The Arrhenius Equation and Temperature Effect

The Arrhenius equation, k = A e^−E_a/RT, shows how the rate constant of a reaction depends on temperature, activation energy and a frequency factor; its logarithmic form, ln k = ln A − E_a/(RT), gives a straight line when ln k is plotted against 1/T. In brief: if k increases from 1.0 × 10⁻³ to 4.0 × 10⁻³ s⁻¹ when the temperature rises from 300 K to 320 K, the activation energy is 55 kJ/mol.

The equation, which Svante Arrhenius proposed in 1889, is the most widely used relationship in chemical kinetics. It makes it possible to predict reaction rates at other temperatures, to find the activation energy from experimental data and to understand why the rates of reactions are so sensitive to temperature. This article explains the equation, shows how to use it with two data points or with a graph, and includes a table of the effect of the activation energy.

What is the Arrhenius equation?

The rate constant k of most reactions increases exponentially with temperature:

k = A · e−Ea / (RT)

Here A is the frequency (pre-exponential) factor, with the same units as k, E_a is the activation energy in J/mol, R = 8.314 J/(mol·K) and T is the absolute temperature in kelvins. The exponential term is the fraction of collisions with sufficient energy. A larger E_a makes k smaller, and a higher T makes k larger.

What forms of the equation are used?

• Exponential form: k = A e^−E_a/RT.
• Logarithmic (linear) form: ln k = ln A − (E_a/R)(1/T).
• Two-point form: ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂), obtained by subtracting the logarithmic equations for two temperatures. A drops out.

How do you find E_a from a graph?

1. Measure k at several temperatures.
2. Calculate ln k and 1/T (in K⁻¹) for each.
3. Plot ln k on the vertical axis against 1/T on the horizontal axis.
4. Draw the best straight line. Its slope is −E_a/R and the intercept is ln A.
5. E_a = −slope × R.

This is called an Arrhenius plot. The line has a negative slope, since k increases as 1/T decreases. A steeper slope means a higher activation energy and a stronger dependence on temperature.

Worked examples

Example 1: finding E_a from two rate constants

The rate constant is 1.0 × 10⁻³ s⁻¹ at 300 K and 4.0 × 10⁻³ s⁻¹ at 320 K.

1. ln(k₂/k₁) = ln 4.0 = 1.386.
2. 1/T₁ − 1/T₂ = 1/300 − 1/320 = 0.003333 − 0.003125 = 2.083 × 10⁻⁴ K⁻¹.
3. E_a = R × 1.386 / (2.083 × 10⁻⁴) = 8.314 × 1.386 / 2.083 × 10⁻⁴ = 55,300 J/mol.

The activation energy is 55 kJ/mol.

Example 2: predicting k at another temperature

A reaction has E_a = 75.0 kJ/mol and k = 2.0 × 10⁻⁴ s⁻¹ at 25 °C. Find k at 50 °C.

1. T₁ = 298.15 K and T₂ = 323.15 K.
2. 1/T₁ − 1/T₂ = 0.0033540 − 0.0030945 = 2.595 × 10⁻⁴ K⁻¹.
3. ln(k₂/k₁) = (75,000 / 8.314) × 2.595 × 10⁻⁴ = 9021 × 2.595 × 10⁻⁴ = 2.341.
4. k₂/k₁ = e^2.341 = 10.4, so k₂ = 2.1 × 10⁻³ s⁻¹.

A rise of 25 °C increases the rate constant by a factor of about 10.

Example 3: the activation energy for “doubling every 10 °C”

Between 298 K and 308 K, a doubling of k means ln 2 = (E_a/8.314)(1/298 − 1/308) = (E_a/8.314)(1.0895 × 10⁻⁴). Hence E_a = 0.693 × 8.314 / 1.0895 × 10⁻⁴ = 52,900 J/mol, or about 53 kJ/mol. The rule of thumb is therefore valid for typical activation energies of about 50 kJ/mol near room temperature.

How much does k change with temperature?

Reactions with higher activation energies are more sensitive to temperature. This is why enzymes with high activation energy for denaturation are destroyed so quickly by a mild fever or when cooking.

What do A and E_a mean?

• E_a: the energy barrier. A catalyst lowers it.
• A: the frequency factor, which includes the collision frequency and the orientation factor. Typical values are 10¹⁰–10¹³ s⁻¹ for first-order reactions and about 10⁸–10¹¹ M⁻¹s⁻¹ for bimolecular gas-phase reactions.
• Limit: as T → ∞, k approaches A.

The equation fits data very well over a modest range of temperature, but a plot may curve over a wide range, because A itself depends slightly on T. For some reactions, such as those with a change in mechanism or enzyme-catalyzed reactions at high temperature, the equation does not apply.

Who was Arrhenius?

Svante Arrhenius (1859–1927) was a Swedish scientist. In 1889 he explained the empirical temperature dependence found by van ‘t Hoff by proposing that reacting molecules must first become “active” by acquiring a minimum energy. He is also known for the theory of electrolytic dissociation (Nobel Prize in Chemistry, 1903) and for an early calculation, in 1896, of the effect of atmospheric carbon dioxide on the temperature of the Earth.

Which mistakes are common?

• Using Celsius: T must be in kelvins.
• Unit conflict for R: if E_a is in kJ/mol, use R = 0.008314 kJ/(mol·K).
• Wrong sign of the slope: the slope of ln k against 1/T is negative, E_a = −slope × R.
• Swapping T₁ and T₂: k₂ goes with T₂.
• Using too few points: two points give an estimate, and several give an average and show the quality of the fit.