The $t\frac{1}{2}$ of a reaction is doubled as the initial concentration of the reactant is doubled. What is the order of the reaction ?

Question

Accepted Answer

The question asks about the order of a reaction where the half-life ($t_{1/2}$) is doubled when the initial concentration of the reactant is doubled. The correct answer is Option 1, which indicates that the reaction is of zero order. Let's explore why this is the case and why the other options are incorrect.

### Explanation for Option 1 (Correct Answer: Zero Order Reaction)

1. **Zero Order Reaction Characteristics:**
   - In a zero order reaction, the rate of reaction is independent of the concentration of the reactant. The rate law is expressed as:
     $$
     \text{Rate} = k
     $$
   - The half-life ($t_{1/2}$) for a zero order reaction is given by:
     $$
     t_{1/2} = \frac{[A]_0}{2k}
     $$
     where $[A]_0$ is the initial concentration of the reactant and $k$ is the rate constant.

2. **Doubling the Initial Concentration:**
   - If the initial concentration $[A]_0$ is doubled, the half-life becomes:
     $$
     t_{1/2} = \frac{2[A]_0}{2k} = \frac{[A]_0}{k}
     $$
   - This shows that the half-life is directly proportional to the initial concentration. Therefore, doubling the initial concentration results in doubling the half-life, which is consistent with the behavior of a zero order reaction.

### Explanation for Why Other Options are Incorrect

2. **Option 2 (First Order Reaction):**
   - For a first order reaction, the rate law is:
     $$
     \text{Rate} = k[A]
     $$
   - The half-life for a first order reaction is independent of the initial concentration and is given by:
     $$
     t_{1/2} = \frac{0.693}{k}
     $$
   - Since the half-life does not depend on the initial concentration, doubling the initial concentration would not affect the half-life. Therefore, Option 2 is incorrect.

3. **Option 3 (Second Order Reaction):**
   - For a second order reaction, the rate law is:
     $$
     \text{Rate} = k[A]^2
     $$
   - The half-life for a second order reaction is given by:
     $$
     t_{1/2} = \frac{1}{k[A]_0}
     $$
   - In this case, the half-life is inversely proportional to the initial concentration. Doubling the initial concentration would result in halving the half-life, not doubling it. Therefore, Option 3 is incorrect.

4. **Option 4 (Third Order Reaction):**
   - For a third order reaction, the rate law is:
     $$
     \text{Rate} = k[A]^3
     $$
   - The half-life for a third order reaction is more complex and typically involves the square of the initial concentration. It is not a simple direct or inverse relationship like the other orders.
   - Doubling the initial concentration would not result in a straightforward doubling of the half-life. Therefore, Option 4 is incorrect.

In conclusion, the behavior described in the question—where the half-life doubles as the initial concentration doubles—is characteristic of a zero order reaction, making Option 1 the correct answer.