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ln and Arrhenius plots

The natural log turns the Arrhenius equation into a straight line, and the line's gradient and intercept give Ea and A. The same ln 2 sets the half-life of a first-order reaction.

ln is log to the base e

A natural logarithm asks the same question as log₁₀ with a different base: e, which is 2.718. ln 2 is 0.693 because e⁰·⁶⁹³ is 2. The reverse is the eˣ key, SHIFT then ln on most calculators. Nothing in A-level chemistry needs you to convert between ln and log; the rule is to use ln wherever the equation says ln, and log wherever it says log. pH, pKa and pKw use log₁₀; everything in rates uses ln.

ln appears in chemistry because rates and rate constants grow exponentially with temperature. The Arrhenius equation k = Ae^(−Ea/RT) has e to a power in it, and the only way to get the power down where it can be measured is to take ln of both sides: ln k = ln A − Ea ÷ RT. The log of a product is the sum of the logs, ln A plus ln of e^(−Ea/RT), and ln of e to anything is that anything.

Written as ln k = (−Ea ÷ R) × (1 ÷ T) + ln A, this is y = mx + c. Plot ln k on the vertical axis against 1 ÷ T in K⁻¹ on the horizontal, and the gradient is −Ea ÷ R, the intercept on the ln k axis is ln A. R is 8.31 J K⁻¹ mol⁻¹, so the gradient has units of K and Ea comes out in J mol⁻¹, to be divided by 1000 for kJ mol⁻¹.

Reading the gradient and the intercept

The gradient is negative, because k rises as T rises and 1 ÷ T falls. A positive Ea from a negative gradient is right; a negative Ea is a sign the minus has been dropped. Take the gradient from two points far apart on the line of best fit, never from two data points, and keep every figure until the end: 1 ÷ T values are small numbers like 3.21 × 10⁻³ K⁻¹ and a rounded one moves the gradient by several per cent.

Ea and A from an Arrhenius plot

A plot of ln k against 1 ÷ T for the decomposition of a compound is a straight line through (3.00 × 10⁻³ K⁻¹, −2.00) and (3.40 × 10⁻³ K⁻¹, −6.50). Find Ea in kJ mol⁻¹ and the pre-exponential factor A, given that k is in s⁻¹.

Show the working

Gradient = change in ln k ÷ change in (1 ÷ T) = (−6.50 − (−2.00)) ÷ (3.40 × 10⁻³ − 3.00 × 10⁻³) = −4.50 ÷ 4.00 × 10⁻⁴ = −1.125 × 10⁴ K.

Gradient = −Ea ÷ R, so Ea = 1.125 × 10⁴ × 8.31 = 9.35 × 10⁴ J mol⁻¹ = 93.5 kJ mol⁻¹.

The intercept is where 1 ÷ T = 0, well off the left of the plotted data, so it is calculated rather than read: ln A = ln k − gradient × (1 ÷ T) = −2.00 − (−1.125 × 10⁴ × 3.00 × 10⁻³) = −2.00 + 33.75 = 31.75.

A = e³¹·⁷⁵ = 6.15 × 10¹³ s⁻¹, with the same units as k because e to a power has none.

When a question gives only two rate constants at two temperatures there is no plot, and the two Arrhenius equations are subtracted: ln k₂ − ln k₁ = (Ea ÷ R)(1 ÷ T₁ − 1 ÷ T₂). ln A cancels because A is the same at both temperatures. ln k₂ − ln k₁ is ln(k₂ ÷ k₁), which is the form worth remembering.

Ea from two rate constants

The rate constant of a reaction is 2.50 × 10⁻³ dm³ mol⁻¹ s⁻¹ at 300 K and 1.20 × 10⁻² dm³ mol⁻¹ s⁻¹ at 320 K. Find the activation energy.

Show the working

ln(k₂ ÷ k₁) = ln(1.20 × 10⁻² ÷ 2.50 × 10⁻³) = ln 4.80 = 1.569.

1 ÷ T₁ − 1 ÷ T₂ = 1 ÷ 300 − 1 ÷ 320 = 3.333 × 10⁻³ − 3.125 × 10⁻³ = 2.083 × 10⁻⁴ K⁻¹. The two reciprocals differ in the fourth figure, which is why they are kept to four.

Ea = R × ln(k₂ ÷ k₁) ÷ (1 ÷ T₁ − 1 ÷ T₂) = 8.31 × 1.569 ÷ 2.083 × 10⁻⁴ = 6.26 × 10⁴ J mol⁻¹ = 62.6 kJ mol⁻¹.

ln 2, and the units of k

A first-order reaction has a constant half-life, t½ = ln 2 ÷ k = 0.693 ÷ k. The ln 2 comes from integrating the first-order rate equation: the concentration falls as e⁻ᵏᵗ, and the time for it to halve is the time at which e⁻ᵏᵗ = ½, so kt = ln 2. A half-life of 60 s gives k = 0.693 ÷ 60 = 1.16 × 10⁻² s⁻¹.

The units of k are whatever makes the rate equation balance, and they change with the overall order. Rate is always in mol dm⁻³ s⁻¹. For a zero-order reaction, rate = k, so k is in mol dm⁻³ s⁻¹. For first order, rate = k[A], and dividing mol dm⁻³ s⁻¹ by mol dm⁻³ leaves s⁻¹. For second order, rate = k[A]² or k[A][B], and dividing by mol² dm⁻⁶ leaves dm³ mol⁻¹ s⁻¹. Third order gives dm⁶ mol⁻² s⁻¹. Each extra order adds dm³ mol⁻¹.

A mark scheme awards the units of k as a separate mark from its value, accepts mol⁻¹ dm³ s⁻¹ in either order, and rejects s⁻¹ for a second-order constant whatever the number beside it.

TRY IT: Logs, a gradient and three sets of units

(a) ln(0.0450).

(b) Ea in kJ mol⁻¹ for an Arrhenius plot of gradient −9.60 × 10³ K.

(d) The value of e⁻³·⁵⁰.

Check your answer

(a) ln(0.0450) = −3.10.

(b) Ea = 9.60 × 10³ × 8.31 = 7.98 × 10⁴ J mol⁻¹ = 79.8 kJ mol⁻¹.

(d) e⁻³·⁵⁰ = 0.0302.

The lessons that use this: The Arrhenius equation and the rate-determining step · Rate equations, orders of reaction and the rate constant · Finding orders from initial rates, half-lives and graphs. Every skill the boards number, and where each is taught: maths skills.