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Chemistry · General chemistry II · Worked example

Find Ea and A from an Arrhenius plot

A first-order reaction has these rate constants at four temperatures. Find the activation energy and the frequency factor.

k=Ae−Ea/⁢R⁢T,lnk=−EaR(1T)+lnA

Transform the data

Arrhenius plots use 1/T in K⁻¹ and ln k, so compute both for every row before fitting.

Rate constants and Arrhenius coordinates
T (K)k (s⁻¹)1/T (K⁻¹)ln k
3001.92 × 10⁻³3.3333 × 10⁻³−6.255
3104.46 × 10⁻³3.2258 × 10⁻³−5.413
3209.81 × 10⁻³3.1250 × 10⁻³−4.624
3302.06 × 10⁻²3.0303 × 10⁻³−3.882

Fit a straight line

The points lie on a straight line (R² = 0.9999998), so one activation energy describes this range. The least-squares slope and intercept are:

slope=−7.83×103 K,intercept=19.84

Activation energy from the slope

The slope equals −Ea/R, so Ea = −R × slope.

Ea=−(8.314 Jmol−1K−1)⁢(−7.83×103 K)=6.51×104 J/mol=65.1 kJ/mol

Frequency factor from the intercept

The intercept is ln A. A has the units of k, and because it is an extrapolation to 1/T = 0, two significant figures are plenty.

A=e19.84≈4.2×108 s−1

Result

Ea = 65.1 kJ/mol and A ≈ 4.2 × 10⁸ s⁻¹.

k=(4.2×108 s−1)e−65.1 kJ/mol⁢/⁢R⁢T

Your turn

Using Ea = 65.1 kJ/mol and A = 4.2 × 10⁸ s⁻¹ from this example, estimate k at 340 K.

Show the answer and explanation

k ≈ 4.2 × 10⁻² s⁻¹.

k = 4.2 × 10⁸ × e^(−65100/(8.314 × 340)) = 4.2 × 10⁸ × e^(−23.03) ≈ 4.2 × 10⁻² s⁻¹. 340 K is just outside the measured range, so treat this as a short extrapolation.

k=4.2×108e−65100⁢/⁢(8.314×340)≈4.2×10−2 s−1

Keep exploring

Fit the same four points in the Kinetics studio, then add a prediction temperature and see when it warns about extrapolation.

Return to the concept →
Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

  • Tro, Chemistry: A Molecular Approach, 4th ed., §14.5 The Effect of Temperature on Reaction Rate, pp. 642–647 (Arrhenius plots, pp. 644–645; two-point form, p. 646)
  • OpenStax Chemistry 2e — Collision theory