Question

The half-life (in seconds) of a first order reaction which takes 4 seconds for 30% completion is: (may use $log_{10}7=0.845$)

• A. 6.65
• B. 5.45
• C. 8.95
• D. 7.77

Hint:

Remember: $t_{1/2}$ for first-order reactions is independent of the initial concentration.

Answer 1

The Correct Option is D

Step 1: Concept
For a first-order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \left(\frac{a}{a-x}\right)$, and the half-life is $t_{1/2} = \frac{0.693}{k}$.

Step 2: Calculation of k
Given $t = 4$ s and 30% completion ($x = 0.3a$), the remaining concentration is $0.7a$.
$k = \frac{2.303}{4} \log \left(\frac{1}{0.7}\right) = \frac{2.303}{4} (\log 10 - \log 7) = \frac{2.303}{4} (1 - 0.845) = \frac{2.303 \times 0.155}{4} \approx 0.08924$ s$^{-1}$.

Step 3: Calculation of Half-life
$t_{1/2} = \frac{0.693}{0.08924} \approx 7.77$ seconds.

Step 4: Conclusion
The correct calculated value matches option (D). Final Answer: (D)