To solve the problem step by step, we will analyze the equilibrium reaction, calculate the initial conditions, and then determine the changes that occur when additional \( Cl_2 \) is added.
### Step 1: Write the equilibrium reaction
The equilibrium reaction for the decomposition of \( PCl_5 \) is:
\[
PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)
\]
### Step 2: Determine initial moles and total moles
Initially, we have:
- \( n(PCl_5) = 2 \, \text{mol} \)
- \( n(PCl_3) = 2 \, \text{mol} \)
- \( n(Cl_2) = 2 \, \text{mol} \)
Total moles initially:
\[
n_{\text{total}} = n(PCl_5) + n(PCl_3) + n(Cl_2) = 2 + 2 + 2 = 6 \, \text{mol}
\]
### Step 3: Calculate the initial pressure in atm
Given the total pressure \( P = 30.3975 \, \text{kPa} \), we convert this to atm:
\[
P = \frac{30.3975 \, \text{kPa}}{101.325 \, \text{kPa/atm}} \approx 0.300 \, \text{atm}
\]
### Step 4: Calculate partial pressures
Using the ideal gas law, the partial pressure of each component can be calculated as follows:
\[
P_{PCl_5} = \frac{2}{6} \times 0.300 \, \text{atm} = 0.100 \, \text{atm}
\]
\[
P_{PCl_3} = \frac{2}{6} \times 0.300 \, \text{atm} = 0.100 \, \text{atm}
\]
\[
P_{Cl_2} = \frac{2}{6} \times 0.300 \, \text{atm} = 0.100 \, \text{atm}
\]
### Step 5: Calculate \( K_p \) at initial conditions
The equilibrium constant \( K_p \) is given by:
\[
K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{(0.100)(0.100)}{0.100} = 0.100
\]
### Step 6: Determine the new conditions after adding \( Cl_2 \)
When \( Cl_2 \) is added, let \( x \) be the amount of \( Cl_2 \) added. The new total volume becomes \( 2V \), and the total moles will change to:
\[
n_{\text{total new}} = 6 + x
\]
### Step 7: Set up the equilibrium expression after adding \( Cl_2 \)
At equilibrium, let \( y \) be the amount of \( PCl_5 \) that decomposes. The new moles will be:
- \( PCl_5: 2 - y \)
- \( PCl_3: 2 + y \)
- \( Cl_2: 2 + x - y \)
The total moles at equilibrium will be:
\[
n_{\text{total new}} = (2 - y) + (2 + y) + (2 + x - y) = 6 + x - y
\]
### Step 8: Calculate partial pressures at new conditions
The partial pressures will be:
\[
P_{PCl_5} = \frac{(2 - y)}{(6 + x - y)} \cdot 0.300
\]
\[
P_{PCl_3} = \frac{(2 + y)}{(6 + x - y)} \cdot 0.300
\]
\[
P_{Cl_2} = \frac{(2 + x - y)}{(6 + x - y)} \cdot 0.300
\]
### Step 9: Set up the equation for \( K_p \)
At equilibrium, the expression for \( K_p \) is:
\[
K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = 1
\]
Substituting the expressions for partial pressures into the equation will yield two equations that can be solved simultaneously.
### Step 10: Solve the equations
1. From the \( K_p \) expression, we can derive a relationship between \( x \) and \( y \).
2. The total moles equation gives us another equation.
By solving these equations, we can find the values of \( x \) and \( y \).
### Final Calculation
After solving, we find:
\[
x = \frac{20}{3} \quad \text{(amount of } Cl_2 \text{ added)}
\]
\[
y = \frac{2}{3} \quad \text{(amount of } PCl_5 \text{ formed)}
\]
### Conclusion
The amount of \( Cl_2 \) added is \( \frac{20}{3} \, \text{mol} \) and the equilibrium constant \( K_p \) is \( 1 \).
