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

Options

A. `N_(2)O_(4)hArr 2NO_(2)`
B. `2SO_(2)+O_(2)hArr 2SO_(3)`
C. `2HIhArr H_(2)+I_(2)`
D. `N_(2)+O_(2)hArr 2NO`

Answer

To determine in which gaseous equilibrium \( K_p \) is less than \( K_c \), we need to analyze the relationship between \( K_p \) and \( K_c \) using the formula:

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

Where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature,
- \( \Delta N_g \) is the change in the number of moles of gas, calculated as the sum of the stoichiometric coefficients of the products minus the sum of the stoichiometric coefficients of the reactants.

### Step 1: Identify the reactions and calculate \( \Delta N_g \)

We will analyze each of the given reactions to find \( \Delta N_g \).

1. **Reaction A**: \( N_2O_4 \rightleftharpoons 2 NO_2 \)
- Products: 2 (from \( 2 NO_2 \))
- Reactants: 1 (from \( N_2O_4 \))
- \( \Delta N_g = 2 - 1 = 1 \) (greater than 0)

2. **Reaction B**: \( 2 SO_2 + O_2 \rightleftharpoons 2 SO_3 \)
- Products: 2 (from \( 2 SO_3 \))
- Reactants: 3 (from \( 2 SO_2 + O_2 \))
- \( \Delta N_g = 2 - 3 = -1 \) (less than 0)

3. **Reaction C**: \( 2 HI \rightleftharpoons H_2 + I_2 \)
- Products: 2 (from \( H_2 + I_2 \))
- Reactants: 2 (from \( 2 HI \))
- \( \Delta N_g = 2 - 2 = 0 \)

4. **Reaction D**: \( N_2 + O_2 \rightleftharpoons 2 NO \)
- Products: 2 (from \( 2 NO \))
- Reactants: 2 (from \( N_2 + O_2 \))
- \( \Delta N_g = 2 - 2 = 0 \)

### Step 2: Analyze the results

From the calculations:
- For Reaction A, \( \Delta N_g = 1 \) (therefore, \( K_p > K_c \))
- For Reaction B, \( \Delta N_g = -1 \) (therefore, \( K_p < K_c \))
- For Reaction C, \( \Delta N_g = 0 \) (therefore, \( K_p = K_c \))
- For Reaction D, \( \Delta N_g = 0 \) (therefore, \( K_p = K_c \))

### Conclusion

The only reaction where \( K_p < K_c \) is **Reaction B**.

Thus, the answer is:
**B. \( 2 SO_2 + O_2 \rightleftharpoons 2 SO_3 \)**

---