`PCl_(5)hArrPCl_(3)+Cl_(2)` in the above reaction the partial pressure of `PCl_(3), Cl_(2) and PCl_(5)` are 0.3, 0.2 and 0.6 atm respectively. If partial pressure of `PCl_(3) and Cl_(2)` was increased twice, partial pressure of `PCl_(5)` will be

Options

A. 0.3
B. 1.2
C. 2.4
D. 0.15

Answer

To solve the problem step by step, we will follow the principles of chemical equilibrium and the use of the equilibrium constant \( K_p \).

### Step 1: Write the reaction and identify the given partial pressures.
The reaction is:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

Given partial pressures:
- \( P_{PCl_3} = 0.3 \, \text{atm} \)
- \( P_{Cl_2} = 0.2 \, \text{atm} \)
- \( P_{PCl_5} = 0.6 \, \text{atm} \)

### Step 2: Calculate the equilibrium constant \( K_p \).
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

Substituting the values:
\[ K_p = \frac{(0.3) \cdot (0.2)}{0.6} \]
\[ K_p = \frac{0.06}{0.6} = 0.1 \]

### Step 3: Determine the new partial pressures after doubling \( P_{PCl_3} \) and \( P_{Cl_2} \).
If the partial pressures of \( PCl_3 \) and \( Cl_2 \) are doubled:
- New \( P_{PCl_3} = 2 \times 0.3 = 0.6 \, \text{atm} \)
- New \( P_{Cl_2} = 2 \times 0.2 = 0.4 \, \text{atm} \)

### Step 4: Set up the equation for \( K_p \) with the new pressures.
Since \( K_p \) remains constant at 0.1, we can write:
\[ K_p = \frac{P_{PCl_3 \, \text{final}} \cdot P_{Cl_2 \, \text{final}}}{P_{PCl_5 \, \text{final}}} \]
Substituting the new values:
\[ 0.1 = \frac{(0.6) \cdot (0.4)}{P_{PCl_5 \, \text{final}}} \]

### Step 5: Solve for \( P_{PCl_5 \, \text{final}} \).
Calculating the numerator:
\[ 0.1 = \frac{0.24}{P_{PCl_5 \, \text{final}}} \]
Rearranging gives:
\[ P_{PCl_5 \, \text{final}} = \frac{0.24}{0.1} = 2.4 \, \text{atm} \]

### Final Answer:
The final partial pressure of \( PCl_5 \) will be \( 2.4 \, \text{atm} \).

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