Verified Solution • 2025–26 Session moderate

QJEEM19Ch3f92: JEE Main 2019 Chemistry - Class 12 JEEM19Ch3f92 Solution: For a reaction scheme A ⟶ k 1 B ⟶ k 2 C , if the rate of…

For a reaction scheme A⟶k1B⟶k2C,

if the rate of formation of B is set to be zero then the concentration of B is given by :

Multiple Choice Options

  • A k1k2[A]{k_1}{k_2}[A]
  • B (k1k2)[A]\left( {{{{k_1}} \over {{k_2}}}} \right)[A]
  • C (k1+k2)[A]({k_1} + {k_2})[A]
  • D (k1−k2)[A]({k_1} - {k_2})[A]

Step-by-Step Verified Solution

Given, the rate of formation of B is set to be zero.

∴\therefore d[B]dt=0{{d\left[ B \right]} \over {dt}} = 0

For this reaction,

A⟶R1k1BA\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{{R_1}}^{{k_1}}} B

Rate of reaction for this reaction (R1) = k1[A]{k_1}\left[ A \right]

For this reaction,

B⟶R2k2CB\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{{R_2}}^{{k_2}}} C

Rate of reaction for this reaction (R2) = k2[B]{k_2}\left[ B \right]

Net rate of formation of B = d[B]dt{{d\left[ B \right]} \over {dt}} = R1 - R2

As d[B]dt=0{{d\left[ B \right]} \over {dt}} = 0

∴\therefore R1 - R2 = 0

⇒\Rightarrow k1[A]{k_1}\left[ A \right] - k2[B]{k_2}\left[ B \right] = 0

⇒\Rightarrow [B]=k1k2[A]\left[ B \right] = {{{k_1}} \over {{k_2}}}\left[ A \right]

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