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 changing the concentrations of reactants A and B affects the rate of the reaction. The rate of the reaction is given by the equation:

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

### Step 1: Write down the initial rate expression
The initial rate expression is:
\[ \text{Rate} = k[A]^2[B] \]

### Step 2: Substitute the new concentrations
We are told that the concentration of A is doubled and the concentration of B is halved. Therefore, we can express the new concentrations as:
- New concentration of A = \( 2[A] \)
- New concentration of B = \( \frac{[B]}{2} \)

### Step 3: Substitute the new concentrations into the rate equation
Now, we substitute these new concentrations into the rate expression:
\[ \text{New Rate} = k(2[A])^2\left(\frac{[B]}{2}\right) \]

### Step 4: Simplify the new rate expression
Now, let's simplify the expression:
\[ \text{New Rate} = k(4[A]^2)\left(\frac{[B]}{2}\right) \]
\[ \text{New Rate} = k \cdot 4[A]^2 \cdot \frac{[B]}{2} \]
\[ \text{New Rate} = k \cdot 2[A]^2[B] \]

### Step 5: Relate the new rate to the old rate
From the original rate expression, we know:
\[ \text{Old Rate} = k[A]^2[B] \]

Thus, we can express the new rate in terms of the old rate:
\[ \text{New Rate} = 2 \cdot \text{Old Rate} \]

### Conclusion
This means that the new rate of the reaction is twice the old rate. Therefore, the rate of reaction will **increase by 2 times**.

### Final Answer
The rate of reaction will increase by 2 times.

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