Rate Equation Calculations Revision Notes | A Level AQA Chemistry
To calculate k, find the orders from experimental data, write the rate equation, rearrange it for k and substitute values. Free notes and 11 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Rate Equation Calculations is part of Rate equations (A-level only) in A Level AQA Chemistry.
5 min
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68
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A Level AQA
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What this lesson covers
To calculate k, find the orders from experimental data, write the rate equation, rearrange it for k and substitute values. Free notes and 11 questions answered.
Perform calculations using the rate equation, including determining the rate constant k with correct units
Explain qualitatively how changes in temperature affect the rate constant k
Apply the Arrhenius equation to relate k, activation energy, and temperature
Calculate activation energy and pre-exponential factors using the logarithmic form of the Arrhenius equation
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Use this lesson to stay inside Rate equations (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Rate equation: A mathematical expression of the form rate = k[A]^m[B]^n that relates the rate of a reaction to the concentrations of reactants raised to their respective orders. Used when determining how concentration affects rate experimentally.
Rate constant (k): The proportionality constant in the rate equation whose value depends on temperature but not on concentration. Its units vary with the overall order of the reaction.
Arrhenius equation: The equation k = Ae^(-Ea/RT) that quantifies how the rate constant k varies with absolute temperature T, activation energy Ea, and the pre-exponential factor A. Given in the AQA data booklet when required.
Activation energy (Ea): The minimum energy that colliding particles must possess for a reaction to occur. Measured in J mol^-1 or kJ mol^-1. A higher Ea means k is more sensitive to temperature changes.
Revision notes
Read the core explanations from Rate Equation Calculations before testing yourself from memory.
1. Why the Rate Equation is the Heart of Kinetics
You already know that increasing concentration generally speeds up a reaction. The rate equation is the tool that makes that relationship precise and quantitative — it tells you exactly how much the rate changes when you change a concentration, and it introduces the rate constant k, which carries all the temperature dependence of the reaction.
Mastering rate equation calculations is one of the highest-yield skills in AQA A Level Physical Chemistry. Questions appear on every paper and reward students who are methodical about units, orders, and the Arrhenius equation.
2. The Rate Equation and the Rate Constant k
The orders m and n must be determined experimentally — you cannot read them from the stoichiometric equation. This is a fundamental point that examiners test repeatedly. Once you know the orders and have a rate measurement with corresponding concentrations, you can calculate k.
The units of k depend on the overall order of the reaction. You derive them by rearranging the rate equation for k and substituting the units of rate (mol dm$^{-3}$ s$^{-1}$) and concentration (mol dm$^{-3}$). This is not something to memorise — it is something to derive each time.
Overall order
Units of k
Example reaction type
Zero order
mol dm$^{-3}$ s$^{-1}$
Rate independent of all concentrations
First order
s$^{-1}$
Rate proportional to one concentration
Second order
mol$^{-1}$ dm$^3$ s$^{-1}$
Rate proportional to two concentrations or one squared
To see where these units come from, consider a second-order reaction: rate = k[A]^2. Rearranging gives $k = \frac{\text{rate}}{[A]^2}$. Substituting units: $k = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})^2} = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{\text{mol}^2 \text{dm}^{-6}} = \text{mol}^{-1} \text{dm}^3 \text{s}^{-1}$. Always show this derivation in full.
Worked example: Calculating k from experimental rate data
The following data were obtained for the reaction A + B → C at constant temperature.\n\nExperiment 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹\nExperiment 2: [A] = 0.20 mol dm⁻³, [B] = 0.10 mol dm⁻³, rate = 4.0 × 10⁻³ mol dm⁻³ s⁻¹\nExperiment 3: [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³, rate = 2.0 × 10⁻³ mol dm⁻³ s⁻¹\n\nDetermine the rate equation and calculate k with units.
Compare experiments 1 and 2: [B] is constant, [A] doubles from 0.10 to 0.20 mol dm$^{-3}$.\n$\frac{\text{rate}_2}{\text{rate}_1} = \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 2$\n$\frac{[A]_2}{[A]_1} = \frac{0.20}{0.10} = 2$\nRate doubles when [A] doubles → first order in A. — [M1] Doubling [A] doubles the rate, so the order with respect to A is 1.
