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. 4

Explanation

To determine the order of the reaction given that the rate of reaction increases two times when the concentration of the reactant is increased four times, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Rate Law**:
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 the reactant, and \( n \) is the order of the reaction.

2. **Set Up Initial Conditions**:
Let the initial concentration of \( A \) be \( [A] \) and the initial rate be \( R_1 \):
\[ R_1 = k [A]^n \]

3. **Set Up New Conditions**:
When the concentration of \( A \) is increased to \( 4[A] \), the new rate \( R_2 \) is given as:
\[ R_2 = k (4[A])^n \]
According to the problem, \( R_2 = 2R_1 \).

4. **Substitute and Rearrange**:
Substitute \( R_1 \) and \( R_2 \) into the equation:
\[ 2R_1 = k (4[A])^n \]
This gives us:
\[ 2(k [A]^n) = k (4[A])^n \]

5. **Cancel Common Terms**:
Since \( k \) and \( [A]^n \) are common on both sides, we can cancel them out (assuming \( k \neq 0 \) and \( [A] \neq 0 \)):
\[ 2 = (4^n) \]

6. **Solve for \( n \)**:
Now we need to solve for \( n \):
\[ 2 = 4^n \]
We can express \( 4 \) as \( 2^2 \):
\[ 2 = (2^2)^n = 2^{2n} \]
Therefore, we have:
\[ 2 = 2^{2n} \]
This implies:
\[ 1 = 2n \quad \Rightarrow \quad n = \frac{1}{2} \]

7. **Conclusion**:
The order of the reaction \( n \) is \( \frac{1}{2} \).

### Final Answer:
The order of the reaction is \( \frac{1}{2} \).

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