`H_(2)O_(2)overset(1st"order")(to)H_(2)O+1/2O_(2)`. Pressure of `O_(2)` is recorded as:
time (min)`" "` 10 min `" "prop`
`P_(O_(2))` (mm)`" "120" "160`
What is half life?

Explanation

To find the half-life of the reaction \( H_2O_2 \rightarrow H_2O + \frac{1}{2}O_2 \) given the pressure of \( O_2 \) at different times, we can follow these steps:

### Step 1: Understand the Reaction and Given Data
The reaction is a first-order reaction, and we are given the pressure of \( O_2 \) at two different times:
- At \( t = 10 \) minutes, \( P(O_2) = 120 \) mm
- At \( t \to \infty \), \( P(O_2) = 160 \) mm

### Step 2: Determine the Change in Pressure
The change in pressure of \( O_2 \) from the initial state to the state at \( t = 10 \) minutes can be calculated as follows:
- Initial pressure of \( O_2 \) (at \( t = \infty \)) = 160 mm
- Pressure of \( O_2 \) after 10 minutes = 120 mm
- Therefore, the amount of \( O_2 \) produced in 10 minutes = \( 160 - 120 = 40 \) mm

### Step 3: Use the First-Order Kinetics Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
Where:
- \( A_0 \) = initial amount (pressure at \( t = \infty \)) = 160 mm
- \( A_t \) = amount at time \( t \) (pressure at \( t = 10 \) min) = 120 mm
- \( t \) = time = 10 min

Substituting the values:
\[ k = \frac{2.303}{10} \log \left( \frac{160}{120} \right) \]

### Step 4: Calculate the Logarithm
Calculate \( \frac{160}{120} = \frac{4}{3} \):
\[ \log \left( \frac{4}{3} \right) \approx 0.1249 \]
Now substituting this back into the equation for \( k \):
\[ k = \frac{2.303}{10} \times 0.1249 \approx 0.0288 \text{ min}^{-1} \]

### 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 \( k \):
\[ t_{1/2} = \frac{0.693}{0.0288} \approx 24.1 \text{ minutes} \]

### Final Answer
The half-life of the reaction is approximately **24.1 minutes**.