When the concentration of a reactant in reaction `A rarr B` is increased by `8` times but rate increases only `2` times, the order of the reaction would be

Options

A. `2`
B. `1//3`
C. `4`
D. `1//2`

Explanation

To determine the order of the reaction \( A \rightarrow B \) given that the concentration of reactant \( A \) is increased by 8 times and the rate increases by only 2 times, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate law for the 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 \( A \).

2. **Define Initial and Final Rates**:
Let the initial concentration of \( A \) be \( [A] \). The initial rate \( R_1 \) can be expressed as:
\[ R_1 = k [A]^n \]
When the concentration of \( A \) is increased by 8 times, the new concentration becomes \( 8[A] \). The new rate \( R_2 \) can be expressed as:
\[ R_2 = k (8[A])^n = k \cdot 8^n \cdot [A]^n \]

3. **Relate the Rates**:
According to the problem, the new rate \( R_2 \) is twice the initial rate \( R_1 \):
\[ R_2 = 2R_1 \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ k \cdot 8^n \cdot [A]^n = 2(k [A]^n) \]

4. **Cancel Out Common Terms**:
Since \( k [A]^n \) appears on both sides, we can cancel it out (assuming \( [A] \neq 0 \)):
\[ 8^n = 2 \]

5. **Solve for \( n \)**:
To solve for \( n \), we can rewrite \( 8 \) as \( 2^3 \):
\[ (2^3)^n = 2^1 \]
This simplifies to:
\[ 2^{3n} = 2^1 \]
Since the bases are the same, we can equate the exponents:
\[ 3n = 1 \]
Solving for \( n \):
\[ n = \frac{1}{3} \]

6. **Conclusion**:
The order of the reaction with respect to \( A \) is \( \frac{1}{3} \).

### Final Answer:
The order of the reaction is \( \frac{1}{3} \).

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