For the reaction, `PCl_(3) (g)+Cl_(2) (g) Leftrightarrow PCl_(5) (g)" at "27^@C, K_p" is 0.41 atm"^(-1).` Then `K_c` is :

Options

A. `6L mol^(-1)`
B. `60 L mol^(-1)`
C. `10.08 L mol^(-1)`
D. `1.008 xx 10^(2) L mol^(-1)`

Explanation

To find the value of \( K_c \) for the reaction \( PCl_3 (g) + Cl_2 (g) \leftrightarrow PCl_5 (g) \) given that \( K_p = 0.41 \, \text{atm}^{-1} \) at \( 27^\circ C \), we can use the relationship between \( K_p \) and \( K_c \):

### Step-by-step Solution:

1. **Identify the Reaction and Given Data:**
- The equilibrium reaction is:
\[ PCl_3 (g) + Cl_2 (g) \leftrightarrow PCl_5 (g) \]
- Given \( K_p = 0.41 \, \text{atm}^{-1} \)
- Temperature \( T = 27^\circ C = 300 \, K \) (convert Celsius to Kelvin by adding 273)

2. **Determine \( \Delta N_g \):**
- \( \Delta N_g \) is calculated as:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- For the reaction:
- Moles of gaseous products = 1 (from \( PCl_5 \))
- Moles of gaseous reactants = 1 (from \( PCl_3 \)) + 1 (from \( Cl_2 \)) = 2
- Therefore:
\[ \Delta N_g = 1 - 2 = -1 \]

3. **Use the Relationship Between \( K_p \) and \( K_c \):**
- The relationship is given by:
\[ K_p = K_c \cdot R^{{\Delta N_g}} \cdot T^{\Delta N_g} \]
- Rearranging for \( K_c \):
\[ K_c = \frac{K_p}{R^{\Delta N_g} \cdot T^{\Delta N_g}} \]

4. **Substitute the Values:**
- Use \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Substitute \( K_p = 0.41 \), \( R = 0.082 \), \( T = 300 \, K \), and \( \Delta N_g = -1 \):
\[ K_c = \frac{0.41}{(0.082)^{-1} \cdot (300)^{-1}} \]

5. **Calculate \( K_c \):**
- Calculate \( (0.082)^{-1} \) and \( (300)^{-1} \):
\[ (0.082)^{-1} = 12.195 \, \text{atm L}^{-1} \text{mol} \]
\[ (300)^{-1} = 0.00333 \, \text{K}^{-1} \]
- Now calculate:
\[ K_c = \frac{0.41}{12.195 \cdot 0.00333} \]
- Calculate the denominator:
\[ 12.195 \cdot 0.00333 \approx 0.0406 \]
- Finally:
\[ K_c \approx \frac{0.41}{0.0406} \approx 10.08 \, \text{L mol}^{-1} \]

### Final Answer:
\[ K_c \approx 10.08 \, \text{L mol}^{-1} \]