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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Half-life (t1/2) at 0.1 M = 5 minutes
- Half-life (t1/2) at 0.01 M = 50 minutes

2. **Recall the Relationship Between Half-life and Concentration:**
The half-life of a reaction can be expressed in terms of the initial concentration \([A_0]\) and the order of the reaction \(n\). For a reaction of order \(n\):
\[ t_{1/2} \propto [A_0]^{(1-n)} \]
This means that the half-life is directly proportional to the initial concentration raised to the power of \(1-n\).

3. **Set Up the Proportionality:**
From the given data, we can write two equations based on the half-lives:
\[ t_{1/2}(0.1) \propto (0.1)^{(1-n)} \quad \text{(1)} \]
\[ t_{1/2}(0.01) \propto (0.01)^{(1-n)} \quad \text{(2)} \]

4. **Express the Proportionality Constants:**
Since both expressions are proportional to the same constant \(k\), we can express them as:
\[ 5 = k \cdot (0.1)^{(1-n)} \quad \text{(3)} \]
\[ 50 = k \cdot (0.01)^{(1-n)} \quad \text{(4)} \]

5. **Divide Equation (3) by Equation (4):**
\[ \frac{5}{50} = \frac{(0.1)^{(1-n)}}{(0.01)^{(1-n)}} \]
Simplifying gives:
\[ \frac{1}{10} = \left(\frac{0.1}{0.01}\right)^{(1-n)} \]
\[ \frac{1}{10} = (10)^{(1-n)} \]

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

7. **Solve for \(n\):**
Rearranging gives:
\[ n = 2 \]

### Conclusion:
The order of the reaction is **2**.

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