The shape of the lesson
By the end of the lesson, learners can
- Going the long way: Build the methane cycle in the simulator.
- Formation data: When you have enthalpies of formation, products minus reactants works because every substance can be made from the same elements.
- Measuring the unmeasurable: Calcium carbonate only decomposes above about 800 °C, so its ΔH cannot be found in a cup calorimeter.
- Formation: products − reactants.
- Combustion: reactants − products.
- Multiply by coefficients; elements have ΔHf = 0.
- Against an arrow: change the sign.
One picture every idea returns to
Hess's law: the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. Build a cycle with the unknown change along the top and a route through measurable changes. With formation data the common point is the elements below: ΔHr = ΣΔHf(products) − ΣΔHf(reactants). With combustion data the common point is the combustion products below: ΔHr = ΣΔHc(reactants) − ΣΔHc(products). Multiply each value by its coefficient and reverse the sign of any arrow you travel against.
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, Any route, same answer.. Some enthalpy changes cannot be measured: you cannot make methane by mixing carbon and hydrogen in a calorimeter. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Going the long way | Build the methane cycle in the simulator. Project tab 4 of the lab (Chemical energetics lab, Hess cycles) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Formation data | When you have enthalpies of formation, products minus reactants works because every substance can be made from the same elements. | Pairs |
| 21 to 30 min | Teach and check: Measuring the unmeasurable | Calcium carbonate only decomposes above about 800 °C, so its ΔH cannot be found in a cup calorimeter. | Numbered Heads Together |
| 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, 13 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 Measuring the unmeasurable, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| The two formula shortcuts have opposite orders: formation is products minus reactants, combustion is reactants minus products. | If unsure, draw the cycle and follow the arrows, changing the sign whenever you travel against one. Never forget to multiply by the coefficients, and remember that elements have ΔH f = 0. |
Hess's law: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
State Hess's law.
Draw a cycle and calculate ΔHf of methane using ΔHc: C −394, H2 −286, CH4 −890 kJ mol−1.
Calculate ΔH for 2NO2(g) → N2O4(g), given ΔHf(NO2) = +33 and ΔHf(N2O4) = +9 kJ mol−1.
Calculate ΔH for C2H4(g) + H2O(l) → C2H5OH(l) using ΔHc: C2H4 −1411, C2H5OH −1367 kJ mol−1.
Anhydrous copper(II) sulfate dissolves with ΔH1 = −66.5 kJ mol−1; hydrated CuSO4·5H2O dissolves with ΔH2 = +11.7 kJ mol−1. Calculate ΔH for CuSO4(s) + 5H2O(l) → CuSO4·5H2O(s) and explain why it cannot be measured directly.
Explain why the enthalpy change of formation of carbon monoxide must be found using Hess's law.
Answer key and marking notes
The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
- Cycle: C + 2H2 → CH4 across the top; both sides go down to CO2 + 2H2O.
- ΔHf + (−890) = −394 + 2(−286) = −966.
- ΔHf = −76 kJ mol−1.
- ΔH = ΣΔHf(products) − ΣΔHf(reactants) = 9 − 2 × 33.
- ΔH = −57 kJ mol−1.
- Water does not burn, so its ΔHc is zero.
- ΔH = ΣΔHc(reactants) − ΣΔHc(products) = −1411 − (−1367) = −44 kJ mol−1.
- Both solids give the same solution, CuSO4(aq).
- ΔH = ΔH1 − ΔH2 = −66.5 − 11.7 = −78.2 kJ mol−1.
- Adding exactly 5 mol of water to the solid cannot be controlled; the solid would partly dissolve and the heat could not be measured cleanly.
- Burning carbon in a limited supply of oxygen always gives a mixture of CO and CO2, never pure CO.
- So ΔHf(CO) is found indirectly from ΔHc(C) and ΔHc(CO).
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.
Which statement is Hess's law?
- Athe total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same
- Bthe enthalpy change for a reaction depends on the route taken from reactants to products
- Cthe enthalpy change of a reaction is equal to its activation energy
- Denergy cannot be created or destroyed, so the enthalpy change of every reaction is zero
Use enthalpies of formation to calculate the standard enthalpy change of combustion of methane: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). (ΔHf: CH4 −75, CO2 −394, H2O(l) −286 kJ mol−1)
- A−891 kJ mol−1
- B+891 kJ mol−1
- C−605 kJ mol−1
- D−1.04 × 103 kJ mol−1
Use enthalpies of combustion to calculate the standard enthalpy change of formation of methane, C(s) + 2H2(g) → CH4(g). (ΔHc: C −394, H2 −286, CH4 −890 kJ mol−1)
- A−76.0 kJ mol−1
- B+76.0 kJ mol−1
- C+210 kJ mol−1
- D−1.86 × 103 kJ mol−1
The enthalpy change for the decomposition CaCO3(s) → CaO(s) + CO2(g) cannot be measured directly, so both solids are reacted separately with hydrochloric acid. Use the cycle to find ΔH for the decomposition.
- A+178 kJ mol−1
- B−178 kJ mol−1
- C−212 kJ mol−1
- D+212 kJ mol−1
Why is the standard enthalpy change of combustion of carbon (graphite) equal to the standard enthalpy change of formation of carbon dioxide?
