The half-life periods of two first-order reactions are in the ratio 3: 2. If t1 is the time required for 25% completion of the first reaction and t2 is the time required for 75% completion of the second reaction, then the ratio, t1 : t2, is (log 3 = 0.48, log 2 = 0.3)

t1t2=t1/21log⁡2log⁡10075t1/22log2⋅log⁡10025=t1/21log⁡43t1/22log⁡(4)=32×0.6−0.480.6=310

The half-life periods of two first-order reactions are in the ratio 3: 2. If t1 is the time required for 25% completion of the first reaction and t2 is the time required for 75% completion of the second reaction, then the ratio, t1 : t2, is (log 3 = 0.48, log 2 = 0.3)

t1t2=t1/21log⁡2log⁡10075t1/22log2⋅log⁡10025=t1/21log⁡43t1/22log⁡(4)=32×0.6−0.480.6=310

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The half-life periods of two first-order reactions are in the ratio 3: 2. If t₁ is the time required for 25% completion of the first reaction and t₂ is the time required for 75% completion of the second reaction, then the ratio, t₁ : t₂, is (log 3 = 0.48, log 2 = 0.3)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

3 : 5

3 : 2

3 : 10

12 : 25

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{t_{1}}{t_{2}} = \frac{\frac{{(t_{1 / 2})}_{1}}{\log 2} \log (\frac{100}{75})}{\frac{{(t_{1 / 2})}_{2}}{\log_{2}} \cdot \log (\frac{100}{25})} = \frac{{(t_{1 / 2})}_{1} \log (\frac{4}{3})}{{(t_{1 / 2})}_{2} \log (4)} \\ = \frac{3}{2} \times \frac{0.6 - 0.48}{0.6} = \frac{3}{10} \end{array}$