Learners' AlleyDesigned by Khow Willard

Chemical Energetics II

Set BAnswer key

Multiple choice

4 marks

CRusting of an iron bar to form iron(III) oxide.[1]

In the process of rusting, O₂(g) reacts with solid iron to form solid iron oxide. Hence there is a decrease in the number of gaseous particles and ΔS is negative.

AThermal decomposition of solid CaCO₃ forms CO₂(g) and increases the number of gaseous particles. Hence ΔS is positive, as there are more ways that particles and the energy can be distributed in gases than in solids. BSublimation causes solid CO₂ to become gaseous CO₂. Hence ΔS is positive. DThe reaction between NaOH and ammonium chloride produces NH₃(g). Hence ΔS is positive.

AN₂O(g) → N₂(g) + ½O₂(g) ΔH = −82 kJ mol⁻¹[1]

ΔG = ΔH − TΔS. For the reaction to be always spontaneous, ΔH needs to be negative (options C and D are incorrect as they have positive ΔH values) and ΔS needs to be positive.

ΔS is positive in option A because there is an increase in the number of moles of gas.

CΔH +, ΔS +, ΔG −[1]

"Cold packs" indicates that the dissolution of NH₄NO₃ in water is an endothermic process, thus ΔH > 0.

Ions in solution are more disordered than in the regular ionic lattice, thus ΔS > 0.

Since the process occurs readily (spontaneously), ΔG < 0.

A−66 J K⁻¹ mol⁻¹[1]

When water freezes at 0 °C, the equilibrium H₂O(l) ⇌ H₂O(s) exists. Hence ΔG = 0.

ΔS = ΔH − ΔGT = −6.0 − 0273 = −0.0220 kJ K⁻¹ mol⁻¹

Amount of water = 54 ÷ 18 = 3 mol. Required entropy change = −0.0220 × 3 = −0.066 kJ K⁻¹ mol⁻¹ = −66 J K⁻¹ mol⁻¹

Structured questions

22 marks

(a)

2NO₂(g) → N₂O₄(g). When NO₂(g) is converted to N₂O₄(g), entropy decreases because there is a decrease in the number of gaseous molecules (from 2 molecules of NO₂ to 1 molecule of N₂O₄). Hence there are fewer ways of arranging the particles, resulting in a decrease in disorder of the system.[1]

(b)

ΔH^⦵ = ΣΔH_f^⦵(products) − ΣΔH_f^⦵(reactants)

= [2(−393.5) + 0 + 4(−241.8)] − [(+48.9) + 2(−19.6)] = −1763.9 kJ mol⁻¹[1]

ΔG^⦵ = ΔH^⦵ − TΔS^⦵ = −1763.9 − [298(+1141.2 × 10⁻³)][1]

= −2.10 × 10³ kJ mol⁻¹[1]

(a)

The standard enthalpy change of hydration, ΔH_hyd^⦵, of an ion is the heat energy evolved when one mole of the free gaseous calcium ions is completely surrounded (hydrated) by water molecules at 298 K and 1 bar.[1]

(b)

Energy cycle: Ca(s) → Ca²⁺(aq) + 2e⁻ directly (+268), or Ca(s) → Ca(g) (+178), then Ca(g) → Ca²⁺(g) + 2e⁻ (1st IE + 2nd IE = +1740), then Ca²⁺(g) → Ca²⁺(aq) (x).[1]

By Hess' Law, ΔH_r = +178 + 1740 + x = +268

x = −1650 kJ mol⁻¹[1]

(c)

ΔS is negative as there is a decrease in the number of moles of gaseous particles when the calcium ion is hydrated, which results in a decrease in disorderliness.[1]

(a)

ΔG = ΔH − TΔS. During melting and boiling, ΔG = 0, so ΔS = ΔH ÷ T

ΔS_melting = +2.3 × 1000 ÷ 234 = +9.83 J mol⁻¹ K⁻¹[1]

(b)

ΔS_vap = +59 × 1000 ÷ 630 = +93.7 J mol⁻¹ K⁻¹[1]

(a)

ΔS^⦵ > 0. This endothermic reaction is thermodynamically feasible when T > ΔH ÷ ΔS.[1]

(b)

ΔS^⦵ < 0. This exothermic reaction is thermodynamically feasible when T < ΔH ÷ ΔS.[1]

(c)

ΔS^⦵ > 0. The reaction is feasible (ΔG < 0 for all temperatures).[1]

(d)

ΔS^⦵ < 0. The reaction is not feasible (ΔG > 0 for all temperatures).[1]

(a)

Sum of ΔG_f^⦵ of reactants = −1012 kJ mol⁻¹. Sum of ΔG_f^⦵ of products = −569 + (−394) = −963 kJ mol⁻¹

So the free energy change, ΔG^⦵ = −963 − (−1012) = +49 kJ mol⁻¹[1]

The reaction will not take place spontaneously because ΔG^⦵ > 0.[1]

(b)

For the reaction to be spontaneous, ΔG = ΔH − TΔS = 101 − T(0.159) < 0, so 101 < T(0.159)[1]

A temperature above 635 K (362 °C) is needed for the reaction to take place.[1]

NoteHere, we assume that ΔH and ΔS do not change with temperature.

(a)

2NI₃(s) → N₂(g) + 3I₂(g) ΔH < 0, ΔS > 0[1]

As 4 moles of gases are produced, there is a very large increase in the entropy of the system. This reaction is also exothermic (formation of the N≡N bond). Hence, highly explosive.[1]

(b)

NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq) ΔH > 0, ΔS > 0[1]

Aqueous ions are formed from ions held fixed in an orderly crystalline lattice, resulting in a large increase in the entropy of the system. Even though the reaction is endothermic, the increase in entropy compensates for the endothermic ΔH, so that ΔG = ΔH − TΔS < 0 even under standard conditions, and NH₄NO₃ dissolves readily.[1]