The decompoistion of `H_(2)O_(2)` can be followed by titration with `KMnO_(4)` and is found to be a first order reaction. The rate constant is `4.5 xx 10^(-2)`. In an experiment, the initial titre value as `25 mL`. The titre value will be `5 mL` after a lapse of

Options

A. `4.5 xx 10^(-2) xx 5 min`
B. `(log_(e) 5)/(4.5 xx 10^(-2)) min`
C. `(log_(e)5//4)/(4.5 xx 10^(-2))min`
D. None of these

Explanation

To solve the problem regarding the decomposition of \( H_2O_2 \) and its titration with \( KMnO_4 \), we need to follow these steps:

### Step 1: Identify the given values
- Initial titre value (\( V_0 \)) = 25 mL
- Final titre value (\( V_t \)) = 5 mL
- Rate constant (\( k \)) = \( 4.5 \times 10^{-2} \, \text{s}^{-1} \)

### Step 2: Use the first-order reaction formula
For a first-order reaction, the relationship between the rate constant, time, and concentrations (or volumes in this case) is given by the equation:

\[ k = \frac{2.303}{T} \log \left( \frac{V_0}{V_t} \right) \]

### Step 3: Rearrange the formula to solve for time \( T \)
We can rearrange the formula to find \( T \):

\[ T = \frac{2.303}{k} \log \left( \frac{V_0}{V_t} \right) \]

### Step 4: Substitute the known values into the equation
Now we can substitute the known values into the equation:

1. Calculate \( \frac{V_0}{V_t} \):

\[ \frac{V_0}{V_t} = \frac{25}{5} = 5 \]

2. Now substitute \( k \) and \( \frac{V_0}{V_t} \) into the equation for \( T \):

\[ T = \frac{2.303}{4.5 \times 10^{-2}} \log(5) \]

### Step 5: Calculate \( \log(5) \)
Using a calculator or logarithm table:

\[ \log(5) \approx 0.699 \]

### Step 6: Substitute \( \log(5) \) back into the equation
Now substitute \( \log(5) \) back into the equation for \( T \):

\[ T = \frac{2.303}{4.5 \times 10^{-2}} \times 0.699 \]

### Step 7: Perform the calculations
1. Calculate \( \frac{2.303}{4.5 \times 10^{-2}} \):

\[ \frac{2.303}{0.045} \approx 51.18 \]

2. Now multiply by \( 0.699 \):

\[ T \approx 51.18 \times 0.699 \approx 35.7 \, \text{s} \]

### Final Answer
The time elapsed for the titre value to change from 25 mL to 5 mL is approximately **35.7 seconds**.

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