Learners' AlleyDesigned by Khow Willard

Chemical Energetics I

Set AAnswer key

From the lesson

13 marks

Reverse the equation: the sign changes, +92

Halve it: ΔH = +46 kJ mol⁻¹[1]

(a)

C(s) + 2H₂(g) + ½O₂(g) → CH₃OH(l)[1]

(b)

C₂H₆(g) + 3½O₂(g) → 2CO₂(g) + 3H₂O(l)[1]

(c)

Mg²⁺(g) + O²⁻(g) → MgO(s)[1]

(d)

½Br₂(l) → Br(g)[1]

q = −(50.0)(4.18)(6.7) = −1400 J[1]

n_H₂O = 0.0250 × 1.00 = 0.02500 mol[1]

ΔH_neut = −1400 ÷ 0.02500 = −56.0 kJ mol⁻¹[1]

ΔH = ΣΔH_f(products) − ΣΔH_f(reactants) = −85 − (+52 + 0) = −137 kJ mol⁻¹[1]

H₂(g) is an element in its standard state, so its ΔH_f^⦵ = 0.[1]

−411 = 107 + 122 + 494 − 349 + LE[1]

LE = −411 − 374 = −785 kJ mol⁻¹[1]

ΔH_at forms one mole of Cl atoms: ½Cl₂(g) → Cl(g). Breaking one mole of Cl–Cl bonds gives two moles of atoms.[1]

Practice

28 marks

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.

(a)

The enthalpy change of combustion is the energy released when 1 mole of a substance is completely burnt in excess oxygen under standard conditions.[1]

(b)

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

n_glucose = 2.18 ÷ (72.0 + 12.0 + 96.0) = 0.01211 mol

ΔH_c(glucose) = −26.45 ÷ 0.01211 = −2184 kJ mol⁻¹ = −2180 kJ mol⁻¹ (3 s.f.)[1]

(a)

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]

(b)(i)

The energy change when 1 mole of a substance is formed from its constituent elements in their standard states at 1 bar, at a specified temperature, usually 298 K.[1]

(ii)

N₂(g) + 52O₂(g) → N₂O₅(g)[1]

(iii)

By Hess' Law, ΔH_f(N₂O₅) = +180 + 2(−57) + (−55)[1]

= +11.0 kJ mol⁻¹[1]

(a)

C₃H₅(NO₃)₃(l) → 32N₂(g) + 14O₂(g) + 3CO₂(g) + 52H₂O(g)[1]

(b)

ΔH_{decomposition} = ΣΔH_f(products) − ΣΔH_f(reactants)

= 3(−394) + 52(−242) − (−364)

= −1423 kJ mol⁻¹[1]

(a)

By Hess' Law, −272 = +416 + ½(+496) + 762 + 1560 + 657 + LE

LE = −3915 kJ mol⁻¹[3]

(b)

|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]

(a)(i)

K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + CO₂(g) + H₂O(l)[1]

(ii)

Heat released = mcΔT = 30.0 × 4.18 × 5.20 = 652.1 J

M_r of K₂CO₃ = 2(39.1) + 12.0 + 3(16.0) = 138.2

n_K₂CO₃ = 2.76 ÷ 138.2 = 0.01997 mol[1]

n_HCl = 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]

(b)(i)

KHCO₃(s) + HCl(aq) → KCl(aq) + CO₂(g) + H₂O(l)[1]

(ii)

Heat absorbed = 30.0 × 4.18 × 3.70 = 464.0 J

M_r of KHCO₃ = 39.1 + 1.0 + 12.0 + 3(16.0) = 100.1

n_KHCO₃ = 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]

(c)

2KHCO₃(s) → K₂CO₃(s) + H₂O(l) + CO₂(g) (2ΔH_r)

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, 2ΔH_r = 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]