Straight-Line Graphs & Linearization — Maths for Chemistry

y = mx + c, the chemist's favourite line

And the art of forcing data onto it

A straight line is the easiest graph to read: a constant gradient m and a y-intercept c.

y = m x + c, m = \frac{rise}{run} = \frac{Δ y}{Δ x}

Most chemistry relationships aren't straight to begin with — so we linearize them. Taking a log or a reciprocal of the right quantity turns a curve into a line whose slope and intercept are physical constants that can be read directly off the graph.

The technique is ubiquitous: the Arrhenius plot ( $ln k$ vs $1/ T$ ) extracts activation energy and pre-exponential factor from the gradient and intercept. The Beer–Lambert plot (A vs c) gives the molar absorptivity. A first-order kinetics plot ( $ln [A]$ vs t) gives the rate constant and confirms the reaction order.

Classic linear plots in physical chemistry

Beer–Lambert

A = ε c l \Rightarrow A vs c, slope = εl

Arrhenius

ln k = - \frac{E _{a}}{R} \cdot \frac{1}{T} + ln A

First-order kinetics

ln [A] = ln [A]_{0} - k t, slope = - k

Clausius–Clapeyron

ln p = - \frac{Δ H _{vap}}{R} \cdot \frac{1}{T} + const

Reaction-order diagnostic plots use linearization to identify kinetics. If you don't know the order, try all three:

Zero order

Plot

[A]

t

. Straight line, slope

= - k

First order

Plot

ln [A]

t

. Straight line, slope

= - k

Second order

Plot

1/ [A]

t

. Straight line, slope

= + k

Langmuir adsorption

1/ θ = 1 + 1/ (K_{L} c)

linearises as

1/ θ

1/ c

with slope

1/ K_{L}

The Clausius–Clapeyron equation $ln p = - Δ H_{vap} / (RT) + const$ linearises as $ln p$ vs $1/ T$ , with slope $= - Δ H_{vap} / R$ . This is structurally identical to the Arrhenius plot — both show how a log-transformed variable depends linearly on reciprocal temperature.

Common pitfalls

Wrong axes for the order. Plotting $1/ [A]$ vs $t$ and calling the slope $- k$ — that is the second-order slope, not first-order. The slope sign and the axes must match the integrated rate law you derived.
Slope sign for Arrhenius/Clausius–Clapeyron. Both give a negative slope on a $ln k$ or $ln p$ vs $1/ T$ plot (assuming activation energy and enthalpy of vaporisation are positive). A positive slope would be unphysical.
Over-extrapolation. Reading off a value far outside the range of the data magnifies any error in the fitted slope. A 1% error in the Arrhenius slope translates to a 1% error in $E_{a}$ — but an extrapolation far outside the data range makes this error dominant.
Forgetting the intercept. The intercept of $ln k$ vs $1/ T$ is $ln A$ (the pre-exponential factor), not zero. Missing it when reading off A is a common error.

Classic linear plots

Beer–Lambert (A vs c, slope εl), first-order kinetics (ln[A] vs t, slope −k), Arrhenius (ln k vs 1/T, slope −Eₐ/R), and the Clausius–Clapeyron plot. Try the line explorer then linearize a curve.

Line explorer

Drag the gradient and intercept — the slope triangle updates in real time to show 'rise over run'.

Change the gradient and intercept

y = 1 x + 1

gradient m1

intercept c1

The green triangle is “rise over run”: go 2 across and 2.0 up, so the gradient is 1. The orange dot marks the intercept c = 1.

Worked example 1Reading a Beer–Lambert calibration graph

A calibration plot of absorbance A against concentration c (mol dm⁻³) gives a straight line through the origin with gradient 48.0 L mol⁻¹ cm⁻¹ when l = 1.00 cm. An unknown gives A = 0.336. Find its concentration.

Linearize a curve

Toggle between the raw exponential decay and its log transform — watch the curve become a straight line.

First-order decay:

[A] = [A]_{0} e^{- k t}

rate constant k0.40 s⁻¹

The raw data curves, so its 'rate' keeps changing — hard to read off a constant.

Worked example 3Arrhenius — extracting Eₐ from an ln k vs 1/T plot

The rate constant for the decomposition of N₂O₅ was measured at two temperatures: $k (318 K) = 4.98 \times 1 0^{- 4} s^{- 1}$ and $k (328 K) = 1.50 \times 1 0^{- 3} s^{- 1}$ . Find $E_{a}$ .

Worked examples

Slope and intercept carry chemistry — two more real problems.

Worked example 4First-order rate constant from a ln[A] vs t plot

Concentration-time data for the decomposition of H₂O₂ gives a straight line when ln[H₂O₂] is plotted against t. The intercept is ln(0.880) and the gradient is −2.31 × 10⁻³ s⁻¹. Find k and the half-life.

Worked example 5Clausius–Clapeyron and enthalpy of vaporisation

Vapour pressures of water: 1.23 kPa at 283 K and 3.17 kPa at 298 K. Use the linearized Clausius–Clapeyron equation to find $Δ H_{vap}$ .

Worked example 6Identifying reaction order — which plot is straight?

Concentration-time data for a decomposition reaction:

t (s): 0, 100, 200, 300, 400

[A] (M): 0.800, 0.400, 0.200, 0.100, 0.050

By plotting (or inspecting) $[A]$ vs t, $ln [A]$ vs t, and $1/ [A]$ vs t, determine the reaction order and find k.

Worked example 7Langmuir adsorption isotherm — linearization

The Langmuir isotherm is $θ = \frac{K _{L} c}{1 + K _{L} c}$ where θ is the surface coverage and c is equilibrium concentration. Show how to linearize it, and state what slope and intercept give $K_{L}$ .

ChallengeChallenge — determine reaction order and k from three concentration-time points

For a reaction A → products, three data points are measured:

t (min): 0, 10, 20

[A] (mol L⁻¹): 0.500, 0.250, 0.125

(a) Determine the reaction order by comparing the three diagnostic plots without a calculator — just by inspection of the ratios. (b) Calculate k. (c) Predict [A] at t = 35 min.

Check yourself

Four questions on gradients, intercepts and linearization in chemistry.

Question 1 of 4 · Score 0

In the equation of a straight line y = mx + c, what is m?

Choose an answer.