The Arrhenius equation and activation energy

Result

The Arrhenius equation and activation energy

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Statement

Let a bimolecular gas-phase step \(\text{A}+\text{B}\to\text{products}\) proceed in a dilute gas held at thermal equilibrium, so that relative translational energy is Maxwell–Boltzmann distributed, and let a pair react whenever the component of relative kinetic energy directed along the line of centres at contact exceeds a fixed threshold \(\varepsilon_0\). Then the thermally averaged rate constant is exactly \(k(T)=N_{\!A}\,\sigma_{\!AB}\sqrt{\dfrac{8RT}{\pi M_\mu}}\;e^{-E_0/RT}\), with \(\sigma_{\!AB}=\pi d_{AB}^{2}\), \(M_\mu\) the molar reduced mass and \(E_0=N_{\!A}\varepsilon_0\). Defining the activation energy by \(E_a\equiv RT^{2}\,\dfrac{d\ln k}{dT}=-R\,\dfrac{d\ln k}{d(1/T)}\) — a definition, valid for any \(k(T)\) whatever — this model gives \(E_a=E_0+\tfrac12RT\). Over any interval narrow enough that \(E_a\) may be treated as constant, integrating the definition returns the Arrhenius equation \(k=A\,e^{-E_a/RT}\), equivalently \(\ln k=\ln A-E_a/(RT)\), with \(A=e^{1/2}N_{\!A}\sigma_{\!AB}\langle v_{\text{rel}}\rangle\) in this model and \(A=p\,e^{1/2}N_{\!A}\sigma_{\!AB}\langle v_{\text{rel}}\rangle\) once a steric factor \(p\le 1\) is admitted.

Why it matters

ideal-gas-law, kinetic-theory-pressure and maxwell-boltzmann-speeds build a complete statistical picture of a gas: how many molecules there are, how fast they move, and how their speeds are distributed. That picture has so far been used only for equilibrium properties — pressure, mean speed, effusion rate (grahams-law-effusion). This result is where it first pays for itself dynamically. The same distribution that fixes \(v_{\text{rms}}\) also fixes what fraction of molecular encounters are violent enough to break bonds, and the answer is an exponential in \(-1/T\). The steep, unmistakable temperature dependence of chemical rate — a factor of two for ten kelvin near room temperature, a factor of \(10^{5}\) between \(300\) and \(600\,\text{K}\) — is therefore not a separate empirical law of kinetics but a direct consequence of the shape of the Maxwell–Boltzmann tail.

The practical payoff is a bridge in both directions. Forwards: given a collision diameter and a barrier height, kinetic theory predicts an absolute second-order rate constant with no fitted parameters at all. Backwards, and far more usefully: two rate measurements at two temperatures deliver \(E_a\), and a single measurement in addition delivers \(A\), so an experimenter converts a table of numbers into a barrier height and a collision efficiency. arrhenius-equation states the fitting law itself from the kinetics side; here it is derived, its activation energy is given an exact definition rather than a slope-of-a-graph one, and the systematic gap between the fitted \(E_a\) and the microscopic threshold \(E_0\) is computed rather than ignored.

Hypotheses

The gas is dilute and remains at thermal equilibrium throughout the reaction.The whole derivation rests on inserting the Maxwell–Boltzmann distribution into a collision integral, which presupposes that elastic collisions repopulate the high-energy tail far faster than reaction drains it. For a reaction with a low barrier, or at a pressure so low that the mean free path approaches the vessel dimensions, the tail is depleted, the effective temperature of the reacting subpopulation falls below \(T\), and the measured rate lies below the equilibrium prediction — the origin of the fall-off behaviour of unimolecular reactions at low pressure.

Molecules are hard spheres of fixed diameter \(d_{AB}=\tfrac12(d_A+d_B)\), so the collision cross-section \(\sigma_{\!AB}=\pi d_{AB}^{2}\) is a temperature-independent constant.Real intermolecular potentials are soft and attractive at long range, so the effective cross-section falls slowly with temperature as fast molecules are deflected less. Ignoring this puts a weak, unmodelled temperature dependence into \(A\); it is small compared with the exponential over ordinary ranges, and it is one of the several reasons the fitted \(A\) is not to be read as a literal collision frequency.

Reaction requires only that the line-of-centres component of relative translational energy reach a fixed threshold \(\varepsilon_0\), independent of temperature.Two separate idealisations sit here. Translational energy alone is counted, so vibrational and rotational energy — which in most real reactions contribute, sometimes dominantly — is treated as a spectator; and orientation is ignored, so every sufficiently energetic collision reacts. Dropping the second alone is what the steric factor \(p\) repairs, empirically. Doing the job properly requires the full potential energy surface, which is transition-state-theory's business.

Motion over the barrier is classical: no tunnelling.A particle with energy below \(\varepsilon_0\) is assigned zero reaction probability. For transfer of a hydrogen atom or a proton at low temperature this is false, and the tunnelling contribution — nearly temperature-independent, since it does not need thermal energy — eventually dominates the exponentially vanishing over-barrier term. The Arrhenius plot then flattens as \(T\) falls and the apparent \(E_a\) tends towards zero.

The measured \(k\) belongs to one elementary step, and the same step remains rate-determining across the whole temperature interval.Every step below concerns a single barrier. A composite rate constant assembled from a pre-equilibrium and a subsequent step still has a perfectly well-defined \(E_a\) by the differential definition, but that \(E_a\) is a signed combination of several energies and can even be negative; it is not a barrier height, and no collision-theory prefactor should be attached to it.

Proof

The argument has three distinct stages, and it is worth keeping them apart. Stage one (Steps 1–2) is pure statistical mechanics: what is the distribution of relative translational energy in a two-component gas. Stage two (Steps 3–7) is the collision integral, evaluated exactly for the line-of-centres model — no appeal is made anywhere to a hand-waved “fraction of molecules with enough energy”. Stage three (Steps 8–10) defines the activation energy properly and shows in what precise sense the Arrhenius form follows. All symbols are rearranged before any number is substituted.

\[ \tfrac12 m_A v_A^{2}+\tfrac12 m_B v_B^{2} \;=\; \tfrac12 M V_{\text{cm}}^{2}+\tfrac12 \mu v_r^{2}, \qquad M=m_A+m_B,\quad \mu=\frac{m_Am_B}{m_A+m_B} \]

The exact kinematic identity obtained by changing variables from \((\mathbf{v}_A,\mathbf{v}_B)\) to the centre-of-mass velocity \(\mathbf{V}_{\text{cm}}\) and the relative velocity \(\mathbf{v}_r=\mathbf{v}_A-\mathbf{v}_B\). The transformation has unit Jacobian, so the product of the two independent Maxwell–Boltzmann factors \(e^{-m_Av_A^{2}/2k_BT}e^{-m_Bv_B^{2}/2k_BT}\) factorises into one Maxwellian in \(\mathbf{V}_{\text{cm}}\) and one in \(\mathbf{v}_r\). B

\[ f(v_r)\,dv_r = 4\pi\left(\frac{\mu}{2\pi k_BT}\right)^{3/2}v_r^{2}\,e^{-\mu v_r^{2}/2k_BT}\,dv_r \quad\Longrightarrow\quad f(\varepsilon)\,d\varepsilon = \frac{2}{\sqrt{\pi}}\,\frac{\varepsilon^{1/2}}{(k_BT)^{3/2}}\,e^{-\varepsilon/k_BT}\,d\varepsilon \]

