Elementary steps, mechanisms and detailed balance

An elementary step is one real collision or one real change, and its order is just its molecularity

Intuition

Think of a recipe. The finished dish tells you the ingredients that went in and what came out. It does not tell you the order of the steps, or how long each one takes. A balanced chemical equation is the dish. The mechanism is the recipe. Only the recipe tells you how fast the whole thing goes.

Last lecture covered a reaction that passes in one trajectory over a single transition state. Only some reactions can be described that way. They include bond stretching and breaking, bond forming and the internal rotation of groups about a bond. Examples: dissociation, A → B + C (like H₂(g) → 2H(g) or HI(g) → H(g) + I(g)), and isomerisation. The lecture asks whether cis–trans isomerisation of azobenzene happens by rotation around the –N=N– bond or by inversion. It does not give the answer.

Elementary processes are unimolecular or bimolecular events between molecules in the reacting system. Termolecular collisions are too rare to be significant. The lecture lists:

• A → B (unimolecular reaction)
• A → B + C (unimolecular dissociation)
• A + B → C (bimolecular reaction)
• A + B → C + D (bimolecular reaction)

The unimolecular cases share one rate law and the bimolecular cases another.

$\frac{d[A]}{dt} = -k[A] \qquad\qquad \frac{d[A]}{dt} = -k[A][B]$

If a step really is elementary, you can write its rate law straight away. The order is set entirely by the molecularity. Other reactions are a sequence of several elementary steps.

Exam detail

Know the four example elementary processes, the two rate laws, and the rule that molecularity sets the order only for a genuinely elementary step.

The balanced equation hides the rate law, so you test mechanisms against data

A balanced equation gives the relative amounts of reactants and products. It does not give the rate law. The gas-phase reaction H₂ + Br₂ → 2HBr has the differential rate law

$\frac{d[\mathrm{HBr}]}{dt} = \frac{k_a[\mathrm{H}_2][\mathrm{Br}_2]^{3/2}}{[\mathrm{Br}_2] + k_b[\mathrm{HBr}]}$

Slide 6 also has a second rate law marked “NOT” and covered by a “BUSTED” stamp. The slide shows a wrong rate law crossed out. The text of that law cannot be read, so it is not reproduced here. A complicated law like the one above is not unusual. Many reactions have rate laws that cannot be described by a simple linear response approach. In general, to find a rate law you need the mechanism, which describes the molecular changes as a series of elementary processes.

How do we know which elementary steps occur? In general that is hard. Some knowledge comes from earlier experiments, or from calculations on elementary steps. But often we just test many schemes built from the simplest elementary steps and see which give the observed kinetics. Several schemes can predict the same overall rate equation. So the best we can say is that a scheme is consistent with the data. If it is not consistent, we can throw that scheme away. The lecture then moves to a worksheet.

Three shortcuts make the algebra manageable

The lecture lists three common approximations.

• Steady state approximation. Assume all reactive intermediates quickly reach an unchanging “steady state” concentration, for example $d[\mathrm{NO}]/dt = 0$.
• Rate-determining step. The slowest bottleneck in a scheme often sets the overall rate. Think of traffic through the narrowest road.
• Pre-equilibrium. Merge any fast reversible reactions into one equilibrium that stays in balance while one product slowly leaks away into later steps. The example is $\mathrm{N_2O_5} \rightleftharpoons \mathrm{NO_2} + \mathrm{NO_3}$.

Pitfall

Do not treat the rate-determining step as “the first step”. It is the slowest one, wherever it falls in the scheme.

At equilibrium every step balances itself, and that gives you K

Intuition

Picture a busy two-way corridor. At equilibrium, as many people walk left as walk right in every stretch of it, not just overall. That is detailed balance. The rule is strict: every single step is balanced, so the whole corridor is.

