The $t_{1/2}$ of a reaction is doubled as the initial concentration of the reactant is doubled, what is the order of the reaction?

Question

Accepted Answer

To determine the order of a reaction based on how the half-life ($t_{1/2}$) changes with the initial concentration of the reactant, we need to understand the relationship between these variables for different reaction orders.

1. **Explanation for Option 4 (Zero Order Reaction):**
   - In a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The half-life for a zero-order reaction is given by the formula:
     $$
     t_{1/2} = \frac{[A]_0}{2k}
     $$
     where $[A]_0$ is the initial concentration and $k$ is the rate constant.
   - From this formula, it is clear that the half-life is directly proportional to the initial concentration $[A]_0$. Therefore, if the initial concentration is doubled, the half-life will also double.
   - This matches the condition given in the question, where the half-life doubles as the initial concentration is doubled, confirming that the reaction is zero-order.

2. **Why Other Options are Incorrect:**

- **Option 1 (Third Order Reaction):**
     - For a third-order reaction, the half-life is inversely proportional to the square of the initial concentration. The formula is:
       $$
       t_{1/2} = \frac{1}{k[A]_0^2}
       $$
     - If the initial concentration is doubled, the half-life would decrease by a factor of four, not double. Therefore, this does not match the condition given in the question.

- **Option 2 (Second Order Reaction):**
     - For a second-order reaction, the half-life is inversely proportional to the initial concentration. The formula is:
       $$
       t_{1/2} = \frac{1}{k[A]_0}
       $$
     - If the initial concentration is doubled, the half-life would be halved, not doubled. This contradicts the condition given in the question.

- **Option 3 (First Order Reaction):**
     - For a first-order reaction, the half-life is independent of the initial concentration. The formula is:
       $$
       t_{1/2} = \frac{0.693}{k}
       $$
     - Doubling the initial concentration would have no effect on the half-life, which remains constant. This does not satisfy the condition that the half-life doubles with the doubling of the initial concentration.

In conclusion, the only scenario where the half-life doubles as the initial concentration is doubled is in a zero-order reaction, which is why Option 4 is the correct answer.