For a reaction `A+ B rarr` products, rate law expression for rate , r=K[A][B]^2`.
If the volume of the container is reduced to half of its original volume, the rate of reaction will become

Options

A. 4 times
B. 8 Times
C. `1/4` times
D. 2 times

Answer

To solve the problem, we need to analyze how the rate of reaction changes when the volume of the container is reduced to half. The rate law expression is given as:

\[ r = K[A][B]^2 \]

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- Let the initial concentrations of A and B be \([A]_1\) and \([B]_1\).
- The rate of reaction initially is given by:
\[ r_1 = K[A]_1[B_1]^2 \]

2. **Effect of Volume Reduction:**
- When the volume of the container is reduced to half, the concentrations of the reactants will change.
- The concentration of a substance is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
- If the volume is halved, the new concentrations will be:
\[ [A]_2 = 2[A]_1 \]
\[ [B]_2 = 2[B]_1 \]

3. **Substituting New Concentrations into Rate Law:**
- Now, we can express the new rate of reaction \( r_2 \) using the new concentrations:
\[ r_2 = K[A]_2[B_2]^2 \]
- Substituting the new concentrations:
\[ r_2 = K(2[A]_1)(2[B]_1)^2 \]

4. **Calculating the New Rate:**
- Expanding the expression:
\[ r_2 = K(2[A]_1)(4[B]_1^2) \]
\[ r_2 = 8K[A]_1[B_1]^2 \]
- Therefore, we can relate \( r_2 \) to \( r_1 \):
\[ r_2 = 8r_1 \]

5. **Conclusion:**
- The rate of reaction becomes 8 times the initial rate when the volume is reduced to half.

### Final Answer:
The rate of reaction will become **8 times** the original rate.