A Data Booklet may be used. Specific heat capacity of water = 4.18 J g−1 K−1 unless stated otherwise.
Source: Learners' Alley, Chemical Energetics I slides
CommonWrite the equation, with state symbols, for each standard enthalpy change.
Source: Learners' Alley, Chemical Energetics I slides
Source: Learners' Alley, Chemical Energetics I slides
Source: Learners' Alley, Chemical Energetics I slides
Source: Learners' Alley, Chemical Energetics I slides
Source: Learners' Alley, Chemical Energetics I slides
Source: Learners' Alley, Chemical Energetics I slides
CommonA student mixes 25.0 cm3 of 0.350 mol dm−3 sodium hydroxide solution with 25.0 cm3 of 0.350 mol dm−3 hydrochloric acid. The temperature increases by 2.5 °C. No heat is lost to the surroundings. The final mixture has a specific heat capacity of 4.2 J g−1 K−1. What is the molar enthalpy change for the reaction?
Source: 2025 ASRJC J2 Prelim P1 Q7
Silane, SiH4, exists as a gas at standard temperature and pressure. Hess' Law can be used to calculate the average Si–H bond energy in gaseous SiH4. Which information is needed to perform the calculation?
Source: 2025 ASRJC J2 Prelim P1 Q8
Source: EJC 2022 JC1 Chemical Energetics Tutorial, A5
In determining the enthalpy change of combustion of glucose, C6H12O6, using a bomb calorimeter, the following results were obtained.
| heat capacity of the calorimeter | 765 J K−1 | mass of glucose burnt | 2.18 g |
|---|---|---|---|
| specific heat capacity of water | 4.2 J g−1 K−1 | mass of water in calorimeter | 1750 g |
| initial temperature of water and calorimeter | 17.84 °C | final temperature of water and calorimeter | 21.10 °C |
Source: EJC 2022 JC1 Chemical Energetics Tutorial, B1
Dinitrogen pentoxide, N2O5, can be produced by the following reaction sequence.
| I | N2(g) + O2(g) → 2NO(g) | ΔH = +180 kJ mol−1 |
| II | NO(g) + ½O2(g) → NO2(g) | ΔH = −57 kJ mol−1 |
| III | 2NO2(g) + ½O2(g) → N2O5(g) | ΔH = −55 kJ mol−1 |
Source: EJC 2022 JC1 Chemical Energetics Tutorial, B4(a)(ii) and (b)
Nitroglycerin, C3H5(NO3)3, is a flammable liquid commonly used to manufacture dynamite. Upon ignition, nitroglycerin decomposes to produce nitrogen, oxygen, carbon dioxide and steam.
| nitroglycerin(l) | H2O(g) | CO2(g) | |
|---|---|---|---|
| ΔHf⦵ / kJ mol−1 | −364 | −242 | −394 |
Source: 2019 CJC J1 Promo P2 Q4(a) and (b)
CommonIron(II) compounds are generally only stable in neutral, non-oxidising conditions. It is difficult to determine the lattice energy of FeO experimentally.
| ΔH / kJ mol−1 | |
|---|---|
| enthalpy change of atomisation of Fe(s) | +416 |
| sum of first and second electron affinity of O(g) | +657 |
| enthalpy change of formation of FeO(s) | −272 |
Data Booklet values: first and second ionisation energies of Fe are +762 and +1560 kJ mol−1, the O=O bond energy is +496 kJ mol−1, and the ionic radii of Fe2+ and Ca2+ are 0.061 nm and 0.099 nm.
Source: 2023 SAJC J2 Prelim P2 Q2(c)(i) and (ii)
ChallengeAssume that the specific heat capacity and density of all solutions are 4.18 J g−1 K−1 and 1.00 g cm−3 respectively.
