`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 these instructions:

### Step 1: Write the equilibrium expression
The reaction given is:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

The equilibrium constant \( K_p \) for this reaction in terms of partial pressures is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

### Step 2: Substitute the initial partial pressures
From the question, the initial partial pressures are:
- \( P_{PCl_3} = 0.3 \, \text{atm} \)
- \( P_{Cl_2} = 0.2 \, \text{atm} \)
- \( P_{PCl_5} = 0.6 \, \text{atm} \)

Now, substituting these values into the equilibrium expression:
\[ K_p = \frac{(0.3) \cdot (0.2)}{0.6} \]

### Step 3: Calculate \( K_p \)
Calculating the value:
\[ K_p = \frac{0.06}{0.6} = 0.1 \]

### Step 4: Determine the new pressures after increasing \( P_{PCl_3} \) and \( P_{Cl_2} \)
If the partial pressures of \( PCl_3 \) and \( Cl_2 \) are increased twice:
- 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 5: Set up the equation for the new equilibrium
Since the temperature remains constant, \( K_p \) will still be \( 0.1 \):
\[ 0.1 = \frac{(0.6) \cdot (0.4)}{P_{PCl_5}'} \]

### Step 6: Solve for \( P_{PCl_5}' \)
Rearranging the equation to solve for \( P_{PCl_5}' \):
\[ P_{PCl_5}' = \frac{(0.6) \cdot (0.4)}{0.1} \]

Calculating the right side:
\[ P_{PCl_5}' = \frac{0.24}{0.1} = 2.4 \, \text{atm} \]

### Final Answer
Thus, the new partial pressure of \( PCl_5 \) is:
\[ P_{PCl_5}' = 2.4 \, \text{atm} \]

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