For which one of the following reactions `K_(p)=K_(c)`?

Options

A. `PCl_(5)hArr PCl_(3)+Cl_(2)`
B. `N_(2)+O_(2)hArr2NO`
C. `N_(2)+3H_(2)hArr 2NH_(3)`
D. `2SO_(3)hArr 2SO_(2)+O_(2)`

Answer

To determine for which reaction \( K_p = K_c \), we can use the relationship between \( K_p \) and \( K_c \):

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

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:

\[ \Delta N_g = \text{(sum of stoichiometric coefficients of products)} - \text{(sum of stoichiometric coefficients of reactants)} \]

For \( K_p \) to equal \( K_c \), \( \Delta N_g \) must be equal to 0. This means that the number of moles of gaseous products must equal the number of moles of gaseous reactants.

Now, let's analyze the given options one by one:

### Option A:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]
- **Products:** 1 (PCl3) + 1 (Cl2) = 2
- **Reactants:** 1 (PCl5) = 1
- **Calculation of \( \Delta N_g \):**
\[ \Delta N_g = 2 - 1 = 1 \quad (\text{not } 0) \]

### Option B:
\[ \text{N}_2 + \text{O}_2 \rightleftharpoons 2 \text{NO} \]
- **Products:** 2 (NO) = 2
- **Reactants:** 1 (N2) + 1 (O2) = 2
- **Calculation of \( \Delta N_g \):**
\[ \Delta N_g = 2 - 2 = 0 \quad (\text{this is } 0) \]

### Option C:
\[ \text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3 \]
- **Products:** 2 (NH3) = 2
- **Reactants:** 1 (N2) + 3 (H2) = 4
- **Calculation of \( \Delta N_g \):**
\[ \Delta N_g = 2 - 4 = -2 \quad (\text{not } 0) \]

### Option D:
\[ 2 \text{SO}_3 \rightleftharpoons 2 \text{SO}_2 + \text{O}_2 \]
- **Products:** 2 (SO2) + 1 (O2) = 3
- **Reactants:** 2 (SO3) = 2
- **Calculation of \( \Delta N_g \):**
\[ \Delta N_g = 3 - 2 = 1 \quad (\text{not } 0) \]

### Conclusion:
The only reaction for which \( K_p = K_c \) is in **Option B**: \( \text{N}_2 + \text{O}_2 \rightleftharpoons 2 \text{NO} \).