Boltzmann Distribution, Temperature and Catalysis

AS · 17 min

A rise of just $10\ ^\circ\text{C}$ can roughly double the rate of many reactions, far more than the extra speed of the molecules alone would explain. The reason is that only the most energetic molecules can react, and the number of those rises steeply with temperature. This note defines activation energy, shows how to sketch and use the Boltzmann distribution of molecular energies, explains the effect of temperature on rate in two ways (energy and collision frequency), and explains how catalysts work, using both the Boltzmann distribution and reaction pathway diagrams, for homogeneous and heterogeneous catalysts. Sketching a labelled Boltzmann distribution is one of the most frequently examined skills in AS Chemistry.

Activation energy

Definition

Activation energy, $E_A$ , is the minimum energy required for a collision to be effective.

Only collisions in which the colliding particles have a combined energy at least equal to $E_A$ can break the necessary bonds and lead to reaction. On a reaction pathway diagram, $E_A$ is the height of the energy barrier between the reactants and the peak. Reactions with a high activation energy are slow at room temperature, because very few collisions have enough energy.

The Boltzmann distribution

In a gas (or a liquid), the molecules do not all have the same energy. They collide constantly, exchanging energy, so at any instant a few are almost stationary, most have moderate energies, and a few have very high energies. The Boltzmann distribution is the graph of the number of molecules with each energy against that energy.

y = 11.28sqrt(x)exp(-x) (3, 0) -- (3, 5) fill 3 7 y = 11.28sqrt(x)exp(-x)

(Horizontal axis: energy; vertical axis: number of molecules with that energy. The vertical line marks the activation energy $E_A$ . The shaded area to its right represents the molecules with energy $\ge E_A$ .)

Key result

Features of the Boltzmann distribution (each one is a marking point when you sketch it):

It starts at the origin: no molecules have zero energy.
It rises to a peak, the most probable energy (the energy possessed by the greatest number of molecules).
It is not symmetrical: it falls more gradually on the high-energy side, with a long tail.
The tail approaches the energy axis but never touches it: there is no maximum energy.
The area under the curve equals the total number of molecules.
Axes are labelled: number of molecules (or proportion of molecules) with a given energy, and energy.

The area under the curve to the right of $E_A$ represents the number of molecules with enough energy to react when they collide. In the sketch above it is only a small fraction of the total, about 11%.

The effect of temperature

When the temperature is raised, the average kinetic energy of the molecules increases, and the whole distribution changes shape. The second sketch shows the same gas at a lower temperature $T_1$ (taller, narrower curve) and a higher temperature $T_2$ (flatter curve).

y = 11.28sqrt(x)exp(-x) y = 5.575sqrt(x)exp(-x/1.6) (3, 0) -- (3, 5)

Key result

At the higher temperature:

the peak moves to the right (higher most probable energy) and is lower;
the curve is broader, with more molecules at high energies;
the area under the curve is the same (the number of molecules has not changed);
the area to the right of $E_A$ is much larger: a greater proportion of molecules have energy $\ge E_A$ .

In this sketch, the fraction of molecules with energy above $E_A$ rises from about 11% to about 29%: more than two and a half times as many.

Method

Explaining the effect of temperature on rate

At a higher temperature, molecules have more kinetic energy (on average).
Energy argument (main reason): the Boltzmann distribution shifts to higher energy, so a greater proportion of molecules have energy greater than or equal to the activation energy. A greater proportion of collisions are effective.
Frequency argument (minor reason): molecules move faster, so they collide more frequently.
Both lead to a greater frequency of effective collisions, so the rate increases.

The energy argument is much more important. A $10\ \text{K}$ rise near room temperature increases the speed of molecules (and so the collision frequency) by under 2%, but it can double the number of molecules with energy above $E_A$ . That is why the rate of many reactions roughly doubles.

Watch out

When drawing two Boltzmann curves at different temperatures, the most common errors are: curves not starting at the origin; the higher-temperature curve with a higher peak (it must be lower); curves crossing the energy axis at high energy (they must not touch it); and areas obviously different (the higher-temperature curve must be lower and broader so that the area stays the same). Draw the higher-temperature curve below the lower-temperature one at low energies and above it at high energies, crossing once.

