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Chemical Kinetics and Nuclear Chemistry question

2019 · 8 Apr · Shift 2 · Q11
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Chemical Kinetics and Nuclear Chemistry question

2019 · 8 Apr · Shift 2 · Q11

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For a reaction scheme A→k1B→k2CA\xrightarrow{{k_1}} B\xrightarrow{{k_2}} CAk1​​Bk2​​C, if the rate of formation of B is set to be zero then the concentration of B is given by :
  1. A
    k1k2[A]{k_1}{k_2}[A]k1​k2​[A]
  2. B
    (k1k2)[A]\left( {{{{k_1}} \over {{k_2}}}} \right)[A](k2​k1​​)[A]
  3. C
    (k1+k2)[A]({k_1} + {k_2})[A](k1​+k2​)[A]
  4. D
    (k1−k2)[A]({k_1} - {k_2})[A](k1​−k2​)[A]
View written solutionFree

Correct answer: B

  1. Write the rate expression for intermediate BBB

For the consecutive reaction A→k1B→k2CA \xrightarrow{k_1} B \xrightarrow{k_2} CAk1​​Bk2​​C

BBB is formed from AAA and consumed to form CCC. So, d[B]dt=k1[A]−k2[B]\frac{d[B]}{dt} = k_1[A] - k_2[B]dtd[B]​=k1​[A]−k2​[B]

  1. Use the given condition

The question says that the rate of formation of BBB is set to be zero, i.e. d[B]dt=0\frac{d[B]}{dt} = 0dtd[B]​=0

Therefore, k1[A]−k2[B]=0k_1[A] - k_2[B] = 0k1​[A]−k2​[B]=0

  1. Solve for [B][B][B]

k2[B]=k1[A]k_2[B] = k_1[A]k2​[B]=k1​[A]

[B]=k1k2[A][B] = \frac{k_1}{k_2}[A][B]=k2​k1​​[A]

  1. Match with the options

The expression obtained is [B]=(k1k2)[A][B] = \left(\frac{k_1}{k_2}\right)[A][B]=(k2​k1​​)[A]

So the correct option is B.

  1. Check against stored answer

Stored correct answer: B

This matches our derived answer.

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