The half-life of a reaction is halved as the initial concentration of the reaction is doubled. The order of the reaction is

Options

A. 0.5
B. 1
C. 2
D. 0

Explanation

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

### Step-by-Step Solution:

1. **Understanding Half-Life in Relation to Reaction Order**:
The half-life (\(t_{1/2}\)) of a reaction is the time required for the concentration of a reactant to decrease to half its initial concentration. The relationship between half-life and concentration depends on the order of the reaction.

2. **Half-Life for Different Orders**:
- For a **zero-order reaction**:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
- For a **first-order reaction**:
\[ t_{1/2} = \frac{0.693}{k} \]
- For a **second-order reaction**:
\[ t_{1/2} = \frac{1}{k[A]_0} \]

3. **Given Condition**:
The problem states that the half-life of the reaction is halved when the initial concentration is doubled. This can be expressed mathematically:
\[ t_{1/2} \text{ (new)} = \frac{1}{2} t_{1/2} \text{ (original)} \]
Let the original concentration be \( [A]_0 \) and the new concentration be \( 2[A]_0 \).

4. **Applying the Condition to the Half-Life Expressions**:
- For a second-order reaction, substituting the new concentration:
\[ t_{1/2} \text{ (new)} = \frac{1}{k(2[A]_0)} = \frac{1}{2k[A]_0} \]
- The original half-life is:
\[ t_{1/2} \text{ (original)} = \frac{1}{k[A]_0} \]
- Thus, we have:
\[ \frac{1}{2k[A]_0} = \frac{1}{2} \left(\frac{1}{k[A]_0}\right) \]

5. **Conclusion**:
Since the relationship holds true, we can conclude that the reaction is second-order. Therefore, the order of the reaction is **2**.

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

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