HSC Exam Practice

Sample response. Hess’s Law states that the total enthalpy change for a chemical reaction is the same regardless of the pathway taken between the initial and final states; ΔH depends only on the identity of reactants and products, not on the intermediate steps.

Marking notes. 1 mark — total ΔH is the same / independent of pathway. 1 mark — depends only on initial and final states (or: enthalpy is a state function). Both required.

Sample response. A state function depends only on the current state of the system — specifically the initial and final conditions — not on the history or route by which those conditions were reached. Enthalpy (H) is a state function because its change ΔH reflects only the difference in bond energies between products and reactants, regardless of how many steps the reaction takes. Hess’s Law is a direct consequence: because ΔH is path-independent, the total ΔH for a reaction is the same whether it occurs in one step or many. This allows chemists to add known ΔH values algebraically to calculate ΔH for an unknown reaction.

Marking notes. 1 mark — correct definition of state function (depends only on initial/final state, not path). 1 mark — enthalpy is a state function because it reflects only the initial and final energy levels (not the route). 1 mark — how this underpins Hess’s Law (path independence allows algebraic addition of known ΔH values to get the unknown ΔH).

Sample response. Manipulation 1 — Reversing the equation: swap reactants and products. Effect: multiply ΔH by −1 (flip the sign). Manipulation 2 — Scaling by a factor n: multiply all coefficients by n. Effect: multiply ΔH by the same factor n. Both operations must be applied to the equation and to ΔH simultaneously — they cannot be separated.

Marking notes. 1 mark per manipulation named. 1 mark per effect on ΔH correctly stated. (4 marks total.)

Sample response. The direct reaction C(s) + ½O₂(g) → CO(g) cannot be performed cleanly in a laboratory calorimeter because burning carbon in a limited oxygen supply always produces a mixture of CO and CO₂ — it is not possible to isolate pure CO as the sole product. Hess’s Law provides an indirect route: two measurable combustion enthalpies (C + O₂ → CO₂, ΔH = −393.5, and CO + ½O₂ → CO₂, ΔH = −283.0) can be combined algebraically. Reversing the second equation and adding it to the first cancels CO₂, leaving C + ½O₂ → CO, with ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹.

Marking notes. 1 mark — burning C in limited O₂ gives a mixture of CO and CO₂; pure CO cannot be isolated. 1 mark — states the indirect principle (algebraic combination of measurable equations). 1 mark — names the two equations used or gives the result ΔH = −110.5 kJ mol⁻¹.

Sample response. An intermediate species appears in the sub-steps of a Hess’s Law calculation but cancels out when the equations are added — it does not appear in the overall target equation. A target species is a reactant or product that appears in the target equation (the overall equation whose ΔH is being determined) and does not cancel; it remains after all intermediates are eliminated.

Marking notes. 1 mark — intermediate: appears in sub-steps but cancels in the combined equation / not in the target. 1 mark — target species: appears in the target equation and does not cancel.

(a) Arrow Q represents the reversed combustion of methane: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g), ΔH = +890.3 kJ mol⁻¹. The sign is positive because this is the reverse of an exothermic combustion reaction — reversing an equation requires the sign of ΔH to flip (from −890.3 to +890.3). On the energy level diagram, the positive sign indicates the arrow goes upward (toward higher enthalpy), representing absorption of energy. [2 marks: 1 for identifying the reaction as the reverse of CH₄ combustion; 1 for explaining why the sign is positive.]

(b) By Hess’s Law, ΔH(R) = ΔH(P) + ΔH(Q). The pathway P + Q takes reactants from the top level to CO₂/H₂O (via P, ΔH = −965.1) then back up to CH₄/O₂ (via Q, ΔH = +890.3). The direct Route R must have the same ΔH total. ΔH(R) = −965.1 + 890.3 = −74.8 kJ mol⁻¹. [3 marks: 1 for writing the Hess’s Law relationship; 1 for correct arithmetic; 1 for correct sign and unit.]

(c) Arrow R represents the reaction C(s) + 2H₂(g) → CH₄(g): the formation of one mole of CH₄ from its constituent elements in their standard states. The standard enthalpy of formation (ΔHₑf) is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions, which is exactly what Arrow R represents. [2 marks: 1 for identifying the reaction is formation from elements in standard states; 1 for stating the definition of ΔHₑf matches this description.]

