In which of the following equilibrium `K_(c) and K_(p)` are not equal?

Options

A. `2C(g)+2O_(2)(g)hArr 2CO_(2)(g)`
B. `2NO(g)hArr N_(2)(g)+O_(2)(g)`
C. `SO_(2)(g)+NO_(2)(g)hArr SO_(3)(g)+NO(g)`
D. `H_(2)(g)+I_(2)(g)hArr 2HI(g)`

Answer

To determine in which of the given equilibria \( K_c \) and \( K_p \) are not equal, we will use the relationship between \( K_p \) and \( K_c \):

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

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 the sum of the stoichiometric coefficients of the products minus the sum of the stoichiometric coefficients of the reactants.

For \( K_p \) to equal \( K_c \), \( \Delta N_G \) must equal 0. Therefore, we need to find the reaction for which \( \Delta N_G \) is not equal to 0.

### Step-by-Step Solution:

1. **Identify the Reactions**: We will analyze each of the given reactions one by one to calculate \( \Delta N_G \).

2. **Reaction A**:
\[ 2C + 2O_2 \rightarrow 2CO_2 \]
- Products: 2 (from \( 2CO_2 \))
- Reactants: 2 (from \( 2C \)) + 2 (from \( 2O_2 \)) = 4
- \( \Delta N_G = 2 - 4 = -2 \)

3. **Reaction B**:
\[ 2NO \rightarrow N_2 + O_2 \]
- Products: 2 (1 from \( N_2 \) + 1 from \( O_2 \))
- Reactants: 2 (from \( 2NO \))
- \( \Delta N_G = 2 - 2 = 0 \)

4. **Reaction C**:
\[ SO_2 + NO_2 \rightarrow SO_3 + NO \]
- Products: 2 (1 from \( SO_3 \) + 1 from \( NO \))
- Reactants: 1 (from \( SO_2 \)) + 1 (from \( NO_2 \)) = 2
- \( \Delta N_G = 2 - 2 = 0 \)

5. **Reaction D**:
\[ H_2 + I_2 \rightarrow 2HI \]
- Products: 2 (from \( 2HI \))
- Reactants: 1 (from \( H_2 \)) + 1 (from \( I_2 \)) = 2
- \( \Delta N_G = 2 - 2 = 0 \)

### Conclusion:
From the calculations, we find:
- For Reaction A, \( \Delta N_G = -2 \) (thus \( K_p \neq K_c \)).
- For Reactions B, C, and D, \( \Delta N_G = 0 \) (thus \( K_p = K_c \)).

Therefore, the answer is **Reaction A** where \( K_c \) and \( K_p \) are not equal.