Reaction mechanisms

A balanced equation is a summary of a journey, and the rate law measured in the laboratory is evidence about the individual steps that journey is made of.

The sixth lesson found orders that bear no relation to the coefficients, and the seventh found an activation energy that belongs to no particular part of the equation. Both are symptoms of the same thing: reactions almost never happen the way they are written. This lesson builds the underlying sequence and shows how to test it.

Elementary steps and molecularity

An elementary step is a reaction that happens exactly as written, in a single encounter, with no intermediate. For an elementary step, and only for an elementary step, the rate law can be written down from the equation. A unimolecular step, one molecule falling apart or rearranging, is first order. A bimolecular step is first order in each of the two species, second order overall. That is not an empirical finding; it follows from what a collision is, since the chance of two particular molecules meeting is proportional to each concentration.

Termolecular steps, requiring three molecules at one point at one time, are possible but rare, because a triple encounter in a gas is orders of magnitude less likely than a double one. Anything requiring four is not seriously proposed. This is the strongest constraint on mechanism building: a balanced equation with large coefficients cannot possibly be elementary, so $2 \to 4 N +$ needs at least three separate steps whatever else is true.

A mechanism is a proposed set of elementary steps that add up to the overall equation. Species that are produced in one step and consumed in another are intermediates: real molecules with real, if short, lifetimes, unlike a transition state, which is a maximum on a path and has none.

The rate determining step

If one step in a sequence is much slower than the rest, the overall rate is that step's rate, and everything else is fast enough to be invisible. This is the rate determining step, and it is the simplest way a mechanism produces a rate law.

Example. Above $500$ K, $N + CO \to NO + C$ is observed to follow $v = k [N]^{2}$ , with carbon monoxide absent from the rate law entirely. Show that the mechanism

N + N \to N + NO (slow)

N + CO \to N + C (fast)

accounts for it.

First check the sum. Adding the two steps and cancelling one $N$ and the $N$ , which appear on both sides, gives $N + CO \to NO + C$ , the observed equation. Then the rate: the first step is bimolecular in nitrogen dioxide, so its rate is $k_{1} [N]^{2}$ , and since it is rate determining the overall rate is that. Carbon monoxide is consumed only in the fast step, which processes nitrate radical as soon as it appears, so its concentration cannot affect anything. The prediction matches, including the surprising absence.

Now you. The observed rate law for $2 N + \to 2 NF$ is $v = k [N] []$ . Is the single-step termolecular mechanism consistent with it? Propose a two step mechanism that is.

Answer

A single termolecular step would give $v = k [N]^{2} []$ , second order in nitrogen dioxide, which is not observed. A consistent mechanism is $N + \to NF + F$ (slow), followed by $N + F \to NF$ (fast). The steps sum correctly and the slow step gives exactly the observed rate law.

Notice what has and has not been shown. The mechanism is consistent with the data, which is the most any mechanism ever is. A rate law can rule mechanisms out; it can never prove one, since another sequence may predict the same law. Mechanisms are supported by detecting the intermediate, by isotopic labelling, by stereochemistry, and by the failure of every proposed alternative.

The steady state approximation

When no step is clearly slowest, the algebra needs a different tool. The steady state approximation assumes that a reactive intermediate, being consumed almost as fast as it is made, sits at a low and nearly constant concentration, so its net rate of change can be set to zero. Not because nothing is happening to it, but because formation and destruction have come into balance.

The classic application answers a question the sixth lesson left hanging: why is a unimolecular decomposition first order at all? A molecule cannot decompose unless something gives it the energy, and the only source is a collision, which is bimolecular. Frederick Lindemann's 1922 answer was that the two events are separate:

A + M rightleftharpoons A^{*} + M (k_{1}, k_{- 1}) A^{*} \to P (k_{2})

where $M$ is any collision partner and $A^{*}$ is an energised molecule. Apply the steady state to $A^{*}$ : it is formed at $k_{1} [A] [M]$ and destroyed at $k_{- 1} [A^{*}] [M] + k_{2} [A^{*}]$ . Setting the two equal and solving,

[A^{*}] = \frac{k_{1} [A] [M]}{k_{- 1} [M] + k_{2}} so v = k_{2} [A^{*}] = \frac{k_{1} k_{2} [A] [M]}{k_{- 1} [M] + k_{2}}

Example. Take the two limits of that expression and say what each predicts.

