What fraction of a reactant remains after 40 min if `t_(1/2)` is 20 min and if Reaction is of first order?

Options

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

Answer

To solve the problem of finding the fraction of a reactant that remains after 40 minutes for a first-order reaction with a half-life (\(t_{1/2}\)) of 20 minutes, we can follow these steps:

### Step 1: Understand the concept of half-life in first-order reactions
In a first-order reaction, the half-life is constant and the amount of reactant decreases by half every half-life period.

### Step 2: Determine the number of half-lives that have passed
Given that the half-life is 20 minutes, we need to calculate how many half-lives fit into 40 minutes:
\[ \text{Number of half-lives} = \frac{\text{Total time}}{t_{1/2}} = \frac{40 \text{ min}}{20 \text{ min}} = 2 \]

### Step 3: Calculate the remaining fraction of the reactant
After each half-life, the amount of reactant remaining is halved:
- After the first half-life (20 minutes):
\[ \text{Remaining fraction} = \frac{1}{2} \]
- After the second half-life (40 minutes):
\[ \text{Remaining fraction} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

### Final Answer
Thus, the fraction of the reactant that remains after 40 minutes is:
\[ \frac{1}{4} \]

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