The relative motion is therefore distributed exactly as maxwell-boltzmann-speeds describes a single species, but with \(\mu\) in place of \(m\) — this is the only structural fact stage one contributes. Converting to the relative translational energy \(\varepsilon=\tfrac12\mu v_r^{2}\) uses \(d\varepsilon=\mu v_r\,dv_r\) and \(v_r=\sqrt{2\varepsilon/\mu}\); the reduced mass cancels identically, which is why the energy distribution is mass-independent. Normalisation checks out, since \(\int_0^\infty x^{1/2}e^{-x}dx=\Gamma(3/2)=\sqrt{\pi}/2\). B

\[ dZ_{AB} = \sigma_{\!AB}\,v_r\,n_An_B\,f(v_r)\,dv_r \]

The collision density resolved by relative speed. In the frame of a B molecule, an A molecule sweeps out a cylinder of cross-section \(\sigma_{\!AB}=\pi d_{AB}^{2}\) and length \(v_r\,dt\) in time \(dt\); the expected number of B centres inside it is \(n_B\sigma_{\!AB}v_r\,dt\), and multiplying by the number density \(n_A\) of A gives collisions per unit volume per unit time. The extra factor of \(v_r\) beyond \(f\) is essential: fast pairs collide more often than their share of the population, and this flux weighting is what makes the final answer differ from a naive population fraction. A

\[ \varepsilon_{\text{loc}} = \varepsilon\left(1-\frac{b^{2}}{d_{AB}^{2}}\right) \ge \varepsilon_0 \quad\Longleftrightarrow\quad b^{2}\le d_{AB}^{2}\left(1-\frac{\varepsilon_0}{\varepsilon}\right) \quad\Longrightarrow\quad \sigma_r(\varepsilon)=\begin{cases} 0, & \varepsilon\lt\varepsilon_0,\\[2pt] \sigma_{\!AB}\left(1-\dfrac{\varepsilon_0}{\varepsilon}\right), & \varepsilon\ge\varepsilon_0. \end{cases} \]

The line-of-centres criterion, made quantitative. For a collision with impact parameter \(b\), the velocity component along the line joining the centres at contact is \(v_r\sqrt{1-b^{2}/d_{AB}^{2}}\), so the energy directed at the barrier is \(\varepsilon(1-b^{2}/d_{AB}^{2})\); a glancing collision (\(b\to d_{AB}\)) delivers nothing however fast it is. The reactive area \(\pi b_{\max}^{2}\) is then the stated energy-dependent cross-section, which rises from zero at threshold towards \(\sigma_{\!AB}\) at high energy. This step, not the Boltzmann factor, is where the chemistry enters. C

\[ -\frac{d[\text{A}]}{dt}=k[\text{A}][\text{B}] \quad\text{with}\quad [\text{A}]=\frac{n_A}{N_{\!A}} \quad\Longrightarrow\quad k = N_{\!A}\,\langle \sigma_r v_r\rangle = N_{\!A}\!\int_0^\infty \sigma_r(\varepsilon)\,v_r(\varepsilon)\,f(\varepsilon)\,d\varepsilon \]

Definition-matching, and the only place where units are at risk. The reactive event density is \(n_An_B\langle\sigma_r v_r\rangle\) molecules per unit volume per unit time; dividing by \(N_{\!A}\) converts to molar concentration per unit time and by \(n_An_B=N_{\!A}^{2}[\text{A}][\text{B}]\) isolates \(k\), leaving one factor of \(N_{\!A}\) upstairs. With SI inputs \(k\) emerges in \(\text{m}^{3}\,\text{mol}^{-1}\text{s}^{-1}\), which is \(10^{3}\) times the customary \(\text{dm}^{3}\,\text{mol}^{-1}\text{s}^{-1}\). A

\[ \begin{aligned} k &= N_{\!A}\int_{\varepsilon_0}^{\infty}\sigma_{\!AB}\left(1-\frac{\varepsilon_0}{\varepsilon}\right)\sqrt{\frac{2\varepsilon}{\mu}}\;\frac{2}{\sqrt{\pi}}\frac{\varepsilon^{1/2}}{(k_BT)^{3/2}}e^{-\varepsilon/k_BT}\,d\varepsilon \\[2pt] &= \frac{2N_{\!A}\sigma_{\!AB}}{\sqrt{\pi}}\sqrt{\frac{2}{\mu}}\;\frac{1}{(k_BT)^{3/2}}\int_{\varepsilon_0}^{\infty}(\varepsilon-\varepsilon_0)\,e^{-\varepsilon/k_BT}\,d\varepsilon \end{aligned} \]

Substituting Steps 2 and 4 into Step 5. The algebraic luck of the model is visible in the second line: the factor \(\varepsilon^{1/2}\) from the density of states and the factor \(\varepsilon^{1/2}\) from the speed \(v_r\) combine with the \((1-\varepsilon_0/\varepsilon)\) of the cross-section to leave the exactly integrable form \((\varepsilon-\varepsilon_0)\). The lower limit is \(\varepsilon_0\) because \(\sigma_r\) vanishes below it. B

\[ \int_{\varepsilon_0}^{\infty}(\varepsilon-\varepsilon_0)e^{-\varepsilon/k_BT}d\varepsilon = (k_BT)^{2}e^{-\varepsilon_0/k_BT} \quad\Longrightarrow\quad k = N_{\!A}\,\sigma_{\!AB}\underbrace{\sqrt{\frac{8k_BT}{\pi\mu}}}_{\textstyle \langle v_{\text{rel}}\rangle}\,e^{-\varepsilon_0/k_BT} = N_{\!A}\sigma_{\!AB}\sqrt{\frac{8RT}{\pi M_\mu}}\;e^{-E_0/RT} \]

The substitution \(x=(\varepsilon-\varepsilon_0)/k_BT\) turns the integral into \((k_BT)^{2}e^{-\varepsilon_0/k_BT}\int_0^\infty xe^{-x}dx\) and \(\int_0^\infty xe^{-x}dx=1\). Collecting the prefactors, \(\tfrac{2}{\sqrt{\pi}}\sqrt{2k_BT/\mu}=\sqrt{8k_BT/\pi\mu}\), which is precisely the mean relative speed of maxwell-boltzmann-speeds with \(\mu\) in place of \(m\). Multiplying numerator and denominator inside the root by \(N_{\!A}\) converts to molar quantities via \(R=N_{\!A}k_B\), \(M_\mu=N_{\!A}\mu\) and \(E_0=N_{\!A}\varepsilon_0\). The result is exact within the model: \(k\propto T^{1/2}e^{-E_0/RT}\). B

\[ E_a \;\equiv\; RT^{2}\,\frac{d\ln k}{dT} \;=\; -R\,\frac{d\ln k}{d(1/T)} \]

