When the concentration of a reactant , A in a reaction `A to ` Products , is doubled the rate of the reaction increases seven times , the order of the reaction is between

Options

A. 0 & 1
B. 1 & 2
C. 2 & 3
D. 3 & 4

Answer

To determine the order of the reaction when the concentration of reactant A is doubled and the rate of the reaction increases seven times, we can follow these steps:

### Step-by-Step Solution:

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

2. **Initial Rate**:
Let the initial concentration of A be \( [A] \). Thus, the initial rate \( R_1 \) can be written as:
\[ R_1 = k[A]^n \]

3. **Final Rate After Doubling Concentration**:
When the concentration of A is doubled, the new concentration becomes \( 2[A] \). The new rate \( R_2 \) can be expressed as:
\[ R_2 = k(2[A])^n = k \cdot 2^n \cdot [A]^n \]

4. **Relate the Rates**:
According to the problem, when the concentration is doubled, the rate increases seven times:
\[ R_2 = 7R_1 \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ k \cdot 2^n \cdot [A]^n = 7(k[A]^n) \]

5. **Cancel Out Common Terms**:
We can cancel \( k[A]^n \) from both sides (assuming \( k \) and \( [A] \) are not zero):
\[ 2^n = 7 \]

6. **Solve for \( n \)**:
To find \( n \), we take the logarithm:
\[ n = \log_2(7) \]
Using the change of base formula:
\[ n = \frac{\log_{10}(7)}{\log_{10}(2)} \]
Calculating the approximate values:
- \( \log_{10}(7) \approx 0.845 \)
- \( \log_{10}(2) \approx 0.301 \)
Thus,
\[ n \approx \frac{0.845}{0.301} \approx 2.81 \]

7. **Determine the Range of the Order**:
Since \( n \approx 2.81 \), we can conclude that the order of the reaction is between 2 and 3.

### Conclusion:
The order of the reaction is between 2 and 3.