The *effective* barrier of a reaction with (fast) pre-equilibrium
Reaction schemes of this form come up all the time:
\[\mathsf{A} \;\underset{k_{-1}}{\overset{k_{1}}{\rightleftharpoons}}\; \mathsf{B} \;\overset{k_{2}}{\longrightarrow}\; \mathsf{C}\]Product C (the species we care about) can be produced from A, but only by way of a reactive intermediate B that is in equilibrium with A. Before going further, here is what each symbol means:
| Symbol | Meaning |
|---|---|
| ${\small k_1,\ k_{-1}}$ | forward / reverse rate constants for A ⇌ B |
| ${\small k_2}$ | rate constant for the B → C step |
| ${\small K = k_1/k_{-1}}$ | equilibrium constant for A ⇌ B (large ${\small K}$ means mostly B; small ${\small K}$ means mostly A) |
| TS₁, TS₂ | transition states connecting A&B and B&C, respectively |
| ${\small \Delta G_1^{\ddagger},\ \Delta G_{-1}^{\ddagger}}$ | activation barriers for the forward / reverse A ⇌ B steps |
| ${\small \Delta G_2^{\ddagger}}$ | activation barrier for the B → C step |
| ${\small \Delta G_{\text{eff}}^{\ddagger}}$ | the one effective barrier that sets how fast C actually forms |
Each barrier ${\small \Delta G_i^{\ddagger}}$ links to its rate constant ${\small k_i}$ through transition-state theory, ${\small k_i = \tfrac{k_B T}{h}\,e^{-\Delta G_i^{\ddagger}/RT}}$.
How does the relative energy of all these species and transition states determine the rate of formation of C? And under what condition can the formation of C be suppressed entirely? This textbook-looking question turns out to be far from trivial. It is exactly what we needed to answer in our recent paper (JACS 2026) to explain why a TAMM-type reaction could stop at the free thiol product (i.e. A) instead of forming the “dead-end” thiazoline (i.e. C).
In short: what is ${\small \Delta G^{\ddagger}_{\text{eff}}}$?
Three tempting guesses
Look at an energy profile for A ⇌ B → C and it is easy to reach for some half-correct answers:
Reading the energies off the reaction profile.
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Guess 1 ${\small \Delta G_{\text{eff}}^{\ddagger}}$ = ${\small \Delta G_2^{\ddagger}}$: Species C is made by climbing ${\small \Delta G_2^{\ddagger}}$ out of B, so a natural guess is that this is the barrier. But since only some of the material is sitting as B at any moment, while the rest is A, a rate that ignored this cannot be right.
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Guess 2 ${\small \Delta G_{\text{eff}}^{\ddagger}}$ = E(TS₂)–E(A): A related guess fixes that halfway. If A is lower in energy than B — which is often the case, and is exactly the picture that invites this guess — then A looks like a resting state (lowest energy point), and TS₂ looks like the one hill left before C. So why not just measure the height of TS₂ above A? This one is closer, but this guess quietly assumes that is always true, and stops working the moment B becomes competitive with A.
Reading the mechanism as two climbs.
- Guess 3 ${\small \Delta G_{\text{eff}}^{\ddagger}}$ = ${\small \Delta G_1^{\ddagger}+\Delta G_2^{\ddagger}}$: A different instinct comes from thinking about the path step by step: first the molecule climbs ${\small \Delta G_1^{\ddagger}}$ to get from A to TS₁, then it climbs ${\small \Delta G_2^{\ddagger}}$ to get from B to TS₂ — so add them, ${\small \Delta G_1^{\ddagger}+\Delta G_2^{\ddagger}}$. This one is, however, never correct, as it treats B as if it were at the same height as TS₁, and never lets the system descend into the valley at B before climbing out.
So: not ${\small \Delta G_2^{\ddagger}}$ alone, not simply the height of TS₂ above A, and not ${\small \Delta G_1^{\ddagger}+\Delta G_2^{\ddagger}}$. What is ${\small \Delta G^{\ddagger}_{\text{eff}}}$, then?
