All subjects · Chemistry · WCH16 Practical Skills in Chemistry II

Advanced graph work: tangents, log graphs and Arrhenius plots

Edexcel IAL Chemistry · WCH16 Practical Skills in Chemistry II

A2 practical data are often processed with tangents and logarithms to turn curves into straight lines. You must read gradients and intercepts correctly and give units.

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Tangents for rates

On a concentration-time graph, the rate at time t is the gradient of the tangent at t (ignore the minus sign). The initial rate is the tangent at t = 0. Draw the tangent touching the curve, make the triangle large, and show the working on the graph (January 2022 paper). Units: mol dm^-3 s^-1.

Turning curves into lines

Order: rate = k[A]ⁿ, so log(rate) = n log[A] + log k. A graph of log(rate) against log[A] has gradient n and intercept log k.
Arrhenius: ln k = -Ea/R x (1/T) + ln A. A graph of ln k against 1/T (T in K) has gradient -Ea/R and intercept ln A (not A).
In a clock reaction, 1/t is proportional to k, so ln(1/t) against 1/T has the same gradient.

Reading the line

Use the line of best fit, not two data points: the gradient method uses all the data and gives more consistent answers than substituting two points into equations (October 2024 report).
Ea = -gradient x R. With R = 8.31 J mol^-1 K^-1, Ea comes out in J mol^-1: divide by 1000 for kJ mol^-1.
The gradient of ln k against 1/T has unit K; a log has no unit.

Worked example

A line on a graph of ln k against 1/T passes through (3.20 x 10^-3 K^-1, -6.10) and (3.40 x 10^-3 K^-1, -8.20). Find Ea.

Gradient = (-8.20 - (-6.10)) / (3.40 x 10^-3 - 3.20 x 10^-3)
= -2.10 / 2.0 x 10^-4 = -10500 K
Ea = -gradient x R = 10500 x 8.31 = 87255 J mol^-1

Answer: 87.3 kJ mol^-1

Key idea: Rate = gradient of a tangent. log(rate) vs log[A]: gradient = order. ln k vs 1/T: gradient = -Ea/R, intercept ln A. Use the best-fit line; give units.

Check yourself

A graph of log(rate) against log[A] passes through (-1.30, -3.60) and (-0.70, -2.40). What is the order with respect to A?
Show answer
2. Gradient = (-2.40 + 3.60) / (-0.70 + 1.30) = 2.0, so second order.
The gradient of a graph of ln k against 1/T is -6200 K. Calculate Ea in kJ mol^-1 (R = 8.31 J mol^-1 K^-1).
Show answer
51.5 kJ mol^-1. Ea = 6200 x 8.31 = 51522 J mol^-1 = 51.5 kJ mol^-1.
A tangent drawn at t = 0 on a graph of [MnO4-] against time passes through (0 s, 0.0200 mol dm^-3) and (50 s, 0 mol dm^-3). Calculate the initial rate.
Show answer
0.00040 mol dm^-3 s^-1. 0.0200 / 50 = 4.0 x 10^-4 mol dm^-3 s^-1.
Which is the correct unit for activation energy calculated from an Arrhenius plot?
- kJ
- kJ mol^-1 K^-1
- kJ mol^-1
- K
Show answer
kJ mol^-1. Ea is an energy per mole. The October 2024 report noted wrong units were common.
What does the intercept on the y-axis of a graph of ln k against 1/T give?
- ln A
- A
- Ea
- -Ea/R
Show answer
ln A. ln k = ln A when 1/T = 0. A must be found as e^intercept.
Why is Ea better found from the gradient of a best-fit line than by substituting two data points into the Arrhenius equation?
- The equation cannot be used with two points
- The gradient method avoids converting to kelvin
- The line uses all the data, reducing the effect of any one poor reading
- The line gives A directly
Show answer
The line uses all the data, reducing the effect of any one poor reading. Two-point methods depend heavily on those two readings.
In a clock reaction, which graph is a straight line through the origin if the reaction is first order in X?
- 1/t against [X]
- t against [X]
- 1/t against [X]²
- log t against [X]
Show answer
1/t against [X]. Rate is proportional to 1/t, and for first order rate is proportional to [X].

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Last updated 8 November 2026. Spotted a mistake? Tell us.