UK Edexcel A-Level Chemistry 9CH0 · Year 13 · Topic 16

Part 4: Arrhenius plots, catalysts and Core Practical 14

Reviewed 9 October 2026.

Obtain activation energy from a fitted slope, recognise the sign change when using clock times, and evaluate the temperature-controlled experiment.

Activation energy is a kinetic barrier

Activation energy is the energy barrier that reacting particles must overcome along a reaction pathway. At a higher temperature, a greater fraction of particles can overcome it, so the rate constant usually increases. Increasing concentration changes encounter frequency, but at fixed temperature and mechanism it does not itself change k.

A homogeneous catalyst is in the same phase as the reacting substances; a heterogeneous catalyst is in a different phase. An aqueous H⁺ catalyst is homogeneous in the aqueous iodination system. A solid metal surface catalysing a gas reaction is heterogeneous. “Same state symbol” is a useful starting point, but phase means a physically uniform region.

A heterogeneous catalyst can adsorb reactants, weaken bonds and provide sites for reaction before products desorb. A homogeneous catalyst can form intermediates and be regenerated. Both provide an alternative pathway with a lower activation barrier; neither changes the reactant/product energy difference or the equilibrium constant at fixed temperature.

Turn an exponential relationship into a straight line

The Arrhenius equation is k = A exp(−Ea/RT). Here A is a pre-exponential factor with the same units as k, R is the gas constant and T is absolute temperature. Edexcel supplies the Arrhenius equation when required, but you must interpret and rearrange it.

Taking natural logarithms gives ln k = ln A − (Ea/R)(1/T). A graph of ln k against 1/T therefore has gradient −Ea/R and intercept ln A. With R = 8.314 J mol⁻¹ K⁻¹, the derived Ea is in J mol⁻¹; divide by 1000 to report kJ mol⁻¹.

An original fitted line passes through (0.00310 K⁻¹, −3.20) and (0.00340 K⁻¹, −5.00). Gradient = [−5.00 −(−3.20)]/(0.00340 −0.00310) = −6000 K. Ea = −gradient ×R = 49 884 J mol⁻¹ = 49.9 kJ mol⁻¹. The negative slope gives a positive activation energy.

Use two well-separated points on the best-fit line, not necessarily two measured points, and retain enough digits in 1/T. If an axis is 1000/T instead of 1/T, its scale factor changes the numerical slope and must be restored in the equation. A correct-looking unit does not rescue a gradient that has not been multiplied by R.

Clock times reverse the slope sign

At fixed initial concentrations, rate is proportional to k. If a clock measures the same sufficiently early reaction extent in each trial, rate is also proportional to 1/t, so ln(1/t) = constant − Ea/(RT). A plot of ln(1/t) against 1/T has negative gradient −Ea/R.

Since ln t = −ln(1/t), plotting ln t instead gives positive gradient +Ea/R. Pearson’s CP14 worksheet uses ln t. Neither graph is intrinsically wrong: read the vertical axis and use the matching sign. Do not label reciprocal time “rate constant” without explaining the fixed-concentration proportionality.

If a ln t versus 1/T fitted slope is +5800 K, Ea = +5800 ×8.314 = 48 221 J mol⁻¹ = 48.2 kJ mol⁻¹. For the equivalent ln(1/t) plot, slope is −5800 K and the same positive energy is obtained.

At two temperatures, subtracting Arrhenius equations gives ln(k₂/k₁) = (Ea/R)(1/T₁ −1/T₂). This is useful for checking a prediction: with Ea = 50.0 kJ mol⁻¹, raising temperature from 298 to 308 K multiplies k by exp[(50000/8.314)(1/298 −1/308)] ≈ 1.93, under the constant-A model.

CP14: the bromide/bromate(V) clock

The main reaction is BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O. A fixed amount of phenol rapidly removes the bromine by electrophilic substitution: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr. Once the phenol is used up, bromine bleaches methyl red. The end point is disappearance of the indicator colour.

Methyl red is not simply reporting an acid–base pH transition here: it is being bleached by bromine after a chemical threshold. The known phenol amount fixes how much bromine must form, so clock times can compare rates at different temperatures when concentrations and volumes are otherwise unchanged.

