Chemistry · General chemistry II · Concept

Reaction mechanisms and rate laws

A mechanism is a list of elementary steps that claims to show how a reaction happens. It is consistent with the data when its steps add up to the overall reaction and its slow step, with any intermediate replaced, predicts the observed rate law.

A concrete case

A reactant that does not count

Below about $225 ° C$ , nitrogen dioxide and carbon monoxide react, ${NO}_{2} + CO \to NO + {CO}_{2}$ . Double the $CO$ and the rate does not change at all. Double the ${NO}_{2}$ and the rate goes up four times. The measured rate law is $rate = k {[{NO}_{2}]}^{2}$ , with no $CO$ in it.

If the reaction were one collision of an ${NO}_{2}$ with a $CO$ , doubling either one would double the rate. It is not one collision. First two ${NO}_{2}$ molecules collide and trade an oxygen atom, ${NO}_{2} + {NO}_{2} \to {NO}_{3} + NO$ . That step is slow. Then the ${NO}_{3}$ meets a $CO$ , ${NO}_{3} + CO \to {NO}_{2} + {CO}_{2}$ , and that step is fast.

Add the two steps and ${NO}_{3}$ cancels, along with one ${NO}_{2}$ , leaving the overall equation. The slow first step sets the pace, and it needs two ${NO}_{2}$ and no $CO$ . $CO$ only has to be there when the fast second step needs it.

The rule

Elementary steps and their rate laws

An elementary step happens exactly as written, in one collision or one rearrangement. Its molecularity is the number of particles that come together in it, one for unimolecular, two for bimolecular and three, rarely, for termolecular.

Because an elementary step is one event, its rate law can be read from its own coefficients. For $A + B \to products$ the rate is $k [A] [B]$ , and for $2 A \to products$ it is $k {[A]}^{2}$ . This works only for an elementary step, never for an overall equation.

A species made in one step and used up in a later one is an intermediate, like ${NO}_{3}$ above. It never appears in the overall equation.

elementary A + B \to products \Rightarrow rate = k [A] [B]

\begin{array}{c} elementary A + B \to products \\ \Rightarrow rate = k [A] [B] \end{array}

The rate-determining step

When one step is much slower than the others, it is the rate-determining step, and the overall rate law is its rate law. If that rate law contains an intermediate, the intermediate must be replaced by species from the overall reaction before it can be compared with the experiment.

A fast, reversible step before the slow one does that. Its forward and reverse rates are equal, so it can be solved for the intermediate.

k_{1} [NO] [{Br}_{2}] = k_{- 1} [{NOBr}_{2}] \Rightarrow [{NOBr}_{2}] = \frac{k_{1}}{k_{- 1}} [NO] [{Br}_{2}]

\begin{array}{c} k_{1} [NO] [{Br}_{2}] = k_{- 1} [{NOBr}_{2}] \\ \Rightarrow [{NOBr}_{2}] = \frac{k_{1}}{k_{- 1}} [NO] [{Br}_{2}] \end{array}

Why it works

Why the slow step sets the rate

Picture two workers on a line. The first can finish $10$ parts a minute, and the second $1000$ . The line turns out $10$ parts a minute, however fast the second worker is. The steps after the slow one can only pass along what it supplies.

Steps before the slow one are fast enough to keep up. If such a step is reversible, its forward and reverse reactions run far faster than the slow step drains it, so it stays close to equilibrium. Setting its forward rate equal to its reverse rate is then a good approximation.

The overall coefficients count what is used in total. The orders count what meets in the slowest collision, together with what a fast equilibrium before it supplies. The two agree only by chance, or when the reaction really is a single step, which is the claim being tested.

The method

Testing a mechanism, one check at a time

1. Add the steps. Cancel every species that appears on both sides. The sum must be the overall equation, or the mechanism describes a different reaction.

2. Find the slow step and write its rate law from its own coefficients.

3. Look for an intermediate in that rate law. If one is there, solve the fast equilibrium before it for the intermediate and substitute.

4. Compare the predicted orders with the observed rate law. If they match, the mechanism is consistent. If they do not, it is ruled out as written.

If–then–else test for a proposed mechanism
Check	If	Then	Else
1. Add the steps	They add up to the overall reaction (intermediates cancel)	Go to check 2	Ruled out: it describes a different reaction
2. Slow step	Exactly one step is slow	Its rate law is $k$ times each reactant’s concentration raised to its coefficient; go to check 3	Mark the rate-determining step before predicting
3. Intermediates	The slow-step rate law has no intermediate	Go to check 4	Replace the intermediate using a fast equilibrium before the slow step, then look again
4. Compare	Predicted orders match the observed rate law	Consistent, possible but not proved	Ruled out as written

Where it breaks

Consistent is not proved. A different mechanism can predict the same rate law. For $2 NO + {Br}_{2}$ , a single collision of three molecules would give $rate = k {[NO]}^{2} [{Br}_{2}]$ too. Detecting the proposed intermediate is the kind of evidence that favors one mechanism over another.

The overall equation is not a step. Reading ${NO}_{2} + CO \to NO + {CO}_{2}$ as one collision predicts $rate = k [{NO}_{2}] [CO]$ , which the experiment rules out.

Three-particle collisions are rare, so a mechanism that needs one in its slow step is less likely than one built from collisions of two.

The fast-equilibrium shortcut assumes the first step runs back far faster than the slow step drains it. When that is not true, chemists use the steady-state approximation instead, which can give a different rate law.

