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QJEEAdv20Chbe85: JEE Advanced 2020 Chemistry - Class 12 JEEAdv20Chbe85 Solution: Which of the following plots is(are) correct for the given…

Which of the following plots is(are) correct for the given reaction?

([P]0 is the initial concentration of P)

JEE Advanced 2020 Paper 2 Offline Chemistry - Chemical Kinetics and Nuclear Chemistry Question 15 English

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/4745cf721b99050a9ebf913daf320e0d896d9f3d18043415b9934237f6c73cea.webp
  • B https://ik.imagekit.io/novatra/photos/55ed3d5c66e109f882f9cc384a555dddd539bfbebec1580042ae5dee815839ce.webp
  • C https://ik.imagekit.io/novatra/photos/c2f60efb1180ee8b3a746d6f4a1be6eb85c9784c2f0c221b3b448fcbf0c53f39.webp
  • D https://ik.imagekit.io/novatra/photos/17e5f52a4759cccff5116c6d5ecd5107dd8a197657b792c39c098db6d78da88a.webp

Step-by-Step Verified Solution

(CH3)3C−Br+NaOH⟶(First order reaction)SN1(CH3)3C−OH+NaBr{(C{H_3})_3}C - Br + NaOH\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{(First\,order\,reaction)}^{{S_{{N^1}}}}} {(C{H_3})_3}C - OH + NaBr

This is first order reaction and for first order reaction t1/2=0.693k{t_{1/2}} = {{0.693} \over k}

So, half-life is independent of initial concentration.

Therefore, the plot (a) correct.

For first order reaction,

(r)=k[(CH3)3C−Br](r) = k[{(C{H_3})_3}C - Br] ;

ln⁡(P0P)=k t\ln \left( {{{{P_0}} \over P}} \right) = k\,t

or ln⁡(PP0)=−k t\ln \left( {{P \over {{P_0}}}} \right) = - k\,t

Hence, plot (b) and (d) are incorrect. For first order reaction,

Q = [P0](1 −- ekt)

Or [Q][P0]=(1−ekt){{[Q]} \over {[{P_0}]}} = (1 - {e^{kt}})

Hence, plot (c) is incorrect.

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