AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/5/5-2)
Unit 5 · Topic 5.2
5.2 Introduction to Rate Law
A rate law has the form rate = k[A]ᵐ[B]ⁿ, where the orders m and n are found by experiment, often by comparing initial rates while changing one concentration at a time. The rate constant k depends on temperature, and its units depend on the overall order.
Key terms
- rate law
- rate constant (k)
- reaction order
- overall order
- method of initial rates
The rate law
A rate law shows how the rate depends on reactant concentrations: rate = k[A]ᵐ[B]ⁿ. k is the rate constant. The exponents m and n are the orders with respect to A and B. The overall order is m + n.
The orders come from experiments, not from the coefficients in the overall balanced equation. For 2NO + O₂ → 2NO₂, the coefficients happen to match the measured rate law, but for many reactions they don't. Orders are usually 0, 1 or 2 in AP problems.
What each order means
Zero order means the rate doesn't depend on that reactant's concentration at all, as long as some is present. It doesn't mean the reactant isn't involved; it may react in a fast step after the slow one.
| Order in A | If [A] doubles, the rate… | If [A] triples, the rate… |
|---|---|---|
| 0 | stays the same | stays the same |
| 1 | doubles | triples |
| 2 | quadruples (×4) | goes up ×9 |
The method of initial rates
- Run the reaction several times, measuring the initial rate each time.
- Find two experiments where only [A] changes. The factor the rate changes by equals (factor [A] changes by)ᵐ. Solve for m.
- Do the same for B using two experiments where only [B] changes.
- Write the rate law, then plug one experiment's data in to calculate k.
- If no pair of experiments changes only one concentration, use the order you already know to divide out that reactant's effect first.
The rate constant
k is constant only at a fixed temperature. Raising the temperature increases k (topic 5.6), which is why reactions go faster when heated even at the same concentrations. Changing concentration changes the rate, but not k.
The units of k make the rate come out in M/s, so they depend on the overall order.
| Overall order | Units of k |
|---|---|
| 0 | M/s (M·s⁻¹) |
| 1 | s⁻¹ |
| 2 | M⁻¹s⁻¹ |
| 3 | M⁻²s⁻¹ |
When the numbers aren't neat
Concentrations don't always double or triple. The rule still works: rate ratio = (concentration ratio)ᵐ. If [A] goes up by a factor of 1.5 and the rate goes up by a factor of 2.25, then 1.5ᵐ = 2.25, so m = 2, because 1.5² = 2.25.
When you can't spot the answer, take logs: m = log(rate ratio) ÷ log(concentration ratio). Round to the nearest whole number (or simple fraction) that fits the data, since experimental rates always have some error.
Once you know the rate law, you can also go the other way and predict the rate for any set of concentrations, as long as the temperature stays the same.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Method of initial rates
For A + B → C, these initial rates were measured. Exp 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ M/s. Exp 2: [A] = 0.20 M, [B] = 0.10 M, rate = 8.0 × 10⁻³ M/s. Exp 3: [A] = 0.10 M, [B] = 0.30 M, rate = 6.0 × 10⁻³ M/s. Find the rate law and k, with units.
Show the solutionHide the solution
- Step 1: Exp 1 → 2: [A] doubles, [B] constant. The rate goes from 2.0 × 10⁻³ to 8.0 × 10⁻³, ×4. 2ᵐ = 4, so m = 2.
- Step 2: Exp 1 → 3: [B] triples, [A] constant. The rate triples. 3ⁿ = 3, so n = 1.
- Step 3: Rate law: rate = k[A]²[B]. Overall order 3.
- Step 4: From Exp 1: k = rate ÷ ([A]²[B]) = 2.0 × 10⁻³ ÷ [(0.10)²(0.10)] = 2.0 M⁻²s⁻¹.
Answer: rate = k[A]²[B], k = 2.0 M⁻²s⁻¹
- Example 2Calculator allowed
When both concentrations change (classic trap)
Using the rate law from the previous example (rate = 2.0 M⁻²s⁻¹ [A]²[B]), a student runs an experiment with [A] = 0.20 M and [B] = 0.20 M and says the rate should double because both concentrations doubled. Find the correct rate.
Show the solutionHide the solution
- Step 1: Doubling [A] multiplies the rate by 2² = 4. Doubling [B] multiplies it by 2¹ = 2.
- Step 2: Combined factor = 4 × 2 = 8 times Exp 1's rate.
- Step 3: Rate = 8 × 2.0 × 10⁻³ = 1.6 × 10⁻² M/s. Check directly: 2.0 × (0.20)² × 0.20 = 0.016 M/s.
Answer: 1.6 × 10⁻² M/s (8 times the original rate, not 2 times)
- Example 3Calculator allowed
Predicting a new rate
For the same reaction, predict the initial rate when [A] = 0.30 M and [B] = 0.20 M.
Show the solutionHide the solution
- Step 1: rate = k[A]²[B] = (2.0 M⁻²s⁻¹)(0.30 M)²(0.20 M).
- Step 2: = 2.0 × 0.090 × 0.20 = 0.036 M/s.
Answer: 3.6 × 10⁻² M/s
Common mistakes
- Taking orders from the coefficients of the overall equation. Orders must come from data (or, for an elementary step, from its coefficients).
- Thinking k changes when concentration changes. Only temperature (or a catalyst) changes k.
- Getting k's units wrong. Work them out from rate ÷ concentration terms.
- Comparing two experiments where both concentrations change without accounting for the one you already know.
On the exam
- Expect a table of initial-rate data on the free-response section: determine each order, write the rate law and calculate k with units. Show which experiments you compared.
- Multiple-choice questions often ask what happens to the rate when a concentration changes by some factor. Raise the factor to the power of the order.
Connected topics
Videos
Check yourself
4 questions on 5.2 Introduction to Rate Law. Pick an answer to see if you got it, and why.
| Experiment | Initial [A] (M) | Initial [B] (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.10 | 0.30 | 6.0 × 10⁻⁴ |
Hypothetical data. Initial-rate data for the reaction A + B → C at constant temperature
Which of the following is the rate law for the reaction?
What is the value of the rate constant, k, including units?
What would the initial rate be if [A] = 0.30 M and [B] = 0.20 M?
The rate law for a reaction is rate = k[X][Y], with the rate measured in M/s. What are the units of k?
0 of 4 answered