A Data Booklet may be used.
CommonUse of the Data Booklet is relevant to this question. In which row are the lattice energy values (kJ mol−1) of the compounds correctly assigned?
| RbBr | CsBr | SrS | BaS | |
|---|---|---|---|---|
| A | −2643 | −2779 | −619 | −647 |
| B | −2779 | −2643 | −647 | −619 |
| C | −647 | −619 | −2779 | −2643 |
| D | −619 | −647 | −2643 | −2779 |
Source: IJC 2017 Promo, via EJC 2022 Chemical Energetics Topical Revision, Q4
Which has a positive enthalpy change of reaction?
Source: NJC 2018 Promo, via EJC 2022 Chemical Energetics Topical Revision, Q5
The following table shows the lattice energies for two silver halides.
| compound | theoretical lattice energy / kJ mol−1 | experimental lattice energy / kJ mol−1 |
|---|---|---|
| AgBr | −759 | −877 |
| AgI | −736 | −867 |
Which statement best explains the difference in the deviation of the experimental lattice energy from the theoretical lattice energy?
Source: 2025 RI J2 Prelim P1 Q11
When heated with oxygen, barium forms barium peroxide, BaO2. The relevant enthalpy changes are shown in the table.
| enthalpy change / kJ mol−1 | |
|---|---|
| Ba(g) → Ba2+(g) + 2e− | p |
| Ba2+(g) + O22−(g) → BaO2(s) | q |
| Ba(s) → Ba(g) | r |
| Ba(s) + O2(g) → BaO2(s) | s |
What is the enthalpy change of the following reaction? O2(g) + 2e− → O22−(g)
Source: 2025 RI J2 Prelim P1 Q12
CommonDiborane, B2H6, is used in rocket fuels. It can be formed from its elements according to the equation 2B(s) + 3H2(g) → B2H6(g). The following data are given.
| 2B(s) + 32O2(g) → B2O3(s) | ΔH1⦵ = −1273 kJ mol−1 |
| B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l) | ΔH2⦵ = −2035 kJ mol−1 |
| H2(g) + ½O2(g) → H2O(l) | ΔH3⦵ = −242 kJ mol−1 |
Source: 2025 CJC J2 Prelim P2 Q1(a)
CommonLithium hydride can be prepared by passing hydrogen gas over heated lithium. The following data will be useful in this question.
| enthalpy change of formation of LiH(s) | −90.5 kJ mol−1 |
|---|---|
| enthalpy change of atomisation of Li(s) | +159.5 kJ mol−1 |
| electron affinity of hydrogen atoms | −73.0 kJ mol−1 |
| enthalpy change of formation of AlCl3(s) | −704 kJ mol−1 |
| enthalpy change of formation of LiCl(s) | −408.5 kJ mol−1 |
Source: 2023 CJC J2 Prelim P3 Q1(a)
Bromine and fluorine react together to give bromine trifluoride. Br2(l) + 3F2(g) → 2BrF3(l)
| standard enthalpy change of formation of BrF3(l) | −301 kJ mol−1 |
|---|---|
| enthalpy change of vaporisation of Br2(l) → Br2(g) | +31 kJ mol−1 |
| enthalpy change of vaporisation of BrF3(l) → BrF3(g) | +44 kJ mol−1 |
Source: A Level 2016 P3 Q2(b), via EJC 2022 JC1 Chemical Energetics Tutorial, B5(a)
| value / kJ mol−1 | |
|---|---|
| standard enthalpy change of atomisation of Ca(s) | +121 |
| standard enthalpy change of formation of Ca2SiO4(s) | −2306 |
| enthalpy change of the reaction Si(s) + 2O2(g) + 4e− → SiO44−(g) | +13872 |
Source: 2023 RI J2 Prelim P3 Q1(c)
Source: RJC H2 Chemistry Tutorial 5a (Chemical Energetics I), Q4(a) and (d)
Environmental chemists burnt a sample of solid sulfur collected near a volcano and measured the temperature change of a fixed mass of water in a calorimeter. Their data are shown below.
| mass of solid sulfur powder burnt / g | 0.76 |
|---|---|
| mass of water / g | 150 |
| initial temperature of water / °C | 29.8 |
| final temperature of water / °C | 39.7 |
In the atmosphere, hydrogen sulfide is oxidised to sulfur dioxide: H2S(g) + 3/2O2(g) → SO2(g) + H2O(l), ΔHr = −562 kJ mol−1. The enthalpy changes of formation of H2S(g) and H2O(l) are −20.6 and −285.8 kJ mol−1.
