The shape of the lesson
By the end of the lesson, learners can
- Why temperature changes k: Raise the temperature in the Boltzmann distribution simulator and watch the shaded area beyond the activation energy grow.
- Finding E a from a graph: Read off the rate constants at several temperatures in the simulator, then convert each into ln k and 1/T.
- k = Ae−Ea/RT; ln k = ln A − Ea/RT.
- Plot ln k against 1/T (T in K).
- Ea = −gradient × 8.31, in J mol−1.
- Only T and catalysts change k.
One picture every idea returns to
The rate constant k increases with temperature because a larger fraction of collisions have energy ≥ Ea. The Arrhenius equation is k = Ae−Ea/RT, where A is the pre-exponential factor (linked to collision frequency and orientation), T is in kelvin and R = 8.31 J K−1 mol−1. Taking logs gives ln k = ln A − Ea/RT, a straight line when ln k is plotted against 1/T: the gradient = −Ea/R and the intercept is ln A. A larger Ea means a steeper line, so the rate is more sensitive to temperature.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A small rise in T, a big jump in k.. Changing concentration changes the rate but not k. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Why temperature changes k | Raise the temperature in the Boltzmann distribution simulator and watch the shaded area beyond the activation energy grow. Project tab 4 of the lab (Rate equations and reaction kinetics lab, Boltzmann distribution) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Finding E a from a graph | Read off the rate constants at several temperatures in the simulator, then convert each into ln k and 1/T. Project tab 5 of the lab (Rate equations and reaction kinetics lab, the Arrhenius plot) and let a student drive it while the class predicts. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Finding E a from a graph, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Always use T in kelvin and R in J K −1 mol −1, so E a comes out in J mol −1: divide by 1000 for kJ. | The gradient is negative; E a = −gradient × R is positive. Plot ln k, not k, and 1/ T, not T. Increasing concentration does not change k; only temperature and catalysts do. |
Temperature, rate constants and the Arrhenius equation: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Two points on an Arrhenius plot are (3.10 × 10−3 K−1, −2.50) and (3.40 × 10−3 K−1, −5.50). Calculate Ea.
Using question 1, calculate ln A and A.
A reaction's rate constant doubles when the temperature rises from 298 K to 308 K. Calculate Ea.
Use the Boltzmann distribution to explain why a small rise in temperature causes a large increase in the rate constant.
A = 1.00 × 1010 dm3 mol−1 s−1 and Ea = 50.0 kJ mol−1. Calculate k at 298 K.
A catalyst lowers Ea from 80 kJ mol−1 to 60 kJ mol−1 at 298 K. Assuming A is unchanged, by what factor does k increase?
Answer key and marking notes
- Gradient = (−5.50 − (−2.50)) / (3.40 × 10−3 − 3.10 × 10−3) = −1.00 × 104 K.
- Ea = −gradient × R = 1.00 × 104 × 8.31 = 83 100 J mol−1.
- Ea = 83.1 kJ mol−1.
- ln A = ln k + Ea/RT = −5.50 + 1.00 × 104 × 3.40 × 10−3 = 28.5.
- A = e28.5 = 2.38 × 1012 (same units as k).
- ln(k2/k1) = ln 2 = 0.693.
- 1/298 − 1/308 = 1.09 × 10−4 K−1.
- Ea = 8.31 × 0.693 / 1.09 × 10−4 = 52 900 J mol−1 = 52.9 kJ mol−1.
- At higher T the distribution flattens and shifts to higher energies.
- The area beyond Ea (molecules with E ≥ Ea) increases by a large proportion.
- So the frequency of successful collisions rises much more than the collision frequency does, and k increases.
- k = 1.00 × 1010 × e−50 000/(8.31 × 298) = 1.00 × 1010 × 1.70 × 10−9.
- k = 17.0 dm3 mol−1 s−1.
- Factor = e(80 000 − 60 000)/(8.31 × 298) = e8.08.
- Factor = 3.22 × 103.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
In the Arrhenius equation k = Ae−Ea/RT, in what units must T and Ea be used with R = 8.31 J K−1 mol−1?
- AT in K and Ea in J mol−1
- BT in °C and Ea in J mol−1
- CT in K and Ea in kJ mol−1
- DT in °C and Ea in kJ mol−1
Why does the rate constant of a reaction increase when the temperature rises?
