g \) for each:

1. **Reaction 1:**
\[ A(g) \rightleftharpoons 2B(g) \]
- Moles of products = 2 (B)
- Moles of reactants = 1 (A)
- \( \Delta N_g = 2 - 1 = 1 \) (Positive)

2. **Reaction 2:**
\[ 3C(g) \rightleftharpoons 2D(g) \]
- Moles of products = 2 (D)
- Moles of reactants = 3 (C)
- \( \Delta N_g = 2 - 3 = -1 \) (Negative)

3. **Reaction 3:**
\[ E(g) \rightleftharpoons F(g) \]
- Moles of products = 1 (F)
- Moles of reactants = 1 (E)
- \( \Delta N_g = 1 - 1 = 0 \) (Zero)

4. **Reaction 4:**
\[ G(g) + H(g) \rightleftharpoons I(g) + J(g) \]
- Moles of products = 2 (I and J)
- Moles of reactants = 2 (G and H)
- \( \Delta N_g = 2 - 2 = 0 \) (Zero)

### Step 4: Conclusion
From the analysis:
- **Reaction 1:** \( K_p > K_c \) (since \( \Delta N_g > 0 \))
- **Reaction 2:** \( K_p < K_c \) (since \( \Delta N_g < 0 \))
- **Reaction 3:** \( K_p = K_c \) (since \( \Delta N_g = 0 \))
- **Reaction 4:** \( K_p = K_c \) (since \( \Delta N_g = 0 \))

Thus, the only reaction where \( K_p < K_c \) is **Reaction 2**.

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

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