- Aboth are represented by the same equation, C(s) + O2(g) → CO2(g), forming 1 mol of CO2
- Bcarbon and carbon dioxide both have enthalpies of formation of zero
- Ccombustion and formation always have equal values for any substance
- Dcarbon dioxide cannot burn, so its enthalpy change of combustion is zero
Use the cycle and these enthalpy changes of combustion to calculate ΔH for C2H4(g) + H2(g) → C2H6(g). (ΔHc: C2H4 −1411, H2 −286, C2H6 −1560 kJ mol−1)
- A−137 kJ mol−1
- B+137 kJ mol−1
- C+149 kJ mol−1
- D−3.26 × 103 kJ mol−1
Calculate the standard enthalpy change of formation of ethanol, 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l). (ΔHc: C −394, H2 −286, C2H5OH −1367 kJ mol−1)
- A−279 kJ mol−1
- B+279 kJ mol−1
- C+687 kJ mol−1
- D+293 kJ mol−1
Carbon monoxide cannot be made cleanly from carbon and oxygen, because some CO2 always forms. Use ΔHc(C) = −394 kJ mol−1 and ΔHc(CO) = −283 kJ mol−1 to find ΔHf(CO), for C(s) + ½O2(g) → CO(g).
- A−111 kJ mol−1
- B+111 kJ mol−1
- C−677 kJ mol−1
- D−505 kJ mol−1
Quiz key
Hess's law states that the total enthalpy change for a reaction is the same whatever route is taken, as long as the starting and finishing states are the same; it follows from conservation of energy. This is why an unknown ΔH can be found from a cycle of known enthalpy changes. If ΔH depended on the route, a cycle could create energy, so the second statement is the opposite of the law. The activation energy is the barrier to reaction and is not the same as ΔH. Energy is conserved overall, but energy can move between the chemicals and the surroundings, so ΔH is not generally zero.
ΣΔHf(products) = −394 + 2(−286) = −966 kJ mol−1. ΣΔHf(reactants) = −75 + 0 = −75 kJ mol−1. ΔH = −966 − (−75) = −891 kJ mol−1, close to the data book value of −890. +891 is reactants − products, the wrong way round. −605 forgets the 2 in front of H2O. −1.04 × 103 adds the ΔHf of methane instead of subtracting it (−966 + (−75) = −1041).
Both sides burn to the same products, CO2 and 2H2O, so ΔHf = ΣΔHc(reactants) − ΣΔHc(products) = [−394 + 2(−286)] − (−890) = −966 + 890 = −76.0 kJ mol−1. +76.0 uses products − reactants, the rule for formation data, which is the wrong way round for combustion data. +210 forgets the 2 in front of H2 (−394 − 286 + 890). −1.86 × 103 adds all three values instead of subtracting the combustion of methane.
By Hess's law the direct route equals the indirect route: ΔH = ΔH1 − ΔH2 = (−17) − (−195) = +178 kJ mol−1. Going against the ΔH2 arrow reverses its sign. The decomposition is endothermic, which is why limestone must be heated strongly. −178 is ΔH2 − ΔH1, the reverse reaction. −212 adds the two values without reversing ΔH2. +212 reverses both arrows, but only the second one is travelled backwards. The direct reaction is not measured because it needs a very high temperature and does not go to completion in a calorimeter.
Burning 1 mol of carbon completely and forming 1 mol of CO2 from its elements in their standard states are the same reaction, C(s) + O2(g) → CO2(g), so both enthalpy changes are −394 kJ mol−1. The same is true for hydrogen and water: ΔHc(H2) = ΔHf(H2O, l) = −286. Only carbon, an element, has ΔHf = 0; CO2 has ΔHf = −394. Combustion and formation are generally different, for example for methane (−890 and −75). It is true that CO2 cannot burn further, but that is not why the two values are equal.
ΔH = ΣΔHc(reactants) − ΣΔHc(products) = [−1411 + (−286)] − (−1560) = −1697 + 1560 = −137 kJ mol−1. +137 subtracts the wrong way round (products − reactants). +149 leaves out the combustion of hydrogen, which is one of the reactants and must also be burned on the left-hand route. −3.26 × 103 adds all three values instead of reversing the ethane arrow. (The bond energy estimate, −124 kJ mol−1, differs because it uses mean bond energies.)
ΔHf = ΣΔHc(reactants) − ΣΔHc(products) = [2(−394) + 3(−286)] − (−1367) = (−788 − 858) + 1367 = −279 kJ mol−1. Oxygen does not burn, so its ΔHc is not included. +279 subtracts the wrong way round. +687 forgets both multipliers, using one C and one H2. +293 doubles the carbon but forgets the 3 in front of H2.
Route 1: C(s) → CO2 directly, −394. Route 2: C(s) → CO(g) → CO2, ΔHf(CO) + (−283). By Hess's law these are equal, so ΔHf(CO) = −394 − (−283) = −111 kJ mol−1. +111 subtracts the wrong way round. −677 adds the two combustion values instead of reversing the CO combustion. −505 doubles the carbon value, but only 1 mol of carbon forms 1 mol of CO.
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.