How to Explain Temperature Increases Using Collision Energy Distributions

Heat two identical samples of magnesium with hydrochloric acid, one at 20°C and the other at 40°C. The warmer sample produces hydrogen gas much faster. Yet, if both samples contain the same amount of magnesium, they can produce exactly the same final volume of hydrogen.

Why does heating make the reaction faster without necessarily producing more product?

The answer lies in the distribution of molecular kinetic energies. Increasing temperature changes how energy is spread across the reacting particles. A greater proportion of particles now have enough energy to overcome the activation energy barrier, so more collisions can result in a reaction.

The important point is that heating increases the fraction of particles with sufficient energy; it does not make every particle react or lower the activation energy.

01What changes when temperature increases?

Imagine a container of gas particles moving around randomly. Some move relatively slowly, while others move much faster. They don’t all have the same kinetic energy.

At a particular temperature, there is a spread of kinetic energies across the particles. This spread is represented by a Maxwell-Boltzmann distribution.

Understanding the distribution

There are three important features to recognise.

1. Particles have a range of kinetic energies.

At the same temperature, some particles have relatively low kinetic energy, while others have much higher kinetic energy.

The distribution curve describes how many particles have each energy. Its peak represents the most probable kinetic energy, not the average kinetic energy.

2. Increasing temperature changes the shape of the curve.

When temperature increases:

• The peak becomes lower.
• The curve becomes broader.
• The most probable kinetic energy increases.
• The high-energy tail extends further to the right.

The total area under the curve stays the same, provided the number of particles remains constant.

This last point matters. Heating doesn’t create extra particles. It changes the way their kinetic energies are distributed.

The two curves also overlap. Some particles at the higher temperature still have less kinetic energy than particles at the lower temperature.

3. A greater fraction of particles now have sufficient energy to react.

This is where activation energy becomes important.

The activation energy, \(E_a\), is the minimum energy required for a collision to result in a reaction through a particular reaction pathway.

On a Maxwell-Boltzmann distribution, we represent it with a vertical line.

Particles with kinetic energies below \(E_a\) cannot overcome the energy barrier. Particles with energies equal to or greater than \(E_a\) satisfy the energy requirement for a successful collision.

When temperature increases, the area to the right of the activation energy line becomes larger.

That area represents a greater fraction of particles with sufficient energy to react.

Why does this increase the reaction rate?

Think about what happens when two reactant particles collide.

A collision isn’t automatically successful. The particles need:

• Sufficient energy to overcome the activation energy barrier.
• A suitable orientation so the necessary bonds can break and form.

Increasing temperature increases the proportion of particles satisfying the energy requirement. It also makes particles move faster, increasing collision frequency.

Consequently, the number of successful collisions per second increases, and the reaction rate increases.

Notice what has not changed: for the same reaction pathway, the activation energy remains constant.

Heating changes the energy distribution, not the activation energy barrier.

For a fuller explanation of the energy requirement, particle orientation, and successful collisions, see Collision Theory and Activation Energy: HSC Chemistry Explained.

02Testing the explanation using experimental data

A Maxwell-Boltzmann distribution explains why temperature should affect reaction rate. But how would you recognise this effect in actual experimental measurements?

Consider an experiment involving magnesium reacting with hydrochloric acid.

The balanced equation is:

\[
\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}
\]

We can measure the volume of hydrogen produced over time to compare the reaction rates at different temperatures.

Setting up a fair comparison

Imagine two identical conical flasks.

• Flask A contains a measured volume of hydrochloric acid at 20°C.
• Flask B contains the same volume and concentration of hydrochloric acid at 40°C.
• Both flasks receive identical magnesium strips.
• The hydrochloric acid is in excess, so magnesium is the limiting reactant.

Each flask is connected to a gas syringe to measure the hydrogen produced.

Before starting, the acid in each flask should reach its target temperature. The magnesium is then added, the flask is sealed, and the stopwatch is started.

We also need to control the following:

• The mass and surface area of magnesium.
• The volume and concentration of hydrochloric acid.
• The pressure and temperature at which the gas volume is measured.
• The apparatus and collection method.

Why control the temperature of the gas syringes?

Because gas expands when heated. If the hydrogen in the warmer experiment is measured at a higher temperature, it could occupy a larger volume even if the same number of gas molecules were produced.

