CP 10 Finding the Activation Energy of a Reaction
A focused revision guide to Core Practical 10 for Edexcel International A Level Chemistry: timing the peroxodisulfate–iodide clock at five temperatures, converting to kelvin, 1/T and ln(1/t), plotting the Arrhenius graph, turning the gradient into an activation energy of 48.4 kJ mol⁻¹, and evaluating the timing and temperature errors.
Finding the Activation Energy: the Arrhenius Plot
Time the iodine clock at five temperatures, turn each time into ln(1/t), plot it against 1/T and read the activation energy from the gradient.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Recap: Rates, Temperature and Kelvin
Three quick questions on the ideas this practical builds on: why temperature changes rate, converting to kelvin, and why 1/t measures the rate of a clock reaction.
What This Practical Is Testing
This practical, Core Practical 10 on the Edexcel International course, times one reaction at five temperatures and turns the results into an activation energy. The reaction is the peroxodisulfate–iodide clock: colourless until a sudden blue-black colour appears.
S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)
I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)
The clock idea is the one used in Core Practicals 9a and 9b (following reaction rates): a small fixed amount of thiosulfate removes the iodine as it forms, and starch turns blue-black when the thiosulfate runs out.
The same amount of iodine forms every run, so rate ∝ 1/t.
On the Edexcel International course this practical is examined through the Unit 6 practical skills paper (WCH16), which asks for methods, apparatus choices, data processing, uncertainty and evaluation.
The specification reference is 11.10 (finding the activation energy from experimental data by calculation and by a graphical method) and 11.13 (this practical), and the data booklet gives the Arrhenius equation k = Ae^(−Eₐ/RT).
Questions ask you to convert temperatures to kelvin, calculate 1/T and ln(1/t), plot the Arrhenius graph, take the gradient from the best-fit line and convert it to Eₐ with its unit, then evaluate the timing and temperature errors.
Key idea: ln k = ln A − Eₐ/(RT). A graph of ln k, or of ln(1/t), against 1/T is a straight line of gradient −Eₐ/R. Multiply the gradient by −R to get Eₐ in J mol⁻¹, then divide by 1000.
Safety and Apparatus
The apparatus is simple; the marks are for using it so that every run is at a known, steady temperature and times the same amount of iodine.
| Item | What it is for | Precision or note |
|---|---|---|
| Thermostatic water bath (or a beaker of water on a tripod) | Brings both solutions to the chosen temperature and holds the mixture there during the run | set between 20 and 60 °C; read the thermometer, not the dial |
| Thermometer (−10 to 110 °C) | Temperature of the mixture at the colour change | ±0.5 °C per reading |
| Two boiling tubes | Hold the iodide mixture and the peroxodisulfate separately while they warm | same tubes every run |
| Two 10 cm³ measuring cylinders | Measure 10.0 cm³ of KI(aq) and 10.0 cm³ of K₂S₂O₈(aq) | ±0.1 cm³ |
| 5 cm³ graduated pipette (or a burette) | Measures 5.0 cm³ of 0.0100 mol dm⁻³ Na₂S₂O₃(aq) | this volume fixes the amount of iodine timed, so it must be identical every run |
| Dropping pipette | Adds about 2 cm³ of starch solution | amount not critical |
| Digital stopwatch | Time from mixing to the first blue-black colour | reads to 0.01 s; the end point is judged to about ±1 s |
The timed-reaction bench: boiling tubes warmed in a water bath with a thermometer, measuring cylinders for the two solutions and a stopwatch. Both solutions are warmed separately and mixed in the bath.
Safety: Potassium peroxodisulfate is an oxidising agent and a skin and respiratory sensitiser: avoid the dust, wear gloves and eye protection. The iodine formed is a low hazard here. Keep the bath below about 60 °C and lift hot tubes with a holder.
