Question

In a first order reaction, concentration of reactant is reduced to (1/8)th of concentration in 23.03 minutes. What is half-life period of reaction?

• A. 25 min
• B. 7.7 min
• C. 15 min
• D. 30 min

Hint:

For first-order reactions, recognizing powers of $\frac{1}{2}$ saves you from doing complex logarithmic calculations! $\frac{1}{2}$ is 1 half-life, $\frac{1}{4}$ is 2 half-lives, $\frac{1}{8}$ is 3 half-lives, $\frac{1}{16}$ is 4 half-lives, and so on.

Answer 1

The Correct Option is B

Step 1: Understand the question.
A first order reaction drops the reactant to $\tfrac{1}{8}$ of its start in $23.03$ minutes. We must find the half-life.

Step 2: Connect the fraction to half-lives.
After each half-life the amount is halved. After $n$ half-lives the fraction left is $\left(\tfrac{1}{2}\right)^n$.

Step 3: Find how many half-lives passed.
We need $\left(\tfrac{1}{2}\right)^n = \tfrac{1}{8}$. Since $\tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3$, we get $n = 3$. So three half-lives have gone by.

Step 4: Relate total time to half-life.
The total time is $n$ times the half-life, so $23.03 = 3 \times t_{1/2}$.

Step 5: Solve for the half-life.
\[ t_{1/2} = \frac{23.03}{3} \approx 7.68\ \text{min} \]
Step 6: Pick the answer.
Rounded, this is $7.7$ minutes, option (B).
\[ \boxed{t_{1/2} \approx 7.7\ \text{min}} \]