In the reaction `2A + B to A_2B`, if the concentration of A is doubled and of B is halved, then the rate of the reaction will:

Options

A. Increase by four times
B. Decrease by two times
C. Increase by two times
D. Remain the same

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the rate law expression
For the reaction \(2A + B \rightarrow A_2B\), we assume it is an elementary reaction. The rate law can be expressed based on the stoichiometric coefficients:
\[ \text{Rate} = k [A]^2 [B]^1 \]
where \(k\) is the rate constant.

### Step 2: Define the initial concentrations
Let the initial concentrations of A and B be:
- \([A] = a\)
- \([B] = b\)

### Step 3: Calculate the initial rate of the reaction
Using the initial concentrations in the rate law:
\[ \text{Initial Rate} (r) = k [A]^2 [B] = k a^2 b \]

### Step 4: Modify the concentrations as per the problem
According to the problem:
- The concentration of A is doubled: \([A] = 2a\)
- The concentration of B is halved: \([B] = \frac{b}{2}\)

### Step 5: Calculate the new rate with modified concentrations
Now, substituting the new concentrations into the rate law:
\[ \text{New Rate} (r') = k [2A]^2 \left[\frac{B}{2}\right] = k (2a)^2 \left(\frac{b}{2}\right) \]
Calculating this gives:
\[ r' = k \cdot 4a^2 \cdot \frac{b}{2} = k \cdot 2a^2 b \]

### Step 6: Relate the new rate to the initial rate
Now, we can relate the new rate \(r'\) to the initial rate \(r\):
\[ r' = 2 \cdot (k a^2 b) = 2r \]

### Conclusion
Thus, the rate of the reaction increases by a factor of 2 when the concentration of A is doubled and the concentration of B is halved.

### Final Answer
The rate of the reaction will **double**.

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