HSC Exam Practice

Sample response. Enthalpy change (ΔH) is the heat energy exchanged between a chemical system and its surroundings at constant pressure. It is calculated as ΔH = H(products) − H(reactants).

Marking notes. 1 mark for definition referencing heat energy exchanged at constant pressure (or heat of reaction); 1 mark for the formula ΔH = H(products) − H(reactants).

Sample response. In an exothermic reaction, ΔH is negative (< 0) and energy flows from the system to the surroundings — the surroundings warm up. In an endothermic reaction, ΔH is positive (> 0) and energy flows from the surroundings into the system — the surroundings cool down.

Marking notes. 1 mark — exothermic: ΔH < 0; 1 mark — endothermic: ΔH > 0; 1 mark — correct direction of energy flow for either type using system/surroundings language. Must not say “the reaction heats up/cools down” without specifying surroundings.

Sample response. (1) Reactants — energy level on the left. (2) Products — energy level on the right. (3) Transition state (activated complex) — peak of the curve. (4) Activation energy (E_a) — arrow from reactant level up to peak. (5) Enthalpy change (ΔH) — arrow from reactant level to product level.

Marking notes. 1 mark per correct feature (max 3 if listing all 5); award 3 marks for all 5 correctly identified. Deduct if “energy” used ambiguously instead of “enthalpy”.

Sample response. The physical state of a substance affects its enthalpy. For water, the liquid state (l) has lower enthalpy than the gaseous state (g) because the change from gas to liquid releases the latent heat of condensation (approx. 44 kJ mol⁻¹ per mole of water). If state symbols are omitted, the ΔH value is ambiguous — it could correspond to either product state — and marks would be lost in an HSC response.

Marking notes. 1 mark — state affects enthalpy (physical state is thermodynamically significant); 1 mark — specific water example (l) vs (g) difference due to latent heat/condensation energy; 1 mark — explains that omitting state symbols makes ΔH ambiguous.

Sample response (a). For 0.500 mol propane: Energy = (0.500/1) × 2220 = 1110 kJ mol⁻¹ released. Thermochemical equation: 0.5 C₃H₈(g) + 2.5 O₂(g) → 1.5 CO₂(g) + 2 H₂O(l)   ΔH = −1110 kJ mol⁻¹.

Sample response (b). Reverse: 3CO₂(g) + 4H₂O(l) → C₃H₈(g) + 5O₂(g)   ΔH = +2220 kJ mol⁻¹.

Marking notes. Part (a): 1 mark for correct ΔH (−1110 kJ mol⁻¹); 1 mark for equation with correct scaling (half coefficients or fraction 0.5). Part (b): 1 mark for correct reversed equation with state symbols; 1 mark for ΔH = +2220 kJ mol⁻¹ (sign flipped). Accept full equation written with all species.

Sample response (a). Endothermic.

Sample response (b). ΔH is positive (> 0) [1]. Energy flows from the surroundings (including the injured ankle) into the system (the dissolving NH₄NO₃), which is why the pack and the area of contact feel cold [1].

Marking notes. 1 mark — endothermic (a); 1 mark — ΔH > 0 (positive sign stated); 1 mark — energy flows from surroundings into system (not “the reaction gets cold”).

Sample response (a). Endothermic [1]. The products sit at a higher enthalpy level (+65 kJ mol⁻¹) than the reactants (0 kJ mol⁻¹), so H(products) > H(reactants), meaning ΔH > 0 [1].

Sample response (b). E_a = height from reactant level to peak = 180 − 0 = 180 kJ mol⁻¹ [1]. ΔH = height from reactants to products = 65 − 0 = +65 kJ mol⁻¹ [1].

Sample response (c). E_a(rev) = E_a(fwd) − ΔH = 180 − 65 = 115 kJ mol⁻¹ [1 for formula; 1 for correct answer]. Reasoning: starting from the products (65 kJ mol⁻¹ above reactants), the energy gap to reach the peak (180 kJ mol⁻¹) is 180 − 65 = 115 kJ mol⁻¹.

Marking notes. Part (a) — 1 mark classification; 1 mark diagram justification (products higher than reactants). Part (b) — 1 mark each; must show values read from diagram. Part (c) — 1 mark for E_a(rev) = E_a(fwd) − ΔH formula or equivalent reasoning; 1 mark for 115 kJ mol⁻¹.

