Verified Solution • 2025–26 Session moderate

QJEEM20Chd157: JEE Main 2020 Chemistry - Class 12 JEEM20Chd157 Solution: A flask contains a mixture of compounds A and B. Both compounds …

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) :
(Use ln 2 = 0.693)

Multiple Choice Options

  • A 180
  • B 120
  • C 300
  • D 900

Step-by-Step Verified Solution

At = A0.e-k1t

Bt = B0.e-k2t

k1 = ln⁡2300{{\ln 2} \over {300}}

k2 = ln⁡2180{{\ln 2} \over {180}}

Given, A0 = B0

and At and Bt are related as [A] = 4[B]

∴\therefore A0.e-k1t = 4B0.e-k2t

⇒\Rightarrow e(ln⁡2180−ln⁡2300)t{e^{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}} = 4

⇒\Rightarrow (ln⁡2180−ln⁡2300)t{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t} = ln 4 = 2ln 2

⇒\Rightarrow (1180−l300)t{\left( {{1 \over {180}} - {{\mathop{\rm l}\nolimits} \over {300}}} \right)t} = 2

⇒\Rightarrow t = 2×180×300120{{2 \times 180 \times 300} \over {120}} = 900 sec

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