In a reaction `A rarr ` products , when start is made from `8.0 xx 10^(-2) M` of A, half life is found to be 120 minute. For the initial concentration `4.0 xx 10^(-2) M`, the half life of the reaction becomes 240 minute. The order of the reaction is `:`

Options

A. zero
B. one
C. two
D. 0.5

Explanation

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

### Step 1: Write down the known values
We have two sets of data:
1. Initial concentration \( [A_1] = 8.0 \times 10^{-2} \, M \) with half-life \( t_{1/2,1} = 120 \, \text{minutes} \)
2. Initial concentration \( [A_2] = 4.0 \times 10^{-2} \, M \) with half-life \( t_{1/2,2} = 240 \, \text{minutes} \)

### Step 2: Use the relationship for half-life and order of reaction
For a reaction of order \( n \), the relationship between half-life (\( t_{1/2} \)) and initial concentration (\( [A] \)) can be expressed as:
\[ t_{1/2} \propto [A]^{(1-n)} \]
This can be rearranged to:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left(\frac{[A_1]}{[A_2]}\right)^{(1-n)} \]

### Step 3: Substitute the known values
Substituting the values we have:
\[ \frac{120}{240} = \left(\frac{8.0 \times 10^{-2}}{4.0 \times 10^{-2}}\right)^{(1-n)} \]
This simplifies to:
\[ \frac{1}{2} = \left(2\right)^{(1-n)} \]

### Step 4: Solve for \( n \)
Taking logarithms on both sides, we can express this as:
\[ \frac{1}{2} = 2^{(1-n)} \]
This implies:
\[ 1-n = -1 \quad \text{(since } 2^{-1} = \frac{1}{2}\text{)} \]
Thus:
\[ n = 2 \]

### Conclusion
The order of the reaction is \( n = 2 \), indicating that it is a second-order reaction.