Rate calculations and graphs

This page covers one Form 4 content standard: rate calculations and graphs. It brings together everything earlier in the chapter and turns it into the numerical and graph-reading skills the exam tests directly. If you can read a curve, calculate an average rate, find an instantaneous rate from a tangent and convert a gas volume into moles, you can answer almost any quantitative rate question.

The shape of a rate graph

A typical experiment records the volume of gas released (or the loss of mass) at regular time intervals, then plots it against time. The curve has a characteristic shape:

• It is steepest at the start, where the rate is fastest and reactants are most concentrated.
• It gradually becomes less steep as reactants are used up and the reaction slows.
• It levels off (becomes flat) when the reaction is complete, a reactant has run out, so no more product forms.

The height of the plateau tells you the total amount of product formed, which depends only on the amount of the limiting reactant, not on the rate.

Average rate from a graph

Average rate = (total change in quantity) ÷ (total time taken).

Read the final volume (or mass change) and the time at which the curve becomes flat, then divide. The unit is cm³ s⁻¹ (for gas volume) or g s⁻¹ (for loss of mass). This gives one number for the whole reaction, or for any interval you choose.

Instantaneous rate from a tangent

The instantaneous rate at a particular time is the gradient of the tangent drawn to the curve at that point.

1. Draw a straight tangent that just touches the curve at the chosen time.
2. Build a right-angled triangle on the tangent and read the change in the y-value (volume) and the change in the x-value (time).
3. Gradient = (change in y) ÷ (change in x) = the instantaneous rate.

The initial rate is the gradient of the tangent at time = 0, and it is the largest instantaneous rate of the whole reaction.

Comparing two curves

When two experiments are drawn on the same axes, compare them in two moves:

• Steepness shows the rate: the steeper curve (at the start) is the faster reaction, caused by higher concentration, higher temperature, larger surface area, or a catalyst.
• Plateau height shows the amount of product: if both curves level off at the same height, the same amount of reactant was used, even if one reached it sooner.

Converting gas volume to moles

You can turn a gas volume into an amount in moles using the molar volume of a gas, 24 dm³ mol⁻¹ at room conditions:

moles of gas = volume of gas ÷ molar volume

Remember that 24 dm³ mol⁻¹ equals 24 000 cm³ mol⁻¹, so if a reaction gives 72 cm³ of gas, that is 72 ÷ 24 000 = 3 × 10⁻³ mol. This lets you connect the graph to the mole calculations from the mole-concept chapter.

Worked example

Question. In a reaction between excess magnesium and dilute hydrochloric acid, hydrogen gas is collected. The volume reaches its maximum of 60 cm³ after 100 s, and the curve is flat after that. From the graph, 36 cm³ has been collected at 20 s. (a) Calculate the average rate over the whole reaction. (b) Calculate the average rate during the first 20 s. (c) State how the rate at 20 s compares with the initial rate.

Step 1, Average rate over the whole reaction. Average rate = total volume ÷ total time = 60 cm³ ÷ 100 s = 0.6 cm³ s⁻¹.

Step 2, Average rate during the first 20 s. Average rate = 36 cm³ ÷ 20 s = 1.8 cm³ s⁻¹.

Step 3, Compare with the initial rate. The reaction is fastest at the very start, so the initial rate is higher than the average over the first 20 s, and the rate at 20 s is lower still because the acid has become less concentrated.

Answer. (a) The average rate over the whole reaction is 0.6 cm³ s⁻¹. (b) During the first 20 s the average rate is 1.8 cm³ s⁻¹, higher than the overall average, because the reaction is faster near the start. (c) The rate at 20 s is lower than the initial rate; the reaction slows continuously as hydrochloric acid is used up.

Practice question

Question. A reaction between a metal carbonate and excess acid produces carbon dioxide. The gas volume levels off at 72 cm³ after 90 s. (a) Calculate the average rate of reaction in cm³ s⁻¹. (b) Using the molar volume of gas, 24 dm³ mol⁻¹ (= 24 000 cm³ mol⁻¹), calculate the number of moles of carbon dioxide produced.

Answer. (a) Average rate = total volume ÷ total time = 72 cm³ ÷ 90 s = 0.8 cm³ s⁻¹. (b) Moles of CO₂ = volume ÷ molar volume = 72 cm³ ÷ 24 000 cm³ mol⁻¹ = 3 × 10⁻³ mol (0.003 mol).

Exam tip

Match the tool to the question. “Rate over” an interval means an average rate, divide change by time. “Rate at” a moment means an instantaneous rate, draw a tangent and find its gradient. Always quote the unit (cm³ s⁻¹ or g s⁻¹). When comparing curves, remember the two-part rule: steepness = rate, plateau height = amount of product. And when a question mixes rate with the mole concept, keep your units consistent, convert dm³ and cm³ carefully before dividing by the molar volume.

Reading graph questions without slipping

The commonest lost marks in this standard come from three slips. First, reading the time when the curve levels off rather than when it reaches a particular value, for an average rate over the whole reaction, use the time the curve becomes flat. Second, forgetting the unit; a rate is never just a number. Third, muddling the two comparisons: a catalysed or hotter reaction is steeper and finishes sooner, but if the amount of limiting reactant is unchanged it reaches the same plateau. If you keep “steeper means faster” and “same height means same amount” separate, you will interpret any pair of curves correctly.

A useful habit is to annotate the graph as you read it: mark the plateau, mark the time it is reached, and mark any point where a tangent is wanted. Then decide whether the question asks for an average or an instantaneous rate before you calculate. That small discipline prevents the most frequent errors and makes multi-part graph questions quick to work through.

Where this fits

This is content standard 7.5 of the Rate of Reaction chapter, the quantitative finish that draws on the concept of rate, the factors, collision theory and catalysts. It is a core Paper 2 and Paper 3 skill in SPM Chemistry (4541). Practise the graph and mole conversions in the worked examples and test yourself with the practice questions. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers drill the tangent method and the “steepness versus plateau” reading until graph questions become straightforward marks.

Quick recap

• Rate graph: steepest at the start, flat when the reaction is complete.
• Average rate = total change ÷ total time; instantaneous rate = gradient of the tangent.
• Steeper = faster; same plateau height = same amount of product.
• moles of gas = volume ÷ molar volume (24 dm³ mol⁻¹ at room conditions).