The definition of activation energy, due in this form to Tolman and adopted by IUPAC. It is a definition and not a law: it assigns an \(E_a\) to every differentiable \(k(T)\), whether or not the Arrhenius form holds, and it makes \(E_a\) in general a function of temperature — the local slope of the Arrhenius plot at the point of interest, times \(-R\). Everything said afterwards about “the” activation energy is a statement about how nearly constant this function is. B

\[ \ln k = \text{const}+\tfrac12\ln T-\frac{E_0}{RT} \quad\Longrightarrow\quad \frac{d\ln k}{dT}=\frac{1}{2T}+\frac{E_0}{RT^{2}} \quad\Longrightarrow\quad E_a = RT^{2}\left(\frac{1}{2T}+\frac{E_0}{RT^{2}}\right)=E_0+\tfrac12 RT \]

Applying Step 8 to Step 7. The activation energy exceeds the microscopic threshold by \(\tfrac12RT\), which is \(1.24\,\text{kJ}\,\text{mol}^{-1}\) at \(298\,\text{K}\) and \(2.5\,\text{kJ}\,\text{mol}^{-1}\) at \(600\,\text{K}\) — a one-to-two per cent effect for a typical barrier, but a systematic one, and the reason that a fitted \(E_a\) is not the same object as a computed barrier height. B

\[ \frac{d\ln k}{d(1/T)}=-\frac{E_a}{R}\ \ \text{with } E_a \text{ constant} \quad\Longrightarrow\quad \ln k = \ln A-\frac{E_a}{RT} \quad\Longrightarrow\quad k = A\,e^{-E_a/RT},\qquad A = p\,e^{1/2}N_{\!A}\sigma_{\!AB}\langle v_{\text{rel}}\rangle \]

Integrating the defining relation of Step 8 with \(E_a\) held constant; the constant of integration is named \(\ln A\). The identification of \(A\) follows from writing \(e^{-E_0/RT}=e^{-(E_a-RT/2)/RT}=e^{1/2}e^{-E_a/RT}\) in Step 7, so the Arrhenius pre-exponential exceeds the collision frequency factor by \(e^{1/2}=1.6487\); the steric factor \(p\le1\) is then appended to account for the orientational requirement that the hard-sphere model omits. Note carefully what has been proved: the Arrhenius form is not exact for this model, it is the result of freezing a slowly varying \(E_a\). A

\[ \langle\varepsilon\rangle_{\text{react}} = \frac{\displaystyle\int_{\varepsilon_0}^{\infty}\varepsilon(\varepsilon-\varepsilon_0)e^{-\varepsilon/k_BT}d\varepsilon}{\displaystyle\int_{\varepsilon_0}^{\infty}(\varepsilon-\varepsilon_0)e^{-\varepsilon/k_BT}d\varepsilon}=\varepsilon_0+2k_BT, \qquad \langle\varepsilon\rangle_{\text{all}}=\tfrac32 k_BT \quad\Longrightarrow\quad \langle\varepsilon\rangle_{\text{react}}-\langle\varepsilon\rangle_{\text{all}} = \varepsilon_0+\tfrac12 k_BT \]

Tolman's interpretation, and an independent confirmation of Step 9. The same substitution \(x=(\varepsilon-\varepsilon_0)/k_BT\) gives \(\int_0^\infty(\varepsilon_0+k_BTx)x e^{-x}dx=\varepsilon_0+2k_BT\) after dividing by the denominator, while \(\langle\varepsilon\rangle\) over the unweighted distribution of Step 2 is \(\Gamma(5/2)/\Gamma(3/2)\cdot k_BT=\tfrac32k_BT\). Multiplying by \(N_{\!A}\) reproduces \(E_a=E_0+\tfrac12RT\) exactly. So the activation energy is not a barrier height at all: it is the excess mean energy of those pairs that react over the mean energy of all pairs, which is a statement that survives even where no barrier exists. C

\[ \ln\frac{k_2}{k_1}=-\frac{1}{R}\int_{1/T_1}^{1/T_2}\!E_a(T)\,d\!\left(\frac1T\right) = -\frac{E_a(T^{*})}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right), \qquad T^{*}=\frac{T_1T_2\ln(T_2/T_1)}{T_2-T_1}\in(T_1,T_2) \]

What the two-point formula actually measures. Integrating Step 8 without assuming \(E_a\) constant and applying the mean value theorem for integrals shows that the two-point value is \(E_a\) evaluated at some interior temperature, never an unweighted average of the endpoints. For the specific model of Step 7, \(E_a=E_0+\tfrac12RT\) is linear in \(T\) and the integral can be done in closed form, which fixes \(T^{*}\) at the value shown — the reciprocal of the mean of \(1/T_1\) and \(1/T_2\) weighted logarithmically. Worked example 2 verifies this numerically to four figures. C

Result

\[ k(T)=p\,N_{\!A}\sigma_{\!AB}\sqrt{\frac{8RT}{\pi M_\mu}}\,e^{-E_0/RT}, \qquad E_a\equiv RT^{2}\frac{d\ln k}{dT}=E_0+\tfrac12RT, \qquad k=A\,e^{-E_a/RT} \]

Reading. Rate constants climb steeply with temperature because the Maxwell–Boltzmann tail above a fixed energy threshold grows exponentially in \(-E_0/RT\), and for no other reason: the collision frequency itself contributes only a feeble \(T^{1/2}\). The activation energy is defined as \(-R\) times the local slope of \(\ln k\) against \(1/T\); it equals the threshold plus \(\tfrac12RT\), and equivalently equals the mean energy of reacting pairs minus the mean energy of all pairs. The Arrhenius equation is the statement that this slope is sensibly constant over the measured interval.

Units check. \(N_{\!A}\sigma\langle v_{\text{rel}}\rangle\) has units \(\text{mol}^{-1}\cdot\text{m}^{2}\cdot\text{m s}^{-1}=\text{m}^{3}\,\text{mol}^{-1}\text{s}^{-1}\), correct for a second-order rate constant; multiply by \(10^{3}\) for \(\text{dm}^{3}\,\text{mol}^{-1}\text{s}^{-1}\). The exponent \(E_0/RT\) is dimensionless with \(E_0\) in \(\text{J}\,\text{mol}^{-1}\) and \(R=8.314\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\); \(M_\mu\) must be in \(\text{kg}\,\text{mol}^{-1}\), not \(\text{g}\,\text{mol}^{-1}\). Constants used throughout: \(R=8.314\,\text{J}\,\text{mol}^{-1}\text{K}^{-1}\), \(N_{\!A}=6.022\times10^{23}\,\text{mol}^{-1}\), \(k_B=R/N_{\!A}=1.381\times10^{-23}\,\text{J}\,\text{K}^{-1}\).

