HSC Exam Practice

Sample response. Activation energy (E_a) is the minimum kinetic energy that colliding reactant particles must have for the collision to result in a chemical reaction. It is measured from the reactant energy level to the peak of the energy profile diagram (the transition state).

Marking notes. 1 mark for "minimum kinetic energy for a collision to result in a chemical reaction" or equivalent. 1 mark for locating E_a as the energy from the reactant level to the transition state peak (not from zero or from the product level).

Sample response. (1) The colliding particles must have kinetic energy equal to or greater than the activation energy (KE ≥ E_a). (2) The particles must collide with the correct orientation so that the reactive sites on each molecule face each other.

Marking notes. 1 mark per condition. Must state both conditions. A response mentioning only energy scores 1/2.

Sample response. On an energy profile diagram for an exothermic reaction, E_a is the energy difference between the reactant energy level and the transition state peak at the top of the curve — it represents the energy barrier particles must overcome for the reaction to proceed [1]. ΔH is the energy difference between the reactant energy level and the product energy level — it represents the net energy change of the reaction (negative for exothermic, so products are lower than reactants) [1]. E_a is always positive (always a barrier) and is typically much larger than the magnitude of ΔH; ΔH can be negative (exothermic) or positive (endothermic) [1].

Marking notes. 1 mark for correctly defining E_a as reactant level to peak. 1 mark for correctly defining ΔH as reactant level to product level. 1 mark for a correct distinguishing statement (e.g. E_a controls rate / is always positive; ΔH controls energy change / can be negative or positive; they are measured between different points on the diagram).

Sample response. The rate of the reaction between CaCO₃ and HCl is fastest at the start and decreases progressively over time until the reaction stops [1]. This occurs because HCl is consumed as the reaction proceeds, so the concentration of H⁺ ions in solution decreases [1]. A lower concentration of H⁺ means fewer H⁺ ions per unit volume colliding with the CaCO₃ surface per second (lower collision frequency); since temperature and E_a are unchanged, the proportion of effective collisions is unchanged, but fewer total collisions means fewer effective collisions per second and a lower reaction rate [1].

Marking notes. 1 mark for correctly outlining the trend (fastest initially, decreases over time). 1 mark for linking the decrease to falling reactant (HCl) concentration. 1 mark for collision theory explanation linking lower concentration to lower collision frequency and thus lower rate. Penalise answers that state the rate is constant or that E_a changes.

Sample response. The combustion of methane (CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890 kJ mol⁻¹) does not occur spontaneously at room temperature despite being thermodynamically favourable because the activation energy is approximately 400 kJ mol⁻¹ [1]. At room temperature, the average kinetic energy of CH₄ and O₂ molecules is far below 400 kJ mol⁻¹, so virtually no collision between them has sufficient energy to reach the transition state [1]. The reaction is thermodynamically favourable (ΔH is very negative) but kinetically controlled — a high E_a prevents the reaction from proceeding without an ignition source (such as a spark) that locally raises the kinetic energy of a small fraction of molecules above E_a [1].

Marking notes. 1 mark for identifying the high E_a as the reason (do not accept ΔH). 1 mark for linking high E_a to insufficient kinetic energy of CH₄/O₂ at room temperature. 1 mark for correctly distinguishing thermodynamic favourability (ΔH) from kinetic feasibility (E_a).

Sample response. (i) When temperature increases, the Maxwell–Boltzmann distribution shifts to higher kinetic energies. The area under the curve to the right of the fixed E_a line is now larger, so a greater proportion of particles have KE ≥ E_a. This means more effective collisions occur per second and the reaction rate increases [2 marks: 1 for "proportion increases", 1 for linking to shift in the MB distribution]. (ii) Collision frequency also increases with temperature — more collisions occur per second. Both effects contribute to the higher reaction rate [1 mark]. Only temperature changes the proportion of particles that exceed E_a (by shifting the MB distribution); increasing concentration increases collision frequency but does NOT change the proportion of collisions that are effective, and a catalyst lowers E_a rather than shifting the distribution [1 mark for correctly distinguishing temperature from catalyst and concentration].

Marking notes. 1 mark for stating that the proportion of particles exceeding E_a increases with temperature. 1 mark for linking this to the shift in the Maxwell–Boltzmann distribution. 1 mark for stating that collision frequency also increases with temperature. 1 mark for correctly explaining that only temperature (not catalyst or concentration) changes the proportion of particles exceeding E_a: concentration changes collision frequency only; a catalyst lowers E_a (moves the threshold) rather than shifting the distribution.

