You want ΔH for the reaction A → C, but only have thermochemical data for: A → B (ΔH₁ = −200 kJ) and C → B (ΔH₂ = −50 kJ). How do you correctly combine these to find ΔH for A → C?
AAdd them directly: ΔH = ΔH₁ + ΔH₂ = −250 kJ
BReverse the second reaction and add: ΔH = ΔH₁ − ΔH₂ = −200 − (−50) = −150 kJ
CAverage the two values: ΔH = (−200 + −50)/2 = −125 kJ
DSubtract the first from the second: ΔH = ΔH₂ − ΔH₁ = −50 − (−200) = +150 kJ
To get A → C, you need B to cancel out. The first equation gives A → B. The second equation gives C → B, but you need B → C (to be consumed, not produced). Reversing C → B gives B → C with ΔH = +50 kJ. Now add: (A → B) + (B → C) = A → C, with ΔH = −200 + (+50) = −150 kJ. The rule: when you reverse a reaction, negate its ΔH. This is because enthalpy is a state function — the same energy that is released going forward must be absorbed going backward. Option A is the most tempting error: adding the raw values without recognizing that the second equation needs to be reversed.
Question 2 Multiple Choice
Why is it valid to reverse a thermochemical equation and simply change the sign of ΔH, rather than recalculating it from scratch?
ABecause all chemical reactions are reversible, so the sign change is always experimentally confirmed
BBecause enthalpy is a state function: ΔH depends only on the initial and final states, not the path. Reversing the reaction swaps initial and final states, so ΔH must flip sign to describe the same state difference in the opposite direction
CBecause the law of conservation of mass requires that energy be conserved when reactions are reversed
DBecause the sign convention for ΔH is arbitrary, so it can be freely changed when convenient
The core reason is the state-function nature of enthalpy. A state function's change depends only on where you start and end, not how you get there. If going from state A to state B releases 100 kJ (ΔH = −100 kJ), then going from B back to A must absorb exactly 100 kJ (ΔH = +100 kJ) — because the difference in enthalpy between the two states is fixed by the states themselves. This is fundamentally different from a path-dependent quantity like heat flow in a non-quasistatic process. Hess's law is entirely a consequence of this state-function property.
Question 3 True / False
Hess's law works because nearly every reaction releases the same total amount of heat regardless of the temperature, pressure, or conditions under which it occurs.
TTrue
FFalse
Answer: False
False. Hess's law works because enthalpy is a state function — the total enthalpy change depends only on the identities of reactants and products, not on the pathway between them. It does NOT say that ΔH is independent of conditions: temperature, pressure, and phase can affect ΔH values (Kirchhoff's law describes temperature dependence). Hess's law applies at a given set of conditions; it says that at those conditions, you can add reactions algebraically. Confusing path-independence (what Hess's law actually claims) with condition-independence (which is false) is a subtle but important error.
Question 4 True / False
If the combustion of 1 mol of propane releases 2,220 kJ (ΔH = −2,220 kJ), then the combustion of 2 mol of propane has ΔH = −4,440 kJ.
TTrue
FFalse
Answer: True
True. One of the manipulation rules underlying Hess's law is that multiplying a balanced thermochemical equation by any coefficient scales ΔH by the same factor. If burning 1 mol of C₃H₈ releases 2,220 kJ, then burning 2 mol releases twice as much: 4,440 kJ. This scaling rule follows directly from enthalpy being an extensive state function — doubling the amount of substance doubles the enthalpy change. This principle is used constantly in Hess's law calculations: you can multiply any step-reaction by the coefficient needed to balance intermediate species.
Question 5 Short Answer
What property of enthalpy makes Hess's law valid, and how does this property differ fundamentally from a path-dependent quantity like the work done against friction?
Think about your answer, then reveal below.
Model answer: Enthalpy is a state function: its value is determined entirely by the current thermodynamic state of the system (composition, temperature, pressure), not by the history of how that state was reached. Therefore, ΔH for any process equals H_final − H_initial, regardless of pathway. Hess's law follows directly: it doesn't matter whether a reaction occurs in one step or ten steps; the total ΔH equals the sum of the individual ΔH values, because the endpoints are the same. Friction work, by contrast, is path-dependent: sliding a book across a rough table a longer path generates more heat than a shorter path between the same endpoints. There is no analog of Hess's law for frictional work because it has no 'state function' equivalent.
The key conceptual move is understanding why state functions enable additive decomposition. Because enthalpy depends only on endpoints, intermediate states are 'invisible' — they can be added and subtracted freely as long as they cancel. Path-dependent quantities cannot be decomposed this way. This also explains why we can reverse reactions and change signs: we are just relabeling which endpoint is 'initial' and which is 'final.'