Half life periods for a reaction at initial concentration of 0.1 M and 0.01 M are 5 and 50 minutes respectively. Then the order of reaction is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial concentration \( [A]_1 = 0.1 \, M \) with half-life \( t_{1/2} = 5 \, \text{minutes} \)
- Initial concentration \( [A]_2 = 0.01 \, M \) with half-life \( t_{1/2} = 50 \, \text{minutes} \)

2. **Analyze the Relationship Between Concentration and Half-Life:**
- The half-life periods are related to the order of the reaction. We can observe that when the concentration decreases from \( 0.1 \, M \) to \( 0.01 \, M \) (which is a factor of 10), the half-life increases from 5 minutes to 50 minutes (which is a factor of 10 as well).

3. **Determine the Order of Reaction:**
- For a **zero-order reaction**, the half-life \( t_{1/2} \) is given by:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
Here, \( t_{1/2} \) is directly proportional to the initial concentration. Thus, if the concentration decreases, the half-life would also decrease, which contradicts our observation.

- For a **first-order reaction**, the half-life is independent of the concentration:
\[ t_{1/2} = \frac{0.693}{k} \]
Therefore, the half-life remains constant regardless of the concentration, which again contradicts our observation.

- For a **second-order reaction**, the half-life is given by:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
This indicates that the half-life is inversely proportional to the initial concentration. Therefore, if the concentration decreases, the half-life increases, which aligns with our observation.

4. **Conclusion:**
- Since the half-life increased by a factor of 10 when the concentration decreased by a factor of 10, we can conclude that the reaction is of **second order**.

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