In a chemical reaction `XtoY`, it is found that the rate of reaction doubles when the concentration of X is incrased four times. The order of the reaction with respect to X is

Options

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

Explanation

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

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

### Step 2: Set up the initial conditions
Let the initial concentration of X be \( [X] \). Therefore, the initial rate of the reaction can be expressed as:
\[ R = k [X]^n \]

### Step 3: Analyze the change in concentration
According to the problem, when the concentration of X is increased four times, the new concentration becomes \( 4[X] \). The new rate of reaction, denoted as \( R' \), can be expressed as:
\[ R' = k (4[X])^n \]

### Step 4: Relate the rates
We know from the problem that the rate doubles when the concentration is increased four times:
\[ R' = 2R \]
Substituting the expressions for \( R \) and \( R' \):
\[ k (4[X])^n = 2(k [X]^n) \]

### Step 5: Simplify the equation
We can simplify this equation:
\[ k (4^n [X]^n) = 2k [X]^n \]
Dividing both sides by \( k [X]^n \) (assuming \( k \) and \( [X] \) are not zero):
\[ 4^n = 2 \]

### Step 6: Solve for \( n \)
Now we need to solve for \( n \):
\[ 4^n = 2 \implies (2^2)^n = 2 \implies 2^{2n} = 2^1 \]
This implies:
\[ 2n = 1 \implies n = \frac{1}{2} \]

### Conclusion
The order of the reaction with respect to X is \( \frac{1}{2} \).

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