Result

The Arrhenius equation

T-040Home CU-202Threads kinetics

Statement

Temperature dependence of rate and the activation energy.

Why it matters

rate-law-order and integrated-rate-laws establish how the rate of a reaction depends on the concentrations of the species present, at fixed temperature; this result supplies the other half of the picture, how the rate constant \(k\) itself depends on temperature. The observation that reaction rates typically rise steeply with temperature, far faster than any simple power-law scaling of molecular speed or diffusion could explain, is one of the central empirical facts of kinetics, and the exponential form derived here is the standard tool for fitting it.

The equation also supplies the practical route to a reaction's activation energy \(E_a\) from nothing more than a handful of rate measurements at different temperatures, a quantity that collision-theory and transition-state-theory then go on to give a physical, mechanistic interpretation of. Later, catalysis-activation-energy revisits exactly this exponential form to explain quantitatively why even a modest reduction in \(E_a\) produces a disproportionately large increase in rate.

Hypotheses

A single, effective activation energy \(E_a\) governs the reaction over the temperature range studied.This holds for an elementary reaction, or for an overall reaction whose rate-determining step does not change across the range; for a genuinely multi-step mechanism (steady-state-approximation), the "activation energy" read off an Arrhenius plot is in general a composite of several steps' barriers, not one microscopic quantity. The pre-exponential factor \(A\) varies only weakly with temperature compared with the exponential term.Collision theory shows \(A\) carries a mild \(\sqrt{T}\) dependence through the mean molecular speed; since the exponential term changes far more steeply with \(T\) whenever \(E_a\) is not tiny, treating \(A\) as constant over a modest range is an excellent approximation, though a modified form \(k=AT^{m}e^{-E_a/RT}\) is used when a wider range or higher precision is needed. Over a genuinely wide temperature range, or for a reaction with an unusually small \(E_a\), a plot of \(\ln k\) against \(1/T\) can show detectable curvature; this signals the single-\(E_a\), constant-\(A\) approximation breaking down, not anomalous chemistry.

Proof

\[ \text{Fraction of collisions with energy} \ge E_a \text{ in a Boltzmann distribution} \propto e^{-E_a/RT} \]

The Boltzmann distribution of molecular kinetic energies (an established result of statistical thermodynamics, cited here) gives the fraction of collisions energetic enough to clear a barrier of height \(E_a\) as an exponential in \(-E_a/RT\); this exponential is the physical origin of the equation's characteristic temperature sensitivity. A

\[ k = A\,e^{-E_a/RT} \]

Writing the rate constant as the product of a collision-frequency-and-orientation factor \(A\) (approximately \(T\)-independent, Hypotheses) and the energetic fraction of Step 1 gives the Arrhenius equation directly; collision-theory develops the physical content of \(A\) in full. A

\[ \ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T} \]

Taking the natural logarithm converts the exponential relationship into a straight line in \(\ln k\) against \(1/T\), with slope \(-E_a/R\) and intercept \(\ln A\) — the standard Arrhenius plot used to extract \(E_a\) from rate data measured at several temperatures. A

\[ \ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]

Subtracting Step 3's expression at two temperatures \(T_1,T_2\) eliminates the unknown \(A\), giving a two-point form that determines \(E_a\) from just two rate-constant measurements without ever needing \(A\) itself. A

Result

\[ k = A\,e^{-E_a/RT} \]

Reading. Rate constants rise exponentially with temperature because an exponentially larger fraction of collisions clear the energy barrier \(E_a\) as \(T\) increases; the two adjustable parameters \(A\) and \(E_a\) are read off the slope and intercept of a plot of \(\ln k\) against \(1/T\).

Scope. Valid over ranges narrow enough that \(E_a\) and \(A\) can be treated as constants (Hypotheses); breaks down for a genuinely non-elementary reaction whose rate-determining step changes with temperature, and for the occasional reaction with a negative apparent \(E_a\).

Corollaries & converses

The two-point form of Step 4 is the standard practical tool: measuring \(k\) at just two temperatures is, in principle, enough to determine \(E_a\), though fitting a full Arrhenius plot from several temperatures gives a more reliable value.
Because \(E_a\) enters the exponent divided by \(RT\), a larger activation energy makes a reaction's rate more sensitive to temperature; this underlies the common laboratory rule of thumb that many reactions' rates roughly double for every \(10\,\text{K}\) rise near room temperature, for \(E_a\) in the typical \(50\text{-}60\,\text{kJ/mol}\) range.
catalysis-activation-energy applies this same exponential sensitivity in reverse: because \(k\) depends exponentially on \(-E_a\), even a modest reduction in barrier height produces a disproportionately large rate increase.

Fails without

Drop the single, effective-\(E_a\) assumption for a genuinely multi-step mechanism: the "activation energy" read from an Arrhenius plot becomes a composite of several steps' barriers, and can even show curvature, or an apparent negative \(E_a\), if the rate-determining step changes across the temperature range studied (steady-state-approximation).
Extrapolate the fitted line far outside the temperature range \(A\) and \(E_a\) were determined over: the mild \(\sqrt{T}\) dependence of \(A\) neglected in the Hypotheses becomes non-negligible over a wide range, so a plot of \(\ln k\) against \(1/T\) can show detectable curvature, making the extrapolated \(k\) unreliable.

