JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Chemical Kinetics

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A flask contains a mixture of compounds A and B . Both compounds decompose by first - order kinetics. The half lives for A and B are 300 s and 180 s , respectively. If the concentrations of

A

and

B

are equal initially, the time required for the concentration of

A

to be four times that of B (in s) is : (Use

ln 2 = 0.693

)

(A)120
(B)300
(C)900✓ Correct
(D)180

Answer: (C)

Solution

Both compounds decay by first-order kinetics, with rate constants

k_{A} = \frac{l n 2}{300}

and

k_{B} = \frac{l n 2}{180}

per second. Starting from equal concentrations,

\frac{[ A ]}{[ B ]} = e^{(k_{B} - k_{A}) t}

, and this ratio is to reach

4

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

. Here

k_{B} - k_{A} = ln 2 (\frac{1}{180} - \frac{1}{300}) = ln 2 \cdot \frac{120}{54000} = \frac{l n 2}{450}

, so

t = \frac{2 l n 2}{l n 2/450} = 900

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Chemical Kinetics

Solution

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