Catalysts

Definition

A catalyst is a substance that increases the rate of a reaction without itself undergoing any permanent chemical change.

Catalysis is the process of increasing the rate of a reaction using a catalyst.

How a catalyst works

In the presence of a catalyst, the reaction takes a different route, with a different mechanism (series of steps) that has a lower activation energy. The catalyst takes part in the reaction but is regenerated at the end, so it is not used up and only small amounts are needed.

Catalysis and the Boltzmann distribution

The catalyst does not change the distribution of molecular energies (the temperature is the same). It lowers the energy barrier, so the line marking $E_A$ moves to the left. Many more molecules now have energy greater than or equal to the new, lower activation energy.

y = 11.28sqrt(x)exp(-x) (3, 0) -- (3, 5) (1.8, 0) -- (1.8, 5) fill 1.8 7 y = 11.28sqrt(x)exp(-x)

(The right-hand vertical line is $E_A$ without a catalyst; the left-hand line is $E_A$ with a catalyst. The shaded area, from the catalysed $E_A$ , is about 31% of the molecules, compared with about 11% beyond the uncatalysed $E_A$ .)

Key result

A catalyst increases rate because:

it provides an alternative reaction pathway with a lower activation energy, so a greater proportion of molecules (and of collisions) have energy $\ge E_A$ ; the frequency of effective collisions increases.

Catalysis and the reaction pathway diagram

On a reaction pathway diagram, the catalysed route has a lower peak. The reactant and product levels are unchanged, so $\Delta H$ is the same.

y = 6 - 4*(1 + tanh(1.2*(x - 5)))/2 + 5*exp(-(x - 5)^2/2) y = 6 - 4*(1 + tanh(1.2*(x - 5)))/2 + 2.5*exp(-(x - 5)^2/2) (4.59, 6) -> (4.59, 9.51) (4.33, 6) -> (4.33, 7.33) (8.5, 6) -> (8.5, 2)

(Upper curve: uncatalysed, with the larger $E_A$ arrow. Lower curve: catalysed, with the smaller $E_A$ arrow. The arrow on the right is $\Delta H$ , the same for both routes. In reality a catalysed route often has more than one peak, one for each step, but a single lower peak is accepted at AS.)

A catalyst lowers the activation energy of the reverse reaction by the same amount, so it speeds up the forward and reverse reactions equally. That is why a catalyst does not change the position of equilibrium; it only helps the system reach equilibrium faster.

Homogeneous and heterogeneous catalysts

Definition

A homogeneous catalyst is in the same phase as the reactants (for example, all in aqueous solution, or all gases).

A heterogeneous catalyst is in a different phase from the reactants (usually a solid catalyst with gaseous or liquid reactants).

catalyst	reaction	type
iron, Fe(s)	Haber process: $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$	heterogeneous
vanadium(V) oxide, $\ce{V2O5(s)}$	Contact process: $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$	heterogeneous
platinum, palladium, rhodium (s)	catalytic converters: $\ce{2CO(g) + 2NO(g) -> 2CO2(g) + N2(g)}$	heterogeneous
nickel, Ni(s)	hydrogenation of alkenes	heterogeneous
manganese(IV) oxide, $\ce{MnO2(s)}$	decomposition of hydrogen peroxide (aq)	heterogeneous
$\ce{H+(aq)}$ (from concentrated $\ce{H2SO4}$ )	esterification: $\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}$	homogeneous
$\ce{Fe^2+(aq)}$ or $\ce{Fe^3+(aq)}$	$\ce{S2O8^2-(aq) + 2I-(aq) -> 2SO4^2-(aq) + I2(aq)}$	homogeneous
enzymes (in solution in cells)	biochemical reactions	homogeneous (usually)

Heterogeneous catalysts work at their surface: reactant molecules are adsorbed onto the surface, bonds within them are weakened, reaction occurs, and the products are released (desorbed). That is why such catalysts are used as powders, fine meshes or thin coatings on a honeycomb support: a large surface area gives more active sites. A homogeneous catalyst usually reacts with one reactant to form an intermediate, which then reacts to form the product and regenerate the catalyst.