Method: Step by Step
Each run uses fresh solutions. The two solutions are warmed apart and mixed in the bath, so the reaction starts at the moment the clock starts and runs at the recorded temperature.
| Step | What you do | Why |
|---|---|---|
| Make up the iodide tube | Put 10.0 cm³ of 0.100 mol dm⁻³ KI(aq), 5.0 cm³ of 0.0100 mol dm⁻³ Na₂S₂O₃(aq) and 2 cm³ of starch in a boiling tube. | The thiosulfate removes the first iodine formed; starch shows the moment free iodine appears. |
| Measure the peroxodisulfate | Put 10.0 cm³ of 0.0400 mol dm⁻³ K₂S₂O₈(aq) in a second boiling tube. | Kept apart until both are at temperature: the reaction must not start early. |
| Warm both tubes | Stand both tubes in the water bath at 20 °C until a thermometer in the iodide tube reads steadily. | Both solutions must be at the bath temperature, or the mixture is cooler than you record. |
| Mix and start the clock | Pour the peroxodisulfate into the iodide tube, start the stopwatch, mix once and return the tube to the bath. | Timing starts when the reaction starts; the bath holds the temperature during the run. |
| Watch for the colour | Stop the stopwatch at the first blue-black colour and record t to the nearest second. | The end point is sudden, but judging it still carries about ±1 s. |
| Record the temperature | Read the thermometer in the mixture at the colour change and record it. | The temperature of the reacting mixture, not the bath setting, is the value you plot. |
| Repeat at four more temperatures | Repeat with fresh solutions at about 30, 40, 50 and 60 °C. | Five points spread over 40 K give a reliable gradient; repeats reduce random error. |
| Process the results | Convert each temperature to kelvin; calculate 1/T and ln(1/t). | The Arrhenius equation is a straight line only in these variables. |
| Plot and measure | Plot ln(1/t) against 1/T, draw the best-fit line and find its gradient. | gradient = −Eₐ/R, so Eₐ = −gradient × R. |
Technique point: Never mix cold solutions and then warm them. The reaction would already be running at an unknown, changing temperature, and the time would belong to no single value of T.
Check: Planning a Temperature Investigation
Order the method for a different reaction investigated in the same way.
Why 1/t Stands In for the Rate Constant
The 5.0 cm³ of 0.0100 mol dm⁻³ thiosulfate removes a fixed amount of iodine before the colour appears: ½ × 0.0100 × 5.0/1000 = 2.50 × 10⁻⁵ mol of I₂, in a total volume of 27 cm³. That is the same in every run.
Δ[I₂] = 2.50 × 10⁻⁵ mol ÷ 0.0270 dm³ = 9.26 × 10⁻⁴ mol dm⁻³, so rate ≈ 9.26 × 10⁻⁴ ÷ t
Only 2.50 × 10⁻⁵ mol of the 4.00 × 10⁻⁴ mol of peroxodisulfate has reacted at the colour change, about 6%, so the concentrations are still close to their starting values and 1/t measures the initial rate.
rate = k[S₂O₈²⁻][I⁻]; the starting concentrations are identical in every run, so rate ∝ k, and k ∝ 1/t
Taking logarithms, ln k = ln(1/t) + a constant. The constant only moves the whole line up or down; the gradient is unchanged, so ln(1/t) can replace ln k without knowing the constant.
Why temperature changes k
Raising the temperature increases the rate constant, not the concentrations. The Boltzmann distribution shows why: a higher temperature flattens the curve and shifts it to higher energy, so the fraction of collisions with energy at least Eₐ grows steeply.
That fraction is the e^(−Eₐ/RT) term of the Arrhenius equation.
The Boltzmann distribution at two temperatures: at the higher temperature a far larger share of the molecules has energy at least equal to Eₐ, which is why a 10 °C rise can double k.
Exam wording: “The same amount of iodine forms in every run and the starting concentrations are the same, so 1/t is proportional to the rate constant; ln(1/t) can therefore be plotted in place of ln k.”
Check: Processing a New Data Set
Convert, take reciprocals and logarithms, and recover a time from ln(1/t) for a different clock reaction.
Results and Processing
The table gives a complete set of results. Times are recorded to the nearest second, the precision of the end point. Temperatures are converted to kelvin, 1/T is kept to four significant figures and ln(1/t) to three decimal places.
