3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 3 · AdvancedSingle correct
Use the following data at 298 K (all in kJ mol−1): ΔfH∘[MgO(s)]=−602; enthalpy of sublimation of Mg(s) =+148; first and second ionisation enthalpies of Mg(g) =+738 and +1451; bond dissociation enthalpy of O2(g)=+498; electron gain enthalpy for O(g)+e−→O−(g) =−141; lattice enthalpy of MgO, defined for MgO(s)→Mg2+(g)+O2−(g), =+3791. The enthalpy change for O−(g)+e−→O2−(g), in kJ mol−1, is ______. Exactly one option is correct.
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Answer: C
Born-Haber cycle (Hess's law): ΔfH∘=ΔsubH+IE1+IE2+21D(O2)+ΔegH1+ΔegH2−ΔlattH.
−602=148+738+1451+249+(−141)+ΔegH2−3791
ΔegH2=−602−148−738−1451−249+141+3791=+744 kJ mol−1.
Insight: adding an electron to the anion O− is strongly endothermic (electron-electron repulsion); MgO is still very stable because the lattice enthalpy of the doubly charged ions more than pays for it.
Traps: using the full 498 for O atoms gives +495; taking ΔegH1 as +141 gives +462; reversing the lattice-enthalpy sign gives nonsense.
Difficulty: Very hardQuestion 2Level 3 · AdvancedSingle correct
Sign convention: ΔU=q+w, where q is heat absorbed by the system and w is work done on the system. 2.00 mol of a monatomic ideal gas (CV,m=23R) at 300 K and 10.0 bar is held in a thermally insulated cylinder. It expands against a constant external pressure of 2.00 bar until its own pressure becomes 2.00 bar. (R=8.314 J K−1 mol−1.) The work w for this process is closest to ______. Exactly one option is correct.
A−3552 J
B−2394 J
C−3991 J
D+2394 J
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Answer: B
Adiabatic: q=0, so ΔU=w. Against constant Pext: w=−Pext(V2−V1) with V=nRT/P.
nCV,m(T2−T1)=−Pext(P2nRT2−P1nRT1), and P2=Pext=2 bar:
23(T2−300)=−(T2−300×0.2)⇒2.5T2=510⇒T2=204 K.
w=ΔU=2×23×8.314×(204−300)=−2394 J (work is done by the gas, so w<0).
Traps: the reversible-adiabatic formula TP(1−γ)/γ = const gives T2=157.6 K and −3552 J (not valid for an irreversible path); treating the expansion as isothermal gives −nRT(1−0.2)=−3991 J; +2394 J is the sign error.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
1 mol of supercooled liquid water at −10∘C (263 K) and 1 bar freezes completely to ice at 263 K, in contact with surroundings that stay at 263 K. Data: ΔfusH=6.01 kJ mol−1 at 273 K; Cp,m[H2O(l)]=75.3 and Cp,m[H2O(s)]=37.7 J K−1 mol−1, both independent of temperature; ln(273/263)=0.0373. Which set gives (ΔSsys, ΔSsurr, ΔSuniv) in J K−1? Exactly one option is correct.
A(−22.0, +22.0, 0)
B(−22.0, +22.9, +0.84)
C(−20.6, +20.6, 0)
D(−20.6, +21.4, +0.81)
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Answer: D
The freezing at 263 K is irreversible, so build a reversible path between the same states (entropy and enthalpy are state functions): liquid 263 K → liquid 273 K → ice 273 K → ice 263 K.
ΔHsys=75.3(10)−6010+37.7(−10)=753−6010−377=−5634 J.
ΔSsys=75.3(0.0373)−2736010−37.7(0.0373)=37.6×0.0373−22.01=−20.6 J K−1.
Surroundings receive −ΔHsys=5634 J reversibly at 263 K: ΔSsurr=5634/263=+21.4 J K−1.
ΔSuniv=−20.6+21.4=+0.81 J K−1>0: spontaneous, as expected for supercooled water.
Insight: neither ΔH at 273 K nor ΔfusH/273 may be used directly at 263 K; the heat actually released at 263 K is 5634 J, not 6010 J. Using 6010 J for both gives (B); assuming equilibrium gives (A) or (C).
Difficulty: Peak challenge