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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2025 · 3 April · Morning Shift · Chemistry · Chemical Kinetics

In a reaction A+B arrow C , initial concentrations of A and B are...

In a reaction A+B→CA+B \rightarrow C, initial concentrations of AA and BB are related as [A]0=8[B]0[A]_0=8[B]_0. The half lives of AA and BB are 10 min and 40 min. respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

Options

  1. A60 min
  2. B80 min
  3. C20 min
  4. D40 min

Correct answer

D. 40 min

Step-by-step solution

Given : [A]0=8[ B]0[\mathrm{A}]_0=8[\mathrm{~B}]_0
\left[\mathrm{t}_{1/2}\right]_{\mathrm{A}}=10 \mathrm{~min}.<br/>[t1/2]B=40 min.<br />\left[\mathrm{t}_{1/2}\right]_{\mathrm{B}}=40 \mathrm{~min}.
Ist \mathrm{I}^{\text{st }} order kinetics
t=\mathrm{t}= ?
[A]t=[B]t[\mathrm{A}]_{\mathrm{t}}=[\mathrm{B}]_{\mathrm{t}}
−kA×t−kB×t-\mathrm{k}_{\mathrm{A}} \times \mathrm{t} \quad-\mathrm{k}_{\mathrm{B}} \times \mathrm{t}
⇒[A]0e=[B]0e\Rightarrow \quad[\mathrm{A}]_0 \mathrm{e}=[\mathrm{B}]_0 \mathrm{e}
⇒[A]0[ B]0=e(kA−kB)t\Rightarrow \quad \frac{[\mathrm{A}]_0}{[\mathrm{~B}]_0}=\mathrm{e}^{\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \mathrm{t}}
⇒8=e(kA−kB)×t\Rightarrow \quad 8=\mathrm{e}^{\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \times \mathrm{t}}
⇒ℓn8=(kA−kB)×t\Rightarrow \quad \ell \mathrm{n} 8=\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \times \mathrm{t}
⇒ln⁡8=ln⁡2(1(ta⁡2)A−1(ta⁡2)B)×t\Rightarrow \quad \ln 8=\ln 2\left(\frac{1}{\left(\operatorname{ta}_2\right)_{\mathrm{A}}}-\frac{1}{\left(\operatorname{ta}_2\right)_{\mathrm{B}}}\right) \times \mathrm{t}
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