The five runs processed for the Arrhenius plot: T in kelvin, the time to the colour change, 1/T and ln(1/t), with the working for the first row.
| T / °C | T / K | t / s | 1/T / K⁻¹ | ln(1/t) |
|---|---|---|---|---|
| 20 | 293 | 120 | 3.413 × 10⁻³ | −4.787 |
| 30 | 303 | 62 | 3.300 × 10⁻³ | −4.127 |
| 40 | 313 | 34 | 3.195 × 10⁻³ | −3.526 |
| 50 | 323 | 19 | 3.096 × 10⁻³ | −2.944 |
| 60 | 333 | 11 | 3.003 × 10⁻³ | −2.398 |
Step 1: T = 20 + 273 = 293 K
Step 2: 1/T = 1 ÷ 293 = 3.413 × 10⁻³ K⁻¹
Step 3: ln(1/t) = ln(1 ÷ 120) = −4.787
Every value of ln(1/t) is negative because every time is longer than 1 s, and the values become less negative as the temperature rises, because the reaction is faster. A positive value, or one that falls with temperature, signals an arithmetic slip.
Significant figures: Keep 1/T to four figures. The rows are only about 0.1 × 10⁻³ K⁻¹ apart (3.413, 3.300, 3.195…), so rounding to two figures shifts the points unevenly: the gradient changes from −5827 K to about −5970 K, roughly 2.5%.
The Arrhenius Equation
The rate constant depends on temperature through the Arrhenius equation. Taking natural logarithms of both sides turns the exponential into a straight-line equation of the form y = mx + c.
k = Ae^(−Eₐ/RT)
ln k = ln A − Eₐ/(RT), so ln k = (−Eₐ/R) × (1/T) + ln A
The Arrhenius equation in both forms, what k, A, Eₐ, R and T mean, and the straight-line plot with gradient −Eₐ/R and intercept ln A.
Eₐ is the activation energy, the minimum energy a collision needs to react. A, the pre-exponential factor, carries the collision frequency and the fraction of collisions with the right orientation.
R is the gas constant, 8.314 J K⁻¹ mol⁻¹, and T is the temperature in kelvin. The term e^(−Eₐ/RT) is the fraction of collisions with energy at least Eₐ.
Why logarithms. A plot of k against T is a curve that gives no simple reading. ln k against 1/T is a straight line, and a gradient is easy to measure and to average over several points.
Why kelvin. 1/T only has meaning on an absolute scale. In °C a reading of zero would give an infinite 1/T and a winter run would give a negative one; the points would not lie on a line.
Units. Because R is in J K⁻¹ mol⁻¹ and the gradient is in K, Eₐ comes out in J mol⁻¹. Divide by 1000 and quote kJ mol⁻¹ to three significant figures.
A catalyst on the same plot
A catalyst provides a route with a lower Eₐ. On an Arrhenius plot the catalysed reaction gives a second line with a smaller (less steep) gradient, lying above the uncatalysed line at every temperature because k is larger.
Exam wording: “gradient = −Eₐ/R, so Eₐ = −gradient × R. A gradient of −5827 K gives Eₐ = 5827 × 8.314 = 48 400 J mol⁻¹ = 48.4 kJ mol⁻¹.”
Check: Terms in the Arrhenius Equation
Place the names of the terms and say what happens to the exponential term as the temperature rises.
Worked Example: the Gradient and Eₐ
The plot below uses the five processed points. Draw the best-fit line first; then read the gradient from two points on the line, as far apart as possible so that reading errors are a small fraction of Δx and Δy.
The Arrhenius plot of ln(1/t) against 1/T for the five runs: a straight line of gradient −5827 K, giving Eₐ = 48.4 kJ mol⁻¹.
Step 1: two points on the line: (3.003 × 10⁻³, −2.398) and (3.413 × 10⁻³, −4.787)
Step 2: gradient = (−4.787 − (−2.398)) ÷ (3.413 × 10⁻³ − 3.003 × 10⁻³) = −2.389 ÷ (4.10 × 10⁻⁴) = −5827 K
Step 3: gradient = −Eₐ/R, so Eₐ = 5827 K × 8.314 J K⁻¹ mol⁻¹ = 48 400 J mol⁻¹
Answer: Eₐ = 48.4 kJ mol⁻¹ (3 s.f.). Textbook values for the peroxodisulfate–iodide reaction lie close to 50 kJ mol⁻¹, so the result agrees within the uncertainty of the timing.