For a fair comparison, both gas volumes should be measured at the same temperature and pressure, or corrected to common conditions.

Predicting the results

Before looking at the measurements, make a prediction.

Which flask should produce hydrogen faster?

The 40°C flask should have a steeper initial gas-volume-versus-time curve. Its particles have a greater fraction of energies above the activation energy, so more successful collisions occur each second.

However, both flasks should eventually produce the same total volume of hydrogen.

Why?

Both contain the same amount of magnesium, which is the limiting reactant. Assuming complete reaction and no gas losses, they must produce the same amount of hydrogen.

The temperature changes how quickly the reaction occurs, not the theoretical amount of hydrogen produced.

Illustrative experimental data

The following is an original illustrative dataset, not a record of an actual experiment.

Notice three things.

1. At 40°C, more hydrogen is produced during the early stages of the reaction.
2. The reaction at 40°C reaches its final volume sooner.
3. Both reactions eventually produce 80 cm³ of hydrogen.

The experimental observations support the collision-theory explanation: increasing temperature increases the reaction rate.

However, there is an important distinction.

The gas-volume measurements are experimental observations. The Maxwell-Boltzmann distribution is the model we use to explain those observations. The gas syringe does not directly measure the kinetic energy of individual particles.

03Average rate and instantaneous rate: Don’t confuse them

When analysing experimental data, you need to distinguish between average reaction rate and instantaneous reaction rate.

They describe different things.

Average reaction rate

Average rate measures the change in a quantity over a specified time interval.

For hydrogen production, we can calculate it using:

\[
\text{Average rate}=
\frac{\Delta V}{\Delta t}
\]

where:

• \(\Delta V\) is the change in hydrogen volume, measured in cm³.
• \(\Delta t\) is the time interval, measured in seconds.

The units are cm³ s⁻¹.

On a volume-time graph, average rate corresponds to the gradient of a straight line joining two points on the curve. This is called a secant.

Instantaneous reaction rate

Instantaneous rate measures how quickly the reaction is proceeding at one particular moment.

For a gas-volume-time graph, it is calculated using the gradient of a tangent drawn to the curve at the required time.

\[
\text{Instantaneous rate}
=
\text{gradient of tangent}
\]

This is important because reaction rates usually decrease as reactants are consumed.

The gradient at 10 seconds may be much steeper than the gradient at 60 seconds, even though both measurements come from the same reaction.

Recognising different types of experimental data

The same distinction applies to other measurements.

The negative signs for reactant concentration and decreasing mass make the reported rates positive.

If you are given a concentration-time graph, use the appropriate concentration gradient. If you are given gas-volume or mass data, use the corresponding change in volume or mass.

One limitation: a gas-volume rate or mass-loss rate is not automatically a molar reaction rate. You may need to convert the measurement into moles using the gas equation or the relevant molar mass.

For more practice with gradients and gas-volume-time graphs, see How to Calculate Reaction Rate from a Gas-Volume-Time Graph.

04Common misconceptions to avoid

05Questions and solutions

Question 1

A chemical reaction is performed at 20°C and then repeated at 50°C. Both experiments use identical reactant quantities and concentrations. The reaction follows the same pathway in both experiments.

Explain, using a Maxwell-Boltzmann distribution, why the reaction rate increases when the temperature is raised.

Your answer should refer to the shape of the distribution, activation energy, and successful collisions.

Solution 1

The reaction rate increases at 50°C because a greater fraction of particles have kinetic energies equal to or greater than the activation energy.

At 50°C, the Maxwell-Boltzmann distribution becomes broader, its peak becomes lower, and the peak shifts towards higher kinetic energies.

The activation energy remains unchanged because the reaction pathway is the same.

Consequently, the area under the higher-temperature curve to the right of the activation energy line increases. This represents a greater proportion of particles with sufficient energy to overcome the activation energy barrier.

More collisions can therefore satisfy the energy requirement for a successful reaction.

Particles also move faster, increasing collision frequency. Together, these effects increase the number of successful collisions per second and therefore increase the reaction rate.

Importantly, not every collision results in a reaction. Suitable orientation is still required.

Question 2

Magnesium reacts with excess hydrochloric acid at two different temperatures. Identical amounts and surface areas of magnesium are used.

The following illustrative gas-volume measurements are obtained:

All gas volumes are measured at the same temperature and pressure.

