Rate Laws and Measurements

Lessons

Senior 6 · Chemistry

Rate Laws and Measurements

5 lessons

Lesson 5 of 5

Molecularity and Reaction Mechanisms

25 min

Carbon monoxide that does not count

Below about 500 K, carbon monoxide reacts with nitrogen dioxide: $\mathrm{NO_2(g) + CO(g) \rightarrow NO(g) + CO_2(g)}$ . If this happened in one collision between an $\mathrm{NO_2}$ molecule and a CO molecule, the rate would be $k[\mathrm{NO_2}][\mathrm{CO}]$ . Experiment gives $\text{rate} = k[\mathrm{NO_2}]^2$ : changing [CO] makes no difference at all. The reaction cannot be a single step. The rate law is evidence about the hidden sequence of steps by which the reaction really happens, its mechanism.

Definition

Elementary step: a single molecular event (one collision or one molecule falling apart) that cannot be broken down further. Example: $\mathrm{NO_2 + NO_2 \rightarrow NO_3 + NO}$ .

Molecularity: the number of particles (molecules, atoms or ions) taking part in an elementary step. Unimolecular: one, e.g. $\mathrm{(CH_3)_3CBr \rightarrow (CH_3)_3C^{+} + Br^{-}}$ . Bimolecular: two, e.g. $\mathrm{OH^{-} + CH_3Br \rightarrow CH_3OH + Br^{-}}$ . Termolecular: three, which is rare because three particles seldom meet at the same instant.

Reaction mechanism: the sequence of elementary steps by which a reaction occurs. Rate-determining step (RDS): the slowest step in the mechanism, which limits the overall rate.

Order and molecularity are different ideas

	Order	Molecularity
Applies to	the overall reaction (or one reactant in it)	one elementary step
How found	by experiment	from the proposed mechanism
Possible values	0, 1, 2, 3 and fractions	1, 2 or (rarely) 3; never 0 or a fraction
Link	For an elementary step only, the rate law can be written from the step: its order in each reactant equals the number of those particles in the step.

Testing a proposed mechanism

A mechanism is acceptable only if:

its steps add up to the overall balanced equation (intermediates cancel);
every step is uni- or bimolecular (termolecular only with good reason);
the rate law it predicts matches the experimental rate law.

To predict the rate law, write the rate equation of the rate-determining step. Species that react only after the slow step do not appear. If an intermediate appears, replace its concentration using a fast equilibrium that comes before the slow step (worked example 3). A mechanism that passes all three tests is consistent with the evidence; it is not proved, because another mechanism might fit as well.

Substitution at carbon: one step or two?

Figure 5.1 compares two halogenoalkanes reacting with hydroxide ions. Bromomethane reacts in one bimolecular step (S_N2, "substitution, nucleophilic, bimolecular"), so both reactants appear in the rate law. 2-Bromo-2-methylpropane first ionises in a slow unimolecular step (S_N1); the hydroxide ion reacts with the carbocation only afterwards, so $[\mathrm{OH^{-}}]$ is absent from the rate law. The three methyl groups both stabilise the carbocation and crowd the back of the carbon atom, which blocks the S_N2 route.

Figure 5.1: Energy profiles for the S_N2 and S_N1 reactions of halogenoalkanes with hydroxide ions.

Worked example 1: a slow first step

Problem. For $\mathrm{NO_2 + CO \rightarrow NO + CO_2}$ , $\text{rate} = k[\mathrm{NO_2}]^2$ . Show that this mechanism fits: step 1 (slow) $\mathrm{NO_2 + NO_2 \rightarrow NO_3 + NO}$ ; step 2 (fast) $\mathrm{NO_3 + CO \rightarrow NO_2 + CO_2}$ .

Solution.

Sum: $\mathrm{2NO_2 + NO_3 + CO \rightarrow NO_3 + NO + NO_2 + CO_2}$ . Cancel $\mathrm{NO_3}$ (intermediate) and one $\mathrm{NO_2}$ : $\mathrm{NO_2 + CO \rightarrow NO + CO_2}$ . ✓

Both steps are bimolecular. ✓

The slow step is elementary with two $\mathrm{NO_2}$ : $\text{rate} = k_1[\mathrm{NO_2}]^2$ . CO reacts only after the slow step, so it is absent. ✓ The mechanism is consistent with the rate law.

Worked example 2: the iodination of propanone

Problem. The experimental rate law is $\text{rate} = k[\mathrm{CH_3COCH_3}][\mathrm{H^{+}}]$ (Lesson 3). Explain how this mechanism accounts for it, and identify the catalyst and the intermediates.

Step 1 (fast): $\mathrm{CH_3COCH_3 + H^{+} \rightleftharpoons CH_3C(OH^{+})CH_3}$
Step 2 (slow): $\mathrm{CH_3C(OH^{+})CH_3 \rightarrow CH_2{=}C(OH)CH_3 + H^{+}}$
Step 3 (fast): $\mathrm{CH_2{=}C(OH)CH_3 + I_2 \rightarrow CH_3COCH_2I + H^{+} + I^{-}}$

Solution. The slow step 2 uses the protonated propanone, whose concentration is set by the fast equilibrium of step 1 and is proportional to $[\mathrm{CH_3COCH_3}][\mathrm{H^{+}}]$ . So $\text{rate} \propto [\mathrm{CH_3COCH_3}][\mathrm{H^{+}}]$ . Iodine reacts only in step 3, after the slow step, so the reaction is zero order in iodine. The intermediates are the protonated propanone and the enol $\mathrm{CH_2{=}C(OH)CH_3}$ . $\mathrm{H^{+}}$ is used in step 1 and given back in step 2, so it is a catalyst; the $\mathrm{H^{+}}$ released in step 3 forms the product HI with $\mathrm{I^{-}}$ . Sum of the steps: $\mathrm{CH_3COCH_3 + I_2 \rightarrow CH_3COCH_2I + HI}$ . ✓

Worked example 3: a fast equilibrium before the slow step

Problem. For $\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$ , $\text{rate} = k[\mathrm{NO}]^2[\mathrm{O_2}]$ . A proposed mechanism is: step 1 (fast equilibrium) $\mathrm{2NO \rightleftharpoons N_2O_2}$ , equilibrium constant $K$ ; step 2 (slow) $\mathrm{N_2O_2 + O_2 \rightarrow 2NO_2}$ . Derive the rate law.