Source: 2025 EJC J2 Prelim P2 Q1(a)(i) to (iv)
ChallengeThe following information is given.
| energy change | / kJ mol−1 |
|---|---|
| atomisation energy of Cu | +339 |
| first electron affinity of O | −141 |
| second electron affinity of O | +791 |
| enthalpy change of formation of copper(I) oxide | −166 |
Source: RJC H2 Chemistry Tutorial 5a (Chemical Energetics I), Q8(a) and (b)
CRbBr −647, CsBr −619, SrS −2779, BaS −2643[1]
Lattice energy ∝ q+q−r+ + r− (ionic radii from the Data Booklet)
Charges: the product of charges is +1 × −1 = −1 for RbBr and CsBr, and +2 × −2 = −4 for SrS and BaS. SrS and BaS will have higher LE than RbBr and CsBr.
Ionic radii: Cs+ has a larger r+ than Rb+. Since Br− is a common anion, RbBr will have a higher LE than CsBr. Ba2+ is bigger than Sr2+. Since S2− is a common anion, SrS will have a larger LE.
Cformation of gaseous O2− ion from gaseous O− ion[1]
The second electron affinity is always positive (endothermic), because energy is required to overcome the electrostatic repulsion between the incoming electron and the singly negatively charged anion.
DThe I− ion is more polarisable than the Br− ion.[1]
Theoretical lattice energy is calculated using |lattice energy| ∝ |q+q−|r+ + r−, which assumes a 100% ionic nature.
The discrepancy between the experimental and theoretical lattice energies shows the presence of covalent character in the bonding in the silver halides, which arises due to substantial polarisation of the anion by the cation.
Since the anionic radius increases down the Group, the polarisability of the anion increases, causing a greater degree of covalency and hence a greater deviation of the experimental from the theoretical value.
Bs − q − p − r[1]
A possible method to solve this question without an energy cycle is to use the algebraic method.
| enthalpy change / kJ mol−1 | |
|---|---|
| Ba(s) + O2(g) → BaO2(s) | s |
| BaO2(s) → Ba2+(g) + O22−(g) | −q |
| Ba2+(g) + 2e− → Ba(g) | −p |
| Ba(g) → Ba(s) | −r |
| O2(g) + 2e− → O22−(g) | s − q − p − r |
Energy cycle: 2B(s) + 3H2(g) → B2H6(g) (ΔHf). With 3O2(g) added, both sides give B2O3(s) + 3H2O(l): directly by −1273 + 3 × (−242), or through B2H6(g) by −2035.[1]
Using Hess' Law, ΔHf(B2H6) = −1273 + 3 × (−242) − (−2035)[1]
= +36 kJ mol−1[1]
| step | ΔH / kJ mol−1 |
|---|---|
| Li(s) + ½H2(g) → LiH(s) | −90.5 |
| Li(s) → Li(g) | +159.5 |
| ½H2(g) → H(g) | ½(+436) |
| Li(g) → Li+(g) + e− | +519 |
| H(g) + e− → H−(g) | −73 |
| Li+(g) + H−(g) → LiH(s) | LE |
Using Hess' Law, −90.5 = +159.5 + ½(+436) + 519 + (−73.0) + LE[1]
LE = −914 kJ mol−1[1]
NoteThe relevant data from the Data Booklet are the 1st IE of Li (+519 kJ mol−1) and the BE of H2 (+436 kJ mol−1).
