The Arrhenius Equation Revision Notes | A Level AQA Chemistry
The Arrhenius Equation: Understand and apply the Arrhenius equation, including calculations and graphical analysis.
The Arrhenius Equation is part of Rate equations (A-level only) in A Level AQA Chemistry.
5 min
Reading time
68
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
The Arrhenius Equation: Understand and apply the Arrhenius equation, including calculations and graphical analysis.
State the Arrhenius equation and identify each variable with correct units
Perform calculations using $k = Ae^{-E_a/RT}$ to find rate constants, activation energies, or temperatures
Rearrange the Arrhenius equation into the linear form $\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$
Use experimental data to plot $\ln k$ against $\frac{1}{T}$ and determine $E_a$ from the gradient
Key ideas to keep in view
Use this lesson to stay inside Rate equations (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Rate constant, $k$: The proportionality constant in a rate equation that relates reaction rate to reactant concentrations. Its value increases with temperature and depends on activation energy.
Activation energy, $E_a$: The minimum energy that colliding particles must possess for a reaction to occur. Measured in J mol$^{-1}$ (or kJ mol$^{-1}$). A larger $E_a$ means $k$ is more sensitive to temperature change.
Arrhenius constant, $A$: A pre-exponential factor in the Arrhenius equation that accounts for collision frequency and the fraction of collisions with the correct orientation. It has the same units as $k$ and is treated as approximately constant over moderate temperature ranges.
Arrhenius equation: The mathematical relationship $k = Ae^{-E_a/RT}$ that quantifies how the rate constant $k$ depends on temperature $T$ (in Kelvin), activation energy $E_a$, and the gas constant $R = 8.314$ J mol$^{-1}$ K$^{-1}$.
Revision notes
Read the core explanations from The Arrhenius Equation before testing yourself from memory.
1. Why Does Heating a Reaction Speed It Up So Dramatically?
You already know from collision theory that increasing temperature gives particles more kinetic energy, leading to more frequent and more energetic collisions. But here is the puzzle: a 10 K rise in temperature often doubles the rate of reaction. A modest energy increase causing such a large rate change cannot be explained by collision frequency alone.
The answer lies in the Boltzmann distribution. At higher temperatures, the tail of the energy distribution shifts significantly — a much larger fraction of molecules now exceed the activation energy $E_a$. The Arrhenius equation captures this relationship mathematically, and it is one of the most powerful tools in physical chemistry.
2. The Arrhenius Equation: Variables and Meaning
The exponential term $e^{-E_a/RT}$ is the fraction of collisions that have energy equal to or greater than $E_a$. When $T$ increases, the exponent $-E_a/RT$ becomes less negative, so $e^{-E_a/RT}$ grows — and $k$ grows with it. When $E_a$ is large, even a small change in $T$ produces a large change in $k$, which is why some reactions are far more temperature-sensitive than others.
3. Calculating with the Arrhenius Equation
Calculations using $k = Ae^{-E_a/RT}$ fall into a few standard patterns: finding $k$ at a given temperature, finding $E_a$ from two $k$–$T$ pairs, or finding $A$. The worked examples below cover the most common exam scenarios. Work through each step before revealing the next — this is where the marks are won.
Worked example: Calculating the Rate Constant $k$ at a Given Temperature
A reaction has an activation energy of 50.0 kJ mol$^{-1}$ and an Arrhenius constant $A = 4.00 \times 10^{11}$ mol$^{-1}$ dm$^3$ s$^{-1}$. Calculate the rate constant $k$ at 350 K. Give your answer to 3 significant figures.
$k = Ae^{-E_a/RT}$ — Always start by stating the equation — this earns the method mark [M1].
$E_a = 50.0 \times 10^3 = 50000 \text{ J mol}^{-1}$ — $R$ is in J mol$^{-1}$ K$^{-1}$, so $E_a$ must also be in J mol$^{-1}$. This conversion is a very common source of errors.
