IB Chemistry HL Topic 7 — Tool 3: Mathematics Paper 1, 2 & IA Practical skill ~13 min read

Graphing Skills

A graph is not decoration at the end of an experiment. It is where the answer lives — in the gradient, the intercept, or the area underneath. Drawing it properly is half the skill; knowing what to pull out of it is the other half.

📚 What you need to know

Rules for drawing a graph

These are the things examiners tick off, one at a time.

The independent variable is the one you chose and controlled. The dependent variable is the one that responded. If you set the concentrations and measured the rates, concentration goes on the x-axis — every time.

Lines of best fit

A line of best fit does not have to be straight, and it does not have to pass through every point. What it has to do is show the trend.

Interpolation and extrapolation

Two words for reading values off a line, and they are not equally safe.

Interpolation and extrapolation Trust the line inside the data, not beyond it extrapolation rate interpolation substrate concentration solid line: inside the data dashed: beyond the data Extrapolation assumes the trend keeps going — which it may not
Extrapolation is not banned. Calorimetry depends on it. But you must be able to justify why the trend should continue into the region you are reading.
The classic legitimate extrapolation. In a calorimetry experiment the mixture starts cooling before it has finished reacting, so you extend the cooling line backwards to the moment of mixing to find the temperature rise that would have happened with no heat loss.

Gradients

The gradient is the answer to a lot of chemistry questions: a rate, an entropy change, a rate constant.

Equation of a straight line y = mx + c   so   gradient m = Δy ÷ Δx

For a straight line, draw a large triangle on the line and divide the vertical side by the horizontal side. Make the triangle as big as the graph allows — a small triangle magnifies your reading errors.

For a curve, the gradient changes from point to point, so you have to draw a tangent first. Line a ruler up against the curve so it is an equal distance from the curve on both sides of the touching point, then draw the tangent and take its gradient.

Taking a gradient off a curve The tangent turns one point on a curve into a straight line you can measure volume of gas / cm³ tangent touches at t = 40 s ∆t = 60 s ∆V = 54 cm³ time / s gradient = 54 ÷ 60 = 0.90 cm³ s⁻¹ Make the triangle large — a small one multiplies your reading errors
The tangent is the red line. It touches the curve at one point only, and its gradient is the rate of reaction at that instant.

🧩 Drawing a tangent that actually works

  1. Mark the point on the curve where you want the gradient.
  2. Lay a ruler against the curve at that point, so the gap between ruler and curve looks equal on both sides.
  3. Draw the tangent long, right across the graph. A short tangent is hard to read.
  4. Build a big triangle on the tangent, not on the curve.
  5. Read both sides off the axes with their units.
  6. Divide, and give the answer a unit made from the two axis units.

Intercepts

An intercept is where the line crosses an axis, and in chemistry it usually carries physical meaning. The best example is a graph of Gibbs energy change against temperature.

Rearranged into y = mx + c ΔG = −ΔS × T + ΔH

Compare that with y = mx + c. Plotting ΔG against T gives a straight line whose gradient is −ΔS and whose y-intercept is ΔH. The x-intercept is the temperature at which ΔG = 0, which is where the reaction stops being feasible.

Three answers from one straight line Gibbs energy against temperature for the ammonia synthesis −60 −40 −20 0 20 40 60 0 200 400 600 800 1000 y-intercept = ∆H x-intercept ≈ 465 K gradient = −∆S ∆G / kJ mol⁻¹ against temperature / K
Below about 465 K the line sits below zero, so ΔG is negative and the reaction is feasible. Above it, ammonia synthesis stops being spontaneous.

Maxima, minima and areas

Two more features worth naming.

Uncertainty bars

An uncertainty bar is drawn above and below a plotted point, showing the absolute uncertainty in that measurement. Usually they are vertical, for the y-values, but they can be horizontal for the x-values too.

Uncertainty bars on a graph The length of the bar is the absolute uncertainty in that reading large uncertainty small uncertainty A best-fit line should pass through most of the bars, not just near the points
A point with a long bar is telling you it was measured badly. If your line misses that point but passes through its bar, the line is still consistent with the data.

Worked examples

WORKED EXAMPLE

On the ΔG against T graph above, the y-intercept is −46.2 kJ mol–1 and the gradient is +0.0994 kJ K–1 mol–1. Find ΔH, ΔS, and the temperature above which the reaction is no longer feasible.

Match the graph to y = mx + c ΔG = −ΔS × T + ΔH, so c is ΔH and m is −ΔS. Read off ΔH ΔH = −46.2 kJ mol–1 (exothermic) Turn the gradient into ΔS −ΔS = +0.0994  →  ΔS = −0.0994 kJ K–1 mol–1 = −99.4 J K–1 mol–1 Find where ΔG = 0 T = 46.2 ÷ 0.0994 = 464.8 Feasible below about 465 K (192°C) both ΔH and ΔS are negative, which is exactly why temperature decides the outcome
WORKED EXAMPLE

A tangent drawn to a volume-time curve at t = 40 s passes through (10 s, 41 cm3) and (70 s, 95 cm3). Calculate the rate of reaction at 40 s.

Use the two points on the tangent, not on the curve ΔV = 95 − 41 = 54 cm3 Δt = 70 − 10 = 60 s Divide 54 ÷ 60 = 0.90 Build the unit from the axes y-axis is cm3, x-axis is s, so the gradient is cm3 per s. Rate at 40 s = 0.90 cm3 s–1 this is the instantaneous rate at that moment, not the average over the whole reaction

💡 Exam tip

⚠️ Common mix-up

That completes Tool 3. Every mark on this page is one you can secure before you even walk into the exam — label the axes, fill the grid, draw a big triangle, and put a unit on the answer.

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