At high pressure $[M]$ is large, so $k_{- 1} [M]$ dominates $k_{2}$ in the denominator and $[M]$ cancels: $v = (k_{1} k_{2} / k_{- 1}) [A]$ , cleanly first order. Energisation is so frequent that it is effectively an equilibrium, and the decomposition of the energised molecule is what limits the rate. At low pressure $k_{2}$ dominates instead, leaving $v = k_{1} [A] [M]$ , second order: now collisions are rare, and every molecule that gets energised decomposes before it can be deactivated.

Now you. The observed first order rate constant is $k_{obs} = v / [A]$ . Show that $1 / k_{obs}$ is linear in $1 / [M]$ , and say why that is a useful test.

Answer

From the expression above, $k_{obs} = k_{1} k_{2} [M] / (k_{- 1} [M] + k_{2})$ , so inverting gives $1 / k_{obs} = k_{- 1} / (k_{1} k_{2}) + 1 / (k_{1} [M])$ . Plotting $1 / k_{obs}$ against $1 / [M]$ should give a straight line whose intercept and slope give the ratios of the rate constants. Real gas phase decompositions do fall off from first order at low pressure and the plot is roughly straight, which is why the mechanism is accepted; it curves at the low pressure end, which is why the modern treatment refines it.

A related shortcut is the pre-equilibrium, where a fast reversible step precedes a slow one. Then the first step stays at equilibrium and the intermediate's concentration is $K_{1}$ times the reactants'. The third order oxidation $2 NO + \to 2 N$ is the standard case: two nitric oxide molecules dimerise reversibly to , which then reacts with oxygen, giving $v = k_{2} K_{1} [NO]^{2} []$ without ever needing a triple collision. It also explains an oddity: this reaction gets slower as the temperature rises, because $K_{1}$ is for an exothermic dimerisation and falls with temperature faster than $k_{2}$ climbs.

Chain reactions

Some mechanisms regenerate their own intermediates, so a single initiation event drives many turnovers. The steps are classified as initiation, which creates the carriers, propagation, which consumes one and makes another, inhibition, which undoes propagation, and termination, which destroys carriers.

The formation of hydrogen bromide is the case that made the method's reputation. Bodenstein's 1906 rate law was

v = \frac{k [] [B]^{1 / 2}}{1 + k' [HBr] / [B]}

which resisted explanation for thirteen years. In 1919 Christiansen, Herzfeld and Polanyi independently proposed a chain: bromine dissociates to atoms, $Br + \to HBr + H$ , then $H + B \to HBr + Br$ regenerating the carrier, with $H + HBr \to + Br$ as inhibition and atom recombination as termination. Applying the steady state to both $H$ and $Br$ reproduces that entire expression, including the half power, which comes from the equilibrium concentration of bromine atoms varying as the square root of $[B]$ , and the inhibition term in the denominator, which comes from the product competing with bromine for hydrogen atoms.

Where termination is inefficient and one propagation step produces two carriers, the chain branches and the rate grows without limit. That is an explosion in the chemical sense, distinct from the merely thermal kind: the hydrogen and oxygen reaction has branching steps and shows explosion limits that depend on pressure in a way no thermal theory predicts. Combustion and polymerisation are both chain chemistry, which is why both are so sensitive to traces of inhibitor.

