The rate of reaction doubles when the concentration of the reactant is increased four times , The order is

Options

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

Answer

To determine the order of the reaction based on the information given, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Rate Law Expression:**
The rate of a reaction can be expressed as:
\[ r = k[A]^n \]
where \( r \) is the rate of reaction, \( k \) is the rate constant, \( [A] \) is the concentration of the reactant, and \( n \) is the order of the reaction.

2. **Set Up the Initial and Changed Conditions:**
Let the initial concentration of the reactant \( [A] \) be \( [A]_0 \). According to the problem, when the concentration is increased four times, the new concentration becomes:
\[ [A] = 4[A]_0 \]
The rate of reaction doubles, so we have:
\[ 2r = k(4[A]_0)^n \]

3. **Express the Initial Rate:**
The initial rate can be expressed as:
\[ r = k[A]_0^n \]

4. **Set Up the Equation:**
From the information given, we can set up the equation:
\[ 2(k[A]_0^n) = k(4[A]_0)^n \]

5. **Cancel Out Common Terms:**
We can cancel \( k \) from both sides (assuming \( k \) is not zero):
\[ 2[A]_0^n = (4^n)[A]_0^n \]

6. **Simplify the Equation:**
Dividing both sides by \( [A]_0^n \) (assuming \( [A]_0 \neq 0 \)):
\[ 2 = 4^n \]

7. **Express \( 4^n \) in Terms of Powers of 2:**
We can express \( 4 \) as \( 2^2 \):
\[ 4^n = (2^2)^n = 2^{2n} \]
Therefore, we have:
\[ 2 = 2^{2n} \]

8. **Equate the Exponents:**
Since the bases are the same, we can equate the exponents:
\[ 1 = 2n \]

9. **Solve for \( n \):**
Dividing both sides by 2 gives:
\[ n = \frac{1}{2} \]

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

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