210 ❯ STEP 4. Review the Knowledge You Need to Score High
For steps 2 and 3, it is necessary to replace the intermediates with reactants. Step 2
gives a rate law matching the one derived in part a.
To see if the stoichiometry is correct, simply add the three steps together and cancel
the intermediates (materials that appear on both sides of the reaction arrow).
Step 1: ClO 2 + ClO 2 → Cl 2 O 4
Step 2: Cl 2 O 4 + OH– → ClO 3 – + HClO 2
Step 3: HClO 2 + OH– → ClO 2 – + H 2 O
Total: 2 ClO 2 + Cl 2 O 4 + 2 OH– + HClO 2 →
Cl 2 O 4 + ClO 3 – + HClO 2 + ClO 2 – + H 2 O
After removing the intermediates (Cl 2 O 4 and HClO 2 ):
2 ClO 2 + 2 OH– → ClO 3 – + ClO 2 – + H 2 O
As this matches the original reaction equation, the mechanism fulfills the overall
stoichiometry requirement.
Give yourself 1 point if you picked step 2, or if you picked a step with a rate law
that matches your wrong answer for part a. Give yourself 1 more point if you explained
the substitution of reactants for intermediates. Give yourself 1 point for summing the
equations and proving the overall equation is consistent.
Total your points. There are 10 points possible. Subtract 1 point if any numerical answer
has an incorrect number of significant figures.❯ Rapid Review
● Kinetics is a study of the speed of a chemical reaction.
● The five factors that can affect the rates of chemical reaction are the nature of the
reactants, the temperature, the concentration of the reactants, the physical state of the
reactants, and the presence of a catalyst.
● The rate equation relates the speed of reaction to the concentration of reactants and has
the form: Rate = k[A ]m[B ]n... where k is the rate constant and m and n are the orders
of reaction with respect to that specific reactant.
● The rate law must be determined from experimental data. Review how to determine the
rate law from kinetics data.
● When mathematically comparing two experiments in the determination of the rate equa-
tion, be sure to choose two in which all reactant concentrations except one remain constant.
● Rate laws can be written in the integrated form.
● If a reaction is first order, it has the rate law of Rate = k[A ]; ln [A ]t – ln [A ] 0 = –kt; a
plot of ln[A] versus time gives a straight line.
● If a reaction is second order, it has the form of Rate=−kk[A];^1 = t
[A]1
[A]
2
t0(integratedrate law); a plot of1
[A]
versus time gives a straight line.● The reaction half-life is the amount of time that it takes the reactant concentration to
decrease to one-half its initial concentration.