Result

Transition state theory

T-042Home CU-202Threads kinetics

Statement

Rate from the activated complex at the barrier top.

Why it matters

collision-theory gives a first, mechanical picture of a reaction rate as a collision frequency multiplied by an energetic and orientational requirement, but it treats molecules as crude hard spheres and can only patch up its systematic failure to predict the pre-exponential factor for complex reactions with an ad hoc "steric factor." Transition state theory (TST) replaces that picture with a proper thermodynamic/statistical treatment of the activated complex sitting at the top of the reaction's energy barrier, giving arrhenius-equation's empirical parameters, \(E_a\) and \(A\), a genuine theoretical foundation.

Because TST expresses the rate constant through thermodynamic-style activation parameters (\(\Delta H^{\ddagger}\), \(\Delta S^{\ddagger}\)) rather than through a purely kinetic collision picture, it also connects naturally to steady-state-approximation: TST supplies the rate constant for each individual elementary step in a proposed mechanism, while the SSA is the separate tool for combining those elementary rate constants into an overall observed rate law.

Hypotheses

An activated complex exists at the potential-energy maximum along the reaction coordinate and is treated as being in quasi-equilibrium with the reactants.The activated complex is not a real, isolable chemical species; treating it as though it were in ordinary equilibrium with the reactants, even though it is not stable enough to isolate, is a formal device that lets ordinary equilibrium thermodynamics be applied to it, exactly as though it were a genuine chemical species. Any activated complex that forms proceeds on to products; it never reverts, or "recrosses," back to reactants.Without this assumption, only some fraction of activated complexes would actually become product, and the simple rate expression built from the quasi-equilibrium concentration alone would overstate the true rate. TST as developed here treats barrier crossing purely classically; it neglects quantum mechanical tunnelling through the barrier, a real, additional contribution to the rate that can be significant for light-particle (proton or hydrogen-atom) transfer reactions, especially at low temperature.

Proof

\[ \text{A}+\text{B} \rightleftharpoons \text{X}^{\ddagger} \to \text{products}, \qquad K^{\ddagger}=\frac{[\text{X}^{\ddagger}]}{[\text{A}][\text{B}]} \]

Postulate a quasi-equilibrium between the reactants and the activated complex \(\text{X}^{\ddagger}\), governed by an equilibrium constant \(K^{\ddagger}\) formally analogous to any ordinary equilibrium constant (Hypotheses). A

\[ \text{rate} = \nu[\text{X}^{\ddagger}], \qquad \nu = \frac{k_BT}{h} \]

The rate is the product of the activated complex's concentration and the universal frequency \(\nu\) at which it crosses over irreversibly to products (Hypotheses' no-recrossing assumption); \(\nu=k_BT/h\) follows from a statistical treatment of the reaction-coordinate motion itself. B

\[ k = \frac{k_BT}{h}K^{\ddagger} \]

Substituting the quasi-equilibrium relation \([\text{X}^{\ddagger}]=K^{\ddagger}[\text{A}][\text{B}]\) (Step 1) into the rate expression (Step 2) gives the Eyring equation, expressing the rate constant directly in terms of the activation quasi-equilibrium constant. A

\[ K^{\ddagger}=e^{-\Delta G^{\ddagger}/RT}, \qquad \Delta G^{\ddagger}=\Delta H^{\ddagger}-T\Delta S^{\ddagger} \]

Ordinary equilibrium thermodynamics relates \(K^{\ddagger}\) to a standard Gibbs energy of activation, which expands into separate enthalpic and entropic activation contributions, exactly as for any other equilibrium constant. A

\[ k=\frac{k_BT}{h}e^{\Delta S^{\ddagger}/R}e^{-\Delta H^{\ddagger}/RT} \quad\text{vs Arrhenius: } k=Ae^{-E_a/RT} \]

Comparing directly against arrhenius-equation's empirical form identifies \(\Delta H^{\ddagger}\) closely with \(E_a\), while the pre-exponential factor \(A\) gains a genuine physical interpretation, \(A\approx(ek_BT/h)e^{\Delta S^{\ddagger}/R}\), an activation-entropy term entirely absent from collision theory's cruder, purely geometric picture. B

Result

\[ k=\frac{k_BT}{h}\,e^{-\Delta G^{\ddagger}/RT}=\frac{k_BT}{h}\,e^{\Delta S^{\ddagger}/R}\,e^{-\Delta H^{\ddagger}/RT} \]

Reading. The rate constant is expressed through the thermodynamic activation parameters of a hypothesised transition state rather than through a purely kinetic collision picture, giving both a barrier height (\(\Delta H^{\ddagger}\)) and an order/probability term (\(\Delta S^{\ddagger}\)) direct physical meaning.

