Question

The time taken for 60% completion of a first order reaction is 13.22 min. What is its half-life ($t_{1/2}$) in min? ($\log(2.5) = 0.398$)

• A. 11
• B. 9
• C. 8
• D. 10

Hint:

Notice that $2.303 \times \log(2) = 0.693$.
For first order reactions, you can relate any two times $t_1$ and $t_2$ directly:
$\frac{t_1}{t_2} = \frac{\log(a/(a-x_1))}{\log(a/(a-x_2))}$.
This bypasses calculating the rate constant $k$ explicitly.

Answer 1

The Correct Option is D

Step 1: Use a shortcut instead of solving for k first.
For a first order reaction, the time taken to reach any fixed leftover fraction is proportional to $\log$ of the reciprocal of that fraction, so the rate constant cancels out if we take a ratio of two such times.
Step 2: Set up the ratio between the two given times.
At 60% completion, 40% remains, so $t_{60}$ involves $\log(100/40) = \log 2.5$. At half life, 50% remains, so $t_{1/2}$ involves $\log 2$. This gives: \[ \frac{t_{60}}{t_{1/2}} = \frac{\log 2.5}{\log 2} \]
Step 3: Plug in the numbers.
$\log 2.5 = 0.398$ and $\log 2 = 0.301$, so the ratio is $0.398/0.301 \approx 1.322$.
Step 4: Solve for the half life.
\[ t_{1/2} = \frac{13.22}{1.322} \]
\[ \boxed{t_{1/2} = 10\text{ min}} \]