HSC Exam Practice

Sample response. Activation energy (E_a) is the minimum energy that colliding reactant molecules must possess for a reaction to occur and products to form. On an energy profile diagram, E_a is represented as the vertical distance from the reactant energy level to the peak of the curve (the transition state), not from the x-axis (zero).

Marking notes. 1 mark for the definition (minimum energy for successful collision / to form transition state). 1 mark for correctly locating E_a as the height from the reactant level to the peak (must specify "from the reactant level" — from the x-axis scores 0 for that component).

Sample response. Activation energy (E_a) is a kinetic quantity — it is the energy barrier that determines the rate of a reaction: how quickly reactants can reach the transition state and form products. Enthalpy change (ΔH) is a thermodynamic quantity — it is the difference between the energy of the products and the energy of the reactants, representing the total energy released or absorbed by the reaction overall. A catalyst lowers E_a by providing an alternative pathway with a lower-energy transition state. It does not change ΔH because the reactants and products (and their energy levels) are identical with or without the catalyst.

Marking notes. 1 mark for distinguishing E_a (kinetic, determines rate / height of barrier) from ΔH (thermodynamic, difference between reactant and product energy). 1 mark for stating the catalyst changes E_a. 1 mark for stating the catalyst does NOT change ΔH, with a reason (reactants and products unchanged / ΔH depends only on start and end energy levels).

Sample response. The iron catalyst is a solid (s), while the reactants N₂ and H₂ are both gases (g). Because the catalyst is in a different physical state from the reactants (solid ≠ gas), Fe is a heterogeneous catalyst.

Marking notes. 1 mark for identifying the physical states of catalyst (solid) and reactants (gas) and noting they differ. 1 mark for the correct classification: heterogeneous. "Heterogeneous" without justification referencing physical states = 1/2 marks.

(a) E_a(reverse, uncatalysed) = E_a(forward) − ΔH = 165 − (−48) = 165 + 48 = 213 kJ mol⁻¹. [1 mark formula/setup, 1 mark correct answer with units = 2 marks]

(b) E_a(forward, catalysed) = 165 − 70 = 95 kJ mol⁻¹ [1 mark]. E_a(reverse, catalysed) = E_a(forward, cat) − ΔH = 95 − (−48) = 95 + 48 = 143 kJ mol⁻¹ [1 mark].

(c) ΔH(catalysed) = −48 kJ mol⁻¹ (unchanged). A catalyst provides an alternative pathway but does not alter the identity or energy of the reactants or products; ΔH depends only on these start and end energy levels. [1 mark for −48 with correct justification]

Sample response. MnO₂(s) is a heterogeneous catalyst (solid catalyst; aqueous H₂O₂ reactant — different phases). It provides an alternative decomposition mechanism with a lower activation energy (E_a drops from 75 kJ mol⁻¹ uncatalysed to 29 kJ mol⁻¹ catalysed). [1 mark] At a given temperature, a greater proportion of H₂O₂ molecules now possess kinetic energy at or above this lower E_a threshold. [1 mark] Therefore more collisions are successful per unit time, increasing the rate of decomposition to H₂O and O₂. The MnO₂ is regenerated and not consumed overall. [1 mark]

Sample response. A homogeneous catalyst is in the same physical state (phase) as the reactants; a heterogeneous catalyst is in a different physical state. [1 mark] Example of homogeneous: H⁺(aq) in acid-catalysed esterification, where both catalyst and reactants are aqueous. [1 mark] Example of heterogeneous: Fe(s) in the Haber process (Incitec Pivot), where the solid iron catalyst contacts gaseous N₂ and H₂; or Pt/Pd/Rh(s) in catalytic converters with gaseous exhaust reactants. [1 mark]

(a) (i) E_a(uncat) = 272 kJ mol⁻¹. (ii) E_a(cat) = 80 kJ mol⁻¹. (iii) ΔH = −197 kJ mol⁻¹. (iv) Reduction = 272 − 80 = 192 kJ mol⁻¹. [2 marks: 1 for E_a values, 1 for ΔH and reduction]

(b) Without V₂O₅, E_a = 272 kJ mol⁻¹. At ~450°C, very few SO₂ and O₂ molecules possess kinetic energy ≥ 272 kJ mol⁻¹, so the uncatalysed rate is extremely slow. [1 mark] V₂O₅ provides an alternative mechanism (oxidation/reduction cycle of vanadium surface) in which the transition state energy is only 80 kJ mol⁻¹ above the reactant level. [1 mark] At 450°C, a much larger fraction of SO₂ and O₂ molecules now meet this lower threshold, dramatically increasing the number of successful collisions per second and hence the rate of SO₃ formation. [1 mark for linking correctly to Boltzmann distribution / proportion of molecules with sufficient energy]