Use accurately measured identical volumes and concentrations of phenol, bromide/bromate and acid for each run, with the same indicator addition. Keep the acid separate until timing begins. Place both portions in a thermostatically controlled or carefully monitored water bath and allow them to reach the selected temperature before mixing.

Mix consistently, start the timer at mixing and keep the reaction mixture in the bath. Record the actual reaction temperature and time to indicator bleaching. Repeat at several temperatures over a suitable range and obtain repeat timings at each. Convert measured °C to K, compute 1/T and ln t, then fit a straight line rather than joining adjacent points.

The experiment uses toxic/corrosive phenol, an oxidising bromate, acid and bromine-producing chemistry, as well as hot water. Use the supervised dilute procedure, eye protection, appropriate protective clothing/gloves and the specified ventilation and waste handling. The technique and interpretation are assessed; a reading exercise does not replace observed practical competence.

Original data and a transparent calculation route

The table is deliberately calculated illustrative data following ln t = 5800/T −15.7, not measured results. Increasing temperature reduces time. Plotting its ln t values against reciprocal kelvin gives a positive gradient of 5800 K and Ea = 48.2 kJ mol⁻¹. Small experimental scatter would normally be expected in real timings.

Keep full calculator precision for transformations. Rounding all reciprocals to 0.003 can erase most of the x-axis variation. Use a sensible display such as 0.0033557 K⁻¹ and enough graph scale to separate points; the eventual energy should reflect experimental uncertainty.

Original idealised CP14-style clock values; not experimental measurements
Temperature / K1/T / K⁻¹ln(t/s)Time / s
2880.003472224.4388984.7
2980.003355703.7630943.1
3080.003246753.1311722.9
3180.003144652.5390012.7
3280.003048781.982937.26

Target the source of slope uncertainty

If the acid is colder than the other reactants when mixed, the actual starting temperature is below the nominal bath temperature and the time is longer than it should be for that label. Equilibrate both portions and measure the reaction mixture where practical. Temperature drift during timing means one trial no longer represents one T.

At high temperatures the clock may be so fast that the fixed delay in mixing and starting/stopping dominates. At low temperatures, long runs may experience more heat exchange. Choose a workable range, repeat measurements and use a consistent objective optical end point when suitable.

A change in phenol amount changes the clock threshold, and a changed total volume changes reactant concentrations. Both can create trends unrelated to temperature. Use the same measured amounts and initial concentration conditions across the series.

Scatter propagates into the fitted gradient; drawing reasonable steepest and shallowest lines or using a regression uncertainty can estimate its effect on Ea. A curved Arrhenius plot may indicate a changing mechanism, experimental bias or a failure of constant-A assumptions. Do not force a high-precision single energy from visibly inconsistent data.

Pearson 9CH0/03 June 2023 Q8(c)(i) reported negative activation-energy answers and inaccurate slopes. The transferable checks are to read the axes, show Δy/Δx, apply the correct ±R factor and report an energy unit; the mark allocation belongs to that particular question.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Q1. A ln k versus 1/T gradient is −7200 K. Find Ea using R = 8.314 J mol⁻¹ K⁻¹.Show answer

Ea = −(−7200) ×8.314 = 59 860.8 J mol⁻¹ = 59.9 kJ mol⁻¹. The negative slope does not mean a negative activation energy.

Q2. Why is the gradient positive for a ln t versus 1/T clock plot?Show answer

At fixed threshold and concentrations, k is proportional to 1/t. Taking ln t instead of ln(1/t) reverses the sign, giving slope +Ea/R.

Q3. Explain the role of phenol and methyl red in CP14.Show answer

Phenol removes bromine until the fixed phenol amount is exhausted. Bromine then bleaches methyl red, giving a timed threshold. Methyl red is not simply detecting neutralisation in this experiment.

Q4. Why must concentrations stay fixed when reciprocal clock time replaces k in an Arrhenius plot?Show answer

Rate = k multiplied by concentration terms. If those terms change across temperatures, variations in 1/t include concentration effects and no longer isolate the temperature dependence of k.

Q5. A catalyst makes a reaction faster at the same temperature. What happens to activation energy, ΔG° and K?Show answer

The catalysed pathway has a lower activation barrier. ΔG° and K for the same overall reaction at the same temperature are unchanged, because the reactant and product thermodynamic states are unchanged.

Sources

Sources and examiner guidance (reviewed 9 October 2026)

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