Consider a certain reaction `A rarr` Products with `k=2.0xx10^(-2)s^(-1)`. Calculate the concentration of `A` remaining after `100s` if the initial concentration of `A` is `1.0 mol L^(-1)`.

Explanation

To solve the problem, we need to determine the concentration of reactant A remaining after 100 seconds given that the reaction follows first-order kinetics. The rate constant \( k \) is given as \( 2.0 \times 10^{-2} \, \text{s}^{-1} \) and the initial concentration of A is \( 1.0 \, \text{mol L}^{-1} \).

### Step-by-Step Solution:

1. **Identify the Order of Reaction**:
The units of the rate constant \( k \) are \( \text{s}^{-1} \), which indicates that the reaction is first-order.

2. **Use the First-Order Kinetics Formula**:
For a first-order reaction, the concentration of A at time \( t \) can be calculated using the formula:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
where:
- \( [A]_0 \) is the initial concentration of A,
- \( [A] \) is the concentration of A at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

3. **Substitute the Known Values**:
We know:
- \( [A]_0 = 1.0 \, \text{mol L}^{-1} \)
- \( k = 2.0 \times 10^{-2} \, \text{s}^{-1} \)
- \( t = 100 \, \text{s} \)

Plugging these values into the equation gives:
\[ \ln \left( \frac{1.0}{[A]} \right) = (2.0 \times 10^{-2}) \times 100 \]

4. **Calculate the Right Side**:
\[ \ln \left( \frac{1.0}{[A]} \right) = 2.0 \]

5. **Exponentiate to Solve for [A]**:
To find \( [A] \), we exponentiate both sides:
\[ \frac{1.0}{[A]} = e^{2.0} \]
Therefore,
\[ [A] = \frac{1.0}{e^{2.0}} \]

6. **Calculate \( e^{2.0} \)**:
Using a calculator, we find:
\[ e^{2.0} \approx 7.389 \]

7. **Final Calculation for [A]**:
\[ [A] = \frac{1.0}{7.389} \approx 0.1353 \, \text{mol L}^{-1} \]

### Final Answer:
The concentration of A remaining after 100 seconds is approximately **0.1353 mol L\(^{-1}\)**.