For the reaction `PCl_(5)hArrPCl_(3)+Cl_(2)`, Supposing at constant temperature, if the volume is increased `16` times the initial volume, the degree of dissociation for this reaction will becomes:

Options

A. `4` times
B. `(1)/(4)` times
C. `2` times
D. `(1)/(2)` times

Answer

To solve the problem regarding the degree of dissociation of the reaction \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) when the volume is increased 16 times, we can follow these steps:

### Step 1: Understand the Relationship Between Degree of Dissociation and Pressure
The degree of dissociation (\( \alpha \)) is inversely proportional to the square root of the pressure. This can be expressed as:
\[ \alpha \propto \frac{1}{\sqrt{P}} \]

### Step 2: Relate Pressure to Volume
According to the ideal gas law, at constant temperature, pressure is inversely proportional to volume:
\[ P \propto \frac{1}{V} \]
This means that if the volume increases, the pressure decreases.

### Step 3: Establish the Relationship Between Initial and Final Conditions
Let:
- \( V_1 \) = initial volume
- \( V_2 \) = final volume = \( 16V_1 \)

Now, since pressure is inversely related to volume, we can express the pressures as:
\[ P_1 \propto \frac{1}{V_1}, \quad P_2 \propto \frac{1}{V_2} = \frac{1}{16V_1} \]

### Step 4: Calculate the Change in Degree of Dissociation
From the relationship established in Step 1:
\[ \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V_1}{16V_1}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

This implies:
\[ \alpha_2 = 4 \alpha_1 \]

### Step 5: Conclusion
Thus, the degree of dissociation becomes four times the initial degree of dissociation when the volume is increased 16 times.

### Final Answer
The degree of dissociation for this reaction will become **four times**.

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