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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2019 · Chemistry · Chemical Kinetics

Given the hypothetical reaction mechanism A I ⟶ B II ⟶ C III ⟶ D IV...

Given the hypothetical reaction mechanism

A⟶IB⟶IIC⟶IIID→IVEA\overset{I}{⟶}B\overset{II}{⟶}C\overset{III}{⟶}D\overset{IV}{→}E and the rate as

Species formed Rate of its information
BB 0.002mol/h0. 002mol / h per mole of AA
CC 0.030mol/h0. 030mol / h per mole of BB
DD 0.011mol/h0. 011mol / h per mole of CC
EE 0.420mol/h0. 420mol / h per mole of DD

The rate determining step is

Options

  1. Astep II
  2. Bstep IIII
  3. Cstep IIIIII
  4. Dstep IVIV

Correct answer

A. step II

Step-by-step solution

Among the formation of B,C,DandEB, C, D and E, the slowest step is the rate determining step. As formation of BB, i.e. 0.002mol/h0 . 002mol / h per mole of AA is the slowest step. Hence, step-(I)\left(I\right) is the rate determining step.
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