Solution. Rate of the slow step: $\text{rate} = k_2[\mathrm{N_2O_2}][\mathrm{O_2}]$ . $\mathrm{N_2O_2}$ is an intermediate, so replace it using step 1: $K = \frac{[\mathrm{N_2O_2}]}{[\mathrm{NO}]^2}$ , so $[\mathrm{N_2O_2}] = K[\mathrm{NO}]^2$ .

\text{rate} = k_2K[\mathrm{NO}]^2[\mathrm{O_2}] = k[\mathrm{NO}]^2[\mathrm{O_2}] \quad\text{with } k = k_2K

The third-order rate law comes from two bimolecular steps; no three-body collision is needed.

Warning

Common mistakes

Writing the rate law from the overall equation. Only an elementary step can be read this way.
Leaving an intermediate in a predicted rate law. Experimental rate laws contain only species you can add to the flask (reactants, catalysts, sometimes products).
Calling a catalyst an intermediate. A catalyst is used in an early step and given back later; an intermediate is formed in one step and used in a later one.
Saying a mechanism is "proved" because it fits the rate law. It is only consistent with the evidence.

Tip

Reading a mechanism: species before or in the slow step appear in the rate law (with intermediates traced back to reactants); species after the slow step do not. Reading a rate law: the species in it, with their orders, tell you what is in the transition state of the slow step.

Allery Chemistry: Rate determining step. Watch how the species in the slow step, and the number of each, are read from the rate equation, and why species that react after it are missing.

✦

Practice

Distinguish between the order of a reaction and the molecularity of a step (three differences).
State the molecularity of each step: (a) $\mathrm{N_2O_4 \rightarrow 2NO_2}$ ; (b) $\mathrm{H + Br_2 \rightarrow HBr + Br}$ ; (c) $\mathrm{2NO + Br_2 \rightarrow 2NOBr}$ .
Explain why termolecular steps are rare.
$\mathrm{2NO_2Cl \rightarrow 2NO_2 + Cl_2}$ goes by step 1 (slow) $\mathrm{NO_2Cl \rightarrow NO_2 + Cl}$ , step 2 (fast) $\mathrm{NO_2Cl + Cl \rightarrow NO_2 + Cl_2}$ . Identify the intermediate and predict the rate law.
For $\mathrm{H_2 + 2ICl \rightarrow I_2 + 2HCl}$ , $\text{rate} = k[\mathrm{H_2}][\mathrm{ICl}]$ . Is this mechanism consistent: step 1 (slow) $\mathrm{H_2 + ICl \rightarrow HI + HCl}$ ; step 2 (fast) $\mathrm{HI + ICl \rightarrow I_2 + HCl}$ ? Give reasons.
The hydrolysis of a halogenoalkane RBr by $\mathrm{OH^{-}}$ has $\text{rate} = k[\mathrm{RBr}]$ . Suggest a two-step mechanism and say which step is slow.
Ozone decomposes, $\mathrm{2O_3 \rightarrow 3O_2}$ , by step 1 (fast equilibrium, constant $K$ ) $\mathrm{O_3 \rightleftharpoons O_2 + O}$ and step 2 (slow) $\mathrm{O + O_3 \rightarrow 2O_2}$ . Derive the rate law and comment on the effect of adding oxygen.

Show answersHide answers

Order is experimental, molecularity theoretical; order applies to the overall reaction, molecularity to one elementary step; order can be zero or fractional, molecularity is 1, 2 or 3.
(a) unimolecular; (b) bimolecular; (c) termolecular.
Three particles must collide at the same instant, with enough energy and the right orientation; this is far less likely than a two-particle collision.
Intermediate: Cl atoms. Slow step contains one $\mathrm{NO_2Cl}$ : $\text{rate} = k[\mathrm{NO_2Cl}]$ , first order.
Yes: the steps add to the overall equation (HI cancels); both are bimolecular; the slow step gives $\text{rate} = k[\mathrm{H_2}][\mathrm{ICl}]$ , matching experiment.
Step 1 (slow): $\mathrm{RBr \rightarrow R^{+} + Br^{-}}$ ; step 2 (fast): $\mathrm{R^{+} + OH^{-} \rightarrow ROH}$ . Only RBr is in the slow step (S_N1).
Rate $= k_2[\mathrm{O}][\mathrm{O_3}]$ with $[\mathrm{O}] = \frac{K[\mathrm{O_3}]}{[\mathrm{O_2}]}$ , so $\text{rate} = \frac{k_2K[\mathrm{O_3}]^2}{[\mathrm{O_2}]}$ . Adding $\mathrm{O_2}$ shifts step 1 to the left, lowering [O] and slowing the reaction: an order of −1 in $\mathrm{O_2}$ .

1Carbon monoxide that does not count

2Order and molecularity are different ideas

3Testing a proposed mechanism

4Substitution at carbon: one step or two?

Carbon monoxide that does not count

Order and molecularity are different ideas

Testing a proposed mechanism

Substitution at carbon: one step or two?