The tangent to the 25°C curve at 30 seconds passes through (15 s, 27 cm³) and (45 s, 45 cm³).

The tangent to the 45°C curve at 30 seconds passes through (15 s, 45 cm³) and (45 s, 75 cm³).

Calculate:

(a) The average rate of hydrogen production during the first 30 seconds at each temperature.

(b) The instantaneous rate at 30 seconds at each temperature.

(c) Explain why the results support collision theory, even though the instantaneous rates are lower than the corresponding average rates.

Solution 2

The 45°C experiment has a higher average and instantaneous rate of hydrogen production than the 25°C experiment.

(a) Average rates

Step 1: Calculate the average rate at 25°C.

\[
\begin{aligned}
r_{\mathrm{avg}}
&=\frac{\Delta V}{\Delta t}\\
&=\frac{36-0}{30-0}\\
&=1.20\text{ cm}^3\text{ s}^{-1}
\end{aligned}
\]

Step 2: Calculate the average rate at 45°C.

\[
\begin{aligned}
r_{\mathrm{avg}}
&=\frac{60-0}{30-0}\\
&=2.00\text{ cm}^3\text{ s}^{-1}
\end{aligned}
\]

(b) Instantaneous rates

Step 3: Calculate the instantaneous rate at 25°C.

Use the gradient of the tangent.

\[
\begin{aligned}
r_{\mathrm{inst}}
&=\frac{45-27}{45-15}\\
&=\frac{18}{30}\\
&=0.60\text{ cm}^3\text{ s}^{-1}
\end{aligned}
\]

Step 4: Calculate the instantaneous rate at 45°C.

\[
\begin{aligned}
r_{\mathrm{inst}}
&=\frac{75-45}{45-15}\\
&=\frac{30}{30}\\
&=1.00\text{ cm}^3\text{ s}^{-1}
\end{aligned}
\]

(c) Explain the difference.

The average rates are greater than the instantaneous rates at 30 seconds because the reaction is slowing down as magnesium is consumed.

The instantaneous rate describes the reaction at exactly 30 seconds, whereas the average rate includes the earlier period when the reaction was proceeding faster.

Both measurements show that the warmer reaction proceeds faster.

This supports collision theory because increasing temperature increases the fraction of particles with sufficient kinetic energy to overcome the activation energy barrier.

Question 3

Calcium carbonate reacts with excess hydrochloric acid to produce carbon dioxide gas.

\[
\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}
\]

Two experiments are performed at 25°C and 45°C. Each uses the same initial amount of calcium carbonate and hydrochloric acid.

The reaction flask is placed on a balance. Carbon dioxide escapes through a loose cotton-wool plug, allowing the decrease in mass to be measured.

The following illustrative measurements are recorded:

Assume that carbon dioxide is the only substance contributing significantly to the mass loss. Neglect evaporation and splashing.

The molar mass of carbon dioxide is 44.0 g mol⁻¹.

(a) Calculate the average mass-loss rate during the first 20 seconds at each temperature.

(b) Calculate the average molar rate of carbon dioxide production during this interval at each temperature.

(c) Explain why the final mass is the same despite the different reaction rates.

Solution 3

The reaction at 45°C has a greater average mass-loss rate and produces carbon dioxide at a higher molar rate. Both experiments have the same final mass because they start with the same limiting amount of calcium carbonate.

(a) Average mass-loss rates

Step 1: Calculate the mass lost at 25°C.

\[
\Delta m=102.40-102.18
\]

\[
\Delta m=0.22\text{ g}
\]

The average mass-loss rate is:

\[
\begin{aligned}
r_m&=\frac{\Delta m}{\Delta t}\\
&=\frac{0.22}{20}\\
&=0.0110\text{ g s}^{-1}
\end{aligned}
\]

Step 2: Calculate the mass lost at 45°C.

\[
\Delta m=102.40-102.05
\]

\[
\Delta m=0.35\text{ g}
\]

Therefore:

\[
\begin{aligned}
r_m&=\frac{0.35}{20}\\
&=0.0175\text{ g s}^{-1}
\end{aligned}
\]

(b) Molar rates of carbon dioxide production

Step 3: Convert the mass of carbon dioxide produced at 25°C into moles.

