Effect of temperature on reaction rates and the concept of activation energy

1. Overview

Raising the temperature speeds up almost every reaction, often by a lot. This section explains why, using two linked ideas: the activation energy, $E_{\text{A}}$, the minimum energy a collision needs to be effective, and the Boltzmann distribution, which shows how energy is shared out among the molecules of a sample. You need to define $E_{\text{A}}$, sketch the distribution, and use it to explain why $E_{\text{A}}$ matters and how temperature affects the rate of a reaction. All of this is explained with the diagram: no calculation is needed. The same diagram comes back in topic 8.3 to explain how catalysts work.

Key Definitions

• Activation energy (EA): the minimum energy required for a collision to be effective, that is, for the colliding particles to react; it is given as a positive value, usually in kJ mol⁻¹.
• Effective collision: a collision between reactant particles that leads to a reaction; the particles must collide with at least the activation energy and, for many reactions, in a suitable orientation.
• Frequency of effective collisions: the number of effective collisions per unit time (for example, per second) in a given volume of the reaction mixture; the rate of a reaction is proportional to it.
• Collision theory: the model in which particles react only when they collide, and only collisions with at least the activation energy (and a suitable orientation) are effective, so the rate depends on the frequency of effective collisions.
• Boltzmann distribution: the way the energies of the molecules in a sample, such as a gas, are spread out at a given temperature, shown as a graph of the number of molecules against energy.
• Most probable energy: the energy at the peak of the Boltzmann distribution; more molecules have energies close to this value than close to any other value.

Content

Activation energy and effective collisions

Particles can react only when they collide. Most collisions, though, are non-effective: the particles bounce apart unchanged. A collision is effective, and leads to a reaction, only when:

1. the particles collide with at least a certain minimum energy, the activation energy; and
2. for many reactions, they meet in a suitable orientation, with the reacting parts of the particles facing each other.

Energy is needed because bonds in the reactants must start to break before new bonds can form, and the electron clouds of two approaching particles repel each other. In a gentle collision the particles simply bounce apart. Only a hard enough collision gets over this energy barrier.

Activation energy, $E_{\text{A}}$, is the minimum energy required for a collision to be effective. It is given as a positive value, usually in kJ mol⁻¹, the same unit as $\Delta H$.

On a reaction pathway diagram (topic 5.1), $E_{\text{A}}$ is the height of the peak above the reactants. Each reaction has its own $E_{\text{A}}$, set by its pathway. Changing the temperature does not change $E_{\text{A}}$: it changes how many particles have enough energy to get over the barrier. A catalyst is different: it provides another route with a lower $E_{\text{A}}$ (topic 8.3).

The Boltzmann distribution

In a gas at a fixed temperature, the molecules do not all have the same energy. Each molecule collides billions of times a second, and each collision shares out the energy differently: one molecule may speed up while the other slows down. A few molecules are slow, most have moderate energies, and a few have very high energies. Each molecule's energy keeps changing, but the overall spread stays the same while the temperature does. This spread is the Boltzmann distribution. The energy plotted is kinetic energy, and particles in a liquid or solution have the same kind of spread.

Feature of the curve	What it tells you
vertical axis: number of molecules; horizontal axis: energy	the height at any energy shows how many molecules have energies close to that value
starts at the origin	almost no molecules have energies close to zero
one peak, towards the left	the most probable energy: more molecules have energies near this value than near any other
not symmetrical: a long tail to the right	a small number of molecules have energies several times the average
the tail gets closer and closer to the energy axis but never meets it	however high the energy, a tiny fraction of molecules has at least that much
area under the whole curve	the total number of molecules in the sample

The mean (average) energy is to the right of the peak, because the long tail pulls the average up. To count the molecules in a range of energies, use the area under that part of the curve, not the height.

How to sketch it:

1. Draw the axes and label them: number of molecules (vertical) and energy (horizontal).
2. Start the curve exactly at the origin.
3. Rise steeply to a single peak, left of centre.
4. Fall more gently into a long tail on the right.
5. Bring the tail closer and closer to the energy axis without touching it, and stop while it is still above the axis.
6. Mark $E_{\text{A}}$ on the energy axis, well to the right of the peak, with a vertical line up to the curve. Shade the area to the right of the line.

Why activation energy matters

Only molecules in the shaded area, those with energy of at least $E_{\text{A}}$, can take part in effective collisions. All the others collide and bounce apart unchanged.

For most reactions at room temperature, $E_{\text{A}}$ lies far out in the tail, so only a tiny fraction of collisions has enough energy. Sketches put $E_{\text{A}}$ much closer to the peak than it really is, so that the shaded area can be seen.

This is the significance of $E_{\text{A}}$: the larger $E_{\text{A}}$, the further right the line, the smaller the area beyond it, so fewer collisions are effective each second and the reaction is slower at the same temperature. A reaction with a large $E_{\text{A}}$ can be so slow that nothing seems to happen, even if it is very exothermic. Methane and air, for example, do not react until a flame or spark gives some molecules enough energy; then the heat released keeps the reaction going.