Scope. Requires the quasi-equilibrium and no-recrossing assumptions (Hypotheses); neglects quantum tunnelling, most significant for light-atom transfer at low temperature.

Corollaries & converses

A negative \(\Delta S^{\ddagger}\) corresponds to a highly ordered, restrictive transition state — typical of a bimolecular association reaction — exactly the situation collision-theory could only capture crudely with its empirical steric factor.
Comparing Steps 4 and 5 shows \(E_a\approx\Delta H^{\ddagger}+RT\) for a reaction in solution, linking arrhenius-equation's empirical activation energy directly to a genuine thermodynamic quantity.
steady-state-approximation and transition-state-theory address complementary aspects of the same kinetic problem: SSA extracts an observed rate law from a proposed multi-step mechanism, while TST supplies the theoretical rate constant for each individual elementary step feeding into that mechanism.

Fails without

Drop the no-recrossing assumption: if a significant fraction of activated complexes revert back to reactants rather than proceeding on to products, Step 2's rate expression, built from the full quasi-equilibrium concentration of \(\text{X}^{\ddagger}\) alone, systematically overestimates the true rate; a transmission coefficient correcting for recrossing is needed to fix this in a more refined treatment.
Ignore quantum tunnelling for a light-particle transfer reaction (Hypotheses' \(t3\)): the classical TST rate can substantially underestimate the true observed rate for hydrogen or proton transfer, especially at low temperature, since tunnelling opens an additional, purely quantum-mechanical pathway through, rather than over, the barrier that the classical treatment cannot capture at all.

Common errors

Treating the activated complex as a real, isolable chemical species, rather than a formal, quasi-equilibrium construct at the potential-energy maximum.
Confusing \(\Delta H^{\ddagger}\) with the reaction's overall thermodynamic \(\Delta H\) — the activation enthalpy concerns only the barrier from reactants to the transition state, independent of whether the overall reaction is exothermic or endothermic.
Assuming a negative \(\Delta S^{\ddagger}\) indicates an error in the analysis, rather than recognising it as the expected, typical sign for essentially any bimolecular association reaction.
Neglecting that classical TST, as developed here, entirely omits quantum tunnelling, which can noticeably raise the true rate above the TST prediction for light-particle transfer.

Discussion

Transition-state theory was developed largely by Henry Eyring, along with Meredith Gwynne Evans and Michael Polanyi, in the mid-1930s, which is why the rate expression derived here is commonly called the Eyring equation. It represented a substantial conceptual advance over collision theory, replacing a purely mechanical, hard-sphere collision picture with a genuinely thermodynamic and statistical treatment grounded in the quantum statistical mechanics then being developed.

Quantum mechanical tunnelling allows a reacting system some non-zero probability of passing through, rather than over, the activation barrier, an effect entirely outside classical TST's scope; it is most pronounced for the lightest nuclei (hydrogen and its isotopes) and becomes proportionally more important as temperature is lowered, since the classical over-the-barrier route becomes increasingly disfavoured while the tunnelling probability is comparatively insensitive to temperature.

Common misconception: that a large activation energy inherently reflects a large activation enthalpy alone, with entropy playing no role. Step 5's expansion of \(\Delta G^{\ddagger}\) shows the observed rate is jointly controlled by both an enthalpic barrier and an entropic term, and a slow reaction can have a comparatively modest \(\Delta H^{\ddagger}\) but a strongly unfavourable, very negative \(\Delta S^{\ddagger}\).

Worked examples

\[ k=3.2\times10^{-4}\,\text{s}^{-1}\ \text{at}\ T=298\,\text{K (illustrative values)}: \quad K^{\ddagger}=\frac{k}{k_BT/h} \]

The universal frequency factor \(k_BT/h\) at \(298\,\text{K}\) is \(\dfrac{(1.381\times10^{-23})(298)}{6.626\times10^{-34}}\approx6.21\times10^{12}\,\text{s}^{-1}\); dividing the given rate constant by this factor gives \(K^{\ddagger}=3.2\times10^{-4}/6.21\times10^{12}\approx5.15\times10^{-17}\). A

\[ \Delta G^{\ddagger}=-RT\ln K^{\ddagger} = -(8.314)(298)\ln(5.15\times10^{-17}) \]