Tip

The detailed mechanisms of homogeneous catalysis (for example $\ce{Fe^2+}$ with peroxodisulfate and iodide) and of adsorption in heterogeneous catalysis are covered at A Level. At AS, you need to define both types, give examples, and explain the effect of a catalyst using the Boltzmann distribution and a reaction pathway diagram.

Worked examples

Routine: sketching and labelling

Sketch the Boltzmann distribution for a sample of gas. Mark the most probable energy and the activation energy, and shade the area representing the molecules that can react.

Solution

Axes: vertical, "number of molecules with energy E"; horizontal, "energy, E". A curve starting at the origin, rising to a single peak, then falling more slowly with a long tail that approaches but never meets the energy axis.

Mark the most probable energy on the energy axis directly below the peak. Draw a vertical line well to the right of the peak and label it $E_A$ . Shade the area under the curve to the right of $E_A$ : these are the molecules with energy $\ge E_A$ , which can react on collision.

Explaining a temperature effect

Use the Boltzmann distribution to explain why the reaction between sodium thiosulfate and hydrochloric acid is faster at $40\ ^\circ\text{C}$ than at $20\ ^\circ\text{C}$ .

Solution

At $40\ ^\circ\text{C}$ the particles have more kinetic energy on average. The Boltzmann distribution moves to the right: the peak is lower and at a higher energy, and the curve has more particles in the high-energy tail. So a greater proportion of particles have energy greater than or equal to the activation energy. In addition, the particles move faster and collide more frequently. Together these produce more frequent effective collisions, so the rate is higher. The first effect is the more important.

Reaction pathway diagram with a catalyst

The decomposition of hydrogen peroxide is exothermic, $\Delta H = -98\ \text{kJ mol}^{-1}$ . Without a catalyst, $E_A = 75\ \text{kJ mol}^{-1}$ ; with the enzyme catalase, $E_A = 23\ \text{kJ mol}^{-1}$ . Sketch the reaction pathway diagram for both routes, and state the activation energy of the reverse reaction for each.

Solution

Sketch: reactants $\ce{2H2O2}$ on the left at a higher level; products $\ce{2H2O + O2}$ on the right, $98\ \text{kJ mol}^{-1}$ lower. Uncatalysed curve rising to a peak $75\ \text{kJ mol}^{-1}$ above the reactants; catalysed curve rising to a peak only $23\ \text{kJ mol}^{-1}$ above the reactants. $\Delta H$ arrow from reactants down to products, the same for both.

Reverse activation energies (from products to peak): uncatalysed $75 + 98 = 173\ \text{kJ mol}^{-1}$ ; catalysed $23 + 98 = 121\ \text{kJ mol}^{-1}$ . The catalyst lowers both by the same amount, $52\ \text{kJ mol}^{-1}$ .

Classifying catalysts

Classify each catalyst as homogeneous or heterogeneous: (a) nickel in the reaction of ethene with hydrogen gas; (b) sulfuric acid in the reaction of ethanoic acid with ethanol; (c) nitrogen dioxide gas in the atmospheric oxidation of sulfur dioxide gas; (d) manganese(IV) oxide powder in the decomposition of aqueous hydrogen peroxide.

Solution

(a) Heterogeneous: solid catalyst, gaseous reactants.

(b) Homogeneous: the acid and the reactants are all in the same liquid phase.

(d) Heterogeneous: solid catalyst, aqueous reactant.

Exam-hard: temperature versus catalyst

A student says: "Raising the temperature and adding a catalyst both increase the rate in the same way, by giving the molecules more energy." Evaluate this statement, using sketches of the Boltzmann distribution to support your answer, and state one other difference between the two changes for a reversible exothermic reaction.

Solution

The statement is incorrect. Both increase the proportion of collisions that are effective, but they do it in different ways.

Raising the temperature does give the molecules more energy. The Boltzmann distribution changes shape: the peak moves to higher energy and becomes lower, and the high-energy tail rises. With $E_A$ unchanged, a greater proportion of molecules now have energy $\ge E_A$ . Collision frequency also increases slightly.
Adding a catalyst does not change the energies of the molecules: the Boltzmann distribution is unchanged. Instead, the catalyst provides an alternative pathway with a lower $E_A$ , so the $E_A$ line moves to the left, and a greater proportion of the same distribution lies beyond it.