Worked example 1: two activation energies at the same temperature

Question: Reactions R and S both take place between gases at 25 °C. R has $E_{\text{A}} = 50$ kJ mol⁻¹ and S has $E_{\text{A}} = 100$ kJ mol⁻¹. In both mixtures the molecules collide equally often, and orientation matters equally. Sketch a Boltzmann distribution, mark both activation energies on it, and use it to explain which reaction is faster.

Step 1: one curve is enough. Both mixtures are at 25 °C, so their molecular energies are spread out in the same way. Draw one curve, following the six sketching steps above.

Step 2: mark both activation energies. Draw a vertical line for $E_{\text{A}}$ of R to the right of the peak. Draw the line for $E_{\text{A}}$ of S further right, because 100 kJ mol⁻¹ is the larger energy.

Step 3: compare the areas. The area to the right of each line is the number of molecules with at least that energy. The area beyond the S line is only part of the area beyond the R line, so it is smaller.

Step 4: link the areas to the rate. With collisions equally frequent, a smaller proportion with at least $E_{\text{A}}$ means fewer effective collisions per second for S.

Answer: R is faster. Its $E_{\text{A}}$ is lower, so a larger area of the distribution lies beyond it: a greater proportion of collisions has at least $E_{\text{A}}$, so the frequency of effective collisions is higher.

For a sense of scale only (you will not be asked to calculate this): at 25 °C both lines lie far out in the tail, so S is not "half as fast" as R. Other things being equal, it is hundreds of millions of times slower.

What a higher temperature does to the distribution

At a higher temperature the molecules have more kinetic energy on average (the average kinetic energy is proportional to the temperature in kelvin). The distribution spreads out towards higher energies:

• The peak is lower and further to the right. The most probable energy is higher, and fewer molecules have energies close to it.
• The area under the curve is the same. The number of molecules has not changed, so a curve that spreads out must also get lower.
• The curves cross once. Beyond the crossing point, the higher-temperature curve stays above the other all along the tail. The tails never meet each other or the axis.
• $E_{\text{A}}$ does not move.

The result is that the area to the right of $E_{\text{A}}$ is larger at the higher temperature, $T_2$, than at $T_1$. A small rise in temperature makes a big difference here, because the far tail is where the number of molecules changes most in proportion.

Lowering the temperature does the opposite: the peak becomes higher and moves left, the tail shrinks towards the axis, and the area beyond $E_{\text{A}}$ gets smaller. That is one reason food keeps longer in a fridge.

A higher temperature and the frequency of effective collisions

Raising the temperature increases the frequency of effective collisions in two ways:

1. Collisions become more frequent. The molecules move faster, so they collide more often. This effect is small (see Worked example 2 (d)).
2. A larger proportion of collisions is effective. More molecules have at least $E_{\text{A}}$, as the larger area beyond $E_{\text{A}}$ shows. This effect is large.

The second effect is the main reason the rate rises. Temperature does not change the fraction of collisions with a suitable orientation.

A rule of thumb. Near room temperature, a 10 °C rise often roughly doubles the rate. It is only a rough guide, not a law: it fits reactions with an $E_{\text{A}}$ of around 50 kJ mol⁻¹. The larger the $E_{\text{A}}$, the bigger the factor, and the same 10 °C rise gives a smaller factor at higher temperatures.

Worked example 2: explaining the effect of a temperature rise

Question: A mixture of two gases reacts slowly at temperature $T_1$. The same mixture, in the same sealed container, is heated to a higher temperature, $T_2$. (a) Sketch the Boltzmann distribution at both temperatures. (b) Use your sketch to explain why the rate of reaction increases. (c) State the effect of the temperature rise on $E_{\text{A}}$. (d) The molecules also collide more often at $T_2$. Explain why this is not the main reason for the faster rate.

(a) On axes labelled number of molecules and energy, draw both curves from the origin, with tails that never meet the axis. Give the $T_2$ curve a lower peak further to the right, crossing the $T_1$ curve once and staying above it along the tail, with about the same area. Mark $E_{\text{A}}$ once and label each curve with its temperature.

(b) The explanation, one link at a time.

1. At $T_2$ the molecules have more kinetic energy on average.
2. So a greater proportion of molecules have energy greater than or equal to $E_{\text{A}}$: the area to the right of $E_{\text{A}}$ is larger under the $T_2$ curve.
3. So a greater proportion of collisions are effective.
4. So the frequency of effective collisions increases, and the rate increases.

(c) $E_{\text{A}}$ does not change. It is set by the reaction pathway. The temperature changes how many molecules can get over the barrier, not the height of the barrier.

(d) Collisions do become more frequent, but only slightly: less than 2% more for a 10 °C rise in a gas at fixed volume. The proportion of collisions with at least $E_{\text{A}}$ rises much more, because $E_{\text{A}}$ lies far out in the tail.

Answer: (a) as in the two-temperature diagram above. (b) At $T_2$ a greater proportion of molecules have at least $E_{\text{A}}$, so effective collisions are more frequent and the rate increases. (c) No effect: $E_{\text{A}}$ is the same at both temperatures. (d) The rise in collision frequency is small compared with the rise in the proportion of collisions with at least $E_{\text{A}}$.

