Calculating Rates of Reaction
The rate of a reaction is how quickly reactants are turned into products. To measure it you track a quantity you can actually see changing — the mass of the reaction mixture (in g) or the volume of gas given off (in cm³) — and divide by the time taken.
mean rate of reaction = quantity of reactant usedtime taken
mean rate of reaction = quantity of product formedtime taken
The quantity can be a mass in g or a volume in cm³, so the units of rate are g/s or cm³/s. Higher Higher Tier students also use the quantity in moles, giving the unit mol/s.
Worked example — mean rate
Magnesium ribbon is added to dilute hydrochloric acid. The reaction produces 48 cm³ of hydrogen gas in the first 60 seconds.
- Write the equation: mean rate = quantity of product formedtime taken
- Substitute with units in place: mean rate = 48 cm³60 s
- Answer: 0.8 cm³/s. The unit does the checking for you — a volume divided by a time must come out in cm³/s.
Worked example — mean rate between two times
Most exam questions don’t start the clock at zero — they ask for the mean rate between two times read off a graph, often with a unit to convert. Reading a volume–time graph, the gas collected is 60 cm³ at 2 minutes and 84 cm³ at 4 minutes. Calculate the mean rate of reaction between these times, in cm³/s.
How the data usually appears in an exam: two readings taken off a volume–time curve. The mean rate between them is the change in volume (24 cm³) divided by the time (120 s).
- Change in quantity = 84 − 60 = 24 cm³ of gas formed in this interval.
- Time taken = 4 − 2 = 2 minutes. The answer is wanted per second, so convert first: 2 × 60 = 120 s.
- mean rate = 24 cm³120 s = 0.20 cm³/s (2 s.f.).
- Forget the minute–second conversion and you’d get 12 cm³/min — right arithmetic, wrong unit, no marks.
Examiners regularly see correct arithmetic lose a mark to a wrong or missing unit. Carry the unit through every line of working: if you divided cm³ by s, the rate is in cm³/s — not “cm³”, not “per second”, and not s/cm³. And before you substitute, check whether the units need converting: compare the units the question gives with the units it asks for — if the time is in minutes but the rate is wanted per second, change it to seconds first.
Rate graphs
A single mean rate hides the most interesting part of the story: the rate changes as the reaction runs. Plot the quantity of product formed (or reactant used) against time and the shape tells you everything:
- Steepest at the start — the rate is fastest at the beginning, when the concentrations of the reactants are highest.
- Gradient decreases — the reaction slows as the reactants are used up.
- Horizontal line — the reaction has finished: one of the reactants has run out, so no more product forms.
The same reaction seen two ways: product formed climbs to a plateau; reactant used up falls to one. The gradient is steepest at the start — the reaction is fastest when the reactant concentrations are highest — and zero when the reaction finishes.
Why the line goes flat — and how high
A rate curve flattens when the reaction finishes — and a reaction finishes the moment one reactant is completely used up. That reactant is the limiting reactant; whatever is left over is in excess. Crucially, the graph carries two separate pieces of information, and exam questions test them one at a time:
- How steep the curve is — the rate — is set by the conditions: concentration, temperature, surface area and any catalyst. Speed the reaction up and the curve rises more steeply (and reaches the plateau sooner).
- How high the plateau sits — the total amount of product — is set by the amount of the limiting reactant. Use more of it and the line levels off higher; use less and it levels off lower — no matter what you did to the rate.
So a faster reaction does not always make more product, and more product does not mean a faster reaction. Reading the steepness and the height as two independent things is one of the most discriminating graph skills in this topic — there’s a predict-the-curve trainer for it in section 2.
When a question asks you to plot the data, marks sit in the mechanics before any chemistry happens:
- Put the axes the right way round: the independent variable (the one you change) goes on the x‑axis, the dependent variable (the one you measure) on the y‑axis.
- Choose scales so the plotted points fill at least half the grid in both directions — pick the scale first, before plotting anything. Use sensible steps (1, 2, 5 or 10 units per square), never awkward ones like 3 or 7.
