Reaction Rates: Study Guide (Chemistry 5.1)

What a reaction rate measures

Some reactions are over in a fraction of a second, like a precipitate forming when two solutions are mixed. Others take years, like iron rusting. Chemical kinetics is the study of how fast reactions happen and what controls that speed.

The rate of a reaction is how quickly reactants are turned into products. You measure it as the change in the amount of a substance divided by the time taken:

rate = change in concentration ÷ change in time = Δ[X] / Δt

Square brackets mean concentration in mol L⁻¹ (M). So the usual unit of rate is M s⁻¹ (mol L⁻¹ s⁻¹). You may also see M min⁻¹, or, for a gas or a solid, mL s⁻¹ or g s⁻¹ when the measurement is a volume or a mass.

Two signs need care:

• A product is formed, so its concentration rises and Δ[product] is positive. Rate of formation = +Δ[product] / Δt.
• A reactant is used up, so its concentration falls and Δ[reactant] is negative. Rate of disappearance = −Δ[reactant] / Δt. The minus sign makes the rate positive.

A rate is always reported as a positive number. If your answer is negative, you have left out the minus sign for a reactant.

Rates of different species: use the coefficients

In one reaction, every species changes at a rate set by the balanced equation. Take the decomposition of hypochlorite ions in warm bleach:

3 ClO⁻(aq) → ClO₃⁻(aq) + 2 Cl⁻(aq)

Each time the reaction happens once, 3 ClO⁻ ions are used and 1 ClO₃⁻ ion and 2 Cl⁻ ions are made. So in any time interval:

• ClO⁻ disappears three times as fast as ClO₃⁻ forms.
• Cl⁻ forms twice as fast as ClO₃⁻ forms.

Chemists remove this ambiguity by defining one rate of reaction: divide each species’ rate by its coefficient. For a general reaction a A + b B → c C + d D:

rate of reaction = −(1/a) Δ[A]/Δt = −(1/b) Δ[B]/Δt = (1/c) Δ[C]/Δt = (1/d) Δ[D]/Δt

Every term gives the same number, so it does not matter which species you measure. When a question asks for “the rate”, check whether it means the rate of one species or the rate of reaction.

Average rate and instantaneous rate

Rates are rarely constant. In most reactions the rate is fastest at the start, when reactant concentrations are highest, and slows as the reactants are used up. Figure 1 shows this for a reaction A → 2 B.

There are two ways to describe the rate from a graph or table like this:

• Average rate over an interval = the change in concentration ÷ the length of the interval. On the graph it is the slope of the straight line (secant) joining two points on the curve. For A from 0 to 100 s: [A] falls from 0.800 M to 0.294 M, so the average rate of disappearance is 0.506 M ÷ 100 s = 0.00506 M s⁻¹.
• Instantaneous rate at one moment = the slope of the tangent to the curve at that moment, with the sign made positive for a reactant. At 50 s the tangent has slope −0.00485 M s⁻¹, so A is disappearing at 0.00485 M s⁻¹ at that instant.

The initial rate is the instantaneous rate at t = 0. It matters in Topic 5.2, because at the very start you know exactly what the concentrations are.

A curve that flattens out tells you the rate is falling. When the reactant curve becomes horizontal, the rate has dropped to zero: the limiting reactant is used up (or the reaction has reached equilibrium).

How rates are measured in the lab

To find a rate you need a property that changes as the reaction goes on and that you can measure many times without stopping the reaction. The choice depends on the reaction.

What changes	What you measure	Typical reaction
A gas is given off	Volume of gas in a gas syringe or upturned measuring cylinder	A metal or a carbonate with acid; catalysed decomposition of hydrogen peroxide
A gas escapes from an open flask	Mass of the flask on a balance	A carbonate with acid
A coloured species forms or is used up	Absorbance in a spectrophotometer, or the time until a colour appears	Fading of a dye; formation of iodine
A precipitate forms	Time until a mark under the flask is hidden	Thiosulfate with acid
Ions are used up or made	Electrical conductivity or pH	Reactions in which H⁺ or OH⁻ is consumed
The number of gas molecules changes	Total pressure in a sealed container	Gas-phase decompositions

In every case you convert the measurement into amounts or concentrations with stoichiometry, then divide by time.

What changes the rate

A reaction happens when reactant particles collide with enough energy and in a suitable orientation. Topic 5.5 builds this collision model in detail. For now, use it to connect what you see in the beaker (the macroscopic scale) to what particles are doing (the particulate scale).

Change	Effect on rate	Particle-level reason
Higher concentration of a reactant in solution (or higher partial pressure of a gas)	Usually faster	More particles in each unit of volume, so collisions happen more often
Higher temperature	Faster	Particles move faster, so they collide more often, and a larger fraction of collisions have enough energy to react
Larger surface area of a solid (powder instead of lumps)	Faster	Only particles at the surface can be hit; breaking a solid up exposes more of them
Adding a catalyst	Faster	The catalyst gives a different route with a lower energy barrier; it is not used up (Topic 5.11)
Other conditions, such as the solvent or, for some reactions, light	Can be faster or slower	They change how easily particles meet or how much energy is available

Two points help in written answers:

• Concentration applies to dissolved substances and gases, not to pure solids or liquids. Adding more lumps of marble does not make each lump react faster. What matters for a solid is how much surface is exposed.
• Temperature has two effects. Faster particles collide more often, but the bigger effect is that many more collisions have enough energy. That is why a 10 °C rise can roughly double the rate of many reactions, far more than the small rise in collision frequency could explain.

