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Chemical Kinetics and Nuclear Chemistry question

2022 · 27 Jul · Shift 2 · Q18
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Chemical Kinetics and Nuclear Chemistry question

2022 · 27 Jul · Shift 2 · Q18

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
[A]→[B]ReactantolimitsProductolimits\begin{matrix} {[A]} & \to & {[B]} \\ {{\mathop{\rm Reactant} olimits} } & {} & {{\mathop{\rm Product} olimits} } \\ \end{matrix}[A]Reactantolimits​→​[B]Productolimits​ If formation of compound [B][\mathrm{B}][B] follows the first order of kinetics and after 70 minutes the concentration of [A][\mathrm{A}][A] was found to be half of its initial concentration. Then the rate constant of the reaction is x×10−6 s−1x \times 10^{-6} \mathrm{~s}^{-1}x×10−6 s−1. The value of xxx is ‾\underline{\hspace{2cm}}​. (Nearest Integer)
Numerical answer
View written solutionFree

Correct answer: 165

  1. For a first-order reaction, the integrated rate law is

[A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}[A]t​=[A]0​e−kt

or equivalently,

k=2.303tlog⁡([A]0[A]t)k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]_t}\right)k=t2.303​log([A]t​[A]0​​)

  1. Given that after 70 minutes, the concentration of [A][A][A] becomes half of its initial value:

[A]t=[A]02[A]_t = \frac{[A]_0}{2}[A]t​=2[A]0​​

So, the time given is actually the half-life of a first-order reaction.

For a first-order reaction,

t1/2=0.693kt_{1/2} = \frac{0.693}{k}t1/2​=k0.693​

Thus,

k=0.693t1/2k = \frac{0.693}{t_{1/2}}k=t1/2​0.693​

  1. Convert time into seconds:

70 min=70×60=4200 s70\text{ min} = 70 \times 60 = 4200\text{ s}70 min=70×60=4200 s

  1. Now calculate kkk:

k=0.6934200=1.65×10−4 s−1k = \frac{0.693}{4200} = 1.65 \times 10^{-4}\,\text{s}^{-1}k=42000.693​=1.65×10−4s−1

  1. Express this in the form x×10−6 s−1x \times 10^{-6}\,\text{s}^{-1}x×10−6s−1:

1.65×10−4=165×10−61.65 \times 10^{-4} = 165 \times 10^{-6}1.65×10−4=165×10−6

Hence,

x=165x = 165x=165

  1. Nearest integer:

165\boxed{165}165​

  1. Comparison with stored answer:

Stored correct answer = 165165165.

Our derived answer matches the stored answer.

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