A catalyst is not an intermediate. A catalyst is used in one step and given back in a later one, so it appears in the rate law but cancels from the overall equation. An intermediate is made first and used up later.

Check your answer

Add the steps yourself. Every species that is not in the overall equation must cancel exactly, with nothing left over on either side.

Check that only overall species are left in the prediction. An intermediate such as ${NOBr}_{2}$ in the final rate law means the substitution is not finished.

Compare every order, including zero. A reactant missing from the observed rate law, like $CO$ , must also be missing from the prediction.

In a Chemistry box, write the overall equation, then each step with the word slow or fast after it, then the observed rate law, such as $rate = k {[NO]}^{2} [{Br}_{2}]$ . The Checker marks the rate-law row. It says “The steps add up to the overall reaction in row 1.” and names the intermediates.

When the orders match, the row reads ✓ with “The mechanism predicts rate = k[Br₂][NO]², which matches this rate law.” When they do not, it reads ✗ and names both rate laws.

Steady states, chains and catalysts

The steady-state approximation assumes an intermediate is used up as fast as it is made, so its concentration stays small and nearly constant. Applied to a mechanism, it gives a rate law that includes the fast-equilibrium result as a special case.

Some reactions run as chains. In the reaction of hydrogen with bromine, a bromine atom attacks $H_{2}$ , and the hydrogen atom formed attacks ${Br}_{2}$ , giving back a bromine atom. One atom can carry the chain through thousands of cycles. The observed rate law has a sum in its denominator, a sign of more than one step competing.

The iodide-catalyzed decomposition of hydrogen peroxide shows a catalyst in a mechanism. $I^{-}$ appears in the slow step’s rate law, $rate = k [H_{2} O_{2}] [I^{-}]$ , yet cancels from the overall equation $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ .

Common mistakes

Using the overall coefficients as the orders. ${NO}_{2} + CO \to NO + {CO}_{2}$ has coefficients of $1$ and $1$ , but below $225 ° C$ the rate law is $k {[{NO}_{2}]}^{2}$ . The orders come from the slow step, not from the overall equation.
Leaving an intermediate in the predicted rate law. Mechanism B for $2 NO + {Br}_{2}$ gives $rate = k_{2} [{NOBr}_{2}] [NO]$ at first. ${NOBr}_{2}$ must be replaced through the fast equilibrium, which turns it into $k {[NO]}^{2} [{Br}_{2}]$ .
Writing the rate law from a fast step when a slow step exists. In the ${NO}_{2} + CO$ mechanism, the fast step ${NO}_{3} + CO$ would predict a $CO$ dependence that the data do not show. The slow first step sets the rate.
Calling a mechanism proven because its rate law matches. A matching rate law makes it possible, not certain. Another mechanism, even a single three-body collision, can predict the same rate law.
Accepting steps that do not add up to the overall reaction. $NO + {Br}_{2} \to {NOBr}_{2}$ then ${NOBr}_{2} \to NOBr + Br$ sum to $NO + {Br}_{2} \to NOBr + Br$ , not $2 NO + {Br}_{2} \to 2 NOBr$ . Such a mechanism is ruled out before any rate law is compared.
Calling a catalyst an intermediate. $I^{-}$ in the hydrogen peroxide mechanism is used in the first step and given back in the second, so it is a catalyst. ${IO}^{-}$ is made in the first step and used in the second, so it is the intermediate.

Key terms

Reaction mechanism: The sequence of simple (elementary) steps that adds up to an overall reaction. The balanced overall equation alone can’t tell you what those steps are.
Elementary step: A single collision or molecular event within a reaction mechanism. Unlike an overall reaction, its rate law can be written straight from its equation.
Rate-determining step: The slowest step in a mechanism, which limits how fast the whole reaction goes. A fast equilibrium before it also shapes the overall rate law.
Reaction intermediate: A substance made in one step of a mechanism and used up in a later step, so it does not appear in the overall equation.
Molecularity: The number of particles that collide in one elementary step: unimolecular (1), bimolecular (2) or, rarely, termolecular (3). It applies only to elementary steps, not to overall reactions.
Bimolecular: Describes an elementary step (one event in a reaction mechanism) in which two particles collide and react. Its rate law is second order overall.
Termolecular: Describes an elementary step (one event in a reaction mechanism) requiring three particles to collide at once. Such steps are rare because three-body collisions are unlikely.
Rate law: An equation linking reaction rate to reactant concentrations, such as $rate = k {[A]}^{2} [B]$ . The units of the rate constant $k$ depend on the overall order.
Catalyst: A substance that speeds up a reaction by giving it a lower-energy pathway and is not used up. It does not change the equilibrium constant or where equilibrium lies.

Work through an example

Nitric oxide and bromine form nitrosyl bromide, $2 NO (g) + {Br}_{2} (g)$ $\to 2 NOBr (g)$ . Experiments give $rate = k {[NO]}^{2} [{Br}_{2}]$ . Which of two mechanisms is consistent with that rate law? Mechanism A is $NO + {Br}_{2} \to {NOBr}_{2}$ (slow), then ${NOBr}_{2} + NO \to 2 NOBr$ (fast). Mechanism B is $NO + {Br}_{2} ⇌ {NOBr}_{2}$ (fast), then ${NOBr}_{2} + NO \to 2 NOBr$ (slow).

Rule a mechanism in or out for

2 NO + {Br}_{2}

→

Explain a negative order with a mechanism →

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Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 9 key terms Test Mechanism B in a Chemistry box See Mechanism B as an energy profile Open worked example on a board Chemistry formulas: kinetics

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