Question

\(60 \%\) of a first order reaction is completed in 60 minute. What is the time taken by reactant to decompose half of its original amount?

• A. 30.6 minute
• B. 20.2 minute
• C. 45.3 minute
• D. 60.1 minute

Hint:

Find the rate constant from 60 percent completion in 60 minutes, then use \(t_{1/2}=0.693/k\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a first order reaction \(k=\dfrac{2.303}{t}\log\dfrac{a}{a-x}\).

Step 2: Rate constant
60 percent is completed, so 40 percent remains: \(k=\dfrac{2.303}{60}\log\dfrac{100}{40}=\dfrac{2.303}{60}\times0.3979\).\n\(k\approx1.527\times10^{-2}\ \mathrm{min^{-1}}\).

Step 3: Half life
\(t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.527\times10^{-2}}\approx45.4\ \text{min}\).

Step 4: Match
The nearest option is 45.3 minute.

Final Answer:
The time for half of the reactant to decompose is about 45.3 minutes. \[ \boxed{\text{(C) }45.3\text{ minute}} \]