Energetics & Thermodynamics

Which statement correctly defines the standard enthalpy of formation, ΔHf°, of a compound?

• AThe enthalpy change when one mole of the compound is formed from its elements in their standard states under standard conditions✓
• BThe enthalpy change when one mole of the compound is burned completely in excess oxygen under standard conditions
• CThe enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions
• DThe enthalpy change when one mole of water is formed in a neutralisation reaction under standard conditions

Answer: Standard enthalpy of formation is the enthalpy change when one mole of a compound is made from its constituent elements, each in their standard states, under standard conditions (100 kPa, stated temperature, usually 298 K). Option B defines enthalpy of combustion; C is atomisation; D is neutralisation.

The standard enthalpy of combustion of carbon is −394 kJ mol⁻¹, of hydrogen is −286 kJ mol⁻¹, and of methane is −890 kJ mol⁻¹. Using Hess's law, calculate the standard enthalpy of formation of methane, CH₄.

• A+76 kJ mol⁻¹
• B−210 kJ mol⁻¹
• C−76 kJ mol⁻¹✓
• D−1570 kJ mol⁻¹

Answer: ΔHf(CH₄) = [ΔHc(C) + 2ΔHc(H₂)] − ΔHc(CH₄) = [(−394) + 2(−286)] − (−890) = (−966) − (−890) = −76 kJ mol⁻¹. The combustion-data cycle subtracts the combustion of the product from the sum of the combustions of the elements.

Using the mean bond enthalpies H–H = 436 kJ mol⁻¹, Cl–Cl = 242 kJ mol⁻¹ and H–Cl = 431 kJ mol⁻¹, calculate the enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g).

• A+184 kJ mol⁻¹
• B−184 kJ mol⁻¹✓
• C−431 kJ mol⁻¹
• D−94 kJ mol⁻¹

Answer: ΔH = Σ(bonds broken) − Σ(bonds formed) = (436 + 242) − (2 × 431) = 678 − 862 = −184 kJ mol⁻¹. The reaction is exothermic, so the sign is negative; the +184 distractor reverses the convention.

Enthalpy changes calculated from mean bond enthalpies often differ slightly from experimentally determined values. What is the main reason for this discrepancy?

• AMean bond enthalpies are measured at 0 K rather than 298 K
• BExperimental values always ignore the energy needed to break bonds
• CMean bond enthalpies only apply to ionic compounds
• DMean bond enthalpies are averaged over many different compounds, so they do not exactly match the bonds in a specific molecule✓

Answer: A mean (average) bond enthalpy is averaged across the same bond type in a range of different molecules and gaseous environments. The actual bond enthalpy in one specific molecule differs from this average, so bond-enthalpy calculations are approximate.

In a calorimetry experiment, burning a fuel raised the temperature of 200 g of water by 25.0 °C. Using the specific heat capacity of water c = 4.18 J g⁻¹ °C⁻¹, calculate the heat energy released to the water (q = mcΔT).

• A5.23 kJ
• B209 kJ
• C2.09 kJ
• D20.9 kJ✓

Answer: q = mcΔT = 200 × 4.18 × 25.0 = 20 900 J = 20.9 kJ. Care with unit conversion: 20 900 J ÷ 1000 = 20.9 kJ, not 2.09 kJ.

When 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid is mixed with 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide, the temperature of the 100 g mixture rises by 6.80 °C. Taking c = 4.18 J g⁻¹ °C⁻¹, calculate the enthalpy of neutralisation per mole of water formed.

• A−2.84 kJ mol⁻¹
• B−56.8 kJ mol⁻¹✓
• C−28.4 kJ mol⁻¹
• D−113.7 kJ mol⁻¹

Answer: q = mcΔT = 100 × 4.18 × 6.80 = 2842 J = 2.842 kJ. Moles of water = 0.0500 dm³ × 1.00 = 0.0500 mol. ΔH = −2.842 / 0.0500 = −56.8 kJ mol⁻¹ (close to the textbook −57 kJ mol⁻¹ for strong acid + strong base).

For the thermal decomposition CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Above approximately what temperature does the reaction become feasible (ΔG ≤ 0)?

• A1106 K✓
• B110 K
• C287 K
• D161 K

Answer: At the feasibility limit ΔG = 0, so T = ΔH/ΔS. Converting ΔS to kJ: T = 178 / 0.161 = 1106 K. Above this temperature the −TΔS term outweighs the positive ΔH, making ΔG negative.

Which set of conditions guarantees that a reaction is feasible (spontaneous, ΔG < 0) at all temperatures?

• AΔH positive and ΔS positive
• BΔH negative and ΔS negative
• CΔH negative and ΔS positive✓
• DΔH positive and ΔS negative

Answer: ΔG = ΔH − TΔS. If ΔH is negative and ΔS is positive, then −TΔS is also negative at all positive temperatures, so ΔG is always negative. Option D is never feasible; A and B are temperature-dependent.