The half-life period of a first order reaction is 20 minutes. The time required for the concentration of the reactant to change from 0.16 M to 0.02 M is :

Options

A. 80 minutes
B. 60 minutes
C. 40 minutes
D. 20 minutes

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Rate Constant (k)
The formula for the rate constant (k) for a first-order reaction is given by:
\[ k = \frac{0.693}{t_{1/2}} \]
where \( t_{1/2} \) is the half-life of the reaction.

Given:
- Half-life \( t_{1/2} = 20 \) minutes

Substituting the values:
\[ k = \frac{0.693}{20} \]
\[ k = 0.03465 \, \text{min}^{-1} \]

### Step 2: Use the First-Order Reaction Formula
For a first-order reaction, the relationship between the rate constant, time, and concentrations is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A_t]} \right) \]
where:
- \( [A_0] \) is the initial concentration
- \( [A_t] \) is the concentration at time \( t \)

Given:
- Initial concentration \( [A_0] = 0.16 \, \text{M} \)
- Final concentration \( [A_t] = 0.02 \, \text{M} \)

### Step 3: Substitute Values into the Formula
Now, substituting the known values into the formula:
\[ t = \frac{2.303}{0.03465} \log \left( \frac{0.16}{0.02} \right) \]

### Step 4: Calculate the Logarithm
First, calculate the ratio:
\[ \frac{0.16}{0.02} = 8 \]

Now, calculate the logarithm:
\[ \log(8) \approx 0.903 \]

### Step 5: Substitute and Calculate Time
Now substitute back into the equation:
\[ t = \frac{2.303}{0.03465} \times 0.903 \]

Calculating this gives:
\[ t \approx \frac{2.303 \times 0.903}{0.03465} \]
\[ t \approx \frac{2.080}{0.03465} \]
\[ t \approx 60.0 \, \text{minutes} \]

### Final Answer
The time required for the concentration of the reactant to change from 0.16 M to 0.02 M is approximately **60 minutes**.

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