Compare experiments 1 and 3: [A] is constant, [B] doubles from 0.10 to 0.20 mol dm$^{-3}$.\n$\frac{\text{rate}_3}{\text{rate}_1} = \frac{2.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 1$\n[B] doubles but rate is unchanged → zero order in B. — [M1] No change in rate when [B] doubles, so the order with respect to B is 0.
rate $= k[A]^1[B]^0 = k[A]$ — [B1] The rate equation is first order overall. [B] does not appear because its order is zero.
$k = \frac{\text{rate}}{[A]} = \frac{2.0 \times 10^{-3}}{0.10} = 2.0 \times 10^{-2}$ — [M1] Rearrange rate = k[A] for k, then substitute values from any experiment.
$k = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}$ — [A1] For a first-order reaction, k has units of s⁻¹. Always derive units — never guess.
3. Temperature and the Rate Constant — a Qualitative Picture
Here is where the rate constant becomes genuinely interesting. When you raise the temperature of a reaction, you are not changing the concentrations — yet the rate increases dramatically. The only way the rate equation can accommodate this is if k itself increases with temperature.
From collision theory, a higher temperature means particles have greater kinetic energy. More particles exceed the activation energy threshold, so a greater proportion of collisions are successful. This increase in the fraction of successful collisions is captured entirely by the increase in k.
The relationship is not linear — k increases exponentially with temperature. A reaction with a large activation energy shows a much steeper increase in k for the same temperature rise than one with a small activation energy. This is the qualitative picture the AQA specification requires you to explain.
4. The Arrhenius Equation — Quantifying the Temperature Dependence of k
The qualitative picture from collision theory is useful, but AQA requires you to work quantitatively. The Arrhenius equation provides the mathematical relationship between k and temperature. You will be given this equation and the gas constant R in the exam, but you need to know how to use it fluently.
Notice the negative sign in the exponent: as $E_a$ increases, the exponent becomes more negative, so $e^{-E_a/RT}$ decreases and k is smaller. As T increases, $E_a/RT$ decreases, the exponent becomes less negative, $e^{-E_a/RT}$ increases, and k gets larger. This is the mathematical expression of the qualitative argument from collision theory.
For most exam calculations, you will use the logarithmic form of the Arrhenius equation, which converts the exponential relationship into a straight-line form. Taking the natural logarithm of both sides gives the linear version shown below.
This linear form is powerful. If you plot $\ln k$ on the y-axis against $1/T$ on the x-axis, you get a straight line with gradient $-E_a/R$ and y-intercept $\ln A$. Examiners frequently ask you to extract $E_a$ from such a graph, or to use two data points to calculate $E_a$ without a graph.
Worked example: Calculating activation energy from two k values at different temperatures
The rate constant for a reaction is 3.46 × 10⁻³ mol⁻¹ dm³ s⁻¹ at 300 K and 2.71 × 10⁻² mol⁻¹ dm³ s⁻¹ at 350 K. Calculate the activation energy Ea in kJ mol⁻¹. (R = 8.314 J mol⁻¹ K⁻¹)
At $T_1 = 300$ K: $\ln k_1 = \ln A - \frac{E_a}{R T_1}$
At $T_2 = 350$ K: $\ln k_2 = \ln A - \frac{E_a}{R T_2}$ — Writing both equations sets up the subtraction that eliminates ln A.
$\ln k_2 - \ln k_1 = -\frac{E_a}{R T_2} + \frac{E_a}{R T_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ — [M1] This rearrangement removes the unknown ln A, leaving only Ea to find.
$\ln k_1 = \ln(3.46 \times 10^{-3}) = -5.666$
$\ln k_2 = \ln(2.71 \times 10^{-2}) = -3.608$
$\ln k_2 - \ln k_1 = -3.608 - (-5.666) = 2.058$ — [M1] Use a calculator for ln values. Keep at least 3 significant figures throughout.