(a) Add the two given equations: CO(g) is a product in the first and a reactant in the second — it cancels. ½O₂ + ½O₂ = O₂ (reactant). Result: C(s) + O₂(g) → CO₂(g). ΔH = −110.5 + (−283.0) = −393.5 kJ mol⁻¹. [2 marks: 1 for showing CO cancellation; 1 for correct ΔH.]

(b) n(C) = 800,000 g ÷ 12.01 g mol⁻¹ = 66,611 mol. Q = 66,611 × 393.5 kJ = 26,211,428 kJ ÷ 1000 = ≈ 26,200 MJ (accept 26,000–26,400 MJ). [3 marks: 1 for n(C) correct; 1 for Q = n × |ΔH|; 1 for converting to MJ with correct magnitude.]

(c) 80% reaches CO₂: Q₁ = 0.80 × 66,611 × 393.5 = 20,968 MJ. 20% stops at CO: Q₂ = 0.20 × 66,611 × 110.5 = 1,470 MJ. Total actual = 20,968 + 1,470 = ≈ 22,440 MJ. Assumption: all reactions go to completion within their respective pathways (no other products formed). [2 marks: 1 for correct splitting and calculation; 1 for a valid stated assumption.]

Sample response. Hess’s Law, published by Swiss-Russian chemist Germain Hess in 1840, is foundational to thermochemistry because it allows chemists to calculate enthalpy changes for reactions that cannot be measured directly. Its significance lies in two related facts: first, it establishes that enthalpy is a state function — ΔH depends only on the identities of reactants and products, not on the pathway taken between them. This is a consequence of the First Law of Thermodynamics (conservation of energy). Second, because ΔH is path-independent, chemists can algebraically combine any set of measurable thermochemical equations that add up to a target reaction, and the sum of their ΔH values gives the ΔH for the unmeasurable reaction. Without this law, an enormous number of important ΔH values — including virtually all standard enthalpies of formation — could not be determined. The clearest example is the standard enthalpy of formation of carbon monoxide, CO(g). In the context of Australian steelmaking at Port Kembla, blast furnace engineers need to know ΔH for C(s) + ½O₂(g) → CO(g). This cannot be measured directly in a calorimeter because burning carbon in limited oxygen always produces a mixture of CO and CO₂. Hess’s Law solves the problem: the measurable ΔH values for C + O₂ → CO₂ (−393.5 kJ mol⁻¹) and CO + ½O₂ → CO₂ (−283.0 kJ mol⁻¹) are combined — the second is reversed (sign changes to +283.0) and added to the first. CO₂ cancels as an intermediate, giving C + ½O₂ → CO, ΔH = −110.5 kJ mol⁻¹ — a value impossible to measure but essential for designing furnace energy budgets. The same principle extends to atmospheric chemistry: the ΔH for SO₂ → SO₃ (a step in acid rain formation in NSW industrial airsheds) is calculated indirectly via Hess’s Law from combustion data for S → SO₂ and S → SO₃. In summary, Hess’s Law transforms thermochemistry from a science limited to directly measurable reactions into one that can quantify the energy changes of any reaction — as long as a set of algebraically combinable measurable reactions can be found. Its significance is not merely historical but remains central to industrial process design and environmental chemistry today.

Marking notes.

• 1 mark — defines Hess’s Law correctly (ΔH is pathway-independent; depends on initial and final states).
• 1 mark — links to enthalpy being a state function and/or First Law of Thermodynamics.
• 1 mark — explains why some ΔH values require Hess’s Law (direct measurement impossible due to mixed products, impractical reactions, etc.).
• 1 mark — describes the mechanism: manipulate and add equations; cancel intermediates; sum ΔH values.
• 1 mark — names and correctly applies a specific Australian industrial or environmental example (Port Kembla CO/CO₂; SO₂/SO₃ acid rain; acceptable alternatives include natural gas pipeline calculations).
• 1 mark — shows the example with specific ΔH values or at minimum correctly identifies the two equations that combine and what cancels.
• 1 mark — reaches an explicit evaluative judgement about the significance of Hess’s Law (not merely descriptive; must state why it is important / what would not be possible without it).