Worked example 1: rates of different species

Question. In a warm bleach solution, hypochlorite decomposes: 3 ClO⁻(aq) → ClO₃⁻(aq) + 2 Cl⁻(aq). At one moment, ClO⁻ is disappearing at 6.0 × 10⁻⁵ M s⁻¹ (an invented value for practice). Find (a) the rate of formation of ClO₃⁻, (b) the rate of formation of Cl⁻ and (c) the rate of reaction.

1. Write the ratio of coefficients: ClO⁻ : ClO₃⁻ : Cl⁻ = 3 : 1 : 2.
2. (a) ClO₃⁻ forms one-third as fast as ClO⁻ disappears: 6.0 × 10⁻⁵ M s⁻¹ × (1/3) = 2.0 × 10⁻⁵ M s⁻¹.
3. (b) Cl⁻ forms two-thirds as fast as ClO⁻ disappears: 6.0 × 10⁻⁵ M s⁻¹ × (2/3) = 4.0 × 10⁻⁵ M s⁻¹.
4. (c) Rate of reaction = −(1/3) Δ[ClO⁻]/Δt = (1/3)(6.0 × 10⁻⁵) = 2.0 × 10⁻⁵ M s⁻¹. Check with Cl⁻: (1/2)(4.0 × 10⁻⁵) = 2.0 × 10⁻⁵ M s⁻¹. The same.

Check. The species with the biggest coefficient (ClO⁻) has the biggest rate. If your Cl⁻ rate came out larger than the ClO⁻ rate, you multiplied by 3/2 instead of 2/3.

Worked example 2: an average rate from mass-loss data

Question. A student adds marble chips (calcium carbonate, in excess) to 50.0 mL of hydrochloric acid in an open flask on a balance. Carbon dioxide escapes, so the mass falls. The invented results are:

Time (s)	0	30	60	120	180
Mass of flask and contents (g)	152.40	151.96	151.64	151.24	151.02

CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

(a) Calculate the average rate of loss of mass over the first 60 s and over the next 60 s. (b) Convert the first rate into the average rate at which HCl is used up, in mol s⁻¹ and in M s⁻¹. (c) Explain why the rate falls.

1. (a) Mass lost from 0 to 60 s = 152.40 − 151.64 = 0.76 g. Average rate = 0.76 g ÷ 60 s = 0.0127 g s⁻¹. From 60 to 120 s: 151.64 − 151.24 = 0.40 g, so 0.40 g ÷ 60 s = 0.0067 g s⁻¹.
2. (b) The mass lost is CO₂ (M = 44.01 g mol⁻¹). Moles of CO₂ per second = 0.01267 g s⁻¹ ÷ 44.01 g mol⁻¹ = 2.88 × 10⁻⁴ mol s⁻¹. From the equation, 2 mol HCl are used for each 1 mol CO₂, so HCl is used at 2 × 2.88 × 10⁻⁴ = 5.76 × 10⁻⁴ mol s⁻¹. In 50.0 mL (0.0500 L) of solution: 5.76 × 10⁻⁴ mol s⁻¹ ÷ 0.0500 L = 0.0115 M s⁻¹. This is the average rate of disappearance of HCl, −Δ[HCl]/Δt.
3. (c) The rate in the second minute is about half the rate in the first. As the acid reacts, [HCl] falls, so H⁺ ions hit the surface of the chips less often. (The chips also get smaller, so a little less surface is exposed.) Fewer successful collisions each second means a lower rate.

Interpretation. The mass readings change least between 120 and 180 s (0.22 g), so the rate is still falling. The total CO₂ lost by 180 s is 1.38 g, or 0.0314 mol. When the mass stops changing, the acid is used up, because the marble is in excess.

Common misconceptions

• “The rate is the same number for every species.” Only the rate of reaction is. The rates of the individual species differ by the ratio of their coefficients.
• “A rate can be negative.” Concentration changes can be negative, but rates are reported as positive. Use −Δ[reactant]/Δt.
• “The rate stays the same all the way through.” For most reactions the rate falls as reactants are used up. That is why an average rate depends on which interval you choose.
• “A faster reaction gives more product.” Rate controls how soon product appears, not how much appears in the end.
• “Adding more of a solid increases its concentration.” A pure solid has no concentration in solution. More lumps add surface, but the rate per unit of surface is unchanged.
• “A catalyst is used up in the reaction.” A catalyst takes part but is regenerated, so the same amount is present at the end.
• “Heating speeds a reaction only because particles collide more often.” The bigger effect is that more collisions have enough energy to react.
• Mixing up the units of the graph with the equation. A rate from a concentration–time graph is in M s⁻¹. Coefficients from the equation are pure numbers that compare rates; they are not part of the units.

Where this leads

This topic links back to the stoichiometry and reaction types of Unit 4, including Topic 4.9, Oxidation–Reduction (Redox) Reactions. Next, Topic 5.2, Introduction to Rate Law, turns “higher concentration means faster” into an equation, rate = k[A]ᵐ[B]ⁿ, and shows how initial rates reveal the powers. Later topics explain the temperature and catalyst effects with the collision model, energy profiles and mechanisms. Try the practice questions now, then use the revision notes and the checklist to consolidate.