$\frac{1}{T_1} = \frac{1}{300} = 3.333 \times 10^{-3}$ K$^{-1}$
$\frac{1}{T_2} = \frac{1}{350} = 2.857 \times 10^{-3}$ K$^{-1}$
$\frac{1}{T_1} - \frac{1}{T_2} = 3.333 \times 10^{-3} - 2.857 \times 10^{-3} = 4.76 \times 10^{-4}$ K$^{-1}$ — [M1] Always use temperatures in Kelvin. Converting 300 K and 350 K is not needed here as they are already in K.
$E_a = \frac{35940}{1000} = 35.9$ kJ mol$^{-1}$ — [A1] Divide by 1000 to convert J mol⁻¹ to kJ mol⁻¹. Give to 3 significant figures to match the data.
Answer: Ea = 35.9 kJ mol⁻¹
5. Using the Arrhenius Equation to Find k at a New Temperature
A common exam question gives you k at one temperature and asks you to find k at another, using a known activation energy. The two-temperature subtraction method from the previous section works perfectly here — you rearrange for $\ln k_2$ rather than for $E_a$.
Worked example: Finding k at a new temperature using the Arrhenius equation
A reaction has an activation energy of 75 000 J mol⁻¹. At 298 K, k = 1.50 × 10⁻⁴ s⁻¹. Calculate k at 320 K. (R = 8.314 J mol⁻¹ K⁻¹)
$\ln k_2 - \ln k_1 = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ — This is the working form derived by subtracting the Arrhenius equation at T₁ from that at T₂.
$\frac{E_a}{R} = \frac{75000}{8.314} = 9021$ K — 75 000 ÷ 8.314 = 9021 K. Keep this value unrounded for the next step.
$\frac{1}{298} = 3.3557 \times 10^{-3}$ K$^{-1}$
$\frac{1}{320} = 3.1250 \times 10^{-3}$ K$^{-1}$
$\frac{1}{T_1} - \frac{1}{T_2} = 3.3557 \times 10^{-3} - 3.1250 \times 10^{-3} = 2.307 \times 10^{-4}$ K$^{-1}$ — [M1] T₁ = 298 K (the temperature where k is known), T₂ = 320 K (the target temperature).
$\ln k_1 = \ln(1.50 \times 10^{-4}) = -8.805$
$\ln k_2 = -8.805 + 2.081 = -6.724$ — [M1] Add the calculated difference to ln k₁ to get ln k₂.
$k_2 = e^{-6.724} = 1.20 \times 10^{-3}$ s$^{-1}$ — [A1] Use e^x on your calculator. The units of k₂ are the same as k₁ (s⁻¹ for a first-order reaction).
Answer: k at 320 K = 1.20 × 10⁻³ s⁻¹
Notice that raising the temperature by only 22 K (from 298 K to 320 K) increased k by a factor of approximately 8. This illustrates why the exponential nature of the Arrhenius equation matters — small temperature changes produce large changes in k, especially when $E_a$ is large.
6. Extracting Ea from a Graph of ln k against 1/T
AQA exam questions often present a table of k values at different temperatures and ask you to plot $\ln k$ against $1/T$ or to use the gradient to find $E_a$. The method is straightforward once you recognise the linear form of the Arrhenius equation.
The gradient of this graph equals $-E_a/R$. To find $E_a$, read off two well-separated points on the line (not data points — points on the drawn line), calculate the gradient as $\Delta(\ln k) / \Delta(1/T)$, and then multiply by $-R$. The negative sign is critical: the gradient is negative, so $E_a$ comes out positive.
Worked example: Calculating Ea from the gradient of a ln k vs 1/T graph
A graph of ln k against 1/T gives a straight line passing through the points (3.10 × 10⁻³ K⁻¹, −4.2) and (3.50 × 10⁻³ K⁻¹, −7.0). Calculate Ea in kJ mol⁻¹. (R = 8.314 J mol⁻¹ K⁻¹)
$\text{gradient} = \frac{\Delta(\ln k)}{\Delta(1/T)} = \frac{-7.0 - (-4.2)}{3.50 \times 10^{-3} - 3.10 \times 10^{-3}}$
$= \frac{-2.8}{0.40 \times 10^{-3}} = \frac{-2.8}{4.0 \times 10^{-4}}$ — [M1] Always use two points that are far apart on the line to minimise reading error. Show the triangle clearly.
