In Q.No .5, if the mixture of gases was allowed to come to quilibrium .The volume of the reaction vessel was then rapidly increased by a factor of two .As a result of the change the reaction quotient `(Q_(c))` would:

Options

A. increase because of the pressure decrease
B. decrease because of the pressure decrease
C. remain the same because the equilibrium constant is indendent of volume
D. increase because the reaction is endothermioc

Answer

To solve the problem step-by-step, we will analyze the effect of increasing the volume of the reaction vessel on the reaction quotient \( Q_c \) for the given reaction.

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]

### Step 2: Understand the Reaction Quotient \( Q_c \)
The reaction quotient \( Q_c \) is defined as:
\[ Q_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \]
where \( [\text{SO}_3] \), \( [\text{SO}_2] \), and \( [\text{O}_2] \) are the molar concentrations of the respective gases.

### Step 3: Initial Conditions
Initially, we have 0.001 moles of each gas in a 1-liter flask:
- \( [\text{SO}_2] = 0.001 \, \text{M} \)
- \( [\text{O}_2] = 0.001 \, \text{M} \)
- \( [\text{SO}_3] = 0 \, \text{M} \)

### Step 4: Calculate Initial \( Q_c \)
Before equilibrium is established, we can calculate \( Q_c \):
\[ Q_c = \frac{[0]^2}{[0.001]^2 \times [0.001]} = 0 \]
Since there are no products initially, \( Q_c = 0 \).

### Step 5: Establish Equilibrium
At equilibrium, let \( x \) be the change in concentration of \( \text{SO}_3 \) produced. The equilibrium concentrations will be:
- \( [\text{SO}_2] = 0.001 - x \)
- \( [\text{O}_2] = 0.001 - \frac{x}{2} \)
- \( [\text{SO}_3] = 2x \)

### Step 6: Effect of Increasing Volume
When the volume of the reaction vessel is doubled, the concentrations of all gases will be halved:
- New \( [\text{SO}_2] = \frac{0.001 - x}{2} \)
- New \( [\text{O}_2] = \frac{0.001 - \frac{x}{2}}{2} \)
- New \( [\text{SO}_3] = \frac{2x}{2} = x \)

### Step 7: Calculate New \( Q_c \)
The new reaction quotient \( Q_c' \) after the volume increase will be:
\[ Q_c' = \frac{[x]^2}{\left[\frac{0.001 - x}{2}\right]^2 \left[\frac{0.001 - \frac{x}{2}}{2}\right]} \]

### Step 8: Analyze the Change in \( Q_c \)
Since the volume increase leads to a decrease in concentration, the reaction will shift towards the side with more moles of gas to re-establish equilibrium. In this case, the forward reaction produces 2 moles of \( \text{SO}_3 \) from 3 moles of reactants. Hence, \( Q_c' \) will increase.

### Conclusion
After the volume of the reaction vessel is increased, the reaction quotient \( Q_c \) will **increase**.