What is the rate of a reaction?
Some reactions are very fast (a firework), some are slow (iron rusting). The rate tells us how fast.
Mean rate = amount of reactant used ÷ time, or amount of product made ÷ time.
Units depend on what you measure: g/s (mass), cm³/s (gas volume) or mol/dm³/s (concentration). In SI, 1 dm³ = 1 L = 10⁻³ m³.
How to measure the rate
- Gas volume: collect the gas in a gas syringe or an upside-down measuring cylinder. Read the volume every 10 s.
- Mass loss: put the flask on a balance. If a gas escapes, the mass goes down. Use cotton wool in the neck so liquid does not spit out.
- Disappearing cross / colour change: time how long until a cross under the flask can no longer be seen (a cloudy solid forms). Rate ∝ 1 ÷ time.
Keep every other condition the same (a fair test): only change one thing at a time.
Reading rate graphs
Plot amount of product (y) against time (x). The curve is steep at the start (fast), gets less steep as reactants are used up, and finally goes flat when the reaction stops.
A steeper line means a faster rate. The mean rate between two times = change in y ÷ change in x.
The rate at one instant = the gradient of the tangent drawn at that point: pick two points on the tangent and divide the rise by the run.
If the same amounts of reactant are used, a faster reaction levels off sooner but at the same final height.
Collision theory and activation energy
For particles to react, they must collide, and the collision must have at least a minimum energy called the activation energy (Eₐ). Weaker collisions just bounce off.
So the rate depends on two things:
- how often particles collide (collision frequency), and
- what fraction of the collisions have enough energy.
Anything that increases either of these makes the reaction faster.
Factors that change the rate
- Concentration (solutions) or pressure (gases): more particles in the same space → more frequent collisions.
- Temperature: particles move faster → more frequent collisions, and a much larger fraction have energy ≥ Eₐ. A rise of about 10 °C often roughly doubles the rate.
- Surface area of a solid: cutting it smaller exposes more particles → more frequent collisions. Powder reacts faster than a lump of the same mass.
- Catalyst: provides another pathway with a lower activation energy → a larger fraction of collisions succeed. It is not used up and is not part of the overall equation.
- Light speeds up some reactions (photosynthesis, silver halides in old photo film).
Catalysts and energy profiles
An energy profile shows energy (y) as the reaction goes on (x). Reactants must climb a hill (Eₐ) before they reach the products. A catalyst makes a lower hill. The start and end energies stay the same.
Examples: iron in making ammonia, platinum and rhodium in a car's catalytic converter, and enzymes, the biological catalysts in living things (like amylase in saliva).
Catalysts save energy and money in industry because reactions can run fast at lower temperatures.
Try it: tablets in water
Drop a fizzy (effervescent) tablet into cold water and another into warm water. Time how long each fizz lasts. Then crush one tablet and compare with a whole one in water at the same temperature. Predict first, then check. In the 3D free play, compare 'lump' with powder and see the graph slope change.
Key formulas and definitions
- Mean rate = amount of reactant used ÷ time taken
- Mean rate = amount of product formed ÷ time taken
- Rate at an instant = gradient of the tangent to the curve
- Rate ∝ 1 ÷ time (for 'time to finish' experiments)
Worked examples
1. A reaction gives 60 cm³ of gas in 30 s. Find the mean rate.
Rate = 60 ÷ 30 = 2 cm³/s.
2. A flask's mass falls from 200.00 g to 198.80 g in 120 s as CO₂ escapes. Find the mean rate.
Mass lost = 1.20 g. Rate = 1.20 ÷ 120 = 0.010 g/s.
3. Gas volume is 30 cm³ at 10 s and 50 cm³ at 30 s. Find the mean rate in this time.
Change = 20 cm³ in 20 s → 1 cm³/s.
4. A tangent to a gas-volume curve passes through (0 s, 10 cm³) and (40 s, 50 cm³). What is the rate at that point?
Gradient = (50 − 10) ÷ (40 − 0) = 40 ÷ 40 = 1 cm³/s.
5. In a disappearing-cross test the cross vanishes after 50 s. Give the rate as 1/time.
Rate = 1 ÷ 50 = 0.02 s⁻¹. A shorter time would mean a faster rate.
6. The concentration of acid falls from 0.12 mol/dm³ to 0.04 mol/dm³ in 40 s. Find the mean rate.
Change = 0.08 mol/dm³. Rate = 0.08 ÷ 40 = 0.002 mol/dm³/s.
7. A reaction takes 80 s at 20 °C. Using 'about double per 10 °C', estimate the time at 40 °C.
20 °C rise → rate × 2 × 2 = ×4. Time ≈ 80 ÷ 4 = 20 s.
Common mistakes
- Saying higher temperature only makes particles collide more often. The bigger reason is that more collisions have enough energy.
- Saying a catalyst gives particles more energy. It lowers the energy needed instead.
- Thinking a faster reaction makes more product. With the same reactants, it makes the same amount, just sooner.
- Mixing up surface area and size: smaller pieces have MORE total surface area.