For the chemical reaction `XrarrY`, it is found that the rate of reaction increases by `2.25` times when the concentration of `X` is increased by `1.5` times, what is the order w.r.t. `X` ?

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To determine the order of the reaction with respect to \( X \) for the reaction \( X \rightarrow Y \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate of the reaction can be expressed as:
\[ \text{Rate} = k [X]^a \]
where \( k \) is the rate constant, \( [X] \) is the concentration of \( X \), and \( a \) is the order of the reaction with respect to \( X \).

2. **Initial Conditions**:
Let the initial concentration of \( X \) be \( [X] = x \). Therefore, the initial rate of the reaction (\( \text{Rate}_1 \)) can be expressed as:
\[ \text{Rate}_1 = k [X]^a = k x^a \]

3. **Change in Concentration**:
When the concentration of \( X \) is increased by 1.5 times, the new concentration becomes:
\[ [X] = 1.5x \]
The new rate of the reaction (\( \text{Rate}_2 \)) can be expressed as:
\[ \text{Rate}_2 = k (1.5x)^a = k (1.5^a) x^a \]

4. **Rate Increase**:
According to the problem, the rate of reaction increases by 2.25 times when the concentration of \( X \) is increased:
\[ \text{Rate}_2 = 2.25 \times \text{Rate}_1 \]
Substituting the expressions for \( \text{Rate}_1 \) and \( \text{Rate}_2 \):
\[ k (1.5^a) x^a = 2.25 \times (k x^a) \]

5. **Cancel Common Terms**:
We can cancel \( k x^a \) from both sides (assuming \( k \) and \( x \) are not zero):
\[ 1.5^a = 2.25 \]

6. **Solve for \( a \)**:
We recognize that \( 2.25 \) can be rewritten as \( 1.5^2 \):
\[ 1.5^a = 1.5^2 \]
This implies:
\[ a = 2 \]

7. **Conclusion**:
Therefore, the order of the reaction with respect to \( X \) is:
\[ \text{Order with respect to } X = 2 \]