For reaction `aA rarr xP`, when `[A] = 2.2 mM`, the rate was found to be `2.4mM s^(-1)`. On reducing concentration of `A` to half, the rate changes to `0.6 mM s^(-1)`. The order of reaction with respect to `A` is

Options

A. `1.5`
B. `2.0`
C. `2.5`
D. `3.0`

Explanation

To determine the order of the reaction with respect to A, we can use the information provided about the rates and concentrations of A.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Initial concentration of A, \([A]_1 = 2.2 \, \text{mM}\)
- Initial rate, \(R_1 = 2.4 \, \text{mM/s}\)
- New concentration of A, \([A]_2 = \frac{2.2}{2} = 1.1 \, \text{mM}\)
- New rate, \(R_2 = 0.6 \, \text{mM/s}\)

2. **Set Up the Rate Law:**
The rate of reaction can be expressed as:
\[ R = k [A]^n \]
where \(k\) is the rate constant and \(n\) is the order of the reaction with respect to A.

3. **Write Two Equations for the Two Conditions:**
For the first condition:
\[ R_1 = k [A]_1^n \implies 2.4 = k (2.2)^n \quad \text{(Equation 1)} \]
For the second condition:
\[ R_2 = k [A]_2^n \implies 0.6 = k (1.1)^n \quad \text{(Equation 2)} \]

4. **Divide Equation 1 by Equation 2:**
\[ \frac{R_1}{R_2} = \frac{k (2.2)^n}{k (1.1)^n} \]
This simplifies to:
\[ \frac{2.4}{0.6} = \frac{(2.2)^n}{(1.1)^n} \]
\[ 4 = \left(\frac{2.2}{1.1}\right)^n \]

5. **Simplify the Fraction:**
\[ \frac{2.2}{1.1} = 2 \]
So we have:
\[ 4 = 2^n \]

6. **Solve for n:**
Since \(4 = 2^2\), we can equate the exponents:
\[ n = 2 \]

### Conclusion:
The order of the reaction with respect to A is **2**.