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S6 Chemistry: Enthalpy Change and Hess Law (Paper 3: challenge)
S6 Chemistry: Enthalpy Change and Hess Law (Paper 3: challenge)
Ten challenge items on the combustion enthalpy of ethanol, hydrogen heating a mass of water, a hydration cycle, a cooling correction, the combustion enthalpy of propane, energy per gram, the air needed for combustion, the measured value as a percentage of the literature value, the formation enthalpy of ethane, and the missing division by moles.
Direct answers Answers in order: minus 922.9, 27.4 K, minus 78.0, minus 876.8, minus 2219.9 kJ per mol, 55.6 and 50.5 kJ per g, 9.52 dm3, 67.5 percent, minus 84.4 kJ per mol, and dividing by the moles gives 890 kJ per mol.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Chemistry: Enthalpy Change and Hess Law |
| Level | S6 |
02 Questions (10 items)
- FoundationWhen 0.500 g of ethanol burns it raises 200 g of water by 12.0 K. Find the combustion enthalpy, taking the molar mass as 46.0 and c as 4.18 J per g per K.
- FoundationWhen 0.200 g of hydrogen burns completely with a combustion enthalpy of minus 286 kJ per mol, all the heat raises 250 g of water. Find the rise, taking the molar mass as 2.0.
- FoundationAnhydrous copper sulphate dissolves with an enthalpy of minus 66.5 kJ per mol and the hydrated salt dissolves with plus 11.5 kJ per mol. Find the enthalpy of hydration.
- IntermediateIn the ethanol experiment above the corrected temperature rise is 11.4 K. Find the corrected combustion enthalpy.
- IntermediateFind the combustion enthalpy of propane from formation enthalpies: propane minus 103.8, carbon dioxide minus 393.5 and liquid water minus 285.8 kJ per mol.
- IntermediateCompare the energy released per gram by methane and propane, with combustion enthalpies of minus 890.3 at molar mass 16.0 and minus 2219.9 at molar mass 44.0 kJ per mol.
- ChallengeWhat volume of air is needed to burn 1.00 dm3 of methane completely at the same temperature and pressure, with air taken as 21 percent oxygen?
- ChallengeThe measured combustion enthalpy of ethanol is 922.9 kJ per mol against a literature value of 1367. What percentage of the literature value was measured?
- ChallengeGiven a combustion enthalpy of minus 1560 for ethane, minus 393.5 for carbon dioxide and minus 285.8 kJ per mol for liquid water, find the formation enthalpy of ethane.
- Error redoA student finds that 0.100 mol gives out 89.0 kJ and writes the combustion enthalpy as 89.0 kJ per mol. What is the mistake?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Minus 922.9 kJ per mol | The water takes in 200 times 4.18 times 12.0 which is 10032 J, and with 0.01087 mol of ethanol 10.032 divided by 0.01087 gives 922.9 kJ per mol. |
| 2 | 27.4 K | The hydrogen is 0.100 mol which releases 28.6 kJ, and dividing by 250 times 4.18 gives 27.4 K. |
| 3 | Minus 78.0 kJ per mol | The cycle gives minus 66.5 minus 11.5 which is minus 78.0 kJ per mol, so energy is given out on hydration. |
| 4 | Minus 876.8 kJ per mol | 200 times 4.18 times 11.4 gives 9530 J, and dividing by 0.01087 mol gives 876.8 kJ per mol. |
| 5 | Minus 2219.9 kJ per mol | Three times minus 393.5 plus four times minus 285.8 minus minus 103.8 gives minus 2219.9 kJ per mol. |
| 6 | 55.6 kJ per g for methane and 50.5 kJ per g for propane, so methane gives more per gram | 890.3 divided by 16.0 gives 55.6 and 2219.9 divided by 44.0 gives 50.5, so methane gives out more energy per gram. |
| 7 | 2.00 dm3 of oxygen, that is about 9.52 dm3 of air | One volume of methane needs two volumes of oxygen, and dividing 2 by 0.21 gives about 9.52 dm3 of air. |
| 8 | 67.5 percent | 922.9 divided by 1367 and multiplied by 100 gives 67.5 percent, and the rest of the energy is lost to the container and the air. |
| 9 | Minus 84.4 kJ per mol | Twice minus 393.5 plus three times minus 285.8 minus minus 1560 gives minus 84.4 kJ per mol. |
| 10 | The moles were not divided out, so it should be 89.0 divided by 0.100 which is 890 kJ per mol | A combustion enthalpy is quoted per mole, so the heat given out must be divided by the number of moles. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Write q = mc(delta T) or the formation enthalpy formula first, divide by the moles only at the end, and check the sign.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.