If the volume of a closed vessel in which the equilibrium `2SO_(2)+O_(2)hArr 2SO_(3)` is set in is halved , the rate of

Options

A. forward reaction will remain same as that of backward reaction
B. forward rection will become double that of backward reaction
C. forward reaction will be halved that of backward reaction
D. all are wrong

Answer

To solve the problem, we need to analyze the effect of halving the volume of a closed vessel on the equilibrium of the reaction:

\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction involves the conversion of sulfur dioxide (SO2) and oxygen (O2) into sulfur trioxide (SO3). The equilibrium expression for this reaction can be written in terms of concentrations.

2. **Equilibrium Constant Expression**:
The equilibrium constant \( K_{eq} \) for the reaction is given by:
\[ K_{eq} = \frac{[SO_3]^2}{[SO_2]^2[O_2]} \]
where \([SO_3]\), \([SO_2]\), and \([O_2]\) are the molar concentrations of the respective gases.

3. **Effect of Halving the Volume**:
When the volume of the closed vessel is halved, the concentrations of all gases will effectively double because concentration is inversely proportional to volume. Therefore:
\[ [SO_2] \text{ and } [O_2] \text{ will double.} \]
This means:
\[ [SO_2] \rightarrow 2[SO_2], \quad [O_2] \rightarrow 2[O_2], \quad [SO_3] \rightarrow 2[SO_3] \]

4. **Substituting into the Equilibrium Expression**:
Substitute the new concentrations into the equilibrium expression:
\[ K_{eq} = \frac{(2[SO_3])^2}{(2[SO_2])^2(2[O_2])} \]
Simplifying this gives:
\[ K_{eq} = \frac{4[SO_3]^2}{4[SO_2]^2 \cdot 2[O_2]} = \frac{2[SO_3]^2}{[SO_2]^2[O_2]} \]

5. **Conclusion on Equilibrium**:
The equilibrium constant \( K_{eq} \) remains unchanged because it is a function of temperature only. However, the system will shift to re-establish equilibrium due to the change in concentration.

6. **Direction of Shift**:
According to Le Chatelier's principle, if the concentration of reactants increases (as it does when volume decreases), the equilibrium will shift to the right to produce more products (SO3) in order to counteract the change.

7. **Rate of Reaction**:
To reach the new equilibrium, the rate of the forward reaction will increase. Since the concentration of reactants has increased, the rate of the forward reaction will increase, potentially doubling to restore equilibrium.

### Final Answer:
The rate of the forward reaction will increase (double) to re-establish equilibrium after the volume is halved.