DDT on exposure to water decomposes according to first order kinetics. Half life = 10 years. How much time it will take for its decomposition to 99%?

Options

A. 50 years
B. 66.6 years
C. 500 years
D. 666 years

Explanation

To solve the problem of how long it will take for DDT to decompose 99% in water, we will follow these steps:

### Step 1: Determine the rate constant (k)
The half-life (t₁/₂) of a first-order reaction is given as 10 years. The formula to calculate the rate constant (k) for a first-order reaction is:

\[ k = \frac{0.693}{t_{1/2}} \]

Substituting the given half-life:

\[ k = \frac{0.693}{10 \text{ years}} = 0.0693 \text{ year}^{-1} \]

### Step 2: Use the first-order kinetics formula
For a first-order reaction, the relationship between the amount of substance remaining and time is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A} \right) \]

Where:
- \(A_0\) is the initial amount of the substance.
- \(A\) is the amount of the substance remaining after time \(t\).

### Step 3: Calculate the amount remaining after 99% decomposition
If 99% of DDT has decomposed, then 1% remains. Therefore:

\[ A = 0.01 A_0 \]

### Step 4: Substitute values into the formula
Now, substituting \(A\) into the first-order kinetics formula:

\[ t = \frac{2.303}{0.0693} \log \left( \frac{A_0}{0.01 A_0} \right) \]

This simplifies to:

\[ t = \frac{2.303}{0.0693} \log \left( \frac{1}{0.01} \right) \]

Calculating the logarithm:

\[ \log \left( \frac{1}{0.01} \right) = \log(100) = 2 \]

### Step 5: Calculate the time
Now substituting back into the equation:

\[ t = \frac{2.303}{0.0693} \times 2 \]

Calculating the value:

\[ t = \frac{2.303 \times 2}{0.0693} \approx \frac{4.606}{0.0693} \approx 66.5 \text{ years} \]

### Conclusion
Thus, the time it will take for DDT to decompose 99% is approximately **66.5 years**.

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