Scope. Exact for the hard-sphere, line-of-centres model in a dilute equilibrium gas. The Arrhenius form that follows from it is an approximation whose quality is quantified by Step 12; the absolute magnitude of \(A\) is reliable only for small, nearly spherical reactants, which is what \(p\) admits.

Corollaries & converses

The two-point formula. \(\ln(k_2/k_1)=-\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\), so \(E_a=R\ln(k_2/k_1)\big/\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\). Two measurements suffice, and \(A\) never enters. By Step 12 the number returned is \(E_a(T^{*})\) for an interior \(T^{*}\), so it should be quoted with the temperature range attached.
The Arrhenius plot. Plotting \(\ln k\) against \(1/T\) gives slope \(-E_a/R\) and intercept \(\ln A\). The intercept sits at \(1/T=0\), i.e. at infinite temperature, so \(A\) is always an extrapolation and is far less precisely determined than \(E_a\): since \(\ln A=\ln k+E_a/RT\), a one per cent error in the slope shifts \(\ln A\) by \(E_a/100RT\), a few tenths for the usual \(E_a/RT\approx20\)–\(40\) — already a factor of \(1.2\)–\(1.5\) in \(A\) itself — and a ten per cent slope error costs an order of magnitude or more.
Temperature sensitivity, quantified. \(\dfrac{d\ln k}{dT}=\dfrac{E_a}{RT^{2}}\), so a fractional rate increase per kelvin of \(E_a/RT^{2}\). Near \(298\,\text{K}\) this is about \(0.07\,\text{K}^{-1}\) for \(E_a=50\,\text{kJ}\,\text{mol}^{-1}\), which integrates to a factor of \(1.93\) over ten kelvin, rising to \(2.20\) for \(E_a=60\,\text{kJ}\,\text{mol}^{-1}\) — the familiar rule of thumb, now carrying the condition \(E_a\approx50\)–\(60\,\text{kJ}\,\text{mol}^{-1}\) that is usually left implicit.
Threshold from slope. \(E_0=E_a-\tfrac12RT\) for this model; for a \(T^{m}\) prefactor generally, \(E_a=E_0+mRT\) (Problems, 3), and transition-state-theory's Eyring form gives \(E_a=\Delta H^{\ddagger}+2RT\) for a gas-phase bimolecular step. Every theory relates \(E_a\) to its own microscopic energy by a small multiple of \(RT\); none of them makes the two identical.
Isotope and mass dependence of \(A\). Since \(A\propto \mu^{-1/2}\), substituting a heavier isotope lowers the prefactor only as the square root of the reduced mass — the same \(\sqrt{\mu}\) scaling that drives grahams-law-effusion. Observed kinetic isotope effects far larger than this are evidence that the barrier itself moved (zero-point energy) or that tunnelling contributes, not that collisions became rarer.
Converse, and it fails. A straight Arrhenius plot does not imply an elementary reaction, a single barrier, or the validity of collision theory. Composite rate constants are usually excellent straight lines, because a sum or product of exponentials in \(1/T\) is dominated by one term over a limited range. Linearity is weak evidence; only the magnitude of \(A\) and independent mechanistic data can distinguish the cases.

Fails without

Thermal equilibrium dropped (tail depletion, low pressure): if reaction removes energetic pairs faster than collisions replenish them, the reacting subpopulation is no longer Maxwell–Boltzmann and Step 2 is simply false. For unimolecular decompositions the same failure appears as the low-pressure fall-off region, where the observed order changes from first to second and the apparent \(E_a\) shifts because the energising collision, not the barrier crossing, has become rate-determining. The correction is a master-equation treatment, not a modified \(E_a\).
Classical over-barrier motion dropped (tunnelling): for hydrogen and proton transfers below roughly \(200\,\text{K}\) the through-barrier contribution, which depends only weakly on temperature, overtakes the exponentially collapsing over-barrier term. The Arrhenius plot bends towards the horizontal, the apparent \(E_a\) falls towards zero as \(T\to0\), and a naive linear fit through the low-temperature points returns a prefactor orders of magnitude below any collision frequency — a signature diagnostic of tunnelling rather than an anomaly of the data.
A single rate-determining step dropped: for a mechanism with a fast pre-equilibrium \(\text{A}+\text{B}\rightleftharpoons\text{C}\) followed by \(\text{C}\to\text{P}\), the composite constant is \(k=K_1k_2\), so by Step 8 and van-t-hoff-equation \(E_a=\Delta U_1^{\circ}+E_{a,2}\). If the association step is sufficiently exothermic this sum is negative and the reaction slows on heating — entirely consistent with the definition of \(E_a\), and entirely inconsistent with reading \(E_a\) as a barrier height. Reactions with bound intermediates routinely show exactly this.
The dilute-gas picture dropped (reaction in solution): encounters in a liquid are not independent collisions but repeated re-collisions inside a solvent cage, and the encounter rate is set by diffusion rather than by \(\sigma\langle v_{\text{rel}}\rangle\). For a diffusion-limited reaction the measured \(E_a\) is the activation energy of viscous flow of the solvent, around \(15\)–\(20\,\text{kJ}\,\text{mol}^{-1}\) in water, and carries no information about the chemical barrier whatsoever. The Arrhenius fit still works; its interpretation does not.
The narrow-range assumption dropped: over hundreds of kelvin the \(T^{1/2}\) of Step 7 and any temperature dependence of \(\sigma\) make \(E_a\) drift, and the plot acquires curvature \(d^{2}\ln k/d(1/T)^{2}=mT^{2}\) for a \(T^{m}\) prefactor (Problems, 3). For \(m=\tfrac12\) the drift is only about \(1.2\,\text{kJ}\,\text{mol}^{-1}\) between \(300\) and \(600\,\text{K}\), which is why the curvature is invisible in ordinary data — but it is not zero, and it sets a floor on how precisely \(E_a\) can be said to exist at all.