Common errors

Using temperature in \(^\circ\text{C}\) rather than kelvin — the Arrhenius equation requires an absolute temperature scale throughout.
Mismatching the units of \(E_a\) and \(R\) (combining \(E_a\) in \(\text{kJ/mol}\) directly with \(R=8.314\,\text{J mol}^{-1}\text{K}^{-1}\)) when evaluating the exponent.
Treating an extracted \(E_a\) as necessarily the barrier height of one elementary step, when the studied reaction is actually multi-step (Hypotheses).
Sign errors when rearranging the two-point form of Step 4, particularly over which temperature is subtracted from which.

Discussion

Svante Arrhenius proposed this temperature dependence in 1889, building on an earlier empirical observation by Jacobus van 't Hoff that equilibrium constants show a closely analogous exponential temperature dependence (the van 't Hoff equation, of the same mathematical form, with \(E_a\) replaced by a reaction enthalpy). At the time Arrhenius's equation was itself a fitted empirical relationship; a genuine physical derivation of it, in terms of molecular collisions and energy barriers, came only later, through collision-theory and, more rigorously still, through transition-state-theory.

Common misconception: that the pre-exponential factor \(A\) is a fixed constant characteristic only of the reactants' identity. Collision theory shows \(A\) packages together the collision frequency and the fraction of collisions with a favourable relative orientation (the steric factor), both of which depend on the specific reaction studied, not on a universal constant.

Worked examples

\[ T_1=298\,\text{K},\ k_1=3.0\times10^{-4}\,\text{s}^{-1}; \qquad T_2=308\,\text{K},\ k_2=1.0\times10^{-3}\,\text{s}^{-1} \]

A first-order rate constant is measured at two temperatures ten kelvin apart; the two-point Arrhenius equation (Step 4) extracts \(E_a\) directly, without needing \(A\). A

\[ E_a = -R\cdot\frac{\ln(k_2/k_1)}{\tfrac{1}{T_2}-\tfrac{1}{T_1}} = -8.314\times\frac{\ln(3.33)}{(1/308-1/298)}\approx 9.1\times10^{4}\,\text{J/mol} \]

Substituting the measured values gives \(E_a\approx91\,\text{kJ/mol}\), a typical magnitude for a reaction whose rate more than triples over a ten-degree rise, consistent with what was observed. A

\[ E_a \approx 91\,\text{kJ/mol} \]

Reading. Two rate measurements ten degrees apart are sufficient to pin down the activation energy via the two-point Arrhenius equation.

Scope. The same procedure, applied to several temperature points and a linear least-squares fit of \(\ln k\) against \(1/T\), is the more statistically reliable version of this calculation used in practice.

Further example. A chemical reaction has an activation energy of \(E_a = 60.0\,\text{kJ/mol}\). The rate constant is measured to be \(k_1 = 2.5 \times 10^{-3}\,\text{s}^{-1}\) at \(T_1 = 298\,\text{K}\). Calculate the rate constant \(k_2\) at \(T_2 = 323\,\text{K}\).

\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

This is the two-point form of the Arrhenius equation, derived by taking the natural logarithm of the ratio of the rate constants at two different temperatures. A

\[ \frac{E_a}{R} = \frac{60000\,\text{J/mol}}{8.314\,\text{J/(mol K)}} \approx 7217\,\text{K} \]

We calculate the pre-factor by dividing the activation energy by the gas constant, ensuring units are consistent (Joules). A

\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{298} - \frac{1}{323} \approx 0.003356 - 0.003096 = 2.60 \times 10^{-4}\,\text{K}^{-1} \]

We compute the difference in the inverse temperatures, which represents the thermal driving force for the change in rate. A

\[ \ln\left(\frac{k_2}{k_1}\right) = 7217 \times 2.60 \times 10^{-4} \approx 1.876 \]

Multiplying the pre-factor by the temperature term gives the natural logarithm of the rate constant ratio. A

\[ \frac{k_2}{k_1} = e^{1.876} \approx 6.53 \]

Exponentiating the result yields the factor by which the rate constant increases. A

\[ k_2 = k_1 \times 6.53 = 2.5 \times 10^{-3}\,\text{s}^{-1} \times 6.53 \approx 1.63 \times 10^{-2}\,\text{s}^{-1} \]

Finally, we multiply the initial rate constant by the calculated factor to find the new rate constant. A

\[ k_2 \approx 1.6 \times 10^{-2}\,\text{s}^{-1} \]

Reading. A 25-degree increase in temperature increases the rate constant by a factor of approximately 6.5 for a reaction with this activation energy.

Scope. This calculation assumes that the activation energy is constant over the temperature range and that the reaction follows Arrhenius behavior.

