The value of `(K_p//K_c)` for the reversible reaction `SO_(2(g))+1//2O_(2(g)) Leftrightarrow SO_(3(g))` at constant temperature T is

Options

A. `(RT)^(1//2)`
B. RT
C. `(RT)^(-1//2)`
D. 1/RT

Answer

To find the value of \( \frac{K_p}{K_c} \) for the reversible reaction

\[ SO_{2(g)} + \frac{1}{2} O_{2(g)} \leftrightarrow SO_{3(g)} \]

at constant temperature \( T \), we will use the relationship between \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction**: The given reaction is:
\[ SO_{2(g)} + \frac{1}{2} O_{2(g)} \leftrightarrow SO_{3(g)} \]

2. **Determine the Change in Moles of Gas (\( \Delta N_g \))**:
- **Products**: There is 1 mole of \( SO_3 \) (gaseous).
- **Reactants**: There is 1 mole of \( SO_2 \) and \( \frac{1}{2} \) mole of \( O_2 \).
- **Total moles of reactants**:
\[ 1 + \frac{1}{2} = \frac{3}{2} \]
- **Calculate \( \Delta N_g \)**:
\[ \Delta N_g = \text{(moles of products)} - \text{(moles of reactants)} = 1 - \frac{3}{2} = 1 - 1.5 = -\frac{1}{2} \]

3. **Use the Relationship Between \( K_p \) and \( K_c \)**:
The relationship is given by:
\[ K_p = K_c (RT)^{\Delta N_g} \]
Substituting \( \Delta N_g = -\frac{1}{2} \):
\[ K_p = K_c (RT)^{-\frac{1}{2}} \]

4. **Rearranging to Find \( \frac{K_p}{K_c} \)**:
\[ \frac{K_p}{K_c} = (RT)^{-\frac{1}{2}} = \frac{1}{\sqrt{RT}} \]

### Final Result:
Thus, the value of \( \frac{K_p}{K_c} \) for the given reaction at constant temperature \( T \) is:
\[ \frac{K_p}{K_c} = \frac{1}{\sqrt{RT}} \]