Use:

\[
n=\frac{m}{M}
\]

where \(n\) is the amount in moles, \(m\) is mass in grams, and \(M\) is molar mass in g mol⁻¹.

\[
n=\frac{0.22}{44.0}
\]

\[
n=0.00500\text{ mol}
\]

Calculate the average molar production rate:

\[
\begin{aligned}
r&=\frac{0.00500}{20}\\
&=2.50\times10^{-4}
\text{ mol s}^{-1}
\end{aligned}
\]

Step 4: Convert the mass of carbon dioxide produced at 45°C into moles.

\[
n=\frac{0.35}{44.0}
\]

\[
n=0.00795\text{ mol}
\]

Calculate the average molar production rate:

\[
\begin{aligned}
r&=\frac{0.00795}{20}\\
&=3.98\times10^{-4}
\text{ mol s}^{-1}
\end{aligned}
\]

The results are:

(c) Explain the final mass.

Both experiments start with the same amount of calcium carbonate, which is the limiting reactant.

The balanced equation shows a 1:1 mole ratio between calcium carbonate and carbon dioxide.

Therefore, both experiments eventually produce the same total amount of carbon dioxide, assuming complete reaction.

The higher temperature increases the rate at which carbon dioxide is produced, but it does not change the theoretical amount produced.

This demonstrates the difference between reaction rate and final product yield.

Question 4

A student investigates the effect of temperature and a catalyst on the same reaction.

Three experiments use identical initial reactant concentrations.

The uncatalysed reaction has an activation energy of 60 kJ mol⁻¹. The catalysed reaction follows an alternative pathway with an activation energy of 35 kJ mol⁻¹.

A student concludes:

“Heating and adding a catalyst both shift the Maxwell-Boltzmann distribution to the right. Since experiments B and C have the same rate, they must have identical energy distributions.”

Evaluate this explanation. Describe what changes in each experiment and explain why the equal reaction rates do not establish that the molecular energy distributions are identical.

Solution 4

The student’s explanation is incorrect. Heating changes the Maxwell-Boltzmann distribution, whereas a catalyst changes the activation energy by providing an alternative reaction pathway.

Experiment A to B: Adding a catalyst

Both experiments occur at 25°C.

Therefore, the Maxwell-Boltzmann distribution remains unchanged.

However, the catalyst provides an alternative reaction pathway with a lower activation energy:

\[
E_a=35\text{ kJ mol}^{-1}
\]

rather than:

\[
E_a=60\text{ kJ mol}^{-1}
\]

The activation energy line moves to the left, increasing the fraction of particles with sufficient energy to react.

More collisions can overcome the energy barrier, so the reaction rate increases.

Experiment A to C: Increasing temperature

Both experiments use the uncatalysed reaction pathway.

The activation energy therefore remains at 60 kJ mol⁻¹.

However, increasing the temperature from 25°C to 45°C changes the Maxwell-Boltzmann distribution.

The curve becomes broader, its peak becomes lower, and a greater fraction of particles have kinetic energies above the unchanged activation energy.

The reaction rate consequently increases.

Why do experiments B and C have the same rate?

Both experiments produce gas at 0.55 cm³ s⁻¹, but this does not mean their energy distributions are identical.

Experiment B has a lower activation energy at 25°C.

Experiment C has a higher temperature but retains the original activation energy.

Different combinations of particle energy, activation energy, collision frequency, and orientation can produce the same overall reaction rate.

The catalyst and the temperature increase achieve a similar measured rate through different changes to the collision process.

Neither treatment changes the theoretical final product amount for the same limiting reactant and reaction.

Question 5

A reaction converts substance A into products.

Two experiments are performed using identical initial concentrations and the same reaction pathway. No catalyst is present.

Independent measurements establish that, over the concentration range considered, the instantaneous reaction rate is directly proportional to the concentration of A:

\[
r=k[A]
\]

Here, \(r\) is the instantaneous reaction rate, \(k\) is a temperature-dependent proportionality constant, and \([A]\) is the concentration of A.

The following illustrative data are obtained:

A student examines the final two rows and argues:

“At 80 seconds, the reaction at 20°C is faster than the reaction at 40°C. Therefore, increasing temperature does not necessarily increase the fraction of successful collisions.”

Answer the following:

(a) Calculate \(k\) for each experiment at 80 seconds.

(b) Using the relationship between rate and concentration, calculate the instantaneous rate of each experiment when both have the same concentration of A, equal to 0.45 mol L⁻¹.

