The half-lives of 2 samples are 0.1 and 0.4 seconds. Their initial conc. are 200 and 50 respectively. What is the order of the reaction ?

Options

A. 0
B. 2
C. 1
D. 4

Explanation

To determine the order of the reaction based on the given half-lives and initial concentrations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Half-life of sample 1, \( t_{1/2,1} = 0.1 \) seconds
- Half-life of sample 2, \( t_{1/2,2} = 0.4 \) seconds
- Initial concentration of sample 1, \( [A]_0 = 200 \)
- Initial concentration of sample 2, \( [B]_0 = 50 \)

2. **Use the Relationship of Half-Life and Concentration:**
The half-life of a reaction is related to the initial concentration and the order of the reaction (n) by the formula:
\[ t_{1/2} \propto [A]_0^{(1-n)} \]
This means:
\[ \frac{t_{1/2,2}}{t_{1/2,1}} = \left(\frac{[B]_0}{[A]_0}\right)^{(1-n)} \]

3. **Substitute the Known Values:**
Substitute the half-lives and initial concentrations into the equation:
\[ \frac{0.4}{0.1} = \left(\frac{50}{200}\right)^{(1-n)} \]

4. **Simplify the Left Side:**
\[ 4 = \left(\frac{1}{4}\right)^{(1-n)} \]

5. **Express the Right Side in Terms of Powers:**
Rewrite \( \frac{1}{4} \) as \( 4^{-1} \):
\[ 4 = 4^{-1(1-n)} \]

6. **Set the Exponents Equal:**
Since the bases are the same, we can set the exponents equal to each other:
\[ 1 = -1(1-n) \]

7. **Solve for n:**
\[ 1 = -1 + n \]
\[ n = 2 \]

8. **Conclusion:**
The order of the reaction is \( n = 2 \).

### Final Answer:
The order of the reaction is 2.

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