Unit 6 · ThermochemistryPractice · 热化学

The pack turning cold means the surroundings lost energy: heat flowed from the surroundings into the reacting substances (the system), so the reaction is endothermic, ΔH > 0, and the products sit higher in enthalpy than the reactants. “Cold” is not something that flows; only heat flows, always from hotter to colder.

Heat lost by the aluminum = heat gained by the water: 60.0 × 0.897 × (88.0 − T) = 120.0 × 4.18 × (T − 20.0), so 53.82(88.0 − T) = 501.6(T − 20.0) and T = 26.6 °C. The water’s m·c (501.6 J/°C) is about 9 times the block’s, so the final temperature lands much closer to the water’s starting temperature.

① Heat absorbed by the solution: m = 100.0 g, q_soln = 100.0 × 4.18 × 5.3 = 2215 J. ② HNO₃ is 0.0600 mol and KOH is 0.0400 mol, so KOH is limiting and 0.0400 mol of water forms; the excess HNO₃ releases no heat. ③ The solution warmed, so the reaction is exothermic: ΔH = −2.215 kJ ÷ 0.0400 mol = −55.4 kJ/mol. Dividing by the moles of HNO₃ would give a too-small −36.9.

Condensation is the reverse of vaporization, so ΔH = −ΔH_vap. q = n × ΔH_vap = 0.750 × 40.7 = 30.5 kJ, released as the molecules come together and form intermolecular attractions. The fusion value is the wrong phase change, and 40.7 kJ forgets to multiply by the moles.

n(C₂H₆) = 4.50 ÷ 30.07 = 0.1497 mol. ΔH is for the reaction as written, which burns 2 mol of C₂H₆, so heat released = 0.1497 ÷ 2 × 3119.4 = 233 kJ. Multiplying 0.1497 × 3119.4 directly doubles it (467 kJ).

Count only the bonds that change. Broken: one C=C and one Cl–Cl, 614 + 242 = 856 kJ. Formed: one C–C and two C–Cl, 348 + 2 × 328 = 1004 kJ. ΔH ≈ 856 − 1004 = −148 kJ/mol. The C=C becomes a C–C, so we break a C=C and form a C–C.

ΔH = ΣnΔH°_f(products) − ΣnΔH°_f(reactants), and H₂ is an element in its standard state with ΔH°_f = 0. ΔH = (−393.5 + 0) − (−110.5 + (−241.8)) = −41.2 kJ/mol, so the reaction is exothermic.

NO₂ is on the left of the target, so reverse ① (flip the sign of ΔH); N₂O₄ is on the right, so use ② as written. Adding them cancels N₂ and 2O₂: ΔH = −ΔH₁ + ΔH₂ = −(+66.4) + (+9.2) = −57.2 kJ/mol.

The new equation is the original reversed and divided by 2: reversing flips the sign, and halving the coefficients halves ΔH. ΔH = −(−1203.4) ÷ 2 = +601.7 kJ/mol. Decomposing MgO absorbs heat.

The recorded “final” temperature is too low. The water’s ΔT is too small, so the calculated heat q is too small, while the metal’s temperature drop (100.0 − T) comes out too large. c_metal = q ÷ (m × ΔT_metal) has a smaller numerator and a larger denominator, so the calculated specific heat is too small.

The heat the warm water gives up does two jobs: melting the ice (0.555 × 6010 = 3336 J) and then warming that meltwater from 0 °C to T. 100.0 × 4.18 × (30.0 − T) = 3336 + 10.0 × 4.18 × (T − 0), so T = 20.0 °C. Forgetting the heat of melting gives a too-high 27.3 °C.

ΔT = heat released ÷ (m·c). Doubling both concentrations at the same volumes doubles the water formed and the heat released while the mass stays the same, so ΔT doubles. Doubling both volumes doubles the heat and the mass, so ΔT is unchanged. Doubling only the HCl leaves KOH limiting, so the same amount of water forms and ΔT is unchanged. The starting temperature doesn’t affect ΔT.