The gradient has the unit K, because ln(1/t) has no unit and 1/T is in K⁻¹; multiplying by R in J K⁻¹ mol⁻¹ leaves J mol⁻¹.
Why the best-fit line, not two data points
Each data point carries its own random error in t and T. A line through all five averages those errors out; a gradient from two raw points inherits both of their errors in full. Draw the line first and read from it.
Why the intercept is unreliable
The intercept, ln A, sits at 1/T = 0, an infinite temperature, far outside the measured range of 3.0 × 10⁻³ to 3.4 × 10⁻³ K⁻¹. Extrapolating about seven times the width of the data magnifies any error in the gradient.
Here ln(1/t) is not ln k in any case, so the intercept is shifted by the unknown constant. The gradient is unaffected, which is why Eₐ from this plot is sound while A is not.
A calculation route from two temperatures
Writing the equation at two temperatures and subtracting gives ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). With k ∝ 1/t, the 293 K and 333 K runs give ln(120/11) = 2.390 and 1/293 − 1/333 = 4.10 × 10⁻⁴ K⁻¹.
Eₐ = 2.390 × 8.314 ÷ (4.10 × 10⁻⁴) = 48 500 J mol⁻¹ = 48.5 kJ mol⁻¹
The two-point value agrees with the graph, but it rests on two measurements only. Use it to check a graph, or when a question gives you just two rate constants.
Sign check: The gradient is negative and Eₐ is positive. If your Eₐ comes out negative you have plotted ln t, or subtracted the points the wrong way round.
Check: Activation Energy From New Data
Turn a gradient into Eₐ, use two rate constants, and explain the kelvin scale and the sign, all on new reactions.
Other Reactions You Can Use
Any reaction that can be timed to a fixed end point at several temperatures gives an Arrhenius plot. Examiners describe one of these and ask you to process the data in the same way.
Thiosulfate and acid. S₂O₃²⁻(aq) + 2H⁺(aq) → S(s) + SO₂(g) + H₂O(l); time for a pencil cross under the flask to disappear behind the sulfur. Simple, but SO₂ is toxic, so keep below about 60 °C.
Magnesium and acid. Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g); time to collect a fixed volume of hydrogen, or for a measured ribbon to disappear. Clean the ribbon with emery paper every run.
Manganate(VII) and ethanedioate. 2MnO₄⁻(aq) + 5C₂O₄²⁻(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l); time for the purple colour to disappear. Watch for autocatalysis by Mn²⁺, which makes the later part of each run faster.
In every version the rule is the same: the same amount of change must be timed in every run, so that 1/t is proportional to the rate constant at that temperature.
Choosing a reaction: Pick one whose times fall between about 30 s and 3 minutes across your temperature range: short times make the ±1 s timing error large, and long runs lose heat and drift in temperature.
Errors, Uncertainty and Improvements
The stopwatch and thermometer are precise; the errors come from the temperature of the mixture not being what you record and from judging the colour change. Each row says which way the result moves.
| Source of error | Effect on the result | Improvement |
|---|---|---|
| Temperature changes during the run | Plotted T is wrong; points scatter about the line | Keep the tube in the bath; record start and end temperatures and plot the mean |
| Solutions mixed before they reach the bath temperature | Reaction runs cooler than recorded: t too long, point below the line | Wait until the thermometer in the tube reads steadily before mixing |
| Judging the first blue-black colour | Random ±1 s; largest effect on the fastest run | Same observer, white background; or a light sensor and data logger |
| Fastest run too short (11 s at 333 K) | ±1 s is 9.1% of the time | Use more dilute peroxodisulfate or a larger volume so every run takes over 30 s |
| Thiosulfate volume varies between runs | Different amount of iodine timed, so 1/t is no longer ∝ rate | Measure the 5.0 cm³ with a graduated pipette or burette, not a dropper |
| Only five temperatures and no repeats | One scattered point changes the gradient | More temperatures across the range and repeats at each |
| 1/T rounded to two significant figures | Points shift unevenly; gradient about 2.5% out | Keep four significant figures in 1/T and three decimal places in ln(1/t) |
Percentage uncertainties. Timing to ±1 s: on the 333 K run, 1 ÷ 11 × 100 = 9.1%; on the 293 K run, 1 ÷ 120 × 100 = 0.83%.