$E_a = \frac{58198}{1000} = 58.2$ kJ mol$^{-1}$ — [A1] Divide by 1000. Round to 3 significant figures to match the precision of the graph data.
Answer: Ea = 58.2 kJ mol⁻¹
7. Exam Technique and Common Mistakes
Rate equation calculations are reliable marks if you follow a consistent method. The most common errors are not conceptual — they are procedural: wrong units for k, temperatures left in Celsius, and sign errors in the Arrhenius calculation. The guidance below addresses each of these directly.
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.
To calculate the rate constant k, determine the orders from experimental data, write the rate equation, rearrange it for k and substitute one experiment's values. For rate = k[A] with a rate of 2.0 × 10⁻³ mol dm⁻³ s⁻¹ and [A] = 0.10 mol dm⁻³, k = 2.0 × 10⁻² s⁻¹.
How does temperature affect the rate constant?
Increasing temperature increases the rate constant k, and the increase is exponential rather than linear. At a higher temperature more particles exceed the activation energy, so a greater proportion of collisions are successful; because concentrations are unchanged, this is captured entirely by the increase in k.
How do you find k at a different temperature?
To find k at a new temperature, use ln k₂ − ln k₁ = (Ea/R)(1/T₁ − 1/T₂). For Ea = 75 000 J mol⁻¹ and k = 1.50 × 10⁻⁴ s⁻¹ at 298 K, ln k₂ = −8.805 + 2.081 = −6.724, giving k = 1.20 × 10⁻³ s⁻¹ at 320 K, about 8 times larger.
Why does a high activation energy make k more sensitive to temperature?
A reaction with a large activation energy shows a much steeper increase in k for the same temperature rise than one with a small activation energy. This follows from the negative exponent in the Arrhenius equation: as T increases, Ea/RT decreases, the exponent becomes less negative and k gets larger.
How do you calculate activation energy from two temperatures?
To calculate activation energy from rate constants at two temperatures, use ln k₂ − ln k₁ = (Ea/R)(1/T₁ − 1/T₂) with temperatures in Kelvin. For k = 3.46 × 10⁻³ at 300 K and 2.71 × 10⁻² at 350 K, Ea = (2.058 × 8.314) ÷ (4.76 × 10⁻⁴) = 35 940 J mol⁻¹, or 35.9 kJ mol⁻¹.
Is the Arrhenius equation given in the A Level Chemistry exam?
The Arrhenius equation and the gas constant R are given in the A Level Chemistry exam, so they do not need to be memorised. What is needed is using them fluently, especially the logarithmic form for finding activation energy or k at a new temperature.
What are the units of k for a zero order reaction?
For a zero order reaction, k has units of mol dm⁻³ s⁻¹, the same as the rate, because rate = k. For first order the units are s⁻¹ and for second order mol⁻¹ dm³ s⁻¹; the units are derived each time by rearranging the rate equation for k.
Why must temperature be in Kelvin in the Arrhenius equation?
Temperature must be in Kelvin in the Arrhenius equation because the gas constant R is in J mol⁻¹ K⁻¹ and the calculation uses 1/T. Leaving temperatures in Celsius is one of the most common procedural errors in rate equation calculations, along with wrong units for k and sign errors.
Why does a small temperature rise cause a large increase in k?
A small temperature rise causes a large increase in k because k increases exponentially with temperature, not linearly. For a reaction with Ea = 75 000 J mol⁻¹, raising the temperature by only 22 K, from 298 K to 320 K, increased k by a factor of approximately 8, from 1.50 × 10⁻⁴ to 1.20 × 10⁻³ s⁻¹.
What are common mistakes in Arrhenius equation calculations?
The most common errors in Arrhenius and rate equation calculations are procedural: wrong units for k, temperatures left in Celsius rather than Kelvin, and sign errors when rearranging. Ea must also be converted to J mol⁻¹ to match R, and the final answer divided by 1000 to give kJ mol⁻¹.
Why does k depend on temperature but not concentration?
The rate constant k carries all the temperature dependence of a reaction. Raising the temperature does not change the concentrations, yet the rate increases dramatically, so the only way the rate equation can accommodate this is if k itself increases, because more particles exceed the activation energy.