Read the passage given below and answer the questions based on it.
The rate of a reaction is concerned with decrease in concentration of reactants or increase in concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants and catalyst affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. Molecularity is defined only for an elementary reaction. Molecularity and order of an elementary reaction are the same.
Which of the following expressions is correct for the rate of the reaction given below?
\(H_2O_2 (aq) + 3I^- (aq) + 2H^+ \longrightarrow 2H_2O (l) + I_3^-\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{\Delta [I^-]}{\Delta t} = \frac{3}{2}\frac{[\Delta H^+]}{\Delta t}\)
When species in a reaction have different stoichiometric coefficients, the rate of change of each concentration is divided by its coefficient, so that the rate of reaction has a single value:
\( \displaystyle \ce{H2O2(aq) + 3I-(aq) + 2H+ -> 2H2O(l) + I3-} \)
\( \displaystyle \text{Rate} = -\frac{\Delta[\ce{H2O2}]}{\Delta t} = -\frac{1}{3}\frac{\Delta[\ce{I-}]}{\Delta t} = -\frac{1}{2}\frac{\Delta[\ce{H+}]}{\Delta t} = \frac{\Delta[\ce{I3-}]}{\Delta t} \)
Taking the \( \ce{I^{-}} \) and \( \ce{H^{+}} \) terms:
\( \displaystyle \frac{1}{3}\frac{\Delta[\ce{I-}]}{\Delta t} = \frac{1}{2}\frac{\Delta[\ce{H+}]}{\Delta t} \;\Rightarrow\; \frac{\Delta[\ce{I-}]}{\Delta t} = \frac{3}{2}\frac{\Delta[\ce{H+}]}{\Delta t} \)
Why the other options are incorrect
Option (1) — Incorrect. It uses only the coefficient of \( \ce{I^{-}} \) and ignores that of \( \ce{H^{+}} \).
Option (2) — Incorrect. The ratio is inverted; \( \ce{I^{-}} \) (3 mol) is used up faster than \( \ce{H^{+}} \) (2 mol), not slower.
Option (4) — Incorrect. It uses only the coefficient of \( \ce{H^{+}} \).