For an elementary reaction `2A+B rarr A_(2)B` if the volume of vessel is quickly reduced to half of it's original volume then rate of reaction will `:`

Options

A. remain unchanged
B. increase four times
C. increase eight times
D. decrease eight times

Answer

To solve the problem, we need to analyze the effect of reducing the volume of the reaction vessel on the rate of an elementary reaction. The reaction given is:

\[ 2A + B \rightarrow A_2B \]

### Step-by-Step Solution:

1. **Write the Rate Law**:
For an elementary reaction, the rate law can be expressed as:
\[ \text{Rate} = k [A]^2 [B]^1 \]
where \( k \) is the rate constant, and \([A]\) and \([B]\) are the concentrations of reactants A and B, respectively.

2. **Understand the Effect of Volume Reduction**:
When the volume of the vessel is reduced to half, the concentration of the reactants will change. The concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
If the volume is halved, the concentration will double.

3. **Calculate New Concentrations**:
Let the initial concentrations of A and B be \([A] = C\) and \([B] = D\). After the volume is halved, the new concentrations will be:
\[ [A]_{\text{new}} = 2C \quad \text{and} \quad [B]_{\text{new}} = 2D \]

4. **Substitute New Concentrations into the Rate Law**:
The new rate of reaction, \( R_{\text{new}} \), can be expressed as:
\[ R_{\text{new}} = k [A]_{\text{new}}^2 [B]_{\text{new}}^1 = k (2C)^2 (2D)^1 \]
Simplifying this gives:
\[ R_{\text{new}} = k \cdot 4C^2 \cdot 2D = 8k [A]^2 [B] \]

5. **Relate New Rate to Original Rate**:
The original rate \( R_{\text{original}} \) is:
\[ R_{\text{original}} = k [A]^2 [B] \]
Thus, we can relate the new rate to the original rate:
\[ R_{\text{new}} = 8 R_{\text{original}} \]

6. **Conclusion**:
Therefore, when the volume of the vessel is quickly reduced to half, the rate of the reaction increases by a factor of 8.

### Final Answer:
The rate of reaction will **increase 8 times**.

---