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Rate Laws and Measurements

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Senior 6 · Chemistry

Rate Laws and Measurements

5 lessons

Lesson 1 of 5

Theories of Reaction Rates

25 min

Two pictures of one event

Picture a reaction in two ways. In the first, molecules are like balls on a crowded field: they react only if they hit each other hard enough and at the right angle. In the second, a reaction is like a journey between two valleys across a ridge: a traveller does not climb the highest summit but crosses at the lowest pass, and the height of that pass decides how many travellers get across each hour.

The first picture is collision theory, which you used in Unit 12. The second is transition state theory, proposed in 1935 by Henry Eyring and, independently, by Meredith Evans and Michael Polanyi. Both explain why a rate depends on $E_\text{a}$ and temperature; the second also says what the molecules look like at the top of the barrier.

Definition

Steric factor, $P$ : the fraction of collisions in which the molecules are oriented so that they can react; $0 < P \le 1$ . Example: $P$ is close to 1 for two atoms, which look the same from every side, and much smaller for large molecules that must meet at one particular group.

Transition state (activated complex), written with the symbol ‡: the arrangement of atoms at the maximum of the energy profile, in which old bonds are partly broken and new bonds partly formed. It lasts about as long as one molecular vibration and cannot be isolated. Example: $\mathrm{[HO\cdots CH_3\cdots Br]^{-\ddagger}}$ .

Reaction intermediate: a species formed in one step of a reaction and used up in a later step. It sits in an energy minimum between two transition states, has complete bonds, and can sometimes be detected. Example: the carbocation $\mathrm{(CH_3)_3C^{+}}$ in the hydrolysis of 2-bromo-2-methylpropane.

Collision theory, quantitatively

Collision theory pictures molecules as hard spheres. The rate constant is the collision frequency factor $Z$ multiplied by the steric factor and by the fraction of collisions with enough energy:

k = P\,Z\,e^{-\frac{E_\text{a}}{RT}}

Comparing with the Arrhenius equation $k = A\,e^{-\frac{E_\text{a}}{RT}}$ shows that $A = P Z$ . Kinetic theory lets you calculate $Z$ from the sizes and masses of the molecules, so measuring $A$ gives $P$ .

Strengths: it explains the effects of concentration, pressure and temperature, and why $A$ is larger for simple species. Limits: it cannot predict $P$ or $E_\text{a}$ ; it treats molecules as structureless spheres; and it says nothing about what happens to the bonds during the collision.

Transition state theory

Transition state theory follows the energy of the atoms as the reaction proceeds. On the way from reactants to products, the system must pass through the transition state at the top of the energy barrier, so

E_\text{a} = E(\text{transition state}) - E(\text{reactants})

The theory treats the transition state as being in a rapid equilibrium with the reactants, and the rate as the rate at which transition states fall apart into products. Figure 1.1 shows the transition state for the reaction of hydroxide ions with bromomethane. The carbon atom is joined to five groups at once: the $\mathrm{C{-}O}$ bond is half made, the $\mathrm{C{-}Br}$ bond half broken, and the negative charge is shared between O and Br. Because the $\mathrm{OH^{-}}$ ion can bond to carbon only from the side opposite the bromine, many collisions fail: this is the origin of a small steric factor.

Figure 1.1: Reactants, transition state and products for $\mathrm{OH^{-} + CH_3Br \rightarrow CH_3OH + Br^{-}}$ .

Reactions in more than one step

Most reactions happen in a series of simple steps. Each step has its own transition state, and an intermediate sits in the valley between them (Figure 1.2). The step whose transition state is highest above the reactants controls the overall rate; it is called the rate-determining step (Lesson 5).

Figure 1.2: Energy profile of a two-step reaction; energies in $\mathrm{kJ\,mol^{-1}}$ , illustrative values.

	Transition state	Intermediate
Position on the profile	An energy maximum	An energy minimum between two maxima
Bonds	Partial bonds being made and broken	Complete bonds
Lifetime	About one vibration (roughly $10^{-13}\,\mathrm{s}$ )	From fractions of a second to much longer
Can it be detected or isolated?	No	Sometimes detected; occasionally isolated

Worked example 1: the steric factor

Problem. For a gas-phase reaction at 500 K, kinetic theory gives a collision frequency factor $Z = 3.0 \times 10^{11}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ . Experiment gives $A = 1.2 \times 10^{9}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ (illustrative values). Find $P$ and interpret it.

Solution.

P = \frac{A}{Z} = \frac{1.2 \times 10^{9}}{3.0 \times 10^{11}} = 4.0 \times 10^{-3}

Only about 4 collisions in every 1000 that have enough energy are correctly oriented. The molecules must meet at one particular atom or bond, so most energetic collisions are wasted.

