U6.9 · Hess's LawPractice · 盖斯定律

The target has 3 C and 4 H₂ on the left and C₃H₈ on the right: take ① × 3, ② × 4, and ③ reversed. Adding them cancels 5O₂, 3CO₂, and 4H₂O(l). ΔH = 3ΔH₁ + 4ΔH₂ − ΔH₃ = 3(−393.5) + 4(−285.8) − (−2219.9) = −103.8 kJ/mol, which is exactly the enthalpy of formation of propane.

H₂O₂ is on the left of the target, so reverse ① (flip the sign); H₂O(l) is on the right, so use ② as written. Adding them cancels H₂ and leaves 1 − ½ = ½ O₂ on the right. ΔH = ΔH₂ − ΔH₁ = (−285.8) − (−187.8) = −98.0 kJ/mol.

② − ① gives 2NO(g) + O₂(g) → 2NO₂(g) (N₂ cancels), ΔH = +66.4 − (+180.6) = −114.2 kJ/mol. The target has half these coefficients, so halve ΔH: −57.1 kJ/mol.

Enthalpy is like altitude: the height gained from the foot of a hill to the top is the same whichever trail you take. With the same initial and final states, one step or several give the same total ΔH, which is why equations can be reversed, scaled, and added to find ΔH. The individual steps generally have different ΔH values.

CaO and CO₂ are on the left of the target, so reverse both ① and ②; CaCO₃ is on the right, so use ③ as written. Adding cancels Ca, C, and 3/2 O₂. ΔH = ΔH₃ − ΔH₁ − ΔH₂ = (−1206.9) − (−635.1) − (−393.5) = −178.3 kJ/mol.

Go from the CO(g) + ½O₂(g) step (−110.5 kJ/mol) down to CO₂(g) (−393.5 kJ/mol): ΔH = final − initial = −393.5 − (−110.5) = −283.0 kJ/mol; going down means exothermic, so it is negative. −393.5 is ΔH for burning C straight to CO₂, −504.0 adds the two numbers, and +283.0 has the wrong sign. This is Hess’s law: ΔH for C → CO₂ equals the sum of C → CO and CO → CO₂.