JEE Main 2022 · 26 Jun · Shift 1 · Chemistry Q84

← JEE Main 2022 · 26 Jun · Shift 1

ChemistryInteger+4 / -1Question 84 of 90

A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______ s.

(Given : $\ln2=0.693$ )

Your answer

Answer and worked solution

Official answer: 200

First order (half-lives independent of concentration): $k_A=\dfrac{0.693}{100}$ , $k_B=\dfrac{0.693}{50}$ .

Ratio. $\dfrac{[A]}{[B]}=e^{(k_B-k_A)t}=4$ , so $t=\dfrac{\ln4}{0.693/100}=200$ s.

Answer: 200

Topics:Half-life Integrated rate equations

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JEE Main 2022 · 26 Jun · Shift 1, Chemistry question 84