Common errors

“\(e^{-E_a/RT}\) is the fraction of molecules with energy greater than \(E_a\).” It is not, and the discrepancy is a factor of several. The fraction of pairs whose relative translational energy exceeds \(\varepsilon_0\) in three dimensions is \(\approx 2\sqrt{\varepsilon_0/\pi k_BT}\,e^{-\varepsilon_0/k_BT}\), larger than the exponential alone by about \(5\) at a typical barrier (Problems, 4). The clean exponential in the Result appears only after the flux weighting of Step 3 and the line-of-centres cross-section of Step 4 have been combined and integrated; it is an outcome, not an input.
Celsius temperatures. \(1/T\) and \(RT\) both demand the absolute scale. Using \(25\) in place of \(298.15\) changes the exponent by a factor of twelve.
Mixing \(\text{kJ}\,\text{mol}^{-1}\) with \(R\) in \(\text{J}\,\text{mol}^{-1}\text{K}^{-1}\). A factor of \(1000\) inside an exponential; the answer is wrong by tens of orders of magnitude and often looks superficially reasonable.
\(\text{g}\,\text{mol}^{-1}\) inside \(\langle v_{\text{rel}}\rangle\). The molar reduced mass must be in \(\text{kg}\,\text{mol}^{-1}\); the error costs a factor of \(\sqrt{1000}\approx31.6\) in \(A\), which is easily mistaken for a steric factor.
Equating \(E_a\) with \(E_0\), or with \(\Delta H^{\ddagger}\), or with a computed barrier height. These differ by multiples of \(RT\) that depend on the model and the molecularity: \(\tfrac12RT\) here, \(2RT\) for gas-phase bimolecular Eyring theory. Quoting four significant figures on an \(E_a\) and then comparing it with a quantum-chemical barrier without the correction is a routine and avoidable inconsistency.
Forgetting the factor \(e^{1/2}\) when comparing a fitted \(A\) with a collision frequency. Omitting it inflates the deduced steric factor by \(1.65\).
The identical-reactant factor of two. For \(\text{A}+\text{A}\), the \(\tfrac12\) that avoids double-counting collision pairs cancels against the two molecules consumed per event, so \(k\) defined by \(-d[\text{A}]/dt=k[\text{A}]^{2}\) is unchanged. But the IUPAC rate of reaction \(v=-\tfrac12 d[\text{A}]/dt\) gives a rate constant smaller by two. State which convention is in use.
“A negative activation energy is impossible.” It is impossible for an elementary step, because \(E_0\ge0\) forces \(E_a\gt0\); it is common for composite constants, and the definition in Step 8 accommodates it without difficulty.

Discussion

Arrhenius proposed the exponential temperature law in 1889, on empirical grounds and by explicit analogy with van 't Hoff's slightly earlier result for equilibrium constants — the two expressions have the same shape, and van-t-hoff-equation is the reason why: for an elementary reversible step \(K=k_f/k_r\), so the temperature derivatives must satisfy \(E_{a,f}-E_{a,r}=\Delta U^{\circ}\). What Arrhenius supplied was the form, together with the physical suggestion that only an activated minority of molecules can react. He did not supply the derivation. That came a generation later, when Trautz and, independently, Lewis assembled the kinetic-theory argument reproduced above; and the interpretation of \(E_a\) as a difference of mean energies, which is the most durable part of the whole subject, is Tolman's.

The structure worth remembering is the competition between a power law and an exponential. Collision frequency scales as \(T^{1/2}\); the reactive fraction scales as \(e^{-E_0/RT}\). Between \(300\) and \(310\,\text{K}\) with \(E_0=60\,\text{kJ}\,\text{mol}^{-1}\), the first factor contributes a rise of \(1.7\) per cent and the second a rise of \(117\) per cent. This is why a chemist may say without much guilt that “\(A\) is temperature-independent”, and it is also why the statement is never exactly true and should not be defended as though it were. The Arrhenius equation is best understood as the leading behaviour of a family of expressions \(k=A'T^{m}e^{-E_0/RT}\), all of which look linear on an Arrhenius plot over any interval a laboratory can conveniently span.

Hard-sphere collision theory is quantitatively poor and conceptually indispensable, and it is worth being precise about which is which. Its prediction of \(E_a\) is not a prediction at all, since \(E_0\) is an input; its prediction of \(A\) is testable, and for reactions between small non-polar molecules it is right to within an order of magnitude, while for reactions between large or highly oriented species the fitted \(p\) can fall to \(10^{-5}\) or below. The failure is structural, not numerical: a hard sphere has no internal coordinates, so it can neither store energy in vibration nor present a reactive face, and both of those are usually decisive. Transition-state theory replaces the sphere by a point on a potential energy surface and recovers the missing factors as an entropy of activation, \(A\propto e^{\Delta S^{\ddagger}/R}\), which is the same information reorganised into a form that can be estimated from molecular structure. The collision picture nevertheless remains the only elementary derivation in which every factor has an unambiguous mechanical meaning, and it is the natural place for the Boltzmann factor to make its first appearance in kinetics.

Common misconceptions. That raising the temperature works chiefly by making molecules move faster: it does that, but the mean speed rises only as \(T^{1/2}\), and essentially all of the rate increase comes from the reshaping of the tail of the distribution. That a catalyst works by increasing \(A\): it works by offering a route with a smaller \(E_0\), and because \(E_0\) sits in an exponent, a reduction of \(20\,\text{kJ}\,\text{mol}^{-1}\) multiplies the rate by \(e^{20000/RT}\approx3\times10^{3}\) at \(298\,\text{K}\). And that \(E_a\) is a property of the reactants alone: it is a property of the reaction path, so the same reactants have different activation energies on different mechanisms — which is precisely what makes catalysis possible.

Worked examples

Example 1. Suppose the second-order rate constant for the gas-phase decomposition \(2\,\text{HI}\to\text{H}_2+\text{I}_2\), with the rate defined by \(-d[\text{HI}]/dt=k[\text{HI}]^{2}\), is measured as \(k_1=3.52\times10^{-7}\,\text{dm}^{3}\text{mol}^{-1}\text{s}^{-1}\) at \(T_1=556\,\text{K}\) and \(k_2=3.02\times10^{-5}\,\text{dm}^{3}\text{mol}^{-1}\text{s}^{-1}\) at \(T_2=629\,\text{K}\). Find \(E_a\) and \(A\); then, taking a representative hard-sphere diameter \(d=3.5\times10^{-10}\,\text{m}\) for HI and the molar mass \(M(\text{HI})=127.91\,\text{g}\,\text{mol}^{-1}\), deduce the steric factor and the threshold energy \(E_0\).

\[ E_a = R\,\frac{\ln(k_2/k_1)}{\dfrac{1}{T_1}-\dfrac{1}{T_2}} = 8.314\times\frac{\ln(85.80)}{1.79856\times10^{-3}-1.58983\times10^{-3}} = 8.314\times\frac{4.4520}{2.0874\times10^{-4}} \]

The two-point form, rearranged for \(E_a\) before any number is inserted. Writing the denominator as \(1/T_1-1/T_2\) rather than \(1/T_2-1/T_1\) removes the leading minus sign and with it the commonest sign error. A

\[ E_a = 8.314\times 2.1328\times10^{4} = 1.773\times10^{5}\,\text{J}\,\text{mol}^{-1} = 177\,\text{kJ}\,\text{mol}^{-1} \]

A large barrier, consistent with a reaction that must break an H–I bond, and consistent with the observed eighty-six-fold rate rise over only \(73\,\text{K}\). A

\[ \ln A = \ln k_1+\frac{E_a}{RT_1} = -14.860+\frac{1.7732\times10^{5}}{8.314\times556} = -14.860+38.360 = 23.500 \quad\Longrightarrow\quad A = 1.61\times10^{10}\,\text{dm}^{3}\text{mol}^{-1}\text{s}^{-1} \]

Rearranging \(\ln k=\ln A-E_a/RT\) at the first data point. Repeating at the second point gives \(\ln A=-10.408+33.908=23.500\), identical to four figures, which is the arithmetic check that the two-point fit is self-consistent. A

\[ T^{*}=\frac{T_1T_2\ln(T_2/T_1)}{T_2-T_1}=\frac{556\times629\times0.123363}{73}=591\,\text{K}, \qquad M_\mu = \tfrac12 M(\text{HI}) = 63.95\,\text{g}\,\text{mol}^{-1}=6.395\times10^{-2}\,\text{kg}\,\text{mol}^{-1} \]