The answer
When the A ⇌ B equilibrium is fast compared to the B → C step (${\small k_1, k_{-1} \gg k_2}$), the build-up of C in fact follows a single exponential, with one effective rate constant (the detailed derivation is provided later in a collapsible section):
\[k_{\text{eff}} = \frac{k_1 k_2}{k_1 + k_{-1}}.\]Writing each ${\small k_i}$ in terms of its barrier and simplifying turns this into a barrier:
\[\Delta G^{\ddagger}_{\text{eff}} = \Delta G^{\ddagger}_{2} + RT\ln\!\left(1 + \frac{1}{K}\right).\]${\small \Delta G_2^{\ddagger}}$ is still in there, but with a non-zero correction term in general.
Seeing it move
Drag the three barrier sliders below. The plot shows the free-energy profile together with all four answers: the three tempting guesses (dotted lines) and the true effective barrier (dashed red). Watch how the true line moves between the guesses as ${\small K}$ changes, and note that it only ever touches two of them — never the third.
Two limits, and why each guess is half right
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Large ${\small K}$ (mostly B): the correction term RTln(1 + 1/K) vanishes and ΔG‡eff → ΔG‡2. This is Guess 1’s regime. When almost everything is already B, waiting for A to become B costs nothing, so the only barrier that matters is the one leaving B.
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Small ${\small K}$ (mostly A): the correction term approximates the energy gap between B and A, so ΔG‡eff approaches ΔG‡2 plus that gap — which is basically the height of TS₂ above A (i.e. Guess 2). This is not a coincidence: when A truly is the resting state, reading the barrier as the highest point above it is the right thing to do. The Guess fails only where it silently assumes this is always true. (It should be noted that the familiar steady-state approximation gives this limiting answer as well, but based on a different assumption.)
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Guess 3 is just wrong. It overshoots the true effective barrier at every value of ${\small K}$. There is no regime where “add both barriers” is the correct move, because the path never actually climbs ${\small \Delta G_1^{\ddagger}}$ and ${\small \Delta G_2^{\ddagger}}$ back to back; it comes back down to B first.
So, while Guess 3 is incorrect, Guesses 1 and 2 are not really wrong so much as incomplete: each is the right answer in its own corner, and the true barrier moves continuously between them as ${\small K}$ changes.
Why this matters for our study
In conclusion, one cannot tell whether C will form appreciably just by comparing ${\small \Delta G_2^{\ddagger}}$ values, or by comparing the height of TS₂ above the starting material. Where the A ⇌ B equilibrium sits matters just as much. This rigorous analysis, instead of simply reading barriers straight off a computed profile, is what let us account for the observed selectivity between the thiol-retaining and cyclized products in N-terminal cysteine chemistry. The full treatment, including the computed barriers it was applied to, is in the paper and its Supporting Information.
For the curious readers: the full derivation from the rate equations
The rate equations
The scheme A ⇌ B → C corresponds to
\[\frac{d[\mathsf{A}]}{dt} = -k_1[\mathsf{A}] + k_{-1}[\mathsf{B}],\] \[\frac{d[\mathsf{B}]}{dt} = k_1[\mathsf{A}] - (k_{-1}+k_2)[\mathsf{B}],\] \[\frac{d[\mathsf{C}]}{dt} = k_2[\mathsf{B}],\]with [A](0) = a0, [B](0) = 0, and [C](0) = 0.