The thermometer, ±0.5 °C on 293 K, contributes only 0.5 ÷ 293 × 100 = 0.17%.
So the hottest run is the weakest point on the graph and timing, not temperature, limits the result. An uncertainty of 9.1% in t is an uncertainty of about ±0.09 in ln(1/t), visibly larger than the scatter of the other points.
Exam focus: For every error, say which way the point moves and what that does to the gradient: “a time that is too short raises ln(1/t), pulls the hot end of the line up and steepens the gradient, so Eₐ comes out too large”.
Check: Evaluating the Experiment
Choose the correct statement about uncertainty, temperature, the best-fit line, improvements and the intercept.
Common Mistakes
The same slips cost marks every year. Each line gives the mistake and the correction that earns the mark.
Celsius. Using °C in 1/T. Convert to kelvin first: T / K = T / °C + 273.
ln t. Plotting ln t instead of ln(1/t), then reporting a negative Eₐ. If you plot ln t the gradient is +Eₐ/R; say so.
The minus sign. Writing Eₐ = gradient × R. The gradient is −Eₐ/R, so Eₐ = −gradient × R, a positive number.
Units. Multiplying by R = 8.314 J K⁻¹ mol⁻¹ gives J mol⁻¹; forgetting to divide by 1000 gives an answer 1000 times too large in kJ mol⁻¹.
Two raw points. Taking the gradient from two data points instead of from the best-fit line.
Rounding early. Rounding 1/T to two figures, or ln(1/t) to one decimal place, before plotting.
Bath temperature. Recording the water-bath setting instead of the temperature of the reacting mixture.
Common Exam Points
Say
“Plot ln(1/t) against 1/T, with T in kelvin.” “The gradient of the best-fit line is −Eₐ/R, so Eₐ = −gradient × R.” “The same amount of iodine forms every run, so 1/t ∝ k.” “The hottest run has the largest percentage uncertainty.”
Do not say
“Plot rate against temperature and read Eₐ from the gradient.” (The Arrhenius plot needs logarithms and reciprocals.) “Eₐ = gradient × R.” (The gradient is negative; Eₐ is positive.) “The temperature was 40 °C, so 1/T = 1/40.” “The intercept gives Eₐ.”
Watch for
Converting °C to K first; gradients with the unit K; Eₐ to three significant figures in kJ mol⁻¹; the two-temperature route ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂); a catalysed line with a smaller gradient; the point with the largest timing uncertainty.
FAQs
Short answers to the questions students most often ask about finding an activation energy.
Why plot ln(1/t) instead of ln k?
Because k is proportional to 1/t when the same amount of iodine forms in every run and the starting concentrations are the same. The proportionality constant only shifts the line up or down; the gradient, and so Eₐ, is unchanged.
Why does the temperature have to be in kelvin?
The Arrhenius equation uses 1/T, which only has meaning on an absolute scale. In °C a reading of zero would give an infinite 1/T and negative temperatures would make no sense, so the points would not lie on a straight line.
How do I get Eₐ from the gradient?
The gradient equals −Eₐ/R, so Eₐ = −gradient × R with R = 8.314 J K⁻¹ mol⁻¹. A gradient of −5827 K gives 5827 × 8.314 = 48 400 J mol⁻¹, which you quote as 48.4 kJ mol⁻¹ to three significant figures.
Why is the run at the highest temperature the least reliable?
The time is shortest there, so the ±1 s needed to judge the colour change is the largest percentage of the reading: 9.1% for an 11 s run against 0.83% for a 120 s run. Using a slower, more dilute mixture fixes this.
What does a catalyst do to the Arrhenius plot?
A catalyst lowers the activation energy, so its line has a smaller, less steep gradient. It also lies above the uncatalysed line at every temperature, because the rate constant is larger at every temperature.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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