In a reaction ` A to B` the rate of reaction increases two times on increasing the concentration of the reactant four times, then order of reaction is

Options

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

Explanation

To determine the order of the reaction \( A \rightarrow B \) based on the given information, we can follow these steps:

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

### Step 2: Set up the initial and final conditions
Let the initial rate of the reaction be \( R \) when the concentration of A is \( [A] \). According to the problem, when the concentration of A is increased four times (i.e., \( [A] \) becomes \( 4[A] \)), the rate of reaction doubles (i.e., \( R \) becomes \( 2R \)).

### Step 3: Write the rate equations for both conditions
1. Initial condition:
\[ R = k [A]^n \quad \text{(1)} \]
2. Final condition with increased concentration:
\[ 2R = k (4[A])^n \quad \text{(2)} \]

### Step 4: Substitute and simplify
From equation (1), we can express \( R \):
\[ R = k [A]^n \]
Substituting this into equation (2):
\[ 2(k [A]^n) = k (4[A])^n \]
This simplifies to:
\[ 2 = k (4^n) \frac{[A]^n}{k [A]^n} \]
Cancelling \( k [A]^n \) from both sides gives:
\[ 2 = 4^n \]

### Step 5: Solve for \( n \)
We can rewrite \( 4^n \) as \( (2^2)^n = 2^{2n} \):
\[ 2 = 2^{2n} \]
This implies:
\[ 2^1 = 2^{2n} \]
Equating the exponents gives:
\[ 1 = 2n \implies n = \frac{1}{2} \]

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

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