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S6 Chemistry: Enthalpy Change and Hess Law (Paper 1: basics)
S6 Chemistry: Enthalpy Change and Hess Law (Paper 1: basics)
Ten basics items on substituting into q = mc(delta T), multiplying moles by combustion enthalpy, bond breaking and making, working back to a temperature rise, the formation enthalpy route, the standard state, a catalyst and enthalpy change, neutralisation enthalpy, a mean rise, and activation energy in exothermic reactions.
Direct answers Answers in order: 2090 J, 222.5 kJ, breaking bonds takes in energy and making bonds gives it out, 32.7 K, minus 890.3 kJ per mol, 100 kPa and 298 K, lower activation energy with unchanged enthalpy, minus 56.8 kJ per mol, 5.3 K, and activation energy is still needed.
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 is needed to raise 100 g of water by 5.0 K? Take c as 4.18 J per g per K.
- FoundationHow much energy is released when 0.25 mol of methane burns completely, with a combustion enthalpy of minus 890 kJ per mol?
- FoundationIn a reaction, are bond breaking and bond making endothermic or exothermic?
- IntermediateWhen 0.050 mol of ethanol burns completely with a combustion enthalpy of minus 1367 kJ per mol, all the energy heats 500 g of water. What is the rise in temperature?
- IntermediateFind the enthalpy of combustion of methane from formation enthalpies: methane minus 74.8, carbon dioxide minus 393.5 and liquid water minus 285.8 kJ per mol.
- IntermediateWhat conditions define the standard state used for standard enthalpy changes?
- ChallengeWhen a catalyst is added, what happens to the activation energy and to the enthalpy change?
- Challenge25.0 cm3 of 1.00 M hydrochloric acid is mixed with 25.0 cm3 of 1.00 M sodium hydroxide, giving 50.0 g of solution and a rise of 6.8 K. Find the enthalpy of neutralisation.
- ChallengeTwo runs give temperature rises of 5.2 K and 5.4 K. What is the mean rise and why is a mean used?
- Error redoA student says an exothermic reaction releases energy by itself and so needs no activation energy. What is wrong with that?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2090 J, that is 2.09 kJ | Energy is mass times specific heat capacity times temperature rise: 100 times 4.18 times 5.0 gives 2090 J. |
| 2 | 222.5 kJ | The energy released is the number of moles times the size of the combustion enthalpy: 0.25 times 890 gives 222.5 kJ. |
| 3 | Breaking bonds takes in energy and making bonds gives out energy | Breaking a bond needs energy input while making a bond releases energy, and the net of the two decides whether the reaction is exothermic. |
| 4 | 32.7 K | The heat released is 68.35 kJ which is 68350 J, and dividing by 500 times 4.18 gives 32.7 K. |
| 5 | Minus 890.3 kJ per mol | Sum the formation enthalpies of the products and subtract those of the reactants: minus 393.5 minus 571.6 plus 74.8 gives minus 890.3. |
| 6 | A pressure of 100 kPa, a temperature usually taken as 298 K, and each substance in its standard state | The standard state fixes the pressure at 100 kPa and the physical state of each substance, while 298 K is the temperature normally quoted. |
| 7 | The activation energy falls while the enthalpy change stays the same | A catalyst provides a route of lower activation energy, and since the starting and final states do not change the enthalpy change stays the same. |
| 8 | Minus 56.8 kJ per mol | The heat given out is 50.0 times 4.18 times 6.8 which is 1421 J, and with 0.0250 mol of water formed that is 56.8 kJ per mol. |
| 9 | 5.3 K, and the mean reduces random error in the readings | Adding 5.2 and 5.4 and halving gives 5.3 K, and averaging repeat readings reduces random error. |
| 10 | An exothermic reaction still needs activation energy to start, just as burning needs a flame to ignite it | Although the overall energy falls, the reactants must still get over the energy barrier, so activation energy is required. |
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 →
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