AQA A-Level Chemistry (7405)

## Learning Outcomes

- Explain why increasing temperature increases reaction rate using collision theory and the Maxwell–Boltzmann distribution
- Interpret and sketch Maxwell–Boltzmann curves for different temperatures
- Distinguish clearly between reaction rate and the rate constant, **k**
- Use the Arrhenius equation and Arrhenius plots to determine **activation energy**
- Process practical data correctly using relative rate, **1/t**
- Identify common exam errors involving kelvin, energy units, and incomplete explanations

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand how temperature affects reaction rate, with a focus on the following syllabus points:

- Know that increasing temperature increases reaction rate because a greater fraction of particles have energy greater than or equal to the activation energy
- Explain temperature effects using the Maxwell–Boltzmann distribution of molecular energies
- Sketch and interpret Maxwell–Boltzmann curves at different temperatures
- Understand that increasing temperature changes the **rate constant, k**, but does not change the order of reaction or the form of the rate equation
- Use the Arrhenius equation, **k = Ae^(-Ea/RT)**, to relate **k**, temperature and activation energy
- Plot **ln k** against **1/T** and determine **Ea** from the gradient
- Use relative rate, **1/t**, in temperature-based practical work where the same endpoint is used each time
- Recognise common practical limitations in Required Practical 3, including temperature drift and subjective endpoints

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. Why does a small increase in temperature often cause a large increase in reaction rate?
2. How does a Maxwell–Boltzmann curve change when temperature increases?
3. What is the difference between reaction rate and the rate constant, **k**?
4. What graph should be plotted to find activation energy from Arrhenius data?
5. Why is rate often taken as proportional to **1/t** in the disappearing cross experiment?

## Introduction

Temperature is one of the most important factors affecting reaction rate. In AQA Chemistry, this topic matters because it joins together ideas from basic collision theory, graphical interpretation of energy distributions, practical skills, and algebraic treatment using the Arrhenius equation.

![Maxwell–Boltzmann energy distribution curves at two temperatures with activation energy marked, showing a greater fraction of molecules above Ea at higher temperature.](https://static.pastpaperhero.com/images/aqa-a-level-chemistry-7405-kinetics-temperature-and-reaction-rate_00.png)

_Higher-temperature Maxwell–Boltzmann distribution is broader and lower, with a larger area beyond the activation energy threshold._

Many students can say that “particles move faster when heated”, but exam questions usually demand more than that. High-mark answers explain how temperature changes the distribution of particle energies, increases the fraction with energy at least equal to the activation energy, and therefore increases the number of **successful collisions** per second.

> **Key Term: reaction rate**
> The change in the amount or concentration of a reactant or product per unit time.

### Collision theory, activation energy and why temperature matters

Collision theory says that particles must collide before they can react. But not every collision leads to reaction. For a collision to be successful, the particles must have enough energy to overcome the energy barrier for reaction.

> **Key Term: activation energy**
> The minimum energy that colliding particles must have for a reaction to occur.

This is the first key point students often miss: increasing temperature does **not** lower activation energy. Only a different route provides an alternative pathway with a lower activation energy. Temperature changes how many particles can reach the existing energy barrier.

As temperature increases, particles have greater kinetic energy on average. That leads to two effects:

- particles move faster, so collisions happen more often
- a greater fraction of particles have energy greater than or equal to **Ea**

The second point is the more important one in exam answers. The collision frequency rises only moderately, but the number of particles able to react rises much more strongly. That is why even a small temperature increase can produce a large increase in rate.

A common rule of thumb is that for many reactions near room temperature, a rise of about **10 K** approximately doubles the rate. This is not a fixed law for all reactions, but it is a useful guide.

Why is the effect so large? Imagine a reaction with an activation energy that sits well into the higher-energy tail of the particle energy distribution. At low temperature, only a small fraction of molecules sit beyond that threshold. A small rightward spread of the distribution can make that fraction much bigger. So the number of successful collisions can rise sharply even if the total number of particles stays the same.

This is also why food keeps longer in a fridge or freezer. At lower temperature, fewer particles have enough energy to react, so decomposition reactions happen more slowly. The activation energy is unchanged, but fewer collisions are successful.

> **Key Term: successful collision**
> A collision that results in reaction because the particles have sufficient energy to overcome the activation energy.

Consider these two reactions at the same temperature:

- **H₂(g) + Cl₂(g) → 2HCl(g)** with a relatively low **Ea**
- **H₂(g) + I₂(g) → 2HI(g)** with a much higher **Ea**

The reaction with chlorine is faster because a larger fraction of collisions can overcome the smaller activation energy barrier. This is an example of how **Ea** and temperature work together to affect rate.

