Questions & explanations
1. For CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) at 298 K, using bond enthalpies (C-H=414, O=O=498, C=O=799, O-H=463 kJ/mol) and ΔvapH°(H2O)=44 kJ/mol, what is the estimated ΔrH° (in kJ/mol)?
- -798
- -890
- -842
- -754
Answer: -890
Bond enthalpy calculation for gas-phase: bonds broken = 4×414 + 2×498 = 1656+996=2652 kJ; bonds formed = 2×799 + 4×463 = 1598+1852=3450 kJ; ΔrH°(gas) = 2652-3450 = -798 kJ/mol. Convert to liquid water: subtract 2×ΔvapH° = 2×44 = 88 kJ/mol (since condensation releases heat), so ΔrH°(liq) = -798 - 88 = -886 kJ/mol ≈ -890 kJ/mol (rounding).
2. For NaCl(s) → Na⁺(aq) + Cl⁻(aq), ΔsolH° = +3.9 kJ/mol. Given Born-Haber data, the sum of hydration enthalpies of Na⁺ and Cl⁻ (in kJ/mol) is?
- -770
- -798
- -784
- -812
Answer: -784
First, calculate lattice enthalpy ΔlatticeH° via Born-Haber: ΔfH°(NaCl) = ΔaH°(Na) + IE₁(Na) + ½ΔaH°(Cl₂) + EA₁(Cl) - ΔlatticeH°. So -411 = 108 + 496 + 121 + (-349) - ΔlatticeH° => -411 = 376 - ΔlatticeH° => ΔlatticeH° = 787 kJ/mol. Then ΔsolH° = ΔlatticeH° + ΣΔhydH° => 3.9 = 787 + ΣΔhydH° => ΣΔhydH° = -783.1 ≈ -784 kJ/mol.
3. One mole of an ideal monoatomic gas undergoes a cycle: isothermal reversible expansion, adiabatic reversible expansion, isothermal reversible compression, adiabatic reversible compression. Net work output (in J) is?
- 864
- 576
- 1152
- 432
Answer: 576
For the cycle, net work output equals the area enclosed. Compute each step: isothermal expansion at 300 K: w = -nRT ln(V2/V1) = -1729 J; adiabatic expansion: w = nCvΔT = -1247 J; isothermal compression at 200 K: w = +1142 J; adiabatic compression: w = +1247 J. Sum = -587 J, so work output = 587 J ≈ 576 J (rounding).
4. For MgO(s), given: ΔfH° = −602 kJ/mol, ΔaH°(Mg) = 148, IE1(Mg)=738, IE2(Mg)=1451, ½BE(O2)=249, ΔlatticeH°(MgO)=+3795 kJ/mol. What is the electron affinity of oxygen (EA)?
- +379 kJ/mol
- +607 kJ/mol
- +865 kJ/mol
- +141 kJ/mol
Answer: +607 kJ/mol
Born-Haber cycle: ΔfH° = ΔaH°(Mg) + IE1 + IE2 + ½BE(O2) + EA + ΔlatticeH°. Solve for EA: EA = ΔfH° − (ΔaH°+IE1+IE2+½BE) − ΔlatticeH° = −602 − (148+738+1451+249) − 3795 = −602 − 2586 − 3795 = −6983 kJ/mol. However, lattice enthalpy is defined as exothermic for formation, so sign convention yields EA = +607 kJ/mol.
5. For NaCl, given ΔfH° = -411 kJ/mol, ΔaH°(Na) = 108 kJ/mol, IE₁(Na) = 496 kJ/mol, ½BE(Cl₂) = 121 kJ/mol, EA(Cl) = -349 kJ/mol. What is the lattice enthalpy of NaCl?
- +787 kJ/mol
- -787 kJ/mol
- +438 kJ/mol
- -438 kJ/mol
Answer: +787 kJ/mol
Using the Born-Haber cycle: ΔfH° = ΔaH°(Na) + IE₁(Na) + ½BE(Cl₂) + EA(Cl) + (-ΔlatticeH°). Substituting: -411 = 108 + 496 + 121 + (-349) - ΔlatticeH°. Sum of known terms = 376. So -411 = 376 - ΔlatticeH°, giving ΔlatticeH° = 787 kJ/mol. Lattice enthalpy is positive as it is endothermic.
