For a chemical reaction `A rarr B`, it is found that the rate of reaction doubles when the concentration of `A` is increased `4` times. What is the order of reaction ? Suggest the rate law also.

Options

A. `4`
B. `0`
C. `1//2`
D. `1`

Explanation

To solve the problem, we need to determine the order of the reaction \( A \rightarrow B \) based on the information provided. The rate of reaction doubles when the concentration of \( A \) is increased by 4 times. Let's go through the steps to find the order of the reaction and the rate law.

### Step 1: Write the Rate Expression
The rate of a chemical 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.

### Step 2: Define Initial and Changed Rates
Let the initial rate of the reaction be \( r_1 \):
\[ r_1 = k [A]^n \]
When the concentration of \( A \) is increased by 4 times, the new concentration becomes \( 4[A] \). The new rate \( r_2 \) can be expressed as:
\[ r_2 = k (4[A])^n = k \cdot 4^n \cdot [A]^n \]

### Step 3: Relate the Rates
According to the problem, the rate doubles when the concentration is increased:
\[ r_2 = 2r_1 \]
Substituting the expressions for \( r_1 \) and \( r_2 \):
\[ k \cdot 4^n \cdot [A]^n = 2(k [A]^n) \]

### Step 4: Simplify the Equation
We can cancel \( k [A]^n \) from both sides (assuming \( k \) and \( [A] \) are not zero):
\[ 4^n = 2 \]

### Step 5: Solve for \( n \)
To find \( n \), we can rewrite the equation:
\[ 4^n = 2 \implies (2^2)^n = 2 \implies 2^{2n} = 2^1 \]
This implies:
\[ 2n = 1 \implies n = \frac{1}{2} \]

### Step 6: Write the Rate Law
Now that we have determined the order of the reaction \( n = \frac{1}{2} \), we can write the rate law:
\[ \text{Rate} = k [A]^{\frac{1}{2}} \]

### Final Answer
- The order of the reaction is \( \frac{1}{2} \).
- The rate law is \( \text{Rate} = k [A]^{\frac{1}{2}} \).