Exact solution via eigenvalues
Because [A] + [B] + [C] = a0, we only need the first two equations. In matrix form,
\[\frac{d}{dt}\begin{bmatrix}[\mathsf{A}]\\ {[\mathsf{B}]}\end{bmatrix} = \mathbf{M}\begin{bmatrix}[\mathsf{A}]\\ {[\mathsf{B}]}\end{bmatrix}, \qquad \mathbf{M}=\begin{bmatrix}-k_1 & k_{-1}\\ k_1 & -(k_{-1}+k_2)\end{bmatrix}.\]The eigenvalues follow from ${\small \det(\mathbf{M}-\lambda\mathbf{I})=0}$:
\[\lambda_{1,2}=\frac{-(k_1+k_{-1}+k_2)\pm\sqrt{(k_1+k_{-1}+k_2)^2-4k_1k_2}}{2}.\]With eigenvectors \(\mathbf{v}_{1,2}=\begin{bmatrix}k_{-1}\\ k_1+\lambda_{1,2}\end{bmatrix}\) and the initial conditions, the coefficients are
\[c_1=\frac{a_0(k_1+\lambda_2)}{k_{-1}(\lambda_2-\lambda_1)},\qquad c_2=-\frac{a_0(k_1+\lambda_1)}{k_{-1}(\lambda_2-\lambda_1)},\]giving the closed-form (biexponential) concentrations
\[[\mathsf{A}](t)=-\frac{a_0}{\lambda_1-\lambda_2}\left[(k_1+\lambda_2)e^{\lambda_1 t}-(k_1+\lambda_1)e^{\lambda_2 t}\right],\] \[[\mathsf{B}](t)=\frac{a_0 k_1}{\lambda_1-\lambda_2}\left[e^{\lambda_1 t}-e^{\lambda_2 t}\right],\] \[[\mathsf{C}](t)=a_0-[\mathsf{A}](t)-[\mathsf{B}](t).\]The rapid pre-equilibrium limit
Let ${\small s=k_1+k_{-1}+k_2}$ and ${\small \Delta=s^2-4k_1k_2}$. Under ${\small k_1,k_{-1}\gg k_2}$, a binomial expansion of ${\small \sqrt{\Delta}}$ gives
\[\lambda_1 \approx -\frac{k_1 k_2}{k_1+k_{-1}} \equiv -k_{\text{eff}}, \qquad \lambda_2 \approx -(k_1+k_{-1}).\]The ${\small e^{\lambda_2 t}}$ term is the fast mode (it decays almost instantly); the ${\small e^{\lambda_1 t}}$ term is the slow mode that governs the observable kinetics. Dropping the fast mode collapses all three concentrations to single exponentials:
\[[\mathsf{C}](t)=a_0\left(1-e^{-k_{\text{eff}} t}\right), \qquad k_{\text{eff}}=\frac{k_1 k_2}{k_1+k_{-1}}.\](The same result follows from the quasi-steady-state approximation applied to B.)
From rate constant to barrier
Substituting ${\small k_i=\tfrac{k_B T}{h}e^{-\Delta G_i^{\ddagger}/RT}}$ into ${\small k_{\text{eff}}}$:
\[k_{\text{eff}}=\frac{k_B T}{h}\cdot\frac{e^{-(\Delta G_1^{\ddagger}+\Delta G_2^{\ddagger})/RT}}{e^{-\Delta G_1^{\ddagger}/RT}+e^{-\Delta G_{-1}^{\ddagger}/RT}}.\]Matching to ${\small k_{\text{eff}}=\tfrac{k_B T}{h}e^{-\Delta G_{\text{eff}}^{\ddagger}/RT}}$ and simplifying:
\[\Delta G^{\ddagger}_{\text{eff}} =\Delta G^{\ddagger}_{2}+RT\ln\!\left(1+e^{-(\Delta G_{-1}^{\ddagger}-\Delta G_1^{\ddagger})/RT}\right) =\Delta G^{\ddagger}_{2}+RT\ln\!\left(1+\tfrac{1}{K}\right),\]since ${\small K=k_1/k_{-1}=e^{-(\Delta G_1^{\ddagger}-\Delta G_{-1}^{\ddagger})/RT}}$.
The two limiting cases discussed above follow directly by taking ${\small K\gg1}$ (mostly B) and ${\small K\ll1}$ (mostly A); in general, since RTln(1 + 1/K) is a positive term, ${\small \Delta G_{\text{eff}}^{\ddagger}}$ is greater than ${\small \Delta G_{\text{2}}^{\ddagger}}$.
The analysis on this page underpins the mechanistic rationalization in JACS 2026, 148, 18020. If you spot an error in the derivation, I’d be glad to hear about it.