2.10 · ThermochemistryPractice · 热化学

ΔH < 0 means exothermic. ΔH from standard enthalpies of formation (equations with the smallest whole-number coefficients): thermite −851.5 kJ; electrolysis of water +571.6 kJ; carbon with steam +131.3 kJ; burning ethanol −1366.7 kJ; H₂O₂ decomposition −196.0 kJ. Combustion, the thermite reaction, and H₂O₂ decomposition release heat; electrolysis needs a steady electric current, and carbon with steam needs steady strong heating, so both are endothermic.

ΔH ≈ (bonds broken) − (bonds formed). Broken: 1 C=C, 4 C–H, and 3 O=O, 614 + 4 × 413 + 3 × 495 = 3751 kJ. Formed: 4 C=O (in 2 CO₂) and 4 O–H (in 2 H₂O), 4 × 799 + 4 × 463 = 5048 kJ. ΔH = 3751 − 5048 = −1297 kJ/mol, exothermic. Enthalpies of formation give −1322.9 kJ/mol; average bond enthalpies only give an estimate.

From q = mcΔT, ΔT = $\dfrac{q}{mc}$ = $\dfrac{25000\ \text{J}}{500\ \text{g} \times 0.449\ \text{J/(g·°C)}}$ = 111.4 °C (convert kJ to J first). Final temperature = 22.0 + 111.4 = 133.4 ≈ 133 °C (q and c have only 3 significant figures). Iron’s specific heat is only about a ninth of water’s, which is why the pan heats up so fast.

q = mcΔT with the same m and ΔT, so the largest specific heat absorbs the most heat. Specific heats (J/(g·°C)) and heat absorbed: aluminum 0.897 → 897 J; water 4.18 → 4180 J; iron 0.449 → 449 J; copper 0.385 → 385 J. Water’s unusually large specific heat is why coastal places have smaller day–night temperature swings than inland ones.

The solution’s heat change is q = mcΔT = 104.0 × 4.18 × (19.0 − 22.0) = −1304 J: the solution lost heat to the dissolving process, so ΔH is positive. n(NH₄NO₃) = 4.00 ÷ 80.05 = 0.04997 mol, and ΔH = +1.304 kJ ÷ 0.04997 mol = +26 kJ/mol (ΔT = 3.0 °C has only 2 significant figures). Instant cold packs work this way.

ΔH = −393.5 kJ/mol means 1 mol of C releases 393.5 kJ. n(C) = 6.00 ÷ 12.01 = 0.4996 mol, so the heat released = 0.4996 × 393.5 = 197 kJ. The question asks for heat released, so give a positive number.

An exothermic reaction has a negative ΔH, so A is wrong. The state of water matters: forming water vapor releases less heat (enthalpies of formation give −1234.7 kJ/mol), so B is wrong. Coefficients are moles, so 2 mol of ethanol releases twice as much, ΔH = −2733.4 kJ/mol, and D is wrong. Only C is correct.

The target has CO and H₂ on the left and CH₃OH on the right: keep the first equation, double the second, and reverse the third (changing the sign of its ΔH). Adding them, O₂, CO₂, and H₂O all cancel, leaving the target. ΔH = ΔH₁ + 2ΔH₂ − ΔH₃ = (−283.0) + 2 × (−285.8) − (−726.4) = −128.2 kJ/mol. This is the industrial route from synthesis gas to methanol.

ΔH = Σ ΔfH(products) − Σ ΔfH(reactants). ΔH = −986.1 − [(−635.1) + (−285.8)] = −65.2 kJ/mol, exothermic.

The system is the dissolving process itself; the solution and calorimeter are the surroundings. The surroundings got colder, so heat flowed from the surroundings into the system: the process is endothermic, ΔH > 0 (about +25.69 kJ/mol in reference tables). A temperature rise would mean an exothermic process, ΔH < 0.