AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/5/5-1)
Unit 5 · Topic 5.1
5.1 Reaction Rates
Reaction rate is how fast reactants are used up or products are made, measured as a change in concentration per unit time. The coefficients in the balanced equation link the rates of the different species, and rates depend on concentration, temperature, surface area and catalysts.
Key terms
- reaction rate
- rate of disappearance
- rate of appearance
- surface area
- catalyst
What reaction rate means
Kinetics is the study of how fast reactions happen. The rate of a reaction is the change in concentration of a reactant or product per unit time, usually in mol/(L·s), written M/s.
Reactant concentrations go down over time, so Δ[reactant] is negative. Rates are reported as positive numbers, so the rate of disappearance of a reactant is written with a minus sign: −Δ[A]/Δt. The rate of appearance of a product is +Δ[P]/Δt.
An average rate is the change over a time interval. An instantaneous rate is the rate at one moment: on a graph of concentration against time, it's the slope of the tangent line at that point. The initial rate is the instantaneous rate at time zero.
Linking rates with coefficients
In 2N₂O₅(g) → 4NO₂(g) + O₂(g), every 2 N₂O₅ that react make 4 NO₂ and 1 O₂. So NO₂ appears twice as fast as N₂O₅ disappears, and O₂ appears half as fast as N₂O₅ disappears.
To get a single rate for the reaction that doesn't depend on which species you watch, divide each species' rate by its coefficient: rate = −½ Δ[N₂O₅]/Δt = ¼ Δ[NO₂]/Δt = Δ[O₂]/Δt.
What changes the rate
The table lists the four main factors. Other conditions, like the solvent or light, can also matter for particular reactions. Topics 5.5 and 5.11 explain the temperature and catalyst effects in detail.
| Factor | Effect | Particle-level reason |
|---|---|---|
| Higher reactant concentration | usually faster | more particles per volume, so more frequent collisions |
| Higher temperature | faster | particles move faster, and a larger fraction of collisions have enough energy to react |
| More surface area of a solid | faster | more particles exposed, so more collisions per second |
| Catalyst | faster | provides a pathway with a lower activation energy |
Measuring a rate
You can't count particles directly, so experiments track a property that depends on concentration: the color of a solution with a spectrophotometer (topic 3.13), the volume or pressure of a gas produced, or the mass lost as a gas escapes. Readings are taken at regular times to build a concentration-versus-time graph.
For most reactions, the rate is highest at the start and slows as the reactants get used up, so the concentration-versus-time curve gets flatter over time.
To estimate an instantaneous rate from a graph, draw a straight line that just touches the curve at the time you care about, then find that line's slope (rise over run). A steeper tangent means a faster rate. Comparing tangents at the start and later on shows the reaction slowing down.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Relating rates with coefficients
For 2N₂O₅(g) → 4NO₂(g) + O₂(g), N₂O₅ is disappearing at 0.0240 M/s at a certain moment. At what rates are NO₂ and O₂ appearing?
Show the solutionHide the solution
- Step 1: NO₂: 4 NO₂ per 2 N₂O₅, so its rate is 0.0240 × 4/2 = 0.0480 M/s.
- Step 2: O₂: 1 O₂ per 2 N₂O₅, so its rate is 0.0240 × 1/2 = 0.0120 M/s.
- Step 3: Check with the single reaction rate: ½(0.0240) = ¼(0.0480) = 0.0120 M/s.
Answer: NO₂ appears at 0.0480 M/s; O₂ appears at 0.0120 M/s.
- Example 2
Average rate from data
The concentration of a reactant falls from 0.500 M to 0.320 M in 60.0 s. What is the average rate of disappearance over that interval?
Show the solutionHide the solution
- Step 1: Δ[A] = 0.320 − 0.500 = −0.180 M.
- Step 2: Rate of disappearance = −Δ[A]/Δt = 0.180 M ÷ 60.0 s = 3.00 × 10⁻³ M/s.
Answer: 3.00 × 10⁻³ M/s
- Example 3
Surface area (classic trap)
Equal masses of calcium carbonate react with the same volume of the same HCl solution. One sample is a single marble chip; the other is a fine powder. Which reacts faster, and does the powder produce more CO₂ in total?
Show the solutionHide the solution
- Step 1: The powder has far more surface exposed, so more acid particles collide with CaCO₃ each second. The powder reacts faster.
- Step 2: The total amount of CO₂ depends on the amount of CaCO₃ (and acid), not on particle size. With equal masses, and enough acid to react with all of it, both samples eventually make the same amount of CO₂.
- Step 3: Rate is about how fast, not how much.
Answer: The powder reacts faster, but both produce the same total amount of CO₂.
Common mistakes
- Giving the rate of disappearance a negative sign. Rates are reported as positive values.
- Assuming every species is used up or formed at the same rate. Use the coefficients.
- Confusing rate with yield: a faster reaction doesn't make more product.
- Saying higher temperature speeds a reaction only because collisions are more frequent. The bigger effect is that more collisions have enough energy.
On the exam
- Expect to relate the rate of one species to another using coefficients, and to explain a rate change at the particle level using collisions.
- Lab questions may ask how changing surface area, concentration or temperature would affect a measured rate. State the direction and the particle-level reason.
Connected topics
Videos
Check yourself
5 questions on 5.1 Reaction Rates. Pick an answer to see if you got it, and why.
For the reaction 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), O₂ is being consumed at a rate of 0.25 M/s at a certain moment. At what rate is H₂O being produced at that moment?
Equal masses of zinc are added to two beakers of 1.0 M HCl at the same temperature. One sample is a single lump and the other is a fine powder. The powder produces hydrogen gas much faster. Which of the following best explains this?
At constant temperature, increasing the concentration of a reactant in solution usually increases the reaction rate. Which of the following best explains this effect?
| Time (s) | [N₂O₅] (M) |
|---|---|
| 0 | 0.0200 |
| 400 | 0.0100 |
| 800 | 0.00500 |
| 1,200 | 0.00250 |
Hypothetical data for the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g) at constant temperature
What is the average rate of disappearance of N₂O₅ between 0 s and 400 s?
What is the average rate of appearance of NO₂ over the same interval?
0 of 5 answered