Sample response (a). From the graph, Trial B collects approximately 15–17 mL of H₂ in the first 20 seconds. Rate = ΔV ÷ Δt ≈ 16 mL ÷ 20 s = 0.80 mL s⁻¹. Accept any answer in the range 0.70–0.90 mL s⁻¹. [2 marks: 1 for reading the graph correctly, 1 for the calculation]

Sample response (b). All three trials use the same mass of zinc (excess Zn) and the same volume of H₂SO₄. The total amount of H₂ produced depends on the total number of moles of reactants that react (stoichiometry: 1 mol Zn yields 1 mol H₂), not on the rate at which they react. Rate affects only how quickly the reaction reaches completion; it does not determine the total quantity of product. All three trials consume the same total amount of Zn and H₂SO₄, so all produce the same total volume of H₂. [2 marks: 1 for identifying same mass of Zn / same moles of limiting reagent, 1 for correctly distinguishing rate from total yield]

Sample response (c). Trial C (2.0 mol L⁻¹) has the steepest initial slope; Trial A (0.5 mol L⁻¹) has the shallowest. A higher concentration of H₂SO₄ means more H⁺ ions per unit volume in solution. This increases the frequency with which H⁺ ions collide with the Zn surface (collision frequency). Since temperature and E_a are identical across all three trials, the proportion of each collision that is effective (has sufficient energy and correct orientation) is unchanged. However, more collisions per second means proportionally more effective collisions per second, so the initial rate is higher for higher concentrations. [3 marks: 1 for correctly identifying collision frequency as the mechanism, 1 for explaining why higher concentration increases collision frequency (more H⁺ per unit volume), 1 for noting that the proportion of effective collisions is unchanged since T and E_a are constant]

Sample response. The statement is partially correct but significantly flawed in its claim that increasing temperature and adding a catalyst work by the same mechanism. Both do increase reaction rate, but they do so through fundamentally different mechanisms, and only one of them alters the energy profile diagram.

A catalyst provides an alternative reaction pathway with a lower activation energy (E_a). For example, in the Haber process (N₂ + 3H₂ ⇌ 2NH₃), an iron catalyst reduces E_a from approximately 335 kJ mol⁻¹ to approximately 163 kJ mol⁻¹. On the energy profile diagram, this is shown as a lower transition state peak, while the reactant and product energy levels remain unchanged — so ΔH is unaffected by the catalyst. On the Maxwell–Boltzmann distribution, the curve itself does not change (temperature is the same), but because E_a is now lower, the fraction of particles to the right of the (shifted) E_a threshold is much larger. More particles can now undergo effective collisions, so the rate increases.

Increasing temperature, by contrast, does NOT lower E_a. The energy profile diagram is unchanged — the transition state peak remains at the same height. What temperature does is change the Maxwell–Boltzmann distribution itself: the curve shifts to higher energies and broadens, so a greater proportion of the existing particles now have KE ≥ the fixed E_a. Additionally, particles move faster at higher temperatures, increasing collision frequency. Both effects increase effective collisions per second, raising the rate.

Therefore the two mechanisms differ fundamentally: a catalyst works by changing the energy profile (lowering E_a) without affecting the distribution of particle energies; increasing temperature works by changing the distribution of particle energies (shifting the Maxwell–Boltzmann curve) without affecting the energy profile diagram. They also differ in their effect on ΔH: a catalyst leaves ΔH unchanged; temperature does not change ΔH either, but the statement incorrectly implies the mechanisms are identical. The statement is correct that both increase rate, but incorrect that they do so "by the same mechanism" — one targets the threshold (E_a), the other targets the distribution of particle energies.

Marking criteria:

• 1 mark — States an overall evaluative judgement (e.g. "partially correct but flawed in claiming the same mechanism").
• 1 mark — Correctly describes the mechanism of a catalyst: lowers E_a by providing an alternative reaction pathway.
• 1 mark — Correctly states that a catalyst does NOT change the Maxwell–Boltzmann distribution, ΔH, or the reactant/product energy levels.
• 1 mark — Correctly describes the mechanism of temperature increase: shifts the Maxwell–Boltzmann distribution to higher energies, increasing the proportion of particles exceeding the fixed E_a.
• 1 mark — Correctly states that increasing temperature does NOT lower E_a (the energy profile peak is unchanged).
• 1 mark — Uses a specific named example (e.g. Haber process with iron catalyst, or glow sticks, or CaCO₃ + HCl) with correct values or context.
• 1 mark — Correctly refers to the Maxwell–Boltzmann distribution in the context of at least one of the two factors.
• 1 mark — Reaches an explicit conclusion that summarises the key mechanistic difference: catalyst lowers E_a (changes the threshold); temperature shifts the energy distribution (changes the proportion of particles reaching the unchanged threshold). Neither changes ΔH.