The temperature at which the fitted \(E_a\) is to be interpreted (Step 12 of the Proof), and the molar reduced mass for two identical HI molecules, \(\mu=m^{2}/2m=m/2\). Converting to \(\text{kg}\,\text{mol}^{-1}\) at this stage rather than later avoids the factor-of-\(31.6\) error listed under Common errors. B

\[ \langle v_{\text{rel}}\rangle=\sqrt{\frac{8RT^{*}}{\pi M_\mu}}=\sqrt{\frac{8\times8.314\times591}{\pi\times6.395\times10^{-2}}}=\sqrt{1.956\times10^{5}}=442\,\text{m}\,\text{s}^{-1}, \qquad \sigma=\pi d^{2}=3.85\times10^{-19}\,\text{m}^{2} \]

The mean relative speed at \(591\,\text{K}\) and the hard-sphere cross-section. For comparison, \(\langle v\rangle\) of a single HI molecule at the same temperature is \(\sqrt{8RT/\pi M}=313\,\text{m}\,\text{s}^{-1}\); the relative speed is larger by \(\sqrt{2}\), as it must be for identical partners. A

\[ A_{\text{coll}} = e^{1/2}N_{\!A}\sigma\langle v_{\text{rel}}\rangle = 1.6487\times(6.022\times10^{23})(3.85\times10^{-19})(442) = 1.69\times10^{8}\,\text{m}^{3}\text{mol}^{-1}\text{s}^{-1} = 1.69\times10^{11}\,\text{dm}^{3}\text{mol}^{-1}\text{s}^{-1} \]

The collision-theory prefactor with \(p=1\), from Step 10 of the Proof. The final conversion is the factor \(10^{3}\) between \(\text{m}^{3}\) and \(\text{dm}^{3}\); omitting it is the single most common source of a spurious steric factor of \(10^{-3}\). B

\[ p = \frac{A_{\text{obs}}}{A_{\text{coll}}} = \frac{1.61\times10^{10}}{1.69\times10^{11}} = 0.095, \qquad E_0 = E_a-\tfrac12RT^{*} = 1.7732\times10^{5}-2.46\times10^{3} = 1.749\times10^{5}\,\text{J}\,\text{mol}^{-1} \]

About one collision in ten with sufficient energy is also correctly oriented — a physically sensible figure for two diatomics that must present their hydrogen ends to one another, and close enough to unity that the hard-sphere picture is doing real work here. The threshold lies \(2.5\,\text{kJ}\,\text{mol}^{-1}\) below the fitted \(E_a\), a \(1.4\) per cent correction that is nevertheless larger than the experimental uncertainty in a good rate determination. B

\[ E_a = 177\,\text{kJ}\,\text{mol}^{-1}\ \ (\text{at }T^{*}=591\,\text{K}), \qquad A = 1.61\times10^{10}\,\text{dm}^{3}\text{mol}^{-1}\text{s}^{-1}, \qquad p = 0.095, \qquad E_0 = 175\,\text{kJ}\,\text{mol}^{-1} \]

Reading. Two rate constants and a table of atomic masses are enough to extract a barrier height, a collision efficiency and an orientational requirement. The barrier accounts for a factor \(e^{-38}\sim10^{-17}\) at \(556\,\text{K}\); everything else in the rate constant is collision frequency.

Scope. The value of \(p\) inherits every crudity of the hard-sphere diameter, which enters as \(d^{2}\): changing \(d\) from \(3.5\times10^{-10}\,\text{m}\) to \(4.0\times10^{-10}\,\text{m}\) would lower \(p\) to \(0.073\). The order of magnitude is meaningful; the second figure is not.

Example 2. A reaction obeys the collision-theory form derived above, \(k=C\,T^{1/2}e^{-E_0/RT}\), with threshold \(E_0=60.00\,\text{kJ}\,\text{mol}^{-1}\). (a) By what factor does \(k\) rise from \(300\,\text{K}\) to \(310\,\text{K}\), and does the “doubles every ten degrees” rule survive? (b) What activation energy would an experimenter extract from those two points, and does it agree with Step 12 of the Proof? (c) Repeat over \(300\)–\(600\,\text{K}\), and say how much curvature this represents.

\[ \frac{k(T_2)}{k(T_1)} = \left(\frac{T_2}{T_1}\right)^{1/2}\exp\!\left[-\frac{E_0}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right], \qquad \frac{E_0}{R}=\frac{60\,000}{8.314}=7216.7\,\text{K} \]

The ratio form, with \(C\) cancelling. Only after the symbolic rearrangement is the numerical value of \(E_0/R\) — a temperature, as the units check requires — computed. A

\[ \frac{1}{310}-\frac{1}{300} = -1.07527\times10^{-4}\,\text{K}^{-1} \quad\Longrightarrow\quad \frac{k(310)}{k(300)} = (1.03333)^{1/2}\,e^{0.77600} = 1.01653\times2.17280 = 2.209 \]

The prefactor contributes \(1.7\) per cent and the exponential contributes \(117\) per cent, so the rule of thumb holds and is essentially entirely due to the Boltzmann factor. A barrier near \(60\,\text{kJ}\,\text{mol}^{-1}\) is exactly the value for which the doubling rule is accurate at room temperature; for \(E_0=100\,\text{kJ}\,\text{mol}^{-1}\) the same ten degrees would multiply \(k\) by \(3.7\). A

\[ E_a^{\text{app}} = R\,\frac{\ln 2.209}{\dfrac{1}{300}-\dfrac{1}{310}} = 8.314\times\frac{0.79241}{1.07527\times10^{-4}} = 6.127\times10^{4}\,\text{J}\,\text{mol}^{-1} = 61.27\,\text{kJ}\,\text{mol}^{-1} \]

What the experimenter reports. It exceeds the threshold \(E_0\) by \(1.27\,\text{kJ}\,\text{mol}^{-1}\), and no amount of experimental care would remove the gap: it is a property of the definition of \(E_a\), not of the measurement. A

\[ T^{*}=\frac{(300)(310)\ln(310/300)}{10}=\frac{93\,000\times0.0327898}{10}=304.95\,\text{K} \quad\Longrightarrow\quad E_0+\tfrac12RT^{*}=60\,000+1267.7=61\,268\,\text{J}\,\text{mol}^{-1} \]

Step 12 of the Proof, tested. The predicted \(61.268\,\text{kJ}\,\text{mol}^{-1}\) reproduces the directly computed \(61.27\,\text{kJ}\,\text{mol}^{-1}\) to five figures, and \(T^{*}\) lies inside \((300,310)\) as the mean value theorem requires — slightly below the arithmetic mean \(305\,\text{K}\), because the weighting is in \(1/T\). C

\[ \frac{k(600)}{k(300)} = \sqrt{2}\;e^{7216.7\times1.66667\times10^{-3}} = 1.41421\times e^{12.0278} = 2.37\times10^{5} \]

