How rate constants depend on temperature

The Arrhenius equation gives the rate constant as

Taking the natural log of both sides gives a straight line when ln k is plotted against 1/T:

So the slope of an Arrhenius plot gives the activation energy, and the intercept gives A. A steeper line means a larger E_a and a rate constant that is more sensitive to temperature. The intercept lies far off the plotted range, which is why A is usually quoted with a large uncertainty.

The temperature coefficient Q₁₀ is the factor by which k increases for a 10 °C rise: Q₁₀ = k(T+10)/k(T) = exp[(Ea/R)(1/T − 1/(T+10))]. It depends only on E_a and T, not on A. The old rule of thumb that "rates double for every 10 °C" corresponds to E_a ≈ 53 kJ mol⁻¹ near room temperature.

What limits A? For a first-order (unimolecular) step, transition-state theory gives k = (k_BT/h) · e^ΔS‡/R · e^−ΔH‡/RT. At 25 °C, k_BT/h ≈ 6 × 10¹² s⁻¹, about the frequency of a molecular vibration, so A is near 10¹³ s⁻¹ when the activation entropy ΔS‡ is close to zero. A large positive ΔS‡ makes A much larger. Protein unfolding, which breaks many contacts at once, has apparent A values of 10⁴⁰ s⁻¹ or more; there A no longer means an attempt frequency.

For a second-order (bimolecular) step, A reflects how often the two reactants meet. In water, molecules diffuse together about 10⁹–10¹⁰ times per second per molar concentration, so no bimolecular rate constant in solution can exceed roughly 10¹⁰ M⁻¹ s⁻¹. The shaded band on the plots marks this diffusion limit. Many reactions fall well below it because only collisions with the right orientation and enough energy react.

The half-life is t_½ = ln 2 / k for a first-order reaction. For a second-order reaction it depends on concentration: t_½ = 1/(k[A]₀) for 2A → products, or for A + B with equal starting concentrations.

Everything here is calculated directly from the equation; no fitting or numerical approximation is involved.