In the reaction `A + B rarr` Products, keeping [A] constant if [B] is doubled , the rate becomes double and keeping [B] constant if [A] is doubled , the rate becomes four times. The order of reaction is :

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the order of the reaction given the information in the question, we can follow these steps:

### Step 1: Write the rate law expression
The rate of the reaction can be expressed as:
\[ \text{Rate} = k[A]^m[B]^n \]
where:
- \( k \) is the rate constant,
- \( [A] \) is the concentration of reactant A,
- \( [B] \) is the concentration of reactant B,
- \( m \) is the order with respect to A,
- \( n \) is the order with respect to B.

### Step 2: Analyze the first condition
Keeping \([A]\) constant and doubling \([B]\):
- Initial rate: \( r_1 = k[A]^m[B]^n \)
- New rate when \([B]\) is doubled:
\[ r_2 = k[A]^m[2B]^n = k[A]^m \cdot 2^n[B]^n = 2^n \cdot r_1 \]

According to the problem, \( r_2 = 2r_1 \). Therefore:
\[ 2^n \cdot r_1 = 2r_1 \]
Dividing both sides by \( r_1 \) (assuming \( r_1 \neq 0 \)):
\[ 2^n = 2 \]
This implies:
\[ n = 1 \]

### Step 3: Analyze the second condition
Keeping \([B]\) constant and doubling \([A]\):
- New rate when \([A]\) is doubled:
\[ r_3 = k[2A]^m[B]^n = k \cdot 2^m[A]^m[B]^n = 2^m \cdot r_1 \]

According to the problem, \( r_3 = 4r_1 \). Therefore:
\[ 2^m \cdot r_1 = 4r_1 \]
Dividing both sides by \( r_1 \):
\[ 2^m = 4 \]
This implies:
\[ 2^m = 2^2 \]
Thus:
\[ m = 2 \]

### Step 4: Determine the overall order of the reaction
The overall order of the reaction is the sum of the individual orders:
\[ \text{Overall order} = m + n = 2 + 1 = 3 \]

### Conclusion
The order of the reaction is **3**.