Other difference: for a reversible exothermic reaction, raising the temperature moves the position of equilibrium to the left and decreases the yield (and the equilibrium constant), whereas a catalyst has no effect on the position of equilibrium or the yield. (Also: the catalyst is not used up, and it allows a lower operating temperature, saving energy.)

Investigating the effect of temperature on rate

Place separate flasks of sodium thiosulfate solution and of dilute hydrochloric acid in a thermostatically controlled water bath at the chosen temperature (for example 20, 30, 40, 50 and $60\ ^\circ\text{C}$ ).
When both have reached the bath temperature, mix them in a flask standing on a cross, and start the stopwatch.
Record the time for the cross to disappear, and the temperature of the mixture at the start and the end; use the mean.
Calculate $1/t$ as a measure of rate and plot $1/t$ against temperature. The curve rises increasingly steeply.

Controlled variables: concentrations and volumes of both solutions, the same cross and observer, the same flask. Errors: temperature falls during the reaction (record mean temperature; use a water bath); judging the end point (use a light sensor); $\ce{SO2}$ is toxic, so work in a well-ventilated room and use dilute solutions.

Watch out

"A catalyst gives the molecules more energy." No: it lowers the activation energy; the energy distribution is unchanged.
"A catalyst is not involved in the reaction." It takes part (it is involved in the mechanism) but is regenerated, so there is no permanent chemical change.
Higher temperature curve with a higher peak, or one that does not start at the origin. See the earlier warning.
Only giving the collision frequency argument for temperature. The main reason is the increased proportion of molecules with energy $\ge E_A$ .
"A catalyst changes $\Delta H$ " or "increases the yield." Neither: $\Delta H$ and the position of equilibrium are unchanged.

Exam tip

Define activation energy exactly: "the minimum energy required for a collision to be effective".
Boltzmann sketch: label both axes, start at the origin, single asymmetric peak, tail not touching the axis, $E_A$ marked to the right of the peak, shaded area beyond $E_A$ . Two-temperature sketches: higher- $T$ curve lower and broader, peak to the right, same area.
Temperature explanation needs three ideas: more molecules with energy $\ge E_A$ (from the distribution); more frequent collisions; therefore more frequent effective collisions.
Catalyst explanation needs: alternative route (different mechanism) with lower $E_A$ ; more molecules with energy $\ge$ the new $E_A$ ; more frequent effective collisions. Draw the new $E_A$ line to the left on the same curve.
Homogeneous: same phase; heterogeneous: different phase. Give the phase of both catalyst and reactants in any example.

Summary

Activation energy: the minimum energy required for a collision to be effective.
Boltzmann distribution: starts at the origin, asymmetric peak (most probable energy), tail never meets the axis, area = total number of molecules.
Area to the right of $E_A$ = molecules able to react.
Higher temperature: peak lower and to the right, same area, many more molecules with energy $\ge E_A$ ; also more frequent collisions; rate increases sharply.
Catalyst: alternative route with a different mechanism and lower $E_A$ ; distribution unchanged; more molecules exceed the lower $E_A$ .
Catalyst lowers $E_A$ of forward and reverse reactions equally: no change to $\Delta H$ or equilibrium position.
Homogeneous catalyst: same phase as reactants ( $\ce{H+}$ in esterification). Heterogeneous: different phase (Fe in Haber, $\ce{V2O5}$ in Contact, Pt/Pd/Rh in catalytic converters).