- Aa greater fraction of collisions have energy equal to or greater than the activation energy
- Bthe activation energy of the reaction decreases
- Cthe concentrations of the reactants increase
- Dthe pre-exponential factor A becomes zero
A graph of ln k against 1/T has a gradient of −7500 K. What is the activation energy?
- A62.3 kJ mol−1
- B0.903 kJ mol−1
- C7.50 kJ mol−1
- D−62.3 kJ mol−1
The graph shows ln k plotted against 1/T for a reaction. What is the activation energy?
- A66.5 kJ mol−1
- B8.00 kJ mol−1
- C0.963 kJ mol−1
- D33.2 kJ mol−1
Which change does NOT change the value of the rate constant k for a reaction?
- Aincreasing the concentration of a reactant
- Bincreasing the temperature
- Cadding a catalyst
- Ddecreasing the temperature
The rate constant of a reaction increases by a factor of 4.00 when the temperature rises from 310 K to 330 K. What is Ea?
- A58.9 kJ mol−1
- B25.6 kJ mol−1
- C0.576 kJ mol−1
- D29.5 kJ mol−1
A reaction has Ea = 75.0 kJ mol−1. By what factor does k increase when the temperature rises from 300 K to 320 K?
- A6.56
- B4.00
- C1.07
- D1.00
For a first-order reaction, A = 4.00 × 1011 s−1 and Ea = 90.0 kJ mol−1. What is k at 350 K?
- A0.0146 s−1
- B3.88 × 1011 s−1
- C3.64 × 10−14 s−1
- D4.00 × 1011 s−1
Quiz key
R is in J K−1 mol−1, so Ea/RT is dimensionless only with Ea in J mol−1 and T in kelvin. Using °C gives the wrong value of 1/T (and division by zero at 0 °C). Using kJ mol−1 makes the exponent 1000 times too small, so k barely depends on temperature.
At higher T the Boltzmann distribution shifts to higher energies, so the area beyond Ea grows; e−Ea/RT increases sharply. Ea is a property of the route and is unchanged by temperature; only a catalyst lowers it. Heating does not change concentrations in solution in any useful way, and concentration does not affect k. A is roughly constant; if it were zero, k would be zero.
Ea = −gradient × R = 7500 × 8.31 = 62 300 J mol−1 = 62.3 kJ mol−1. 0.903 comes from dividing 7500 by 8.31 instead of multiplying. 7.50 forgets R. Activation energy is always positive; the negative sign belongs to the gradient.
Gradient = (−5.0 − (−1.0)) / (3.50 × 10−3 − 3.00 × 10−3) = −4.0 / 5.0 × 10−4 = −8000 K. Ea = 8000 × 8.31 = 66 500 J mol−1 = 66.5 kJ mol−1. 8.00 ignores R (and the ×10−3 scale). 0.963 divides by R. 33.2 halves the gradient by misreading the axis.
Concentration changes the rate through the concentration terms in the rate equation, but k stays the same. Raising or lowering T changes e−Ea/RT. A catalyst provides a route with a lower Ea, which increases k.
ln 4.00 = 1.386; 1/310 − 1/330 = 1.96 × 10−4 K−1. Ea = 8.31 × 1.386 / 1.96 × 10−4 = 58 900 J mol−1 = 58.9 kJ mol−1. 25.6 uses log10 4 instead of ln 4. 0.576 divides by the temperature difference (20 K) instead of the difference in 1/T. 29.5 uses ln 2 (treating the factor as 2).
(75 000 / 8.31) × (1/300 − 1/320) = 9025 × 2.08 × 10−4 = 1.88, and e1.88 = 6.56. 4.00 applies the rough rule that the rate doubles every 10 K, which only holds for Ea near 50 kJ mol−1. 1.07 is just 320/300, assuming k is proportional to T. 1.00 comes from leaving Ea in kJ, which makes the exponent almost zero.
Ea/RT = 90 000 / (8.31 × 350) = 30.9; e−30.9 = 3.64 × 10−14; k = 4.00 × 1011 × 3.64 × 10−14 = 0.0146 s−1. 3.88 × 1011 leaves Ea in kJ, so the exponential is almost 1. 3.64 × 10−14 is only the exponential factor, forgetting to multiply by A. k equals A only at infinite temperature or when Ea is zero.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.