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Chemical Kinetics and Nuclear Chemistry question

2025 · 3 Apr · Shift 1 · Q19
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Chemical Kinetics and Nuclear Chemistry question

2025 · 3 Apr · Shift 1 · Q19

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
In a reaction A+B→CA+B \rightarrow CA+B→C, initial concentrations of AAA and BBB are related as [A]0=8[B]0[A]_0=8[B]_0[A]0​=8[B]0​. The half lives of AAA and BBB 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?
  1. A
    20 min
  2. B
    40 min
  3. C
    80 min
  4. D
    60 min
View written solutionFree

Correct answer: B

  1. Write first-order decay expressions

For a first-order disappearance, [X]t=[X]0e−kt[X]_t=[X]_0 e^{-kt}[X]t​=[X]0​e−kt where kkk is the rate constant.

So for AAA and BBB: [A]t=[A]0e−kAt,[B]t=[B]0e−kBt[A]_t=[A]_0 e^{-k_A t}, \qquad [B]_t=[B]_0 e^{-k_B t}[A]t​=[A]0​e−kA​t,[B]t​=[B]0​e−kB​t

  1. Use half-life relation for first-order reaction

For first order kinetics, t_{1/2}= rac{\ln 2}{k}

Hence, kA=ln⁡210k_A=\frac{\ln 2}{10}kA​=10ln2​ and kB=ln⁡240k_B=\frac{\ln 2}{40}kB​=40ln2​

  1. Use the given initial concentration relation

Given, [A]0=8[B]0[A]_0=8[B]_0[A]0​=8[B]0​

We need the time when [A]t=[B]t[A]_t=[B]_t[A]t​=[B]t​

Substitute: [A]0e−kAt=[B]0e−kBt[A]_0 e^{-k_A t}=[B]_0 e^{-k_B t}[A]0​e−kA​t=[B]0​e−kB​t

Using [A]0=8[B]0[A]_0=8[B]_0[A]0​=8[B]0​, 8[B]0e−kAt=[B]0e−kBt8[B]_0 e^{-k_A t}=[B]_0 e^{-k_B t}8[B]0​e−kA​t=[B]0​e−kB​t

Cancel [B]0[B]_0[B]0​: 8e−kAt=e−kBt8e^{-k_A t}=e^{-k_B t}8e−kA​t=e−kB​t

So, 8=e(kA−kB)t8=e^{(k_A-k_B)t}8=e(kA​−kB​)t

Taking natural log, ln⁡8=(kA−kB)t\ln 8=(k_A-k_B)tln8=(kA​−kB​)t

  1. Substitute rate constants
=\ln 2\left(\frac{4-1}{40}\right) =\frac{3\ln 2}{40}$$ Also, $$\ln 8=3\ln 2$$ Thus, $$t=\frac{3\ln 2}{(3\ln 2)/40}=40\text{ min}$$ 5. **Check options** The required time is $$\boxed{40\text{ min}}$$ So the correct option is **B**.
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