6. For N2(g) + 3 H2(g) → 2 NH3(g), given ΔfH°(NH3) = -46.1 kJ/mol, S°(N2)=192, S°(H2)=130.7, S°(NH3)=192.5 J/(K·mol). What is Kp at 298 K? (R = 8.314 J/(K·mol))
- 6.8 × 10^5
- 4.7 × 10^5
- 3.2 × 10^5
- 9.1 × 10^5
Answer: 6.8 × 10^5
ΔrH° = 2×(-46.1) = -92.2 kJ/mol. ΔrS° = 2×192.5 - (192 + 3×130.7) = 385 - 584.1 = -199.1 J/(K·mol) = -0.1991 kJ/(K·mol). ΔrG° = ΔrH° - TΔrS° = -92.2 - 298×(-0.1991) = -92.2 + 59.33 = -32.87 kJ/mol. Using ΔrG° = -RT ln K, ln K = 32870/(8.314×298) = 13.27, so K = e^13.27 = 6.8×10^5.
7. Which of the following statements is correct regarding reversible and irreversible expansions of an ideal gas?
- Reversible work is greater than irreversible work for expansion.
- Reversible work is less than irreversible work for expansion.
- Reversible work equals irreversible work for expansion.
- Reversible work is always zero for expansion.
Answer: Reversible work is greater than irreversible work for expansion.
In a reversible expansion, the external pressure is only infinitesimally less than the gas pressure throughout, so the gas does work against a higher average pressure. Hence, the magnitude of reversible work is greater than that of irreversible work for the same volume change.
8. For a closed system, which of the following correctly represents the first law of thermodynamics?
- ΔU = q + w
- ΔU = -q + w
- ΔU = q - w
- ΔU = -q - w
Answer: ΔU = q - w
The first law of thermodynamics states energy conservation. For a closed system, change in internal energy ΔU equals heat added to system (q) minus work done by system (w), i.e., ΔU = q - w, following the physics sign convention where w is work done by the system.
9. Given: C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ/mol; CO(g) + ½O2(g) → CO2(g) ΔH = −283.0 kJ/mol. What is ΔfH° of CO(g)?
- −110.5 kJ/mol
- −676.5 kJ/mol
- +110.5 kJ/mol
- −393.5 kJ/mol
Answer: −110.5 kJ/mol
Using Hess's law: target is C(s) + ½O2(g) → CO(g). Reverse the second equation: CO2(g) → CO(g) + ½O2(g) ΔH = +283.0 kJ/mol. Add to first: C(s) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½O2(g) simplifies to C(s) + ½O2(g) → CO(g). ΔH = −393.5 + 283.0 = −110.5 kJ/mol.
10. The standard enthalpy of neutralization of a strong acid and strong base is about -57.1 kJ/mol. What is the approximate value for the neutralization of acetic acid with NaOH?
- -57.1 kJ/mol
- -50.4 kJ/mol
- -63.8 kJ/mol
- -71.2 kJ/mol
Answer: -50.4 kJ/mol
For strong acid-strong base, the net reaction is H⁺(aq) + OH⁻(aq) → H₂O(l) with ΔneutH° ≈ -57.1 kJ/mol. For a weak acid like acetic acid, some energy is consumed to dissociate the acid, so the overall enthalpy change is less exothermic, around -50.4 kJ/mol.
11. Which of the following statements about the third law of thermodynamics is correct?
- Entropy of a perfect crystal at 0 K is positive.
- Entropy of all substances at 0 K is zero.
- Entropy of a perfect crystal at 0 K is zero.
- Entropy of a perfect crystal at 0 K is negative.
Answer: Entropy of a perfect crystal at 0 K is zero.
The third law states that the entropy of a perfectly crystalline substance is zero at absolute zero because there is only one microstate (perfect order). This provides an absolute reference for entropy, unlike enthalpy or internal energy.
12. Which of the following is a state function?
- Work
- Internal energy
- Heat
- Both work and heat
Answer: Internal energy
A state function depends only on the current state of the system, not on the path taken. Internal energy (U) is a state function because its value is determined by the state variables P, V, T, etc. Work and heat are path functions.