At `700` K, the equilibrium constant, `K_p` for the reaction `2SO_3(g) hArr 2SO_2(g) +O_2 (g)` is `1.8 xx10^(-3)` atm. The value of `K_c` for the above reaction at the same temperature in moles per litre would be

Options

A. `1.1 xx 10^(-7)`
B. `3.1 xx 10^(-5)`
C. `6.2 xx 10^(-7)`
D. `9.3 xx 10^(-7)`

Answer

To find the value of \( K_c \) for the reaction \( 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \) at \( 700 \, K \), given that \( K_p = 1.8 \times 10^{-3} \, \text{atm} \), we can follow these steps:

### Step-by-Step Solution

1. **Write the expression for \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \cdot (RT)^{\Delta n} \]
where:
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas.

2. **Determine \( \Delta n \)**:
For the reaction:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
- The number of moles of gaseous products (\( n_p \)) = 2 (from \( SO_2 \)) + 1 (from \( O_2 \)) = 3.
- The number of moles of gaseous reactants (\( n_r \)) = 2 (from \( SO_3 \)).
- Therefore, \( \Delta n = n_p - n_r = 3 - 2 = 1 \).

3. **Substitute the values into the equation**:
We know:
- \( K_p = 1.8 \times 10^{-3} \, \text{atm} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 700 \, K \)
- \( \Delta n = 1 \)

Now substituting these values into the equation:
\[ K_p = K_c \cdot (RT)^{\Delta n} \]
becomes:
\[ 1.8 \times 10^{-3} = K_c \cdot (0.0821 \cdot 700)^1 \]

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

5. **Rearrange to find \( K_c \)**:
\[ K_c = \frac{K_p}{RT} = \frac{1.8 \times 10^{-3}}{57.47} \]

6. **Perform the calculation**:
\[ K_c = \frac{1.8 \times 10^{-3}}{57.47} \approx 3.13 \times 10^{-5} \, \text{mol/L} \]

### Final Answer
Thus, the value of \( K_c \) at \( 700 \, K \) is approximately:
\[ K_c \approx 3.1 \times 10^{-5} \, \text{mol/L} \]