Over three hundred kelvin the rate constant rises by five orders of magnitude, of which the \(T^{1/2}\) prefactor supplies a factor of only \(1.41\). A

\[ T^{*}_{300\to600}=\frac{(300)(600)\ln 2}{300}=600\ln 2=415.9\,\text{K} \quad\Longrightarrow\quad E_a^{\text{app}}=60\,000+\tfrac12(8.314)(415.9)=61.73\,\text{kJ}\,\text{mol}^{-1} \]

The apparent activation energy drifts by \(61.73-61.27=0.46\,\text{kJ}\,\text{mol}^{-1}\), i.e. \(0.75\) per cent, between a ten-kelvin window at room temperature and a three-hundred-kelvin window. That drift is the entire observable content of the curvature of the Arrhenius plot for this model, and it is comparable with the scatter of ordinary rate data — which is why the plot looks straight. B

\[ \frac{k(310)}{k(300)}=2.21; \qquad E_a^{\text{app}}\big|_{300\text{-}310}=61.27\,\text{kJ}\,\text{mol}^{-1}, \qquad E_a^{\text{app}}\big|_{300\text{-}600}=61.73\,\text{kJ}\,\text{mol}^{-1}, \qquad E_0=60.00\,\text{kJ}\,\text{mol}^{-1} \]

Reading. The rule that rates double for a ten-degree rise is a statement about barriers near \(60\,\text{kJ}\,\text{mol}^{-1}\), not a general law. The activation energy an experiment returns is systematically above the true threshold by \(\tfrac12RT^{*}\), and its dependence on the chosen temperature range is a fraction of a per cent — small, but exactly the size of the effect that makes “the” activation energy an approximate concept.

Scope. Purely a property of the model \(k\propto T^{1/2}e^{-E_0/RT}\); a real reaction may show far larger curvature from tunnelling, a changing rate-determining step, or a temperature-dependent cross-section (Fails without).

