Question

What is the half-life of a first order reaction if time required to decrease concentration of reactant from 1.0 M to 0.25 M is 10 hour?

• A. 12 hour
• B. 4 hour
• C. 5 hour
• D. 10 hour

Hint:

For first order reactions, each half-life reduces concentration to half. Two half-lives reduce to \(1/4\)th.

Answer 1

The Correct Option is C

Step 1: Understand the question.
A first order reaction drops from 1.0 M to 0.25 M in 10 hours. We need its half-life.
Step 2: Recall the half-life idea.
A half-life is the time for the amount to fall to half. After $n$ half-lives the fraction left is: \[ \frac{[A]}{[A]_0} = \left(\frac{1}{2}\right)^n \]
Step 3: Find the fraction left.
\[ \frac{0.25}{1.0} = \frac{1}{4} \]
Step 4: Write the fraction as a power of one half.
\[ \frac{1}{4} = \left(\frac{1}{2}\right)^2 \] So $n = 2$ half-lives have passed.
Step 5: Find one half-life.
Two half-lives take 10 hours, so one half-life is: \[ t_{1/2} = \frac{10}{2} = 5 \text{ hours} \]
Step 6: Choose the answer.
The half-life is 5 hours, which is option 3. \[ \boxed{5\ \text{hours}} \]