Sample response (a). Photosynthesis is the reverse of glucose combustion. Equation: 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)   ΔH = +2803 kJ mol⁻¹. Sign is reversed because the reaction direction is reversed.

Sample response (b). ΔH = +2803 kJ mol⁻¹ (positive) means photosynthesis is endothermic [1]. Energy must be absorbed from the surroundings — in this case, from sunlight — to drive the reaction and convert CO₂ and H₂O into glucose. The energy is stored as chemical potential energy in the bonds of glucose [1].

Sample response (c). One assumption is that the same pathway and standard conditions (25°C, 100 kPa) apply to photosynthesis as to combustion. In practice, photosynthesis occurs via a multi-step enzyme-catalysed pathway in chloroplasts under variable temperature and light conditions, so the actual enthalpy profile may differ. Accept also: assumes all water is in liquid state (l).

Marking notes. (a) 1 mark for correct reversed equation with state symbols; 1 mark for ΔH = +2803 kJ mol⁻¹. (b) 1 mark for identifying endothermic + energy absorbed from surroundings (sunlight); 1 mark for linking to energy stored in glucose bonds. (c) 1 mark for a valid assumption (standard conditions, same pathway, or state symbol assumption).

Sample response. An energy profile diagram is a graph of enthalpy (y-axis) versus reaction coordinate (x-axis) that simultaneously communicates both thermodynamic properties (where the energy ends up — the enthalpy change ΔH) and kinetic properties (how easily the reaction starts — the activation energy E_a). These are independent features that must be read separately from the diagram.

The thermodynamic property is ΔH: the vertical distance from the reactant level to the product level. If products sit lower than reactants, ΔH < 0 (exothermic); if products sit higher, ΔH > 0 (endothermic). An Australian example of an exothermic reaction is the combustion of methane at the Karratha LNG facility: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ mol⁻¹. On the diagram, the product level is 890 kJ mol⁻¹ below the reactant level, and the ΔH arrow points downward. An Australian example of an endothermic reaction is the dissolution of ammonium nitrate in instant cold packs used in Australian sports medicine, ΔH > 0; the product level sits above the reactant level and the ΔH arrow points upward.

The kinetic property is E_a: the vertical distance from the reactant level to the transition state peak. E_a represents the minimum energy that colliding particles must possess for the reaction to proceed. A larger E_a means fewer collisions are effective, so the reaction is slower. Crucially, E_a is measured from the reactant level to the peak — not from the x-axis — because the position of the reactant level on the y-axis is arbitrary; only differences between levels are meaningful.

When the reaction equation is reversed, the reactants and products swap positions on the diagram. ΔH flips sign (if combustion is −890, decomposition is +890 kJ mol⁻¹). The activation energy for the reverse reaction becomes E_a(rev) = E_a(fwd) − ΔH, measured from the product level up to the same peak.

When a catalyst is added, the energy profile changes only in the height of the peak — E_a decreases because the catalyst provides an alternative lower-energy pathway. Crucially, ΔH is unchanged: reactant and product levels remain at the same positions, so the overall energy released or absorbed is unaffected. A catalyst speeds the reaction without altering the thermodynamics.

Marking criteria (9 marks):

• 1 mark — Correctly identifies the two independent properties communicated: thermodynamic (ΔH) and kinetic (E_a).
• 1 mark — Defines ΔH from the diagram: vertical distance from reactant to product level; sign determined by relative positions.
• 1 mark — Defines E_a from the diagram: vertical distance from reactant level to transition state peak (not from x-axis).
• 1 mark — Correctly describes exothermic diagram features and names an Australian exothermic example (e.g. LNG/methane combustion at Karratha, hand warmer).
• 1 mark — Correctly describes endothermic diagram features and names an Australian endothermic example (e.g. cold pack/NH₄NO₃ dissolution, photosynthesis).
• 1 mark — Correctly explains what happens when the equation is reversed: ΔH flips sign; E_a(rev) = E_a(fwd) − ΔH; reactant/product levels swap.
• 1 mark — Correctly explains what a catalyst does: lowers E_a only; ΔH is unchanged because reactant and product energy levels are unaffected.
• 1 mark — Uses precise thermochemical language throughout: enthalpy change, activation energy, transition state, exothermic/endothermic with correct sign conventions, system/surroundings distinction.
• 1 mark — Produces a coherent, structured response that explicitly links thermodynamic and kinetic information as distinct but simultaneously readable from the same diagram.