In the reaction `2A+B to A_2B`, Rate = k[A]^2 [ B ] if the concentration of A is doubled and that of B is halved, then the rate of reaction will:

Options

A. Decrease 2 times
B. Increase2 times
C. Increase 4 times
D. Remain the same

Answer

To solve the problem, we need to analyze how the rate of the reaction changes when the concentrations of the reactants are altered. The reaction given is:

\[ 2A + B \rightarrow A_2B \]

The rate law expression is:

\[ \text{Rate} = k[A]^2[B] \]

where:
- \( k \) is the rate constant,
- \( [A] \) is the concentration of A,
- \( [B] \) is the concentration of B.

### Step-by-Step Solution:

1. **Identify the Initial Rate Expression**:
The initial rate of the reaction can be expressed as:
\[ R = k[A]^2[B] \]

2. **Change the Concentrations**:
According to the problem, the concentration of A is doubled and the concentration of B is halved:
- New concentration of A: \( [A]_{new} = 2[A] \)
- New concentration of B: \( [B]_{new} = \frac{1}{2}[B] \)

3. **Substitute the New Concentrations into the Rate Expression**:
The new rate \( R' \) with the altered concentrations becomes:
\[ R' = k[2A]^2\left[\frac{1}{2}B\right] \]

4. **Calculate the New Rate**:
Now, substituting the new concentrations into the rate expression:
\[ R' = k(2[A])^2\left(\frac{1}{2}[B]\right) \]
Simplifying this:
\[ R' = k \cdot 4[A]^2 \cdot \frac{1}{2}[B] \]
\[ R' = 2k[A]^2[B] \]

5. **Relate the New Rate to the Initial Rate**:
Since the initial rate \( R = k[A]^2[B] \), we can relate the new rate to the initial rate:
\[ R' = 2R \]

6. **Conclusion**:
The new rate \( R' \) is twice the initial rate \( R \). Therefore, when the concentration of A is doubled and the concentration of B is halved, the rate of reaction will increase by a factor of 2.

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

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