`H_(2)O_(2)toH_(2)O+1/2O_(2)`(1storder)
Time to time the `H_(2)O_(2)` solutioni is titrated with standard acidified `KMnO_(4)`. Thedata is
Time (min)`" "0" "15`
`KMnO_(4) ("consumed")" "16ml" "2ml`
What is half life in minutes?

Explanation

To find the half-life of the reaction \( H_2O_2 \rightarrow H_2O + \frac{1}{2}O_2 \) which follows first-order kinetics, we will follow these steps:

### Step 1: Understand the Reaction and Data
We have the reaction:
\[ H_2O_2 \rightarrow H_2O + \frac{1}{2}O_2 \]
The data provided is:
- At time \( t = 0 \) min, \( 16 \, \text{ml} \) of \( KMnO_4 \) is consumed.
- At time \( t = 15 \) min, \( 2 \, \text{ml} \) of \( KMnO_4 \) is consumed.

### Step 2: Determine Initial and Remaining Concentrations
From the data:
- Initial concentration of \( H_2O_2 \) (denoted as \( A_0 \)) is represented by the \( KMnO_4 \) consumed at \( t = 0 \):
\[ A_0 = 16 \, \text{ml} \]
- At \( t = 15 \) min, the amount of \( KMnO_4 \) consumed is \( 2 \, \text{ml} \), which indicates the remaining concentration of \( H_2O_2 \) (denoted as \( A_t \)):
\[ A_t = 16 - 2 = 14 \, \text{ml} \]

### Step 3: Apply the First-Order Kinetics Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
Substituting the values:
- \( t = 15 \, \text{min} \)
- \( A_0 = 16 \, \text{ml} \)
- \( A_t = 14 \, \text{ml} \)

We have:
\[ k = \frac{2.303}{15} \log \left( \frac{16}{14} \right) \]

### Step 4: Calculate the Logarithm
Calculating the logarithm:
\[ \log \left( \frac{16}{14} \right) = \log(1.142857) \approx 0.05799 \]
Now substituting this back into the equation for \( k \):
\[ k = \frac{2.303}{15} \times 0.05799 \]
Calculating \( k \):
\[ k \approx \frac{2.303 \times 0.05799}{15} \approx 0.0083 \, \text{min}^{-1} \]

### Step 5: Calculate the Half-Life
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{0.0083} \approx 83.5 \, \text{min} \]

### Final Answer
The half-life of the reaction is approximately:
\[ t_{1/2} \approx 83.5 \, \text{min} \]