Problems

A first-order rate constant is measured as \(k=4.60\times10^{-4}\,\text{s}^{-1}\) at \(350\,\text{K}\) and \(8.80\times10^{-3}\,\text{s}^{-1}\) at \(400\,\text{K}\). Find \(E_a\) and \(A\), and check the value of \(A\) at both data points.
Solution
Rearrange before substituting: \(E_a=R\ln(k_2/k_1)\big/\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)\).
\(k_2/k_1=8.80\times10^{-3}/4.60\times10^{-4}=19.130\), so \(\ln(k_2/k_1)=2.9512\).
\(\dfrac{1}{350}-\dfrac{1}{400}=2.857143\times10^{-3}-2.500000\times10^{-3}=3.57143\times10^{-4}\,\text{K}^{-1}\).
\(E_a=8.314\times2.9512/3.57143\times10^{-4}=8.314\times8263.4=6.870\times10^{4}\,\text{J}\,\text{mol}^{-1}=68.7\,\text{kJ}\,\text{mol}^{-1}\).
For \(A\), use \(\ln A=\ln k+E_a/(RT)\). At \(350\,\text{K}\): \(\ln A=-7.6843+68\,703/(8.314\times350)=-7.6843+23.610=15.926\). At \(400\,\text{K}\): \(\ln A=-4.7330+68\,703/(8.314\times400)=-4.7330+20.659=15.926\). The two agree, as they must for a two-point fit.
Hence \(A=e^{15.926}=8.25\times10^{6}\,\text{s}^{-1}\). Note that this is many orders of magnitude below the \(\sim10^{13}\,\text{s}^{-1}\) of a molecular vibration frequency, which for a unimolecular reaction signals a strongly negative entropy of activation — a tight, ordered transition state.
Take the reaction of Problem 1 to be bimolecular after all, with the \(E_a=68.7\,\text{kJ}\,\text{mol}^{-1}\) determined at a mean temperature of \(375\,\text{K}\). Find the threshold energy \(E_0\), then compute the reactive fraction \(e^{-E_0/RT}\) at \(375\,\text{K}\) and at \(425\,\text{K}\), and state the factor by which raising the temperature by \(50\,\text{K}\) increases it.
Solution
From Step 9 of the Proof, \(E_0=E_a-\tfrac12RT\) evaluated at the temperature the fit refers to:
\(E_0=68\,703-\tfrac12(8.314)(375)=68\,703-1559=67\,144\,\text{J}\,\text{mol}^{-1}=67.1\,\text{kJ}\,\text{mol}^{-1}\).
At \(375\,\text{K}\): \(RT=8.314\times375=3117.8\,\text{J}\,\text{mol}^{-1}\), so \(E_0/RT=67\,144/3117.8=21.536\) and the reactive fraction is \(e^{-21.536}=4.4\times10^{-10}\).
At \(425\,\text{K}\): \(RT=3533.5\,\text{J}\,\text{mol}^{-1}\), so \(E_0/RT=19.002\) and the fraction is \(e^{-19.002}=5.6\times10^{-9}\).
The ratio is \(5.6\times10^{-9}/4.4\times10^{-10}=12.7\), or directly \(e^{21.536-19.002}=e^{2.534}=12.6\) (the difference is rounding). Fewer than one collision in a hundred million is reactive even at the higher temperature, yet a \(50\,\text{K}\) rise multiplies that minute fraction by thirteen. This is the whole content of the Arrhenius equation in one number: reaction lives entirely in the far tail of the distribution, where small shifts in \(T\) have enormous proportional effects.
For the general form \(k=A'T^{m}e^{-E_0/RT}\) with \(A'\), \(m\) and \(E_0\) all constant, show that \(E_a=E_0+mRT\) and that the Arrhenius plot has curvature \(\dfrac{d^{2}\ln k}{d(1/T)^{2}}=mT^{2}\). Evaluate the drift in \(E_a\) between \(300\) and \(600\,\text{K}\) for the collision-theory value \(m=\tfrac12\) and for the transition-state value \(m=1\).
Solution
Activation energy. \(\ln k=\ln A'+m\ln T-\dfrac{E_0}{RT}\), so \(\dfrac{d\ln k}{dT}=\dfrac{m}{T}+\dfrac{E_0}{RT^{2}}\), and by the definition in Step 8, \(E_a=RT^{2}\left(\dfrac{m}{T}+\dfrac{E_0}{RT^{2}}\right)=mRT+E_0\).
Curvature. Put \(u=1/T\), so \(\ln T=-\ln u\) and \(\ln k=\ln A'-m\ln u-\dfrac{E_0}{R}u\). Then \(\dfrac{d\ln k}{du}=-\dfrac{m}{u}-\dfrac{E_0}{R}\) and \(\dfrac{d^{2}\ln k}{du^{2}}=\dfrac{m}{u^{2}}=mT^{2}\).
The second derivative is positive for \(m\gt0\), so the plot is convex when viewed against \(1/T\): the local slope becomes less steep as \(1/T\) increases, i.e. the apparent \(E_a\) falls as the temperature falls, exactly as \(E_a=E_0+mRT\) states.
Numbers. The drift is \(\Delta E_a=mR\,\Delta T=mR(300)\). For \(m=\tfrac12\): \(\Delta E_a=0.5\times8.314\times300=1.25\,\text{kJ}\,\text{mol}^{-1}\). For \(m=1\): \(2.49\,\text{kJ}\,\text{mol}^{-1}\).
Against a typical \(E_a\) of \(60\)–\(180\,\text{kJ}\,\text{mol}^{-1}\) these are drifts of one to four per cent across a three-hundred-kelvin range. Since the curvature term \(mT^{2}\) enters a plot whose slope is of order \(E_0/R\sim10^{4}\,\text{K}\), detecting it requires rate data of better than one per cent precision over a wide range — which is why experiments almost never distinguish \(m=\tfrac12\) from \(m=1\) from \(m=0\).
The naive account of the Arrhenius factor claims that \(e^{-E_0/RT}\) is the fraction of pairs whose relative translational energy exceeds \(E_0\). Compute that fraction exactly from Step 2 of the Proof, obtain its leading behaviour for \(E_0\gg RT\), and evaluate the resulting discrepancy for \(E_0=67.1\,\text{kJ}\,\text{mol}^{-1}\) at \(375\,\text{K}\).
Solution
Write \(x=\varepsilon/k_BT\) and \(x_0=\varepsilon_0/k_BT=E_0/RT\). From \(f(\varepsilon)\,d\varepsilon=\dfrac{2}{\sqrt{\pi}}\dfrac{\varepsilon^{1/2}}{(k_BT)^{3/2}}e^{-\varepsilon/k_BT}d\varepsilon\),
\(F=\displaystyle\frac{2}{\sqrt{\pi}}\int_{x_0}^{\infty}x^{1/2}e^{-x}\,dx\).
Exact evaluation. Integrate by parts with \(u=x^{1/2}\), \(dv=e^{-x}dx\): \(\displaystyle\int_{x_0}^{\infty}x^{1/2}e^{-x}dx = x_0^{1/2}e^{-x_0}+\frac12\int_{x_0}^{\infty}x^{-1/2}e^{-x}dx\), and the remaining integral is \(\sqrt{\pi}\,\operatorname{erfc}(\sqrt{x_0})\). Hence
\(F = \dfrac{2}{\sqrt{\pi}}\sqrt{x_0}\,e^{-x_0}+\operatorname{erfc}(\sqrt{x_0})\).
Asymptotics. Repeated integration by parts gives \(\displaystyle\int_{x_0}^{\infty}x^{-1/2}e^{-x}dx = x_0^{-1/2}e^{-x_0}\left(1-\dfrac{1}{2x_0}+\cdots\right)\), so \(F\simeq \dfrac{2}{\sqrt{\pi}}\sqrt{x_0}\,e^{-x_0}\left(1+\dfrac{1}{2x_0}\right)\) and the leading ratio to the bare exponential is \(2\sqrt{x_0/\pi}\).
Numbers. \(x_0=67\,144/(8.314\times375)=21.536\). The leading factor is \(2\sqrt{21.536/\pi}=2\sqrt{6.855}=5.24\), with a correction of \(1+1/(2\times21.536)=1.023\), giving \(5.36\). So the naive fraction, \(2.4\times10^{-9}\), exceeds the exponential \(e^{-21.536}=4.4\times10^{-10}\) by a factor of about five.
The moral is that the exponential in the Result does not come from counting energetic pairs. It emerges only after the \(v_r\) flux weighting of Step 3 (which favours fast pairs, pushing the fraction up) is combined with the line-of-centres cross-section of Step 4 (which discards glancing collisions, pushing it down); the two effects together convert \(2\sqrt{x_0/\pi}\,e^{-x_0}\) into exactly \(e^{-x_0}\). Getting the right answer from the naive argument is a coincidence of shape, not of magnitude.
Prove Tolman's interpretation for the line-of-centres model — that \(E_a\) equals the mean molar energy of reacting pairs minus the mean molar energy of all pairs — and evaluate both means, and their difference, at \(500\,\text{K}\) for \(E_0=100.0\,\text{kJ}\,\text{mol}^{-1}\). Comment on why this makes a negative \(E_a\) intelligible.
Solution
Mean energy of reacting pairs. The reactive flux at energy \(\varepsilon\) is \(\sigma_r(\varepsilon)v_r(\varepsilon)f(\varepsilon)\propto(\varepsilon-\varepsilon_0)e^{-\varepsilon/k_BT}\) for \(\varepsilon\ge\varepsilon_0\), as established in Step 6. With \(\varepsilon=\varepsilon_0+k_BTx\),
\(\langle\varepsilon\rangle_{\text{react}}=\dfrac{\int_0^{\infty}(\varepsilon_0+k_BTx)\,x\,e^{-x}dx}{\int_0^{\infty}x\,e^{-x}dx}=\dfrac{\varepsilon_0\Gamma(2)+k_BT\,\Gamma(3)}{\Gamma(2)}=\varepsilon_0+2k_BT\),
using \(\Gamma(2)=1\) and \(\Gamma(3)=2\).
Mean energy of all pairs. Over the unweighted distribution of Step 2, \(\langle\varepsilon\rangle_{\text{all}}=k_BT\,\Gamma(5/2)/\Gamma(3/2)=\tfrac32k_BT\), the equipartition value for three degrees of relative translational freedom.
Difference. \(\langle\varepsilon\rangle_{\text{react}}-\langle\varepsilon\rangle_{\text{all}}=\varepsilon_0+2k_BT-\tfrac32k_BT=\varepsilon_0+\tfrac12k_BT\), which on multiplication by \(N_{\!A}\) is \(E_0+\tfrac12RT\) — exactly the \(E_a\) obtained by differentiation in Step 9. The two routes agree, which is the content of Tolman's theorem.
Numbers at \(500\,\text{K}\). \(RT=8.314\times500=4157\,\text{J}\,\text{mol}^{-1}\). Mean energy of reacting pairs: \(E_0+2RT=100\,000+8314=108.3\,\text{kJ}\,\text{mol}^{-1}\). Mean energy of all pairs: \(\tfrac32RT=6.24\,\text{kJ}\,\text{mol}^{-1}\). Difference: \(102.1\,\text{kJ}\,\text{mol}^{-1}\), and the check \(E_0+\tfrac12RT=100\,000+2079=102.1\,\text{kJ}\,\text{mol}^{-1}\) agrees.
Negative activation energies. Nothing in the difference-of-means statement requires the reacting subpopulation to be the hotter one. If a mechanism reacts preferentially through pairs that are colder than average — as happens when a fast, exothermic pre-equilibrium forms a bound complex whose population falls with temperature — then \(\langle E\rangle_{\text{react}}\lt\langle E\rangle_{\text{all}}\) and \(E_a\lt0\). A barrier height can never be negative; a difference of mean energies obviously can. Tolman's reading is therefore the one that generalises, and it is why Step 8 is stated as a definition rather than as a consequence of the Arrhenius form.