S6 Chemistry: Energetics (Paper 2: Intermediate)

S6 Chemistry: Energetics (Paper 2: intermediate)

2026-11-12

S6 Chemistry: Energetics (Paper 2: intermediate)

Ten intermediate items on the symbols in q equals m c delta T, the definition of standard enthalpy of combustion, Hess law, a neutralisation calculation, the bond energy method, the formation enthalpy method, ethanol combustion, combining equations, a bond energy estimate and the difference between the enthalpy change and activation energy.

Direct answers　Answers in order: mass, specific heat capacity and temperature rise, one mole burning completely, independent of route, -57.1 kJ per mole, -185 kJ per mole, -891 kJ per mole, -580 kJ per mole, -111 kJ per mole, -802 kJ per mole, 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: Energetics
Level	S6

02　Questions (10 items)

FoundationIn q equals m c delta T, what do m, c and delta T stand for, and what are their units?
FoundationWhat is the definition of the standard enthalpy of combustion?
FoundationState Hess law.
Intermediate50.0 cm3 of 0.50 M hydrochloric acid is mixed with 50.0 cm3 of 0.50 M sodium hydroxide and the temperature rises by 3.4 degrees Celsius. Find the enthalpy of neutralisation, taking density 1.0 g per cm3 and c = 4.2.
IntermediateUse bond energies to find the enthalpy change of H2 plus Cl2 giving 2HCl, with H-H 436, Cl-Cl 243 and H-Cl 432 in kJ per mole.
IntermediateUse enthalpies of formation to find the enthalpy change of CH4 plus 2O2 giving CO2 plus 2H2O, with dHf CH4 -75, CO2 -394 and H2O -286.
ChallengeBurning 2.00 g of ethanol, M = 46.0, raises 200 g of water by 30.0 degrees Celsius. Find the experimental enthalpy of combustion.
ChallengeGiven that C plus O2 gives CO2 with dH -394 and CO plus half O2 gives CO2 with dH -283, find dH for C plus half O2 giving CO.
ChallengeEstimate the enthalpy change of CH4 plus 2O2 giving CO2 plus 2H2O using bond energies: C-H 413, O=O 498, C=O 799 and O-H 463.
Error redoA student writes that a negative enthalpy change means the reaction needs no energy to start. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	m is the mass heated in grams, c is the specific heat capacity in J per g per K and delta T is the temperature rise in K	Multiplying the three in a consistent set of units gives q in joules.
2	The enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions	The definition fixes one mole of substance and complete combustion, and since burning releases heat the value is negative.
3	The total enthalpy change depends only on the initial and final states, not on the route taken	Enthalpy is a state function, so the enthalpy changes of several steps can be added or subtracted to give that of a single step.
4	about -57.1 kJ per mole	The total mass is 100 g and q is 100 times 4.2 times 3.4 which is 1428 J; the limiting amount is 0.025 mol, and 1428 divided by 25 gives 57.1 kJ per mole with a negative sign.
5	-185 kJ per mole	Breaking bonds absorbs 436 plus 243 which is 679 while forming bonds releases 2 times 432 which is 864, and the difference is -185.
6	-891 kJ per mole	The sum for products minus the sum for reactants gives -394 plus 2 times -286 which is -966, and subtracting -75 gives -891.
7	about -580 kJ per mole	q is 200 times 4.2 times 30 which is 25200 J, the ethanol is 0.0435 mol, and 25.2 divided by 0.0435 gives 580 with a negative sign.
8	-111 kJ per mole	Subtracting the second equation from the first gives -394 minus -283 which is -111.
9	about -802 kJ per mole	Breaking gives 4 times 413 plus 2 times 498 which is 2648 while forming gives 2 times 799 plus 4 times 463 which is 3450, and the difference is -802.
10	A negative enthalpy change only means the products lie lower in energy; the reaction still needs activation energy to start	The overall energy change and the barrier to start are two different things, and an exothermic reaction may still need heating to begin.

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 and balance the equation first, then use q = mcDT or add formation enthalpies to find the enthalpy change.Tutor Registration →

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