Verified Solution • 2025–26 Session moderate

QJEEM21Ch46ac: JEE Main 2021 Chemistry - Class 12 JEEM21Ch46ac Solution: A and B decompose via first order kinetics with half-lives 54.0 …

A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer).

Step-by-Step Verified Solution

Initially : [A0]=[B0]=a\left[ {{A_0}} \right] = \left[ {{B_0}} \right] = a

After time 't' min : [A]=16[B]\left[ A \right] = 16\left[ B \right]

[A]=[A0]e−kAt\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}

[B]=[B0]e−kBt\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}

⇒a . e−kAt=16ae−kBt\Rightarrow a\,.\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}

⇒e−(kA−kB)t=16\Rightarrow {e^{ - \left( {{k_A} - {k_B}} \right)t}} = 16

⇒(kB−kA)t=ln⁡16\Rightarrow \left( {{k_B} - {k_A}} \right)t = \ln 16

⇒ln⁡2(118−154)t=4ln⁡2\Rightarrow \ln 2\left( {{1 \over {18}} - {1 \over {54}}} \right)t = 4\ln 2

⇒t=54×18×436=108\Rightarrow t = {{54 \times 18 \times 4} \over {36}} = 108 min

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