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

Options

A. 0
B. 2
C. `3//2`
D. 4

Answer

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

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
For a reaction A → B, the rate law can be expressed as:
\[ \text{Rate} = k [A]^n \]
where \( k \) is the rate constant and \( n \) is the order of the reaction.

2. **Set Up the Initial Conditions**:
Let the initial concentration of A be \( [A] \). Therefore, the initial rate of the reaction can be expressed as:
\[ \text{Rate}_1 = k [A]^n \]

3. **Change the Concentration**:
When the concentration of A is increased four times, the new concentration becomes \( 4[A] \). The new rate of the reaction can be expressed as:
\[ \text{Rate}_2 = k [4A]^n = k (4[A])^n = k \cdot 4^n \cdot [A]^n \]

4. **Relate the Two Rates**:
According to the problem, the rate increases eight times when the concentration is increased four times:
\[ \text{Rate}_2 = 8 \cdot \text{Rate}_1 \]
Substituting the expressions for the rates:
\[ k \cdot 4^n \cdot [A]^n = 8 \cdot (k [A]^n) \]

5. **Cancel Common Terms**:
Since \( k \) and \( [A]^n \) are common in both sides, we can cancel them:
\[ 4^n = 8 \]

6. **Express Both Sides with the Same Base**:
We can express 8 as \( 2^3 \) and 4 as \( 2^2 \):
\[ (2^2)^n = 2^3 \]
This simplifies to:
\[ 2^{2n} = 2^3 \]

7. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ 2n = 3 \]

8. **Solve for n**:
Dividing both sides by 2 gives:
\[ n = \frac{3}{2} \]

### Conclusion:
The order of the reaction is \( \frac{3}{2} \) or 1.5.

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