How do we link a complicated mechanism, which may include species that are neither reactants nor products, to the equilibrium state? The lecture’s example is $2\mathrm{NO} + 2\mathrm{H_2} \rightleftharpoons \mathrm{N_2} + 2\mathrm{H_2O}$, believed to occur in three elementary steps:

1. $\mathrm{NO} + \mathrm{NO} \rightleftharpoons \mathrm{N_2O_2}$ (coefficients $k_1$, $k_{-1}$)
2. $\mathrm{N_2O_2} + \mathrm{H_2} \rightleftharpoons \mathrm{N_2O} + \mathrm{H_2O}$ (coefficients $k_2$, $k_{-2}$)
3. $\mathrm{N_2O} + \mathrm{H_2} \rightleftharpoons \mathrm{N_2} + \mathrm{H_2O}$ (coefficients $k_3$, $k_{-3}$)

The principle of detailed balance says that at equilibrium, the forward and backward rates of each elementary process are equal.

Try it: worked example

Use detailed balance on the three steps for 2NO + 2H2 to get the equilibrium constant K in terms of the rate coefficients.

Show answer and steps

Answerk1 k2 k3 / (k-1 k-2 k-3) = [N2]eq [H2O]eq^2 / ([NO]eq^2 [H2]eq^2) = K. The intermediates N2O2 and N2O cancel.

1. Step i at equilibrium, forward rate equals backward rate: $k_1[\mathrm{NO}]^2_{eq} = k_{-1}[\mathrm{N_2O_2}]_{eq}$. Rearrange to $\dfrac{k_1}{k_{-1}} = \dfrac{[\mathrm{N_2O_2}]_{eq}}{[\mathrm{NO}]^2_{eq}}$.
2. Step ii the same way: $k_2[\mathrm{N_2O_2}]_{eq}[\mathrm{H_2}]_{eq} = k_{-2}[\mathrm{N_2O}]_{eq}[\mathrm{H_2O}]_{eq}$. Rearrange to $\dfrac{k_2}{k_{-2}} = \dfrac{[\mathrm{N_2O}]_{eq}[\mathrm{H_2O}]_{eq}}{[\mathrm{N_2O_2}]_{eq}[\mathrm{H_2}]_{eq}}$.
3. Step iii the same way: $k_3[\mathrm{N_2O}]_{eq}[\mathrm{H_2}]_{eq} = k_{-3}[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]_{eq}$. Rearrange to $\dfrac{k_3}{k_{-3}} = \dfrac{[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]_{eq}}{[\mathrm{N_2O}]_{eq}[\mathrm{H_2}]_{eq}}$.
4. Multiply the three ratios: $\dfrac{k_1k_2k_3}{k_{-1}k_{-2}k_{-3}} = \dfrac{[\mathrm{N_2O_2}]_{eq}}{[\mathrm{NO}]^2_{eq}}\cdot\dfrac{[\mathrm{N_2O}]_{eq}[\mathrm{H_2O}]_{eq}}{[\mathrm{N_2O_2}]_{eq}[\mathrm{H_2}]_{eq}}\cdot\dfrac{[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]_{eq}}{[\mathrm{N_2O}]_{eq}[\mathrm{H_2}]_{eq}}$.
5. Cancel the intermediates. $[\mathrm{N_2O_2}]_{eq}$ is on the top of the first fraction and the bottom of the second. $[\mathrm{N_2O}]_{eq}$ is on the top of the second and the bottom of the third.
6. What is left is $\dfrac{[\mathrm{N_2}]_{eq}[\mathrm{H_2O}]^2_{eq}}{[\mathrm{NO}]^2_{eq}[\mathrm{H_2}]^2_{eq}}$, which the slide labels $K$, the equilibrium constant.

So the equilibrium constant is fixed by the rate coefficients of all the steps, and the intermediates never show up in it.

Optional aside

Slide 9 gives the elementary steps and the detailed balance equations, but does not work through the rate law for the forward reaction. Slide 7 of this lecture mentions a worksheet, but the worksheet itself is not in the slides.

Recall

Why do the intermediates N2O2 and N2O vanish from the equilibrium constant?

Each one appears once on top and once on the bottom when the three detailed-balance ratios are multiplied, so they cancel.