What fraction of a reactant remains unreacted after 40 min if `t_(1//2)` is of the reaction is 20 min? consider the reaction is first order

Options

A. `1//4`
B. `1//2`
C. `1//8`
D. `1//6`

Answer

To solve the problem of determining the fraction of a reactant that remains unreacted after 40 minutes for a first-order reaction with a half-life of 20 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Half-life (\( t_{1/2} \)) = 20 minutes
- Time elapsed (t) = 40 minutes
- Reaction order = First order

2. **Calculate the Rate Constant (k):**
For a first-order reaction, the relationship between the half-life and the rate constant is given by:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the given half-life:
\[ k = \frac{0.693}{20 \text{ min}} = 0.03465 \text{ min}^{-1} \]

3. **Use the First-Order Kinetics Formula:**
The first-order kinetics formula relates the concentration of reactant at time \( t \) to the initial concentration:
\[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \]
Here, \( [A_0] \) is the initial concentration, and \( [A] \) is the concentration at time \( t \).

4. **Calculate the Natural Logarithm:**
Substitute \( k \) and \( t \) into the equation:
\[ \ln \left( \frac{[A_0]}{[A]} \right) = 0.03465 \text{ min}^{-1} \times 40 \text{ min} = 1.386 \]

5. **Exponentiate to Find the Ratio:**
To find the ratio of the initial concentration to the concentration at time \( t \):
\[ \frac{[A_0]}{[A]} = e^{1.386} \approx 4 \]

6. **Determine the Fraction Remaining:**
The fraction of the reactant that remains unreacted can be calculated as:
\[ \text{Fraction remaining} = \frac{[A]}{[A_0]} = \frac{1}{4} \]

### Final Answer:
The fraction of the reactant that remains unreacted after 40 minutes is \( \frac{1}{4} \).

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