Learners' AlleyDesigned by Khow Willard

Chemical Energetics I

Set BAnswer key

Multiple choice

4 marks

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.

ANeutralisation is an exothermic reaction. BFormation of CO₂ gas is the same as combustion of solid carbon (graphite), which is an exothermic reaction. DThis reaction is exothermic, as a solid ionic compound is formed from gaseous ions (i.e. lattice energy).

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

Structured questions

29 marks

(a)

Standard enthalpy change of formation of B₂O₃(s)[1]

(b)

Energy cycle: 2B(s) + 3H₂(g) → B₂H₆(g) (ΔH_f). 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, ΔH_f(B₂H₆) = −1273 + 3 × (−242) − (−2035)[1]

= +36 kJ mol⁻¹[1]

(a)

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).

(b)

4LiH(s) + AlCl₃(s) → LiAlH₄(s) + 3LiCl(s)[1]

ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants)

−276 = [ΔH_f + 3(−408.5)] − [4(−90.5) + (−704)][1]

ΔH_f = −116.5 kJ mol⁻¹[1]

Energy cycle: Br₂(l) + 3F₂(g) → 2BrF₃(l) directly (2ΔH_f), 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]

2ΔH_f(BrF₃) = ΔH_vap(Br₂) + BE(Br−Br) + 3BE(F−F) − 6BE(Br−F) − 2ΔH_vap(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)	ΔH_f[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]

(a)

Enthalpy change of neutralisation is the enthalpy change when an amount of acid or alkali is neutralised to form 1 mole of water, the reaction being carried out using dilute aqueous solutions.[1]

(b)

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]

(a)

S(s) + O₂(g) → SO₂(g)[1]

(b)

q = −mcΔT

q = −(150)(4.18)(39.7 − 29.8)

q = −6207 J[1]

Amount of S = 0.76 ÷ 32.1 = 0.02368 mol

ΔH_c = −62070.02368 = −262 000 J mol⁻¹

ΔH_c = −262 kJ mol⁻¹[1]

(c)

S(s) + O₂(g) → SO₂(g) represents ΔH_f(SO₂), and also ΔH_c(S).

ΔH_r = [ΔH_f(SO₂) + (−285.8)] − (−20.6)

−562 = [ΔH_f(SO₂) + (−285.8)] − (−20.6)[1]

ΔH_c(S) = ΔH_f(SO₂) = −297 kJ mol⁻¹[1]

(d)

In (b), the heat lost to the surroundings was not accounted for or the experiment was not 100% efficient or the sulfur sample from the volcano was not pure.[1]

(a)

step	ΔH / kJ mol⁻¹
2Cu(s) + ½O₂(g) → Cu₂O(s)	ΔH_f[Cu₂O] = −166
2Cu(s) → 2Cu(g)	ΔH_atom[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]

(b)

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]