> **Test Tip:** If a question asks why temperature increases rate, do not stop at “particles move faster”. Add: “a greater proportion of particles have energy greater than or equal to the activation energy, so there are more successful collisions per second.”

### Maxwell–Boltzmann distributions and exam interpretation

The Maxwell–Boltzmann distribution shows the spread of molecular energies in a sample of gas. It does not mean all molecules have the same energy. In fact, most have moderate energies, some have low energies, and some have much higher energies.

> **Key Term: Maxwell–Boltzmann distribution**
> A graph showing the distribution of particle energies in a sample at a given temperature.

On a Maxwell–Boltzmann graph:

- the x-axis shows energy
- the y-axis shows number of molecules, or fraction of molecules, with that energy
- the total area under the curve is constant for a fixed number of particles

When temperature increases:

- the peak becomes lower
- the peak shifts to the right
- the curve becomes broader
- the total area stays the same
- the area to the right of **Ea** increases

That last point is the one linked directly to reaction rate. The area beyond **Ea** represents the fraction of molecules with enough energy to react when they collide. At higher temperature, this area becomes much larger.

Students often describe the curve vaguely. Be precise. Do not say “the whole curve moves right”. The shape changes as well: it becomes lower and broader.

Also avoid saying that all particles now have enough energy. Even at higher temperature, only a **greater proportion** do.

Suppose you are shown two curves, A and B, for the same gas. Curve B has a lower peak and is spread further right. B is the higher-temperature curve. If the activation energy line is drawn, the higher-temperature curve will have the larger area to the right of that line.

This helps explain why small temperature changes can have large effects. The area under the tail beyond **Ea** can increase by a large percentage even when the average energy does not increase by much.

One subtle point that stronger students include: the Maxwell–Boltzmann graph shows energies of individual particles, but reaction requires colliding particles to provide enough collision energy overall. In practice, the graph is still used to explain the increased chance of collisions having enough energy to react.

> **Key Term: fraction of molecules with energy ≥ Ea**
> The proportion of particles in a sample that have enough energy to overcome the activation energy.
>
> **Test Tip:** In a sketch question, keep the area under both curves the same. A very common error is drawing the higher-temperature curve with a bigger total area, which would wrongly suggest more particles.

### Rate constant, Arrhenius equation and worked calculations

A reaction rate is the measured speed of reaction in a particular experiment. The **rate constant**, **k**, is the constant in the rate equation and depends on temperature.

> **Key Term: rate constant (k)**
> The constant in a rate equation that links reactant concentrations to reaction rate at a particular temperature.

For example, if the rate equation is:

rate = k[A][B]

then increasing temperature increases **k**, so the rate increases. But the reaction order does not change just because temperature changes. Students often mix these ideas up.

The Arrhenius equation shows how **k** depends on temperature and activation energy:

k = Ae^(-Ea/RT)

where:

- **k** = rate constant
- **A** = Arrhenius constant or frequency factor
- **Ea** = activation energy in **J mol⁻¹**
- **R** = gas constant = **8.31 J K⁻¹ mol⁻¹**
- **T** = temperature in **K**
  > **Key Term: Arrhenius equation**
  > An equation that relates the rate constant to temperature and activation energy: k = Ae^(-Ea/RT).

As temperature increases, the value of the exponent becomes less negative, so **k** increases. As **Ea** increases, the exponent becomes more negative, so **k** decreases.

The linear form is:

ln k = ln A - Ea/RT

or

ln k = -(Ea/R)(1/T) + ln A

This has the form **y = mx + c**, so:

- y = **ln k**
- x = **1/T**
- gradient = **-Ea/R**
- intercept = **ln A**

That means activation energy can be found from the gradient:

Ea = -(gradient) × R

#### Worked example: finding Ea from an Arrhenius plot

Suppose a graph of **ln k** against **1/T** gives a gradient of **-9600 K**.

Step 1: write the relationship

gradient = -Ea/R

So:

-Ea/R = -9600

Step 2: rearrange

Ea = 9600 × R

Step 3: substitute **R = 8.31 J K⁻¹ mol⁻¹**

Ea = 9600 × 8.31  
Ea = 79776 J mol⁻¹

Step 4: convert to **kJ mol⁻¹**

Ea = 79.8 kJ mol⁻¹

Answer: **Ea = 79.8 kJ mol⁻¹**

#### Worked example: using two temperatures

A useful rearrangement is:

ln(k₂/k₁) = (Ea/R)[(1/T₁) - (1/T₂)]

Suppose:

- **Ea = 50.0 kJ mol⁻¹**
- **T₁ = 298 K**
- **T₂ = 308 K**
- find **k₂/k₁**

Step 1: convert **Ea** to **J mol⁻¹**

50.0 kJ mol⁻¹ = 50000 J mol⁻¹

Step 2: substitute

ln(k₂/k₁) = (50000/8.31)[(1/298) - (1/308)]

Step 3: calculate bracket first

(1/298) - (1/308)  
= 0.003356 - 0.003247  
= 0.000109 K⁻¹

Step 4: calculate multiplier

50000/8.31 = 6016.85

Step 5: multiply

ln(k₂/k₁) = 6016.85 × 0.000109  
ln(k₂/k₁) = 0.656

Step 6: remove natural log

k₂/k₁ = e^0.656  
k₂/k₁ = 1.93

So **k₂** is about **1.9 times** **k₁**. That fits the common idea that a 10 K rise near room temperature often nearly doubles rate.

Common student errors in Arrhenius work:

- using temperature in **°C** instead of **K**
- leaving **Ea** in **kJ mol⁻¹** when **R** is in **J K⁻¹ mol⁻¹**
- plotting **k** against **T** instead of **ln k** against **1/T**
- forgetting that the gradient is negative
- writing that temperature lowers **Ea**

### Required Practical 3 and handling temperature data

In Required Practical 3, a common method is the reaction between sodium thiosulfate and hydrochloric acid. Sulfur forms, making the mixture cloudy, and the time for a marked cross to disappear is measured.

A typical equation is:

Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)

The same visible endpoint is used each time: the cross just disappears. Because the same amount of sulfur is needed to reach that point, rate is taken as proportional to **1/time**.

> **Key Term: relative rate (1/t)**
> A value proportional to reaction rate when the same fixed endpoint is used in each experiment.

If time falls from 200 s to 100 s when temperature is raised, the rate has not decreased: it has doubled, because **1/100** is twice **1/200**.

#### Worked example: relative rate from practical data

Results:

- 25 °C: time = 220 s
- 35 °C: time = 110 s

Step 1: convert to relative rates

At 25 °C: rate ∝ 1/220 = 0.00455 s⁻¹  
At 35 °C: rate ∝ 1/110 = 0.00909 s⁻¹

Step 2: compare

0.00909 / 0.00455 = 2.00

So the reaction at 35 °C is approximately twice as fast.

For an Arrhenius-style analysis of practical data, you may use **ln(1/t)** in place of **ln k**, because **1/t** is proportional to rate and, for the same endpoint, proportional to **k**.

This is useful, but only if the endpoint is truly consistent.

Practical limitations:

- the disappearing cross endpoint is subjective
- there may be heat loss to the surroundings
- the reactants may cool before mixing or during reaction
- there is a delay between mixing and starting the timer
- volumes and temperature measurements may be imprecise

Ways to improve the method:

- use a water bath to keep temperature constant before mixing
- insulate the reaction vessel
- use a thermometer or temperature probe for each run
- mix in the same way each time
- use colorimetry instead of judging the cross by eye

When evaluating, do not just list errors. Explain their effect. For example, if temperature falls during the reaction, the measured time will be longer than it should be, so the calculated rate will be lower than the true initial rate.

> **Test Tip:** If a practical question asks for a graph, think carefully about what is being measured. If the experiment records time to a fixed endpoint, plot **1/t** or discuss relative rate, not raw time as though it were rate.

## Key Point Checklist

_This article has covered the following key knowledge points:_

- Increasing temperature increases reaction rate mainly because a greater fraction of particles have energy greater than or equal to **Ea**
- Collision frequency also increases with temperature, but this is usually the smaller effect
- **Activation energy** is the minimum energy needed for a successful reaction
- Temperature does not lower **Ea**; enzymess lower **Ea**
- A higher-temperature Maxwell–Boltzmann curve is lower, broader and shifted to the right
- The total area under a Maxwell–Boltzmann curve stays constant for the same number of particles
- The area to the right of **Ea** represents the fraction of molecules able to react
- Temperature increases the **rate constant, k**, but does not change reaction order
- The Arrhenius equation is **k = Ae^(-Ea/RT)**
- A graph of **ln k** against **1/T** gives a straight line with gradient **-Ea/R**
- In practical work with a fixed endpoint, rate is proportional to **1/t**
- Common errors include using °C instead of K, mixing J and kJ, and confusing time with rate

## Key Terms and Concepts

- reaction rate
- activation energy
- successful collision
- Maxwell–Boltzmann distribution
- fraction of molecules with energy ≥ Ea
- rate constant (k)
- Arrhenius equation
- relative rate (1/t)


Kinetics - Temperature and reaction rate