(c) V₂O₅ is a solid; SO₂ and O₂ are gases. Solid ≠ gas → V₂O₅ is a heterogeneous catalyst. [1 mark state + 1 mark justification = 2 marks]

(a) ΔH = +62 kJ mol⁻¹ (positive) → endothermic: the products are at higher energy than the reactants; the system absorbs heat from the surroundings. [1 mark sign interpretation] E_a(reverse, uncat) = 248 − 62 = 186 kJ mol⁻¹. [1 mark formula, 1 mark answer = 2 marks for (a) total; sign: endothermic, so products are higher than reactants, and the reverse reaction starts from a higher level, climbing less far to the same peak — hence E_a(reverse) < E_a(forward).]

(b) 40% reduction: E_a(forward, cat) = 248 × (1 − 0.40) = 248 × 0.60 = 148.8 kJ mol⁻¹ (accept 149 kJ mol⁻¹). [1 mark] E_a(reverse, cat) = 148.8 − 62 = 86.8 kJ mol⁻¹ (accept 87 kJ mol⁻¹). [1 mark] Both marks require working shown.

(c) ΔH(catalysed) = +62 kJ mol⁻¹ (unchanged). [1 mark] This means the reaction system absorbs 62 kJ of thermal energy from the surroundings per mole of product formed. The reactants and products are identical with or without the catalyst, so the energy stored in the product bonds relative to the reactant bonds is unchanged — the reaction remains endothermic regardless of the catalyst's effect on the rate.

(d) Acceptable assumptions: the catalyst lowers E_a of the reverse reaction by the same 40% (i.e. the catalyst lowers the peak proportionally for both forward and reverse reactions from the respective starting levels); OR the reduction applies to the activation energy measured from the reactant level (not from an arbitrary reference). [1 mark for any valid, clearly stated assumption about the catalyst's uniform effect on the energy diagram]

Sample response. Catalysts are central to modern industrial chemistry because they increase reaction rates at lower operating temperatures and pressures, reducing energy costs and infrastructure demands. Their role is kinetic — they lower activation energy (E_a) by providing an alternative mechanism — not thermodynamic: they do not change ΔH, the identity of reactants or products, and are not consumed overall. Two key examples illustrate both the power and the limitations of industrial catalysts. First, the Haber process at Incitec Pivot (Brisbane) uses Fe(s) as a heterogeneous catalyst (different phase from gaseous N₂ and H₂ reactants). The Fe surface reduces E_a from ~335 to ~150 kJ mol⁻¹, enabling commercial ammonia production at ~450°C and ~200 atm rather than the prohibitively high temperatures that would be needed uncatalysed. However, the catalyst does not change ΔH (−92 kJ mol⁻¹) or shift the equilibrium position. Crucially, a catalyst cannot increase the equilibrium yield of NH₃ — it only increases the rate at which equilibrium is approached. To maximise yield, engineers must manipulate temperature and pressure (Le Chatelier), not the catalyst. Second, the Contact process uses V₂O₅(s) — again heterogeneous — to reduce E_a for SO₃ synthesis from ~272 to ~80 kJ mol⁻¹ at ~450°C. The same limitation applies: ΔH = −197 kJ mol⁻¹ is unchanged; the catalyst cannot drive more SO₃ formation at equilibrium. The key distinction is therefore: catalysts resolve kinetic limitations (too slow to be commercially viable at safe temperatures) but cannot overcome thermodynamic limitations (unfavourable equilibrium position). In industrial practice, both must be addressed together — the catalyst accelerates the approach to equilibrium, while conditions (T, P, concentration) are adjusted to set the equilibrium position where it is needed.

Marking notes. 1 mark — correctly distinguishes kinetic (E_a, rate) from thermodynamic (ΔH, equilibrium position, yield) effects of a catalyst. 1 mark — correctly names and classifies the catalyst in example 1 (Haber / Fe(s), heterogeneous, solid ≠ gas) with justification. 1 mark — correctly names and classifies the catalyst in example 2 (Contact / V₂O₅(s), heterogeneous; or catalytic converter Pt/Pd/Rh, heterogeneous) with justification. 1 mark — states that the catalyst lowers E_a and explains the mechanism of rate increase (lower E_a → greater proportion of molecules with sufficient energy → more successful collisions). 1 mark — correctly states that ΔH is unchanged by the catalyst in both examples. 1 mark — identifies and explains the limitation: a catalyst does not change the equilibrium position or yield; it only changes the rate at which equilibrium is achieved (does not alter ΔG or K_eq). 1 mark — reaches an explicit evaluative statement: catalysts are essential for kinetic viability (commercially viable rate) but cannot substitute for thermodynamic optimisation (yield requires manipulation of T, P, concentration).