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QJEEM21Ch19cf: JEE Main 2021 Chemistry - Class 12 JEEM21Ch19cf Solution: For the first order reaction A 2B, 1 mole of reactant A gives…

For the first order reaction A →\to 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is __________ min. (Round off to the nearest integer).

[Use : ln 2 = 0.69, ln 10 = 2.3]

Properties of logarithms : ln xy = y ln x;

ln⁡(xy)=ln⁡x−ln⁡y\ln \left( {{x \over y}} \right) = \ln x - \ln y

(Round off to the nearest integer)

Step-by-Step Verified Solution

JEE Main 2021 (Online) 27th July Evening Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 88 English Explanation
Now, t=t1/2ln⁡2×[A0][At]t = {{{t_{1/2}}} \over {\ln 2}} \times {{[{A_0}]} \over {[{A_t}]}}

100=t1/2ln⁡2×ln⁡10.9⇒t1/2=690100 = {{{t_{1/2}}} \over {\ln 2}} \times \ln {1 \over {0.9}} \Rightarrow {t_{1/2}} = 690 min (taking ln 3 = 1.11)

Ans. 600 to 700

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