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: Understanding the Question:
We are given the time it takes for a first-order reaction to reduce its initial concentration to $\frac{1}{8}$ of its original value. We need to find the half-life ($t_{1/2}$) of the reaction.

Step 2: Key Formula or Approach:
For a first-order reaction, the concentration remaining after $n$ half-lives is given by:
$$[\text{A}]_t = [\text{A}]_0 \times \left(\frac{1}{2}\right)^n$$
Where the total time elapsed ($t$) is $n \times t_{1/2}$.

Step 3: Detailed Explanation:
The problem states that the final concentration is $\frac{1}{8}$ of the initial concentration.
$$\frac{[\text{A}]_t}{[\text{A}]_0} = \frac{1}{8}$$
We know that $\frac{1}{8} = \left(\frac{1}{2}\right)^3$.
This means exactly 3 half-lives have passed ($n = 3$).
We are given the total time $t = 23.03\ \text{minutes}$.
$$t = 3 \times t_{1/2}$$
$$23.03 = 3 \times t_{1/2}$$
$$t_{1/2} = \frac{23.03}{3} \approx 7.676\ \text{min}$$
Rounding to the nearest decimal given in the options yields $7.7\ \text{min}$.

Step 4: Final Answer:
The half-life period is approximately 7.7 min, matching option (B).