Substituting \(\ln(5.15\times10^{-17})\approx-37.5\) into Step 4's relation gives \(\Delta G^{\ddagger}\approx-(2478)(-37.5)\approx92{,}900\,\text{J/mol}\approx92.9\,\text{kJ/mol}\), the activation Gibbs energy consistent with the given rate constant and temperature. A

\[ \Delta G^{\ddagger}\approx92.9\,\text{kJ/mol from } k=3.2\times10^{-4}\,\text{s}^{-1}\text{ at }298\,\text{K} \]

Reading. A single measured rate constant, together with temperature, converts directly through the Eyring equation into an activation Gibbs energy, without needing to separately measure \(\Delta H^{\ddagger}\) and \(\Delta S^{\ddagger}\) individually.

Scope. Separating \(\Delta G^{\ddagger}\) into \(\Delta H^{\ddagger}\) and \(\Delta S^{\ddagger}\) individually requires measuring \(k\) at several temperatures and analysing the resulting Eyring plot, analogous to how an Arrhenius plot separately extracts \(E_a\) and \(A\).

Further example. A first-order decomposition reaction has a measured rate constant of \(k = 5.0\times10^{-3}\,\text{s}^{-1}\) at \(T = 350\,\text{K}\). Determine the activation Gibbs energy \(\Delta G^{\ddagger}\) for this process, assuming the standard transition state theory (Eyring equation) applies with a transmission coefficient \(\kappa = 1\).

\[ k = \frac{k_B T}{h} K^{\ddagger} \implies K^{\ddagger} = \frac{k h}{k_B T} \]

The Eyring equation relates the rate constant to the equilibrium constant of the activated complex. Rearranging isolates \(K^{\ddagger}\), which is dimensionless. A

\[ K^{\ddagger} = \frac{(5.0\times10^{-3}\,\text{s}^{-1})(6.626\times10^{-34}\,\text{J}\cdot\text{s})}{(1.381\times10^{-23}\,\text{J/K})(350\,\text{K})} \]

Substituting the given values and fundamental constants. The units of \(\text{s}^{-1}\) and \(\text{s}\) cancel, leaving a dimensionless quantity as required for an equilibrium constant. A

\[ K^{\ddagger} = \frac{3.313\times10^{-36}}{4.8335\times10^{-21}} \approx 6.854\times10^{-16} \]

Performing the division yields the equilibrium constant for the transition state. The value is very small, indicating the transition state is much less populated than the reactants. A

\[ \Delta G^{\ddagger} = -RT \ln K^{\ddagger} \]

The thermodynamic relationship between the standard Gibbs free energy of activation and the equilibrium constant is used. A

\[ \Delta G^{\ddagger} = -(8.314\,\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1})(350\,\text{K}) \ln(6.854\times10^{-16}) \]

Substituting \(R\), \(T\), and the calculated \(K^{\ddagger}\). Note that \(\ln(6.854\times10^{-16}) \approx -34.91\). A

\[ \Delta G^{\ddagger} = -(2910) (-34.91) \approx 101{,}588\,\text{J/mol} \approx 101.6\,\text{kJ/mol} \]

Multiplying the terms gives the activation Gibbs energy. The positive sign confirms the barrier is energetically unfavorable relative to the ground state. A

\[ \Delta G^{\ddagger} \approx 101.6\,\text{kJ/mol} \]

Reading. This value represents the total free energy barrier the system must overcome at 350 K. It combines both enthalpic and entropic contributions to the rate limitation.

Scope. This calculation assumes the reaction follows first-order kinetics and that the transition state is in quasi-equilibrium with the reactants. It does not distinguish between \(\Delta H^{\ddagger}\) and \(\Delta S^{\ddagger}\).

