JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Chemical Kinetics

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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

: ln 2 = 0.693

)

Answer: 200

Solution

Since half-lives are independent of concentration, both A and B follow first-order kinetics.

k_{A} = \frac{l n 2}{100}

and

k_{B} = \frac{l n 2}{50}

For first-order reactions:

[A]_{t} = [A]_{0} e^{- k_{A} t}

and

[B]_{t} = [B]_{0} e^{- k_{B} t}

Given

[A]_{0} = [B]_{0}

and we need

[A]_{t} = 4 [B]_{t}

[A]_{0} e^{- k_{A} t} = 4 [B]_{0} e^{- k_{B} t}

e^{- k_{A} t} = 4 e^{- k_{B} t}

e^{(k_{B} - k_{A}) t} = 4

(k_{B} - k_{A}) t = ln 4 = 2 ln 2

(\frac{l n 2}{50} - \frac{l n 2}{100}) t = 2 ln 2

\frac{l n 2}{100} \times t = 2 ln 2

t = 200

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Chemical Kinetics

Solution

More JEE Main questions on Chemical Kinetics