Worked example 2: reading a two-step energy profile

Problem. Use Figure 1.2 to find (a) the activation energy of each step, (b) the overall activation energy, (c) $\Delta H$ for the reaction, (d) the activation energy of the reverse reaction, and (e) the rate-determining step.

Solution.

(a) Step 1: $85 - 0 = 85\,\mathrm{kJ\,mol^{-1}}$ . Step 2: $55 - 30 = 25\,\mathrm{kJ\,mol^{-1}}$ .

(b) The highest point above the reactants is TS₁, so the overall $E_\text{a} = 85\,\mathrm{kJ\,mol^{-1}}$ .

(d) Going backwards, the products must climb to TS₁: $E_\text{a}(\text{reverse}) = 85 - (-40) = 125\,\mathrm{kJ\,mol^{-1}}$ . In general $E_\text{a}(\text{reverse}) = E_\text{a}(\text{forward}) - \Delta H$ .

(e) Step 1, because its transition state is the highest point on the path; once the intermediate forms, it crosses the lower second barrier quickly.

Warning

Common mistakes

Calling the transition state an intermediate. A transition state is at a peak and cannot be isolated; an intermediate is in a valley.
Choosing the rate-determining step as "the step with the larger barrier measured from its own start". Compare the transition states with the starting reactants: the highest one on the whole profile marks the slow step.
Thinking that every collision with $E \ge E_\text{a}$ succeeds. The orientation must also be right; $P$ can be far below 1.

Tip

On any energy profile: peaks = transition states (one per step); valleys between peaks = intermediates; number of peaks = number of steps; $E_\text{a}(\text{reverse}) = E_\text{a}(\text{forward}) - \Delta H$ .

Professor Dave Explains: Energy Diagrams, Catalysts, and Reaction Mechanisms. Watch for the difference between a transition state (a peak) and an intermediate (a valley), and how a multi-step profile is read.

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Practice

State two limitations of simple collision theory.
Give two differences between a transition state and a reaction intermediate.
A reaction has $Z = 5.0 \times 10^{10}$ and $A = 2.0 \times 10^{8}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ . Calculate the steric factor.
Explain why the reaction of $\mathrm{OH^{-}}$ with bromomethane has a steric factor much less than 1.
A one-step reaction has $E_\text{a} = 60\,\mathrm{kJ\,mol^{-1}}$ and $\Delta H = +25\,\mathrm{kJ\,mol^{-1}}$ . Sketch its energy profile and calculate the activation energy of the reverse reaction.
Sketch an energy profile for an endothermic two-step reaction in which the second step is rate-determining. Label the intermediate and both transition states.
A profile has reactants at 0, TS₁ at +40, intermediate at +10, TS₂ at +70 and products at +20 (all in $\mathrm{kJ\,mol^{-1}}$ ). Identify the rate-determining step, the overall $E_\text{a}$ and $\Delta H$ .

Show answersHide answers

Any two: it cannot predict the steric factor; it cannot predict $E_\text{a}$ ; it treats molecules as hard, structureless spheres; it says nothing about bond changes during the collision.
A transition state is at an energy maximum, has partial bonds and cannot be isolated; an intermediate is at a minimum between two maxima, has complete bonds and can sometimes be detected.
$P = \frac{2.0 \times 10^{8}}{5.0 \times 10^{10}} = 4.0 \times 10^{-3}$ .
The $\mathrm{OH^{-}}$ ion can form a bond to carbon only by approaching from the side opposite the bromine atom; collisions at the bromine end or the hydrogen side do not react, however energetic.
Reactants low, products 25 higher, one peak 60 above the reactants; $E_\text{a}(\text{reverse}) = 60 - 25 = 35\,\mathrm{kJ\,mol^{-1}}$ .
Two peaks with a valley between them; products above reactants; the second peak (TS₂) higher than the first.
TS₂ (+70) is the highest point, so step 2 is rate-determining; overall $E_\text{a} = 70\,\mathrm{kJ\,mol^{-1}}$ ; $\Delta H = +20\,\mathrm{kJ\,mol^{-1}}$ (endothermic).

Sources and credits

Video: Energy Diagrams, Catalysts, and Reaction Mechanisms, Professor Dave Explains, embedded from YouTube.
The text, tables, diagrams, illustrative values, worked examples and Practice items were written and drawn by Ganzaa.

1Two pictures of one event

2Collision theory, quantitatively

3Transition state theory

4Reactions in more than one step

Two pictures of one event

Collision theory, quantitatively

Transition state theory

Reactions in more than one step