Problems

A reaction has \(k=1.0\times10^{-2}\,\text{s}^{-1}\) at \(T=310\,\text{K}\). Compute \(K^{\ddagger}\) using \(k_BT/h\approx6.46\times10^{12}\,\text{s}^{-1}\) at this temperature.
Solution
\(K^{\ddagger}=k/(k_BT/h)=1.0\times10^{-2}/6.46\times10^{12}\approx1.55\times10^{-15}\).
Two bimolecular reactions have similar \(\Delta H^{\ddagger}\) but very different rates at the same temperature. Explain, in terms of \(\Delta S^{\ddagger}\), how this is possible.
Solution
By Step 5's Eyring expression, the rate constant depends on both \(\Delta H^{\ddagger}\) (through the exponential barrier term) and \(\Delta S^{\ddagger}\) (through the pre-exponential entropy term). If the two reactions have similar \(\Delta H^{\ddagger}\) but very different rates, the difference must arise predominantly from \(\Delta S^{\ddagger}\): a reaction requiring a much more restrictive, precisely ordered transition state (more negative \(\Delta S^{\ddagger}\)) will be markedly slower than one with a more loosely constrained transition state, even at an identical activation enthalpy.
Explain qualitatively why a sterically demanding bimolecular reaction (requiring precise mutual orientation of two large reactants) is expected to have a strongly negative \(\Delta S^{\ddagger}\), and relate this to how collision theory would instead describe the same effect.
Solution
Forming a highly specific, precisely oriented transition state from two independently tumbling reactant molecules represents a large loss of rotational and translational freedom, giving a strongly negative \(\Delta S^{\ddagger}\) (Corollaries). Collision theory, lacking any thermodynamic activation-entropy concept, can only capture this same steric demand indirectly, through an empirically fitted "steric factor" reducing the predicted rate below the simple collision-frequency estimate — TST's \(\Delta S^{\ddagger}\) gives this same physical effect a proper theoretical basis instead.
A unimolecular isomerization reaction has a rate constant of \(k = 2.5\times10^{-2}\,\text{s}^{-1}\) at \(T = 400\,\text{K}\). Calculate the activation Gibbs energy \(\Delta G^{\ddagger}\) in kJ/mol. Use \(k_B = 1.381\times10^{-23}\,\text{J/K}\) and \(h = 6.626\times10^{-34}\,\text{J}\cdot\text{s}\).
Solution
First, calculate the equilibrium constant \(K^{\ddagger}\): \[ K^{\ddagger} = \frac{k h}{k_B T} = \frac{(2.5\times10^{-2})(6.626\times10^{-34})}{(1.381\times10^{-23})(400)} \] \[ K^{\ddagger} = \frac{1.6565\times10^{-35}}{5.524\times10^{-21}} \approx 2.999\times10^{-15} \] Next, calculate \(\Delta G^{\ddagger}\): \[ \Delta G^{\ddagger} = -RT \ln K^{\ddagger} = -(8.314)(400) \ln(2.999\times10^{-15}) \] \[ \ln(2.999\times10^{-15}) \approx -33.43 \] \[ \Delta G^{\ddagger} = -(3325.6)(-33.43) \approx 111{,}176\,\text{J/mol} \] Rounding to three significant figures: \[ \Delta G^{\ddagger} \approx 111\,\text{kJ/mol} \]
Two reactions, A and B, have identical activation enthalpies \(\Delta H^{\ddagger}\) at a given temperature. Reaction A has a rate constant 1000 times larger than Reaction B. Explain this difference using the concept of activation entropy \(\Delta S^{\ddagger}\). Which reaction has the more negative \(\Delta S^{\ddagger}\)?
Solution
The Eyring equation is given by: \[ k = \frac{k_B T}{h} e^{\Delta S^{\ddagger}/R} e^{-\Delta H^{\ddagger}/RT} \] Since \(\Delta H^{\ddagger}\) and \(T\) are identical for both reactions, the ratio of their rate constants depends solely on the exponential term involving entropy: \[ \frac{k_A}{k_B} = \frac{e^{\Delta S_A^{\ddagger}/R}}{e^{\Delta S_B^{\ddagger}/R}} = e^{(\Delta S_A^{\ddagger} - \Delta S_B^{\ddagger})/R} \] Given \(k_A/k_B = 1000\): \[ 1000 = e^{(\Delta S_A^{\ddagger} - \Delta S_B^{\ddagger})/R} \] Taking the natural log: \[ \ln(1000) = \frac{\Delta S_A^{\ddagger} - \Delta S_B^{\ddagger}}{R} \] \[ 6.908 = \frac{\Delta S_A^{\ddagger} - \Delta S_B^{\ddagger}}{8.314} \] \[ \Delta S_A^{\ddagger} - \Delta S_B^{\ddagger} \approx 57.4\,\text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \] Since \(\Delta S_A^{\ddagger} > \Delta S_B^{\ddagger}\), Reaction B has the more negative activation entropy. A more negative \(\Delta S^{\ddagger}\) indicates a transition state that is more ordered or constrained relative to the reactants, which reduces the probability of forming the activated complex and thus lowers the rate.