Source: RJC H2 Chemistry Tutorial 5a (Chemical Energetics I), Q3
Reverse the equation: the sign changes, +92
Halve it: ΔH = +46 kJ mol−1[1]
q = −(50.0)(4.18)(6.7) = −1400 J[1]
nH2O = 0.0250 × 1.00 = 0.02500 mol[1]
ΔHneut = −1400 ÷ 0.02500 = −56.0 kJ mol−1[1]
ΔH = ΣΔHf(products) − ΣΔHf(reactants) = −85 − (+52 + 0) = −137 kJ mol−1[1]
H2(g) is an element in its standard state, so its ΔHf⦵ = 0.[1]
−411 = 107 + 122 + 494 − 349 + LE[1]
LE = −411 − 374 = −785 kJ mol−1[1]
ΔHat forms one mole of Cl atoms: ½Cl2(g) → Cl(g). Breaking one mole of Cl–Cl bonds gives two moles of atoms.[1]
B−60 kJ mol−1[1]
Limiting reagent is hydrochloric acid = (0.35 × 0.025) = 0.00875 mol
q = (25.0 + 25.0)(4.2)(2.5) = 525 J
ΔH = −(525) ÷ (0.00875) = −60 kJ mol−1
BΔH⦵atomisation(Si), ΔH⦵atomisation(H), ΔH⦵formation(SiH4)[1]
Using |LE| ∝ |q+q−|r+ + r−:
Mg2+ has a higher charge than Na+, and O2− has a higher charge than Cl−. Mg2+ has a smaller ionic radius than Na+, and O2− has a smaller ionic radius than Cl−.[1]
MgO would thus have the largest lattice energy in magnitude (most exothermic).[1]
NoteCharge is the more dominant factor affecting lattice energies if the ions concerned are from the same period.
Heat absorbed by water = mcΔT = (1750)(4.2)(21.10 − 17.84) = 23960 J = 23.96 kJ[1]
Heat absorbed by calorimeter = CΔT = (765)(21.10 − 17.84) = 2494 J = 2.494 kJ[1]
Heat change of combustion of glucose = −(23.96 + 2.494) = −26.45 kJ
nglucose = 2.18 ÷ (72.0 + 12.0 + 96.0) = 0.01211 mol
ΔHc(glucose) = −26.45 ÷ 0.01211 = −2184 kJ mol−1 = −2180 kJ mol−1 (3 s.f.)[1]
More energy is required to break the bonds in the reactants than is given out when the products are formed, due to the much stronger N≡N bond in the reactants. Hence, the reaction is endothermic.[1]
By Hess' Law, ΔHf(N2O5) = +180 + 2(−57) + (−55)[1]
= +11.0 kJ mol−1[1]
ΔHdecomposition = ΣΔHf(products) − ΣΔHf(reactants)
= 3(−394) + 52(−242) − (−364)
= −1423 kJ mol−1[1]
By Hess' Law, −272 = +416 + ½(+496) + 762 + 1560 + 657 + LE
LE = −3915 kJ mol−1[3]
|lattice energy| ∝ q+ × q−r+ + r−
Since the anions are the same, and Fe2+ and Ca2+ have the same charge but Fe2+ has a smaller radius, there are stronger ionic bonds or greater attraction between Fe2+ and O2−.[1]
Hence, the lattice energy of FeO is more exothermic or more negative than that of CaO.[1]
Heat released = mcΔT = 30.0 × 4.18 × 5.20 = 652.1 J
Mr of K2CO3 = 2(39.1) + 12.0 + 3(16.0) = 138.2
nK2CO3 = 2.76 ÷ 138.2 = 0.01997 mol[1]
nHCl = 301000 × 2 = 0.06 mol, hence HCl is in excess.
ΔH1 = −652.1 ÷ 0.01997 = −32650 J mol−1 = −32.7 kJ mol−1[1]
Heat absorbed = 30.0 × 4.18 × 3.70 = 464.0 J
Mr of KHCO3 = 39.1 + 1.0 + 12.0 + 3(16.0) = 100.1
nKHCO3 = 2.00 ÷ 100.1 = 0.01998 mol, and HCl is in excess.[1]
ΔH2 = +464.0 ÷ 0.01998 = +23220 J mol−1 = +23.2 kJ mol−1[1]
2KHCO3(s) → K2CO3(s) + H2O(l) + CO2(g) (2ΔHr)
Both sides react with 2HCl(aq) to give 2KCl(aq) + 2H2O(l) + 2CO2(g): 2ΔH2 from 2KHCO3(s), ΔH1 from K2CO3(s).
By Hess' Law, 2ΔHr = 2ΔH2 − ΔH1 = +79.1 kJ mol−1[1]
Heat change for the decomposition of 1 mole of KHCO3 = +39.6 kJ mol−1 (3 s.f.)[1]