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

Options

A. `2NO_((g))hArrN_(2(g))_O_(2)(g)`
B. `SO_(2(g))+NO_(2(g))hArrSO_(3(g))+NO((g))`
C. `H_(2(g))+I_(2(g))hArr2HI_((g))`
D. `2C_((s ))+O_(2(g))hArr2CO_(2(g))`

Answer

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

\[ K_p = K_c \cdot R^{\Delta N_g} \]

where \( \Delta N_g \) 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-by-Step Solution

1. **Identify the Equilibria**: We need to analyze each equilibrium reaction provided in the question.

2. **Calculate \( \Delta N_g \) for Each Reaction**:
- **First Reaction**: \( 2 \text{NO} \rightarrow \text{N}_2 + \text{O}_2 \)
- Moles of gaseous products = 1 (N2) + 1 (O2) = 2
- Moles of gaseous reactants = 2 (NO)
- \( \Delta N_g = 2 - 2 = 0 \)
- Therefore, \( K_p = K_c \).

- **Second Reaction**: \( \text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO} \)
- Moles of gaseous products = 1 (SO3) + 1 (NO) = 2
- Moles of gaseous reactants = 1 (SO2) + 1 (NO2) = 2
- \( \Delta N_g = 2 - 2 = 0 \)
- Therefore, \( K_p = K_c \).

- **Third Reaction**: \( \text{H}_2 + \text{I}_2 \rightarrow 2 \text{HI} \)
- Moles of gaseous products = 2 (HI)
- Moles of gaseous reactants = 1 (H2) + 1 (I2) = 2
- \( \Delta N_g = 2 - 2 = 0 \)
- Therefore, \( K_p = K_c \).

- **Fourth Reaction**: \( \text{C} (s) + \text{O}_2 \rightarrow 2 \text{CO}_2 \)
- Moles of gaseous products = 2 (CO2)
- Moles of gaseous reactants = 1 (O2) (Note: Solid carbon is not counted)
- \( \Delta N_g = 2 - 1 = 1 \)
- Therefore, \( K_p \neq K_c \) because \( K_p = K_c \cdot R^1 \).

3. **Conclusion**: The equilibrium in which \( K_p \) and \( K_c \) are not equal is the fourth reaction: \( \text{C} (s) + \text{O}_2 \rightarrow 2 \text{CO}_2 \).

### Final Answer
In the fourth reaction, \( K_p \) and \( K_c \) are not equal.