Activation Energy: Why Temperature Speeds Up Reactions

The energy barrier on a reaction diagram

A reaction diagram plots energy against the progress of a reaction. Reactants sit on the left and products on the right, and between them the curve rises to a maximum. The arrangement of atoms at that maximum is the transition state, or activated complex: a short-lived structure in which old bonds are partly broken and new ones partly formed.

The activation energy, Ea, is the height of the barrier measured from the reactants. The reverse reaction has its own barrier, measured from the products. The two barriers and the enthalpy change are linked: ΔH = Ea(forward) − Ea(reverse). For example, if a reaction has Ea(forward) = 75 kJ/mol and ΔH = −40 kJ/mol, the reverse barrier is 75 − (−40) = 115 kJ/mol. An exothermic reaction always has a higher reverse barrier than forward barrier.

The size of ΔH does not tell you the size of Ea. A strongly exothermic reaction can still be extremely slow if its barrier is high. The combustion of a hydrocarbon in air at room temperature is an everyday example.

Collisions, energy and orientation

Collision theory says that molecules must collide to react, and that a collision leads to reaction only if two conditions are met. The molecules must collide with at least the activation energy, and they must be oriented so that the right atoms meet.

At any temperature, molecular kinetic energies are spread over a wide range. The fraction of collisions with energy of at least Ea is roughly exp(−Ea/RT), where exp(x) means e raised to the power x. That fraction is small but very sensitive to temperature. For Ea = 50.0 kJ/mol, Ea/RT is 20.17 at 298.15 K and 19.52 at 308.15 K, so the fraction rises from 1.7 × 10⁻⁹ to 3.3 × 10⁻⁹, a factor of 1.9. A 10 °C rise, only about 3 percent more in kelvins, nearly doubles the rate.

The familiar rule of thumb that rates double for every 10 °C rise holds only near room temperature and for barriers of about 50 kJ/mol. Between 25 °C and 35 °C, a 20 kJ/mol barrier gives a factor of only 1.3, and a 100 kJ/mol barrier gives a factor of 3.7.

Three forms of the Arrhenius equation

The Arrhenius equation is k = A exp(−Ea/RT). Here A is the frequency factor, which reflects how often molecules collide and how many collisions have a suitable orientation. R is the gas constant, 8.314 J mol⁻¹ K⁻¹ (NIST CODATA value, rounded), and T is in kelvins. Ea must be in joules per mole when R is in joules.

Taking natural logarithms gives a straight-line form: ln k = −(Ea/R)(1/T) + ln A. A plot of ln k against 1/T has slope −Ea/R and intercept ln A. This is the best method when you have rate constants at several temperatures.

Writing the linear form for two temperatures and subtracting gives the two-point form: ln(k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁). With any four of the quantities k₁, k₂, T₁, T₂ and Ea known, it gives the fifth. The frequency factor cancels, so you do not need to know it.

Worked example: Ea from two rate constants

OpenStax section 12.5 lists second-order rate constants for the decomposition 2 HI(g) → H₂(g) + I₂(g) at five temperatures. Two of them are k₁ = 1.22 × 10⁻⁶ L mol⁻¹ s⁻¹ at T₁ = 575 K and k₂ = 1.16 × 10⁻³ L mol⁻¹ s⁻¹ at T₂ = 700 K.

Rearrange the two-point form: Ea = R ln(k₂/k₁) / (1/T₁ − 1/T₂). The ratio k₂/k₁ is 951, and ln 951 = 6.857. The reciprocal temperatures are 1/575 K = 1.7391 × 10⁻³ K⁻¹ and 1/700 K = 1.4286 × 10⁻³ K⁻¹, which differ by 3.106 × 10⁻⁴ K⁻¹. So Ea = (8.314 J mol⁻¹ K⁻¹)(6.857) / (3.106 × 10⁻⁴ K⁻¹) = 1.84 × 10⁵ J/mol, or 184 kJ/mol.

Keep extra digits in 1/T. The difference between two reciprocals is much smaller than either one, so rounding them to two significant figures first can shift Ea by several percent.

Worked example: a five-point Arrhenius plot and a prediction

Using all five data points is better than using two, because it averages out the error in each measurement. The OpenStax values are k = 3.52 × 10⁻⁷, 1.22 × 10⁻⁶, 8.59 × 10⁻⁵, 1.16 × 10⁻³ and 3.95 × 10⁻² L mol⁻¹ s⁻¹ at 555, 575, 645, 700 and 781 K. Convert each pair to 1/T and ln k, then fit a straight line by least squares, using a spreadsheet or graphing calculator.

The fitted slope is −2.229 × 10⁴ K, and the points lie very close to the line (R² = 0.9996). Then Ea = −slope × R = 2.229 × 10⁴ K × 8.314 J mol⁻¹ K⁻¹ = 1.85 × 10⁵ J/mol, or 185 kJ/mol, in close agreement with the two-point result. The intercept is 25.20, so A = e raised to the power 25.20 = 8.8 × 10¹⁰ L mol⁻¹ s⁻¹.

Use the fitted equation to predict k at 625 K, a temperature between the measured points: k = (8.8 × 10¹⁰) × e raised to the power −(1.853 × 10⁵)/(8.314 × 625) = 2.9 × 10⁻⁵ L mol⁻¹ s⁻¹. The prediction sits between the measured values at 575 K and 645 K, as it should. Predictions far outside the measured range are less reliable, because Ea and A can drift with temperature.

How a catalyst changes the barrier

A catalyst provides a different mechanism with a lower activation energy and is regenerated at the end. It lowers the barrier for the forward and reverse reactions by the same amount, so it speeds both directions equally. It does not change ΔH, and it does not change the position of equilibrium. It only helps the system reach equilibrium sooner.

Worked example. Suppose an enzyme lowers Ea by 20.0 kJ/mol at body temperature, 37.0 °C or 310.15 K, and A stays the same. The rate constant increases by a factor of e raised to the power (20 000 J/mol)/(8.314 J mol⁻¹ K⁻¹ × 310.15 K) = e raised to the power 7.756 = 2.3 × 10³. A modest-looking change in the barrier gives a rate more than two thousand times faster, because Ea sits in an exponent.

Arrhenius calculation errors

Before reporting an activation energy, check that it is positive and of a sensible size. Most activation energies for reactions studied in general chemistry are tens to a few hundred kilojoules per mole.

• Mixing kJ and J: with R = 8.314 J mol⁻¹ K⁻¹, Ea must be in J/mol. A value a thousand times too small or too large usually means this slip.
• Using Celsius temperatures. 1/T must use kelvins.
• Swapping k₁ and k₂ relative to T₁ and T₂, which flips the sign of Ea.
• Using log instead of ln. If you plot log₁₀ k against 1/T, the slope is −Ea/(2.303R), not −Ea/R.
• Rounding 1/T values early, which magnifies error in their small difference.
• Applying the 10 °C doubling rule as if it were exact.