Temperature in rate experiments

Rate depends so strongly on temperature that a rate experiment must hold it steady, usually in a thermostatically controlled water bath (with ice to work below room temperature). If a 10 °C rise doubles the rate, even a 1 °C drift changes it by several per cent (about 7%). Bring each solution to the bath temperature before mixing.

To quench a reaction in a sample taken from a mixture, run the sample into a much larger volume of iced water. Quenching does not stop the reaction completely: it slows it so much that the amounts hardly change while you analyse the sample (for example, by titration), so analyse it straight away. Two separate effects slow it. The lower temperature means a smaller proportion of collisions has at least $E_{\text{A}}$. The extra water dilutes the sample, so collisions between reactant particles are less frequent (topic 8.1).

Key Equations

There is no formula to calculate with in this section, only relationships to use in explanations.

Rate and effective collisions: the rate of reaction is proportional to the frequency of effective collisions (the number of effective collisions per second in a given volume).

What sets the frequency of effective collisions: how often the particles collide, and the proportion of those collisions with at least $E_{\text{A}}$ (and a suitable orientation). Heating increases both, the second far more.

Areas on the Boltzmann distribution: area under the whole curve = total number of molecules (the same at every temperature for a fixed sample); area to the right of $E_{\text{A}}$ = number of molecules with energy of at least $E_{\text{A}}$.

Temperature ($T$ in K, $t$ in °C): $T = t + 273$. The average kinetic energy of the molecules is proportional to $T$.

Rule of thumb (approximate, near room temperature, for $E_{\text{A}}$ around 50 kJ mol⁻¹): a 10 °C rise roughly doubles the rate.

Common Mistakes to Avoid

• Mistake: sketching the Boltzmann distribution like a reaction pathway diagram, starting part way up the vertical axis or labelling the axes "enthalpy" and "progress of reaction". Fix: start at the origin and label the axes number of molecules and energy. A curve starting a third of the way up the axis claims that many molecules have zero energy.
• Mistake: drawing the higher-temperature curve with a higher peak, or letting a tail touch the energy axis or merge into the other curve. Fix: the area stays the same, so a curve that spreads to the right must be lower, as the $T_2$ peak is in the two-temperature diagram. Its tail stays above the $T_1$ tail to the end, and both stay above the axis.
• Mistake: saying that a higher temperature lowers (or raises) the activation energy. Fix: $E_{\text{A}}$ is set by the reaction pathway. A reaction with $E_{\text{A}} = 50$ kJ mol⁻¹ at 15 °C still has 50 kJ mol⁻¹ at 45 °C, so the $E_{\text{A}}$ line on the sketch does not move. What changes is the shaded area beyond it. Only a catalyst offers a route with a lower $E_{\text{A}}$ (topic 8.3).
• Mistake: writing "more collisions" or "more successful collisions" with no mention of time. Fix: write "more frequent effective collisions" or "more effective collisions per second". If each effective collision uses up one reactant particle, using up 0.010 mol takes $6.0 \times 10^{21}$ effective collisions at any temperature; heating only makes them happen in less time.
• Mistake: explaining the effect of temperature only as "the particles move faster, so they collide more often". Fix: give the main effect first: a greater proportion of molecules have at least $E_{\text{A}}$. For a gas at fixed volume, a 10 °C rise near room temperature makes collisions less than 2% more frequent, yet the rate often roughly doubles.
• Mistake: reading the height of the curve at $E_{\text{A}}$ as the number of molecules that can react. Fix: that height shows only how many molecules have energies close to $E_{\text{A}}$. The number with at least $E_{\text{A}}$ is the whole shaded area to the right of the line.
• Mistake: thinking that a higher temperature slows down an exothermic reaction. Fix: heating speeds up both exothermic and endothermic reactions, and both the forward and the reverse reaction of an equilibrium. Magnesium reacts faster with warm dilute hydrochloric acid than with cold, although the reaction is exothermic. $\Delta H$ decides which way an equilibrium shifts (topic 7.1), not whether the rate rises.

Exam Tips

• Define activation energy with the key word minimum: "the minimum energy required for a collision to be effective". "The energy needed to start a reaction" is too vague.
• Sketch with labelled axes and each curve labelled with its temperature. A second, hotter curve has a lower peak further right, crosses the first once and keeps its tail above it.
• Explain the effect of temperature as the chain in Worked example 2 (b): more energy on average, so a greater proportion of molecules with at least $E_{\text{A}}$, so more frequent effective collisions, so a faster rate.
• When a question says "use the Boltzmann distribution", refer to the diagram: say that the area to the right of $E_{\text{A}}$ is larger at the higher temperature. "The molecules have more energy" on its own does not use the diagram.
• Answer the question asked. If a question asks for the effect of temperature on $E_{\text{A}}$, the answer is "none" ($E_{\text{A}}$ does not change); an explanation of why the rate rises does not answer it.