For a first order reaction with half life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately

Options

A. 1500 s
B. 500 s
C. 900 s
D. 600 s

Answer

To solve the problem, we need to determine the time taken for the concentration of a reactant in a first-order reaction to decrease from \( \frac{M}{10} \) to \( \frac{M}{100} \). Given that the half-life of the reaction is 150 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship of half-life and rate constant (k)**:
For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
We know \( t_{1/2} = 150 \) seconds. We can rearrange this to find \( k \):
\[ k = \frac{0.693}{150} \]

2. **Calculate the rate constant (k)**:
\[ k = \frac{0.693}{150} \approx 0.00462 \, \text{s}^{-1} \]

3. **Use the first-order rate equation**:
The integrated rate law for a first-order reaction is:
\[ \ln\left(\frac{[A_0]}{[A]}\right) = kt \]
Here, \( [A_0] = \frac{M}{10} \) and \( [A] = \frac{M}{100} \).

4. **Substitute the concentrations into the equation**:
\[ \ln\left(\frac{\frac{M}{10}}{\frac{M}{100}}\right) = kt \]
Simplifying the left side:
\[ \ln\left(\frac{M/10}{M/100}\right) = \ln\left(\frac{100}{10}\right) = \ln(10) \]

5. **Set up the equation with k**:
\[ \ln(10) = kt \]
We know \( k \approx 0.00462 \, \text{s}^{-1} \):
\[ \ln(10) \approx 2.303 \]
Therefore:
\[ 2.303 = 0.00462 \cdot t \]

6. **Solve for time (t)**:
\[ t = \frac{2.303}{0.00462} \approx 498.7 \, \text{seconds} \]

7. **Round the result**:
The time taken for the concentration to fall from \( \frac{M}{10} \) to \( \frac{M}{100} \) is approximately 500 seconds.

### Final Answer:
The time taken for the concentration of the reactant to fall from \( \frac{M}{10} \) to \( \frac{M}{100} \) is approximately **500 seconds**.