Problems

A reaction's rate constant increases from \(1.0\times10^{-5}\,\text{s}^{-1}\) at \(20^\circ\text{C}\) to \(2.5\times10^{-4}\,\text{s}^{-1}\) at \(50^\circ\text{C}\). Find \(E_a\).
Solution
Convert to kelvin: \(T_1=293\,\text{K}\), \(T_2=323\,\text{K}\). \(\ln(k_2/k_1)=\ln(25)=3.22\). \(\tfrac{1}{T_1}-\tfrac{1}{T_2}=\tfrac{1}{293}-\tfrac{1}{323}=3.17\times10^{-4}\,\text{K}^{-1}\). \(E_a=R\times3.22/3.17\times10^{-4}=8.314\times3.22/3.17\times10^{-4}\approx8.4\times10^{4}\,\text{J/mol}\approx84\,\text{kJ/mol}\).
Given \(A=5.0\times10^{13}\,\text{s}^{-1}\) and \(E_a=100\,\text{kJ/mol}\), estimate \(k\) at \(400\,\text{K}\).
Solution
\(E_a/RT = 100000/(8.314\times400)=30.1\). \(e^{-30.1}\approx8.5\times10^{-14}\). \(k=A e^{-E_a/RT}\approx5.0\times10^{13}\times8.5\times10^{-14}\approx4.3\,\text{s}^{-1}\).
Explain qualitatively why doubling \(E_a\) at fixed \(T\) reduces \(k\) far more drastically than doubling \(T\) at fixed \(E_a\) increases it, for a typical \(E_a\) around \(80\,\text{kJ/mol}\) near room temperature.
Solution
\(k\) depends on \(E_a\) through the ratio \(E_a/RT\) inside an exponential: at \(298\,\text{K}\), \(RT\approx2.5\,\text{kJ/mol}\), so an \(E_a\) of \(80\,\text{kJ/mol}\) corresponds to an exponent of roughly \(-32\); doubling \(E_a\) to \(160\,\text{kJ/mol}\) roughly squares an already tiny factor, crashing \(k\) by many orders of magnitude. Doubling \(T\) instead only halves the exponent's magnitude (to about \(-16\), since \(RT\) itself scales with \(T\)), a comparatively mild change; because the exponent scales as \(1/T\) rather than linearly, the two operations are far from symmetric in their effect on \(k\).
A certain enzyme-catalyzed reaction has an activation energy of \(E_a = 50.0\,\text{kJ/mol}\). By what factor does the rate constant increase when the temperature is raised from \(25^\circ\text{C}\) to \(35^\circ\text{C}\)?
Solution
First, convert temperatures to Kelvin: \(T_1 = 25 + 273.15 = 298.15\,\text{K}\) \(T_2 = 35 + 273.15 = 308.15\,\text{K}\) Use the two-point Arrhenius equation: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substitute the values (\(E_a = 50000\,\text{J/mol}\), \(R = 8.314\,\text{J/(mol K)}\)): \[ \frac{E_a}{R} = \frac{50000}{8.314} \approx 6014\,\text{K} \] Calculate the temperature term: \[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{298.15} - \frac{1}{308.15} \approx 0.003354 - 0.003245 = 1.09\times10^{-4}\,\text{K}^{-1} \] Calculate the natural log of the ratio: \[ \ln\left(\frac{k_2}{k_1}\right) = 6014 \times 1.09\times10^{-4} \approx 0.656 \] Solve for the factor \(k_2/k_1\): \[ \frac{k_2}{k_1} = e^{0.656} \approx 1.92 \] The rate constant increases by a factor of approximately 1.92.
For a reaction with an activation energy of \(E_a = 75.0\,\text{kJ/mol}\), the rate constant is \(k_1 = 0.050\,\text{s}^{-1}\) at \(300\,\text{K}\). Calculate the rate constant \(k_2\) at \(350\,\text{K}\).
Solution
Given: \(E_a = 75000\,\text{J/mol}\) \(T_1 = 300\,\text{K}\) \(T_2 = 350\,\text{K}\) \(k_1 = 0.050\,\text{s}^{-1}\) \(R = 8.314\,\text{J/(mol K)}\) Use the Arrhenius relation: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Calculate the pre-factor: \[ \frac{E_a}{R} = \frac{75000}{8.314} \approx 9021\,\text{K} \] Calculate the temperature difference term: \[ \frac{1}{300} - \frac{1}{350} = 0.003333 - 0.002857 = 4.762\times10^{-4}\,\text{K}^{-1} \] Calculate the exponent: \[ \ln\left(\frac{k_2}{k_1}\right) = 9021 \times 4.762\times10^{-4} \approx 4.296 \] Exponentiate to find the ratio: \[ \frac{k_2}{k_1} = e^{4.296} \approx 73.4 \] Calculate \(k_2\): \[ k_2 = k_1 \times 73.4 = 0.050 \times 73.4 \approx 3.67\,\text{s}^{-1} \] The rate constant at 350 K is approximately \(3.7\,\text{s}^{-1}\).