4LiH(s) + AlCl3(s) → LiAlH4(s) + 3LiCl(s)[1]
ΔHr = ΣΔHf(products) − ΣΔHf(reactants)
−276 = [ΔHf + 3(−408.5)] − [4(−90.5) + (−704)][1]
ΔHf = −116.5 kJ mol−1[1]
Energy cycle: Br2(l) + 3F2(g) → 2BrF3(l) directly (2ΔHf), or by Br2(l) → Br2(g), then Br2(g) + 3F2(g) → 2Br(g) + 6F(g), then 2Br(g) + 6F(g) → 2BrF3(g), then 2BrF3(g) → 2BrF3(l).[1]
2ΔHf(BrF3) = ΔHvap(Br2) + BE(Br−Br) + 3BE(F−F) − 6BE(Br−F) − 2ΔHvap(BrF3)
2(−301) = 31 + 193 + 3(158) − 6BE(Br−F) − 2(44)[1]
6BE(Br−F) = 1212
BE(Br−F) = 1212 ÷ 6 = +202 kJ mol−1[1]
| step | ΔH / kJ mol−1 |
|---|---|
| 2Ca(s) + Si(s) + 2O2(g) → Ca2SiO4(s) | ΔHf[Ca2SiO4] = −2306 |
| 2Ca(s) → 2Ca(g) | 2(+121) |
| 2Ca(g) → 2Ca2+(g) + 4e− | 2 × (1st IE[Ca] + 2nd IE[Ca]) |
| Si(s) + 2O2(g) + 4e− → SiO44−(g) | +13872 |
| 2Ca2+(g) + SiO44−(g) → Ca2SiO4(s) | LE[Ca2SiO4] |
Using Hess' Law, −2306 = 2(+121) + 2(+590 + 1150) + 13872 + LE[Ca2SiO4][2]
LE[Ca2SiO4] = −19900 kJ mol−1 (3 s.f.)[1]
HCN is a weak acid and energy is required to ionise it to produce one mole of H+ ions before neutralisation. NH3 is a weak base and energy is required to ionise it to produce one mole of OH− ions before neutralisation can take place.[1]
The ionisation of the weak acid is endothermic, which results in a less exothermic enthalpy change of neutralisation for the weak acid–weak base than for the strong acid–strong base.[1]
q = −mcΔT
q = −(150)(4.18)(39.7 − 29.8)
q = −6207 J[1]
Amount of S = 0.76 ÷ 32.1 = 0.02368 mol
ΔHc = −62070.02368 = −262 000 J mol−1
ΔHc = −262 kJ mol−1[1]
S(s) + O2(g) → SO2(g) represents ΔHf(SO2), and also ΔHc(S).
ΔHr = [ΔHf(SO2) + (−285.8)] − (−20.6)
−562 = [ΔHf(SO2) + (−285.8)] − (−20.6)[1]
ΔHc(S) = ΔHf(SO2) = −297 kJ mol−1[1]
| step | ΔH / kJ mol−1 |
|---|---|
| 2Cu(s) + ½O2(g) → Cu2O(s) | ΔHf[Cu2O] = −166 |
| 2Cu(s) → 2Cu(g) | ΔHatom[Cu] × 2 = 2(+339) |
| 2Cu(g) → 2Cu+(g) + 2e− | 1st IE[Cu] × 2 = 2(+745) |
| ½O2(g) → O(g) | ½BE[O2] = ½(+496) |
| O(g) + e− → O−(g) | 1st EA[O] = −141 |
| O−(g) + e− → O2−(g) | 2nd EA[O] = +791 |
| 2Cu+(g) + O2−(g) → Cu2O(s) | LE[Cu2O] |
By Hess' Law, −166 = 2(+339) + 2(+745) + ½(+496) + (−141) + (+791) + LE[Cu2O][2]
Lattice energy of copper(I) oxide, LE[Cu2O] = −3232 kJ mol−1[1]
The LE of CuO is more exothermic (−4143 kJ mol−1), so stronger ionic bonds are formed.[1]
For CuO, the higher charge of Cu2+ compared with Cu+ gives a larger q+, and the smaller cationic radius (r+), as the electron cloud is attracted more strongly to the nucleus, gives a smaller interionic distance (r+ + r−).[1]