Unit 5 · Topic 5.2 Beta

Introduction to Rate Law

A rate law, rate = k[A]^m[B]^n, gives how the rate depends on each reactant concentration.

Practice 5: Mathematical Routines

Part 1 · Hook

Why this matters

Double the amount of bleach in a stain remover and it might work twice as fast, four times as fast, or no faster at all. Which one happens cannot be read from the label or from the balanced equation. Chemists find out by experiment, and they write the answer as a rate law.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If 2ᵐ = 4, what is m?

Show the answer

2² = 4, so m = 2.

Correct: 2:
4:
1:
8:

2. For 2 SO₂ + O₂ → 2 SO₃, SO₂ is used up at 0.40 M/s. How fast is O₂ used up?

0.20 M/s
0.40 M/s
0.80 M/s
0.10 M/s

Show the answer

One O₂ reacts for every 2 SO₂, so O₂ is used half as fast: 0.20 M/s.

Correct: 0.20 M/s:
0.40 M/s:
0.80 M/s:
0.10 M/s:

Part 4 · See it

See it first

Three small graphs of rate against the concentration of a reactant A. Zero order: a flat line, so doubling [A] leaves the rate the same. First order: a straight line through the origin, so doubling [A] doubles the rate. Second order: an upward curve, so doubling [A] multiplies the rate by four. Dashed lines mark one concentration and twice that concentration on each graph. — Doubling a concentration leaves a zero-order rate unchanged, doubles a first-order rate and quadruples a second-order rate. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Each trial changes only one starting concentrationany change in the initial rate is caused by that reactant alone
The rate changes by (concentration factor) raised to the ordercomparing the two factors gives the order for that reactant
With every order knownany one trial gives the rate constant k, whose units follow from the overall order
k is fixed at a given temperaturethe rate law predicts the rate for any new set of concentrations

Part 6 · Key ideas

Key ideas

A rate law has the form rate = k[A]^m[B]ⁿ. The orders m and n come from experiment, not from the coefficients.
Order 0: rate unchanged when [A] changes. Order 1: rate changes by the same factor. Order 2: rate changes by the factor squared.
Method of initial rates: compare two trials where only one concentration changes.
k depends on the reaction and the temperature, not on concentration. Its units are M/s divided by M raised to the overall order.

Part 7 · Misconception

A common mistake

The wrong idea: The orders in a rate law are the coefficients of the balanced equation.

What actually happens: Orders are measured. The balanced equation gives the total amounts that react, not how the rate depends on each concentration.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Initial rates for making nitrosyl chloride

Nitrogen monoxide and chlorine react at a fixed temperature:

2 NO(g) + Cl₂(g) → 2 NOCl(g)

Three trials start with different concentrations, and the initial rate of formation of NOCl is measured in each.

Initial concentrations and initial rates
Trial	[NO] (M)	[Cl₂] (M)	Initial rate (M/s)
1	0.0100	0.0100	1.80 × 10⁻⁴
2	0.0200	0.0100	7.20 × 10⁻⁴
3	0.0100	0.0200	3.60 × 10⁻⁴

1. What is the order of the reaction with respect to NO?

Zero order
First order
Second order
Third order

Show the answer

Compare trials 1 and 2, where only [NO] changes: [NO] doubles and the rate goes from 1.80 × 10⁻⁴ to 7.20 × 10⁻⁴ M/s, a factor of 4. Since 2ᵐ = 4, m = 2.

Zero order: Zero order would leave the rate unchanged when [NO] changes.
First order: First order would double the rate when [NO] doubles; here it quadruples.
Correct: Second order: Right: doubling [NO] multiplies the rate by 4 = 2².
Third order: Third order would multiply the rate by 2³ = 8.

2. What is the order of the reaction with respect to Cl₂?

Zero order
Half order
First order
Second order

Show the answer

Compare trials 1 and 3, where only [Cl₂] changes: it doubles and the rate doubles (1.80 → 3.60 × 10⁻⁴ M/s). 2ⁿ = 2 gives n = 1.

Zero order: Zero order would leave the rate unchanged; it doubled.
Half order: Half order would multiply the rate by about 1.4, not 2.
Correct: First order: Right: the rate changes by the same factor as [Cl₂].
Second order: Second order would quadruple the rate when [Cl₂] doubles.

3. Using trial 1, calculate the rate constant k. Include units.

Type a number and its unit.

Show the answer

rate = k[NO]²[Cl₂], so k = rate / ([NO]²[Cl₂]) = (1.80 × 10⁻⁴ M/s) / ((0.0100 M)²(0.0100 M)) = (1.80 × 10⁻⁴ M/s) / (1.00 × 10⁻⁶ M³) = 1.80 × 10² M⁻² s⁻¹. The units are M/s ÷ M³ = M⁻² s⁻¹.

Answer: 180 M^-2 s^-1

Data table

A reaction that ignores one reactant

Two dissolved reactants, X and Y, react at 25 °C. Initial rates of disappearance of X are measured.

Initial concentrations and rates for X + Y → products
Trial	[X] (M)	[Y] (M)	Initial rate (M/s)
1	0.15	0.15	2.4 × 10⁻⁵
2	0.30	0.15	4.8 × 10⁻⁵
3	0.15	0.45	2.4 × 10⁻⁵

4. Which rate law agrees with the data?

rate = k[X]
rate = k[X][Y]
rate = k[Y]
rate = k[X]²

Show the answer

Trials 1 and 2: [X] doubles, rate doubles, so first order in X. Trials 1 and 3: [Y] triples, rate is unchanged, so zero order in Y ([Y]⁰ = 1). rate = k[X].

Correct: rate = k[X]: Right: first order in X, zero order in Y.
rate = k[X][Y]: Tripling [Y] would triple the rate; the data show no change.
rate = k[Y]: The rate does not change with [Y] but does change with [X].
rate = k[X]²: Doubling [X] would quadruple the rate; it only doubles.

5. For rate = k[A][B]², what are the units of k when concentrations are in M and time in s?

M s⁻¹
s⁻¹
M⁻¹ s⁻¹
M⁻² s⁻¹

Show the answer

The overall order is 1 + 2 = 3. k = rate/([A][B]²) has units (M/s)/M³ = M⁻² s⁻¹.

M s⁻¹: These are the units of a rate, which is also the units of k for zero order.
s⁻¹: These are the units for a first-order rate law.
M⁻¹ s⁻¹: These are the units for a second-order rate law.
Correct: M⁻² s⁻¹: Right: third order overall gives M⁻² s⁻¹.

6. For the rate law rate = k[NO₂]², the concentration of NO₂ is doubled at constant temperature. Predict each change.

Variable	Change
Initial rate	—
Rate constant k	—
Overall order of the reaction	—

Show the answer

The rate goes up by 2² = 4. The rate constant and the order stay the same: k changes only with temperature (or a new pathway).

Initial rate: increases. The rate depends on [NO₂]², so doubling [NO₂] multiplies the rate by 4.
Rate constant k: no change. k depends on temperature and the reaction, not on concentration.
Overall order of the reaction: no change. The order is a property of the rate law (2 here); it does not change with concentration.

Part 9 · Summary

Summary

A rate law, rate = k[A]^m[B]^n, gives how the rate depends on each reactant concentration. The orders come from initial-rate data by changing one concentration at a time. The rate constant k is fixed at a given temperature, and its units depend on the overall order.

Part 10 · Up next

What comes next

Part 11 · Connections