- Label each axis with the quantity and its unit (“Volume of gas / cm³”, “Time / s”).
- Plot points accurately as small neat crosses — no “blobs” bigger than half a small square.
- Draw a single, smooth line (or curve) of best fit — not feathery sketching, not dot-to-dot, and never forced through an anomalous point.
The rate at a single moment — tangents
The mean rate over a whole experiment is one number, but the graph shows the rate constantly changing. To find the rate at one particular time, draw a tangent to the curve at that time — a straight line, ruled so it just touches the curve at that point and follows its direction — then measure the tangent’s gradient (slope). The steeper the tangent, the faster the reaction at that moment.
gradient = change in ychange in x = ΔyΔx
Everyone must be able to draw a tangent and use its slope to compare rates. Higher Tier students must also calculate the gradient to give a numerical rate at that time — with the unit taken straight from the axes (cm³/s if the axes are cm³ and s).
Two habits separate full marks from dropped ones when measuring any gradient:
- Draw the largest triangle you reasonably can on the tangent — ideally spanning at least half the drawn line. Reading errors of half a square matter far less on a big triangle than on a tiny one, so a large triangle makes your gradient more accurate. Where the line crosses gridline intersections (or the axes) cleanly, those are ideal corner points — easy to read exactly.
- Annotate the coordinates you used. Mark the two points and write their values on the graph. Examiner reports note students who calculated a gradient but never showed which coordinates they read — making the working impossible to credit if the final answer slips.
📐 Tangent explorer — the rate at any moment
Magnesium + dilute hydrochloric acid, with the hydrogen collected in a gas syringe. Slide along the curve and watch the tangent, the triangle and the gradient calculation update. Notice when the reaction is fastest — and what the gradient becomes once it finishes.
Worked example — rate in mol/s Higher
Higher Tier questions sometimes want the rate in mol/s instead of cm³/s. You won’t be expected to know the conversion — the question gives you a factor — but you do need to draw (or read) a tangent first, then convert its gradient.
A tangent drawn to a graph of volume of oxygen against time has a gradient of 0.84 cm³/s. Under the conditions used, 1 cm³ of oxygen = 4.0 × 10⁻⁵ mol. Calculate the rate of reaction at that moment, in mol/s.
- The gradient of the tangent is the rate in cm³/s: 0.84 cm³/s.
- Convert each cm³ to moles with the factor given: rate = 0.84 × (4.0 × 10⁻⁵) mol/s.
- = 3.4 × 10⁻⁵ mol/s (2 s.f.). The unit follows the working: (cm³/s) × (mol/cm³) = mol/s.
- Mean rate — one number for a whole period: total quantitytotal time. Use it when the question says “calculate the mean rate between…” or gives a table of start/end values.
- Rate at an instant — the gradient of the tangent at that time. Use it when the question says “at 30 seconds” or “at this point”.
- Quoting a mean rate when the question asked for the rate at a time is a classic way to lose all the marks while doing correct arithmetic.
🧪 Exam-style questions
Calcium carbonate reacts with dilute hydrochloric acid. 36 cm³ of carbon dioxide is collected in the first 45 seconds. Calculate the mean rate of reaction over this period, in cm³/s.
Show answer
- mean rate = quantity of product formedtime taken = 3645. 1 mark
- = 0.8 cm³/s. 1 mark
- The unit comes straight from the working: cm³ divided by s gives cm³/s.
On a graph of volume of gas produced against time, when is the rate of reaction greatest? Tick (✓) one box.
A reaction loses mass as a gas escapes. Which set of units is correct for the rate of this reaction? Tick (✓) one box.
A tangent is drawn to a volume–time curve at t = 15 s. The tangent passes through the points (5, 18) and (35, 54), where x is in seconds and y in cm³. Calculate the rate of reaction at 15 s.
Show answer
- Δy = 54 − 18 = 36 cm³. 1 mark
- Δx = 35 − 5 = 30 s. 1 mark
- gradient = 3630 = 1.2 cm³/s. 1 mark
- Note how the two points span a wide run of the tangent — a big triangle — and both coordinates are stated in the working.