$-\frac{E_a}{RT} = -\frac{50000}{8.314 \times 350}$
$= -\frac{50000}{2909.9}$
$= -17.18$ — Keep at least 4 significant figures here to avoid rounding errors in the final answer. [M1]
$e^{-17.18} = 3.44 \times 10^{-8}$ — Use the $e^x$ function on your calculator. A common mistake is using $\log$ instead of $\ln$ — they are not the same.
$k = 1.38 \times 10^{4}$ mol$^{-1}$ dm$^3$ s$^{-1}$ (3 s.f.) — Units of $k$ match units of $A$ for a second-order reaction. Always include units — omitting them loses the accuracy mark.
4. The Linear Form: Plotting $\ln k$ Against $1/T$
The exponential form $k = Ae^{-E_a/RT}$ is powerful for single-point calculations, but it is not straightforward to use when you have a table of $k$ values at different temperatures. The solution is to take the natural logarithm of both sides, which converts the equation into the form of a straight line: $y = mx + c$.
Compare this to $y = mx + c$: here $y = \ln k$, $x = 1/T$, gradient $m = -E_a/R$, and intercept $c = \ln A$. Once you have the gradient from your graph, rearranging gives $E_a = -\text{gradient} \times R$. Because the gradient is negative, multiplying by $-R$ gives a positive value for $E_a$, as expected.
The graph above shows a typical Arrhenius plot. Notice the straight-line relationship with a negative gradient — this is the hallmark of Arrhenius behaviour. The steeper the gradient (more negative), the larger the activation energy, and the more sensitive the rate constant is to temperature changes.
Worked example: Determining $E_a$ from an Arrhenius Plot
A student plots $\ln k$ against $1/T$ and obtains a straight line. Two points on the line are: $(2.50 \times 10^{-3}$ K$^{-1}$, $3.20)$ and $(3.30 \times 10^{-3}$ K$^{-1}$, $-0.40)$. Calculate the activation energy $E_a$ in kJ mol$^{-1}$.
$\text{gradient} = \frac{\Delta(\ln k)}{\Delta(1/T)} = \frac{y_2 - y_1}{x_2 - x_1}$ — Always use two points that are as far apart as possible on the line to minimise percentage error. [M1]
$\text{gradient} = \frac{-0.40 - 3.20}{(3.30 \times 10^{-3}) - (2.50 \times 10^{-3})}$
$= \frac{-3.60}{0.80 \times 10^{-3}}$ — Subtract carefully: $-0.40 - 3.20 = -3.60$ and $3.30 - 2.50 = 0.80$, so the denominator is $0.80 \times 10^{-3}$ K$^{-1}$. [M1]
$\text{gradient} = \frac{-3.60}{8.0 \times 10^{-4}} = -4500 \text{ K}$ — $-3.60 \div 8.0 \times 10^{-4}$: divide $3.60$ by $8.0$ to get $0.45$, then account for $10^{-4}$ in the denominator: $0.45 \times 10^{4} = 4500$. With the negative sign: gradient $= -4500$ K.
$E_a = \frac{37413}{1000} = 37.4 \text{ kJ mol}^{-1}$ (3 s.f.) — Divide by 1000 to convert J to kJ. Check: a value in the range 30–100 kJ mol$^{-1}$ is physically reasonable for most reactions.
Answer: $E_a = 37.4$ kJ mol$^{-1}$
5. Finding $\ln A$ and $A$ from the Graph
Once you have the gradient, the y-intercept of the $\ln k$ vs $1/T$ graph gives you $\ln A$. To recover $A$ itself, raise $e$ to the power of the intercept: $A = e^{\text{intercept}}$. In practice, the y-intercept is often far off the plotted region, so you may need to use the equation of the line with a known point to calculate it.