(c) Explain why the warmer experiment can have a lower instantaneous rate at 80 seconds without contradicting the Maxwell-Boltzmann distribution.

(d) Identify the fairest comparison for determining the effect of temperature on reaction rate, and explain why comparing the two experiments at the same elapsed time can be misleading.

Solution 5

The student’s conclusion is incorrect. The warmer reaction has a greater temperature-dependent rate constant, but it has consumed more reactant by 80 seconds. Its lower remaining concentration causes its instantaneous rate to be slightly lower at that particular time.

This is a more subtle comparison than simply checking which reaction is faster at a particular moment.

(a) Calculate \(k\) for each experiment.

Step 1: Rearrange the supplied rate relationship.

\[
r=k[A]
\]

Therefore:

\[
k=\frac{r}{[A]}
\]

The units of \(k\) are s⁻¹.

Step 2: Calculate \(k\) at 20°C.

\[
\begin{aligned}
k_{20}
&=\frac{0.0045}{0.45}\\
&=0.010\text{ s}^{-1}
\end{aligned}
\]

Step 3: Calculate \(k\) at 40°C.

\[
\begin{aligned}
k_{40}
&=\frac{0.0040}{0.20}\\
&=0.020\text{ s}^{-1}
\end{aligned}
\]

The proportionality constant at 40°C is twice that at 20°C.

This indicates that the warmer conditions produce a greater reaction rate at the same reactant concentration.

(b) Compare the rates at the same concentration.

Step 4: Calculate the rate at 20°C when \([A]=0.45\text{ mol L}^{-1}\).

\[
\begin{aligned}
r_{20}&=k_{20}[A]\\
&=0.010\times0.45\\
&=0.0045\text{ mol L}^{-1}\text{ s}^{-1}
\end{aligned}
\]

Step 5: Calculate the rate at 40°C at the same concentration.

\[
\begin{aligned}
r_{40}&=k_{40}[A]\\
&=0.020\times0.45\\
&=0.0090\text{ mol L}^{-1}\text{ s}^{-1}
\end{aligned}
\]

At the same concentration, the reaction at 40°C is twice as fast.

(c) Explain the apparent contradiction.

The important distinction is that reaction rate depends on both temperature and the concentration of reactants.

At 80 seconds, the two experiments are not in the same chemical state.

The 20°C experiment still contains 0.45 mol L⁻¹ of A, while the 40°C experiment contains only 0.20 mol L⁻¹.

The warmer experiment has consumed more reactant because it proceeded faster during the earlier stages.

Even though its temperature-dependent rate constant is greater, fewer reactant particles remain available to collide at 80 seconds.

This is why its instantaneous rate can be slightly lower.

The Maxwell-Boltzmann distribution has not stopped working. It explains how temperature affects the fraction of particles with sufficient energy. It does not, by itself, account for differences in the amount of reactant remaining.

(d) Identify the fairest comparison.

The fairest comparison uses identical reactant concentrations and compares the instantaneous rates at those concentrations.

This controls the availability of reactant particles, allowing the effect of temperature to be examined more clearly.

Comparing rates at the same elapsed time can be misleading because the warmer reaction may have progressed much further.

The key lesson is that a higher temperature increases the proportion of particles with sufficient energy, but the observed reaction rate also depends on how many reactant particles are still available.

06What to remember when explaining a temperature increase

When an HSC Chemistry question asks you to explain why increasing temperature increases reaction rate using collision energy distributions, build your answer in this order:

1. Increasing temperature increases the average kinetic energy of the particles.
2. The Maxwell-Boltzmann distribution becomes broader, its peak becomes lower, and it shifts towards higher kinetic energies.
3. The activation energy remains unchanged for the same reaction pathway.
4. A greater fraction of particles now have kinetic energies equal to or greater than the activation energy.
5. More collisions can satisfy the energy requirement for a successful reaction. Particles also move faster, increasing collision frequency.
6. Therefore, the number of successful collisions per second increases, and the reaction rate increases.

Keep the distinction between temperature and activation energy clear.

Temperature changes the distribution of particle energies. A catalyst changes the available reaction pathway and can lower the activation energy.

Once you can explain these two effects separately, you’re in a much better position to tackle questions comparing temperature, catalysts, concentration, and reaction-rate graphs.