Key Concepts and Summary
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.
Practice Problems: Reaction Mechanisms
Why are elementary reactions involving three or more reactants very uncommon?
Solution
Since each elementary step describes a collision (or decomposition) of molecules, a termolecular elementary step (indicating third-order kinetics) would mean having 3 molecules colliding simultaneously – and with the correct orientation and sufficient energy. Since the molecules in a mixture are moving about randomly, this is unlikely to happen spontaneously, and it would take a very long time to have even one successful collision, let alone enough collisions to produce a measurable amount of product. Therefore, termolecular and higher orders of reaction are rare, since unimolecular and bimolecular reactions will proceed much more quickly and be the dominant reactions.In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction $A+B⟶C$? Can we predict the effect if the reaction is known to be an elementary reaction?
Solution
In general, for the overall reaction we cannot predict the effect of changing the concentration without knowing the rate law (or the elementary steps of the reaction mechanism).
If the reaction is known to be an elementary reaction, then we can use the stoichiometry of the elementary step to build a rate law: in this case it would be first-order in [A] and doubling the concentration of A doubles the rate.
Define these terms:
(a) unimolecular reaction
(b) bimolecular reaction
(c) elementary reaction (elementary step)
(d) overall reaction
.Solution
(a) A unimolecular reaction is a reaction (specifically an elementary step in a reaction mechanism) that only involves one molecule of reactant. For example, a reaction of the type $A \rightarrow B + C$ would be a unimolecular reaction. These reactions exhibit first-order kinetics.
(b) A bimolecular reaction is a reaction (specifically an elementary step in a reaction mechanism) that involves two molecules of reactant. For example, a reaction of the type $2A\rightarrow B$ or $A+B\rightarrow C$ would each be bimolecular reactions. These reactions exhibit overall second-order kinetics (though they may be first-order in two individual reactants).
(c) An elementary step in a reaction mechanism is a chemical reaction that describes a single collision between reacting molecules (or the decomposition of a single molecule). An elementary step may begin with the reactants from the overall reaction, or may involve intermediate species produced in an earlier step or catalysts that may be directly involved in a reaction step. All elementary steps in a valid reaction mechanism must combine to produce the overall reaction.
(d) The overall reaction in a mechanism describes the stoichiometry of the reactants and products of a chemical reaction. It does not include any intermediate species (that may be produced in one elementary step and consumed in a later step), nor any catalysts (that may be consumed in an elementary step but are regenerated later) – though use of a catalyst may be noted.
What is the rate law for the elementary termolecular reaction $A+2B\rightarrow \text{products}$ ?
For $3A \rightarrow \text{products}$ ?
Solution
For $A+2B \rightarrow \text{products}$: $Rate=k[A][B]^2$
For $3A \rightarrow \text{products}$ : $Rate=k[A]^3$
Given the following overall reactions and their corresponding rate laws, which of these reactions are likely to proceed via a single-step reaction mechanism?
(a)$Cl_2+CO⟶Cl_2CO\qquad rate=k[Cl_2]^{3/2}[CO]$
(b) $PCl_3+Cl_2⟶PCl_5\qquad rate=k[PCl_3][Cl_2]$
(c) $2NO+H_2⟶N_2+H_2O_2\qquad rate=k[NO][H_2]$
(d) $2NO+O_2⟶2NO_2\qquad rate=k[NO]^2[O_2]$
(e) $NO+O_3⟶NO_2+O_2 \qquad rate=k[NO][O_3]$
Solution
Since elementary steps in a reaction mechanism describe a single collision event, their stoichiometry determines the rate law for that step in the reaction. Overall reactions may describe a more complex action, so the molecularity of the rate law does not necessarily match the stoichiometry of an “overall” reaction (the rate limiting step in the mechanism is probably not the same stoichiometry as the overall reaction).
For the reactions above, if they are “overall” reactions and the rate law does match, it is possible that this is a one-step reaction, and the overall reaction is the same as its’ elementary step. If the rate law doesn’t match, then there must be at least one other step in the mechanism that influences the rate law, so the overall reaction must proceed via a multi-step mechanism.
Following this logic, the reactions above that could be one-step reactions are (b), (d), and (e)
Write the rate law for each of the following elementary reactions:
(a) $O_3\; ^{\underrightarrow{\;\small{{\text{sunlight}\;}}}}\;O_2+O$
(b) $O_3+Cl⟶O_2+ClO$
(c) $ClO+O⟶Cl+O_2$
(d) $O_3+NO⟶NO_2+O_2$
(e) $NO_2+O⟶NO+O_2$
Solution
(a) Rate = k[O3]
(b) Rate = k[O3][Cl]
(c) Rate = k[ClO][O]
(d) Rate = k[O3][NO]
(e) Rate = k[NO2][O]
Nitrogen monoxide, NO, reacts with hydrogen, H2, according to the following equation: $$2NO+2H_2⟶N_2+H_2O$$
What would the rate law be if the mechanism for this reaction were: $$2NO+H_2⟶N_2+H_2O_2\quad \text{(slow)}$$ $$H_2O_2+H_2⟶2H_2O_2\quad \text{(fast)}$$
Solution
The order of the rate law will match the stoichiometry of the rate-determining (slow) step. Therefore: $$\text{rate} = k[NO]^2 [H_2 ]$$
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises: $$Cl_2\;(g)\rightarrow 2Cl\;(g)\quad \text{(fast)}$$ $$CO\;(g)+Cl\;(g)\rightarrow COCl (g)\quad \text{(slow}) $$ $$COCl\;(g)+Cl\;(g)\rightarrow COCl_2\, (g)\quad \text{(fast})$$
(a) Write the overall reaction.
Solution
The overall reaction should be the sum of all the elementary steps: $$ Cl_2\, (g) + CO\, (g) \rightarrow COCl_2 \, (g)$$
(b) Identify all intermediates and catalysts.
Solution
Intermediates are species that are formed as products of one elementary step of a mechanism, and consumed in a later step (so that they do not appear in the overall reaction).
$Cl\, (g)$ and $CoCl\, (g)$ are intermediates in this reaction.
Catalysts can be identified as species that are added to a reaction (they first appear as a reactant) but are regenerated in a later step, so that they do not appear in the overall reaction. Some heterogeneous catalsts are also written over the reaction arrow (e.g. Pt in $A\: ^{\underrightarrow{\small{\; Pt\; }}}\: B$). There are no catalysts in this reaction mechanism.
(c) Which step is the rate-determining step?
Solution
The second step (slowest reaction) will be the rate determining step in this mechanism.
(d) Write the rate law for each elementary reaction.
Solution
- $\text{rate}= [Cl_2]$
- $\text{rate}= [CO][Cl] $
- $\text{rate}= [COCl][Cl]$