The mathematical expression that relates the rate of a reaction to the concentrations of the reactants is known as the rate expression, rate equation or rate law.
For the reaction \(A + B \rightarrow \text{Products}\), the rate equation is
\[\text{Rate} \propto [A][B]\]
\[\text{Rate} = k[A][B]\]
k is known as the rate constant or specific reaction rate. The units of the rate constant are \((\text{mol})^{1-n}\,(\text{L})^{n-1}\,\text{s}^{-1}\), where n is the order of the reaction.
The rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically, but must be determined experimentally.
Several chemical reactions take place in a sequence of steps, and the overall rate of the reaction is governed by the slowest step, called the rate determining step.
Consider the following statements about the decomposition of hydrogen peroxide catalysed by iodide ion in an alkaline medium, \( \ce{2H2O2 -> 2H2O + O2} \), which takes place in two steps.
(A) The iodide-catalysed decomposition of H2O2 is first order with respect to both H2O2 and I−.
(B) In the iodide-catalysed decomposition of H2O2, the first step is a unimolecular elementary reaction.
(C) In the iodide-catalysed decomposition of H2O2, IO− is an intermediate, as it is formed during the course of the reaction but does not appear in the overall balanced equation.
(D) For a complex reaction, the molecularity of the slowest step is the same as the order of the overall reaction.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
The decomposition of H2O2 catalysed by I− is a complex reaction that takes place in two bimolecular elementary steps:
\( \displaystyle \ce{H2O2 + I^- -> H2O + IO^-} \quad \text{(slow)} \)
\( \displaystyle \ce{H2O2 + IO^- -> H2O + I^- + O2} \quad \text{(fast)} \)
The slow first step decides the rate, so the rate law is Rate = k[H2O2][I−].
(A) The iodide-catalysed decomposition of H2O2 is first order with respect to both H2O2 and I−. — True. Each concentration appears with the power one in the rate law.
(B) In the iodide-catalysed decomposition of H2O2, the first step is a unimolecular elementary reaction. — False. In the first step one H2O2 molecule and one I− ion collide, so the step is bimolecular. Both the steps are bimolecular elementary reactions.
(C) In the iodide-catalysed decomposition of H2O2, IO− is an intermediate, as it is formed during the course of the reaction but does not appear in the overall balanced equation. — True. IO− is made in the first step and used up in the second, so it cancels out when the two steps are added.
(D) For a complex reaction, the molecularity of the slowest step is the same as the order of the overall reaction. — True. Here the slow step is bimolecular (H2O2 and I−), and the order of the whole reaction is 1 + 1 = 2.
NCERT (2026-27 edition):
3.2.4 Molecularity of a Reaction – “The rate equation for this reaction is found to be”
\( \displaystyle \text{Rate} = \frac{-d[\ce{H2O2}]}{dt} = k[\ce{H2O2}][\ce{I^-}] \)
“This reaction is first order with respect to both H2O2 and I–. Evidences suggest that this reaction takes place in two steps”
\( \displaystyle \text{(1)}\ \ce{H2O2 + I^- -> H2O + IO^-} \)
\( \displaystyle \text{(2)}\ \ce{H2O2 + IO^- -> H2O + I^- + O2} \)
“Both the steps are bimolecular elementary reactions. Species IO- is called as an intermediate since it is formed during the course of the reaction but not in the overall balanced equation. The first step, being slow, is the rate determining step. Thus, the rate of formation of intermediate will determine the rate of this reaction.”
“(iii) For complex reaction, order is given by the slowest step and molecularity of the slowest step is same as the order of the overall reaction.”