At particular concentration , the half life of the reaction is 100 minutes. When the concentration of the reactant become double half life becomes , 25 minutes , then what will be the order of the reaction ?

Options

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

Answer

To determine the order of the reaction based on the given half-lives at different concentrations, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The half-life (T₁/₂) at a certain concentration (A) is 100 minutes.
- When the concentration is doubled (2A), the half-life becomes 25 minutes.

2. **Use the Relationship Between Half-Life and Concentration**:
- The relationship between half-life (T₁/₂), concentration (A), and the order of the reaction (n) is given by:
\[ T_{1/2} \propto A^{(1 - n)} \]
- This means that the half-life is directly proportional to the concentration raised to the power of (1 - n).

3. **Set Up the Equations**:
- For the first concentration (A):
\[ T_{1/2} = 100 \text{ minutes} \implies 100 \propto A^{(1 - n)} \quad \text{(Equation 1)} \]
- For the doubled concentration (2A):
\[ T_{1/2} = 25 \text{ minutes} \implies 25 \propto (2A)^{(1 - n)} \quad \text{(Equation 2)} \]

4. **Express the Proportional Relationships as Equations**:
- From Equation 1:
\[ 100 = k \cdot A^{(1 - n)} \]
- From Equation 2:
\[ 25 = k \cdot (2A)^{(1 - n)} = k \cdot 2^{(1 - n)} \cdot A^{(1 - n)} \]

5. **Divide Equation 1 by Equation 2**:
- This gives:
\[ \frac{100}{25} = \frac{k \cdot A^{(1 - n)}}{k \cdot 2^{(1 - n)} \cdot A^{(1 - n)}} \]
- Simplifying this:
\[ 4 = \frac{1}{2^{(1 - n)}} \]
- Rearranging gives:
\[ 2^{(1 - n)} = \frac{1}{4} = 2^{-2} \]

6. **Equate the Exponents**:
- Since the bases are the same, we can equate the exponents:
\[ 1 - n = -2 \]
- Solving for n:
\[ n = 3 \]

7. **Conclusion**:
- The order of the reaction is 3.

### Final Answer:
The order of the reaction is **3**.