`K_p` for the following reaction at 700 K is `1.3 xx 10^(-3) "atm"^(-1)` . The `K_c` at same temperature for the reaction `2SO_2 +O_2 hArr 2SO_3` will be

Options

A. `1.1 xx10^(-2)`
B. `3.1 xx10^(-2)`
C. `5.2 xx10^(-2)`
D. `7.4 xx10^(-2)`

Explanation

To find the value of \( K_c \) for the reaction \( 2SO_2 + O_2 \rightleftharpoons 2SO_3 \) given that \( K_p \) at 700 K is \( 1.3 \times 10^{-3} \, \text{atm}^{-1} \), we can use the relationship between \( K_p \) and \( K_c \).

### Step-by-step Solution:

1. **Identify the Reaction and Calculate \( \Delta n_g \)**:
- The balanced reaction is:
\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]
- Count the moles of gaseous products and reactants:
- Products: \( 2 \, \text{moles of } SO_3 \)
- Reactants: \( 2 \, \text{moles of } SO_2 + 1 \, \text{mole of } O_2 = 3 \, \text{moles} \)
- Calculate \( \Delta n_g \):
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 3 = -1 \]

2. **Use the Relationship Between \( K_p \) and \( K_c \)**:
- The relationship is given by:
\[ K_p = K_c (RT)^{\Delta n_g} \]
- Rearranging this gives:
\[ K_c = \frac{K_p}{(RT)^{\Delta n_g}} \]

3. **Substitute the Known Values**:
- Given:
- \( K_p = 1.3 \times 10^{-3} \, \text{atm}^{-1} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 700 \, \text{K} \)
- Substitute \( R \), \( T \), and \( \Delta n_g \) into the equation:
\[ K_c = \frac{1.3 \times 10^{-3}}{(0.0821 \times 700)^{-1}} \]

4. **Calculate \( RT \)**:
- Calculate \( RT \):
\[ RT = 0.0821 \times 700 = 57.47 \, \text{L atm} \]

5. **Calculate \( (RT)^{-1} \)**:
- Calculate \( (RT)^{-1} \):
\[ (RT)^{-1} = \frac{1}{57.47} \approx 0.0174 \, \text{atm}^{-1} \]

6. **Final Calculation for \( K_c \)**:
- Now substitute back into the equation for \( K_c \):
\[ K_c = 1.3 \times 10^{-3} \times 0.0174 \]
- Calculate \( K_c \):
\[ K_c \approx 2.26 \times 10^{-5} \, \text{mol}^{-1} \text{L} \]

### Final Answer:
The value of \( K_c \) at 700 K for the reaction \( 2SO_2 + O_2 \rightleftharpoons 2SO_3 \) is approximately \( 2.26 \times 10^{-5} \, \text{mol}^{-1} \text{L} \).