Enzymes and the saturating rate law

Biological catalysts bind their substrate first and react afterwards, which gives a rate law of a shape that turns up wherever a catalyst can be saturated. Write

E + S rightleftharpoons ES \to E + P

and apply the steady state to the complex $ES$ , using the fact that the total enzyme is conserved. The result, from Leonor Michaelis and Maud Menten in 1913 in the form Briggs and Haldane later justified, is

v = \frac{v_{\max} [S]}{K_{M} + [S]}

with $v_{\max} = k_{2} [E]_{0}$ and $K_{M} = (k_{- 1} + k_{2}) / k_{1}$ . The shape has two limits and a useful midpoint. When $[S] ≪ K_{M}$ the rate is first order in substrate; when $[S] ≫ K_{M}$ the enzyme is saturated and the rate is $v_{\max}$ , independent of substrate, which is the zero order behaviour of the sixth lesson. And when $[S] = K_{M}$ the rate is exactly half of $v_{\max}$ , which is how $K_{M}$ is measured.

Example. An enzyme has $K_{M} = 5.0 \times 10^{- 5}$ mol dm⁻³ and $v_{\max} = 2.0 \times 10^{- 3}$ mol dm⁻³ s⁻¹. Find the rate at $[S] = 1.0 \times 10^{- 5}$ and at $1.0 \times 10^{- 3}$ mol dm⁻³.

At the low concentration, $v = 2.0 \times 10^{- 3} \times 1.0 \times 10^{- 5} / (6.0 \times 10^{- 5}) = 3.3 \times 10^{- 4}$ mol dm⁻³ s⁻¹, which is $17$ per cent of the maximum. At the high one, $v = 2.0 \times 10^{- 3} \times 1.0 \times 10^{- 3} / (1.05 \times 10^{- 3}) = 1.9 \times 10^{- 3}$ , which is $95$ per cent. Raising the substrate concentration a hundredfold has raised the rate by less than a factor of six, because the enzyme was already running out of capacity.

Now you. For the same enzyme, at what substrate concentration is the rate three quarters of $v_{\max}$ ?

Answer

Set $[S] / (K_{M} + [S]) = 0.75$ , giving $[S] = 3 K_{M} = 1.5 \times 10^{- 4}$ mol dm⁻³. The general result is that reaching a fraction $f$ of the maximum needs $[S] = K_{M} f / (1 - f)$ , which diverges as $f$ approaches one: saturation is approached and never reached.

The turnover numbers involved are extraordinary. Catalase processes about $4 \times 10^{7}$ molecules of hydrogen peroxide per enzyme molecule per second, and carbonic anhydrase around $10^{6}$ , close to the limit set by how fast substrate can diffuse to the active site.

Kinetics and thermodynamics must agree

The two halves of this course have run in parallel, and one relation ties them together. Consider an elementary step at equilibrium. Equilibrium is dynamic: forward and reverse continue at equal rates, so $k_{f} [A] = k_{r} [B]$ , and therefore

K = \frac{[B]}{[A]} = \frac{k_{f}}{k_{r}}

The equilibrium constant of an elementary step is the ratio of its rate constants. This is not an extra assumption but a requirement, the principle of detailed balance: every individual step must be separately balanced at equilibrium, not merely the overall process. If a step measured $k_{f} = 1.6 \times 10^{- 2}$ and its reverse $2.95 \times 10^{- 4}$ in the same units, thermodynamics is obliged to report $K = 54.2$ for it, and if it reports something else, one of the measurements is wrong.

The same requirement forbids a cycle of three reactions all of which run preferentially in the same direction, which would be a chemical perpetual motion machine. And it connects to the previous lesson: since $K$ depends on temperature through $Δ H$ and each rate constant through its $E_{a}$ , taking logarithms of $K = k_{f} / k_{r}$ and differentiating recovers $E_{a, f} - E_{a, r} = Δ H$ for the step, which was drawn from the energy profile there and is now derived.

The general machinery is complete. The remaining lessons apply it to the two families of reaction that account for most of the chemistry anyone actually does, starting with the transfer of a proton.