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S6 Chemistry: Enthalpy Change and Hess Law (Paper 4: misconception drill)
S6 Chemistry: Enthalpy Change and Hess Law (Paper 4: misconception drill)
Ten drill items on q = mc(delta T), combustion enthalpy per mole, the sign on reversing an equation, temperature scales, zero formation enthalpy of elements, average bond energy, percentage error, fuel mass from the heat needed, adding two Hess steps, and the calorimeter heat that makes a result too small.
Direct answers Answers in order: 1463 J, minus 890, plus 572 kJ, no conversion needed, zero, 415 kJ per mol, 9.9 percent, about 0.188 g, minus 393.5 kJ per mol, and the calculated value is too small.
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)
- FoundationHow much energy does 100 g of water take in when it warms by 3.5 K, taking c as 4.18 J per g per K?
- Foundation0.100 mol of a fuel gives out 89.0 kJ on burning. What is its combustion enthalpy?
- FoundationIf the enthalpy change for two moles of hydrogen burning to liquid water is minus 572 kJ, what is it for the reverse reaction?
- IntermediateWhen using q = mc(delta T), must the temperature difference be changed from Celsius to kelvin first?
- IntermediateWhat is the standard enthalpy of formation of an element in its standard state?
- IntermediateMethane has four C to H bonds with a total bond energy of 1660 kJ per mol. What is the average energy of one C to H bond?
- ChallengeA measured combustion enthalpy is 654 kJ per mol against a literature value of 726. What is the percentage error?
- ChallengeWhat mass of methane must burn to raise 250 g of water by 10.0 K, with a combustion enthalpy of minus 890 kJ per mol, a molar mass of 16.0 and c of 4.18 J per g per K?
- ChallengeGiven that carbon burning to carbon monoxide has an enthalpy change of minus 110.5 and carbon monoxide burning to carbon dioxide has minus 283.0 kJ per mol, find the value for carbon burning straight to carbon dioxide.
- Error redoIgnoring the heat taken up by the calorimeter itself makes the enthalpy change of an exothermic reaction too large or too small?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 1463 J | 100 times 4.18 times 3.5 gives 1463 J. |
| 2 | Minus 890 kJ per mol | 89.0 divided by 0.100 gives 890, and the value is negative because energy is given out. |
| 3 | Plus 572 kJ | Reversing a reaction keeps the size of the enthalpy change but flips its sign. |
| 4 | No, because a temperature difference has the same value in Celsius or in kelvin | The two scales have different zero points but the same size of degree, so the difference is unchanged. |
| 5 | Zero | Formation enthalpy is for one mole of a substance made from its elements, and an element made from itself has no change, so it is zero. |
| 6 | 415 kJ per mol | 1660 divided by 4 gives 415 kJ per mol, and an average bond energy is a mean value. |
| 7 | 9.9 percent | Taking 726 minus 654, dividing by 726 and multiplying by 100 gives 9.9 percent, and the measured value is low through heat loss. |
| 8 | About 0.188 g | The heat needed is 250 times 4.18 times 10.0 which is 10450 J, dividing by 890000 J per mol gives 0.01174 mol, and multiplying by 16.0 gives 0.188 g. |
| 9 | Minus 393.5 kJ per mol | Adding the two steps cancels the carbon monoxide in between, and minus 110.5 minus 283.0 gives minus 393.5 kJ per mol. |
| 10 | Too small, because the heat given out is underestimated | Part of the heat goes into the calorimeter instead of the water, so the rise is too small and the calculated heat given out is too small. |
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.