Worked example: Finding the Arrhenius Constant $A$ from Graph Data
Using the same graph as before (gradient $= -4500$ K, and the point $(2.50 \times 10^{-3}$ K$^{-1}$, $3.20)$ lies on the line), find the value of $A$.
$\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$ — This is the straight-line equation. We know the gradient and one point, so we can find the intercept $\ln A$.
$\ln A = 3.20 + 11.25 = 14.45$ — Rearrange by adding 11.25 to both sides.
$A = e^{14.45} = 1.90 \times 10^{6}$ — $e^{14.45} \approx 1.90 \times 10^{6}$. The units of $A$ match the units of $k$ for this reaction. [A1]
Answer: $A = 1.90 \times 10^{6}$ (same units as $k$)
6. Using Two Data Points Without a Graph
Sometimes an exam question gives you $k$ at two different temperatures and asks for $E_a$ without requiring a graph. You can derive a two-point formula directly from the linear Arrhenius equation by writing it for each temperature and subtracting. This is a compact and reliable method.
Worked example: Finding $E_a$ from Two Rate Constants
At 300 K, $k_1 = 1.50 \times 10^{-3}$ s$^{-1}$. At 320 K, $k_2 = 6.20 \times 10^{-3}$ s$^{-1}$. Calculate the activation energy $E_a$ in kJ mol$^{-1}$.
$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ — This is derived by writing $\ln k_2 - \ln k_1$ from the linear form. [M1]
$E_a = \frac{56600}{1000} = 56.6 \text{ kJ mol}^{-1}$ (3 s.f.) — Divide by 1000 to convert J mol$^{-1}$ to kJ mol$^{-1}$. A value around 50–60 kJ mol$^{-1}$ is physically reasonable.
Answer: $E_a \approx 56.6$ kJ mol$^{-1}$
7. Molecular Interpretation: What the Equation Tells Us
The Arrhenius equation is not merely a mathematical tool — it encodes real physical chemistry. The term $e^{-E_a/RT}$ represents the fraction of molecular collisions that have sufficient energy to overcome the activation energy barrier. As $T$ increases, this fraction grows exponentially, which is why rate constants are so sensitive to temperature.
A catalyst lowers $E_a$, which makes the exponent $-E_a/RT$ less negative. This increases $e^{-E_a/RT}$ and therefore increases $k$ — the reaction proceeds faster at the same temperature. The Arrhenius equation quantifies exactly how much faster: halving $E_a$ does not halve $k$; it raises $k$ by a factor of $e^{E_a/2RT}$, which can be enormous.
The Arrhenius constant $A$ accounts for collision frequency and the fraction of collisions with the correct geometric orientation. Even if two molecules collide with enough energy, they must approach each other in the right orientation for bonds to break and form. $A$ captures both of these factors simultaneously.
8. Exam Technique and Common Pitfalls
Arrhenius equation questions appear on AQA Paper 1 and Paper 2 at A Level. They typically carry 3–6 marks and test calculation, graph interpretation, and explanation. The mark scheme rewards precise method, correct unit handling, and accurate use of the logarithmic form. Here is what separates full-mark answers from near-misses.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
You already know from collision theory that increasing temperature gives particles more kinetic energy, leading to more frequent and more energetic collisions. But here is the puzzle: a 10 K rise in temperature often doubles the rate of reaction. A modest energy increase causing such a large rate change cannot be explained by collision frequency alone.
What should I learn from this Chemistry lesson?
By the end of this lesson, you should be able to State the Arrhenius equation and identify each variable with correct units; Perform calculations using $k = Ae^{-E_a/RT}$ to find rate constants, activation energies, or temperatures; Rearrange the Arrhenius equation into the linear form $\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$; Use experimental data to plot $\ln k$ against $\frac{1}{T}$ and determine $E_a$ from the gradient.
How should I revise The Arrhenius Equation?
Read each section, then close the notes and explain the idea from memory. Check your explanation against the page and finish with an exam question from Physical chemistry.