Practice

Question

Define activation energy.
Sketch (or describe) a Boltzmann distribution, labelling the axes, the most probable energy and the activation energy.
On the same axes, sketch distributions for a gas at two temperatures, $T_1$ and a higher $T_2$ , and state three differences between the curves.
Explain why a small increase in temperature can cause a large increase in rate.
Define a catalyst, and explain, using the Boltzmann distribution, how a catalyst increases the rate of a reaction.
Sketch a reaction pathway diagram for an endothermic reaction with and without a catalyst, labelling $\Delta H$ and both activation energies.
Explain the difference between homogeneous and heterogeneous catalysis, with one example of each.
Explain why catalytic converters contain the metal catalysts as a thin layer on a ceramic honeycomb.
For a reaction, $E_A = 120\ \text{kJ mol}^{-1}$ uncatalysed and $85\ \text{kJ mol}^{-1}$ catalysed; $\Delta H = +40\ \text{kJ mol}^{-1}$ . Calculate the activation energy of the reverse reaction with and without the catalyst, and use your answer to explain why a catalyst does not change the position of equilibrium.
In the Boltzmann sketches in this note, about 11% of molecules had energy above $E_A$ at the lower temperature and about 29% at the higher temperature, while the catalysed $E_A$ gave about 31% at the lower temperature. A chemist wants to double the rate of an industrial reaction that is exothermic and reversible. Compare the use of a higher temperature with the use of a catalyst, considering rate, yield and cost.

Answers

The minimum energy required for a collision to be effective.
Vertical axis: number of molecules with energy E; horizontal axis: energy. Curve starts at the origin, rises to a peak (most probable energy marked below it on the axis), falls more gradually with a long tail that approaches but does not touch the energy axis. $E_A$ marked as a vertical line on the high-energy side, well to the right of the peak.
The $T_2$ curve: (1) has a lower peak; (2) its peak is at a higher energy (moved to the right); (3) has more molecules at high energy, a larger area beyond $E_A$ ; the total area under both curves is the same.
At a higher temperature the Boltzmann distribution shifts to higher energies. Because $E_A$ is in the tail of the distribution, even a small shift greatly increases the proportion of molecules with energy $\ge E_A$ , so the proportion of effective collisions rises sharply; collisions are also slightly more frequent. The frequency of effective collisions increases greatly.
A catalyst is a substance that increases the rate of a reaction without itself undergoing any permanent chemical change. It provides an alternative pathway (different mechanism) with a lower activation energy. On the Boltzmann distribution, which is unchanged, the $E_A$ line moves to the left, so a greater proportion of molecules have energy $\ge E_A$ ; the frequency of effective collisions increases.
Reactants at a lower level, products higher (endothermic); $\Delta H$ arrow pointing up from reactants to products, positive. Uncatalysed curve with a high peak; catalysed curve with a lower peak; both $E_A$ arrows drawn from the reactants up to their peaks.
Homogeneous: catalyst in the same phase as the reactants, for example $\ce{H+(aq)}$ in the esterification of ethanoic acid with ethanol (all in the liquid phase). Heterogeneous: catalyst in a different phase, for example solid iron in the Haber process with gaseous nitrogen and hydrogen.
Heterogeneous catalysts act at their surface (reactants are adsorbed onto active sites). A thin layer on a honeycomb gives a very large surface area for a small mass of expensive metal, maximising the number of active sites and the rate, and allowing the exhaust gases to flow through.
Reverse $E_A = E_A(\text{forward}) - \Delta H$ for an endothermic reaction. Uncatalysed: $120 - 40 = 80\ \text{kJ mol}^{-1}$ . Catalysed: $85 - 40 = 45\ \text{kJ mol}^{-1}$ . The catalyst lowers both forward and reverse activation energies by the same amount ( $35\ \text{kJ mol}^{-1}$ ), so it speeds up both reactions; the equilibrium is reached faster but its position is unchanged.
Higher temperature: increases the proportion of molecules with energy $\ge E_A$ (from 11% to 29% in the sketch) and the collision frequency, so the rate rises; but because the reaction is exothermic, the position of equilibrium moves to the left, reducing the yield, and heating costs energy. Catalyst: gives a similar increase in the proportion able to react (31% in the sketch) at the same temperature, with no effect on the yield and lower energy costs; the catalyst is not used up, although it has an initial cost and can be poisoned. The catalyst is the better choice; industrial processes usually combine a catalyst with a compromise temperature.

Activation energy

The Boltzmann distribution

The effect of temperature

Catalysts

How a catalyst works

Catalysis and the Boltzmann distribution

Catalysis and the reaction pathway diagram

Homogeneous and heterogeneous catalysts

Worked examples

Practice

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