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Drivers of reactions review
Summary of Module 4: energy changes, calorimetry, energy profiles, bond energies, Hess's law, enthalpies of formation, entropy and Gibbs free energy, with key formulas and a mixed HSC-style test.
Every idea in this module answers one of two questions: how much energy does a reaction move, and will it go at all? The first is answered by enthalpy, measured with calorimeters or calculated with Hess's law. The second needs entropy as well, combined in Gibbs free energy. Use this page to pull them together before a test.
Module summary
| Topic | The big idea | Key tool |
|---|---|---|
| 4A Exo and endo | Heat flows out of (ΔH −) or into (ΔH +) the system | ΔH = H(products) − H(reactants) |
| 4B Calorimetry | Measure heat from the temperature change of a known mass of solution | q = mcΔT; ΔH = −q/n; extrapolate |
| 4C Combustion | ΔcH: 1 mol burns completely; bomb calorimeters are calibrated | C = q/ΔT; q = CΔT; kJ g⁻¹ = |ΔcH|/M |
| 4D Energy profiles | Eₐ is the barrier; catalysts give a lower-Eₐ pathway, same ΔH | ΔH = Eₐ(fwd) − Eₐ(rev) |
| 4E Bond energies | Breaking absorbs, forming releases | ΔH ≈ Σ broken − Σ formed |
| 4F Hess's law | ΔH is independent of route | Reverse: change sign; multiply: multiply ΔH; add |
| 4G Formation | Elements in standard states are zero | ΔH° = ΣnΔfH°(products) − ΣnΔfH°(reactants) |
| 4H Entropy | Systems tend to more dispersed arrangements | ΔS° = ΣnS°(products) − ΣnS°(reactants) |
| 4I Gibbs | Spontaneous when ΔG is negative | ΔG° = ΔH° − TΔS°; T = ΔH/ΔS |
Key formulas and units
- $q = mc\Delta T$: q in J, m in g, c = 4.18 J g⁻¹ K⁻¹ for water and dilute solutions.
- $\Delta H = -\dfrac{q}{n}$: convert q to kJ first; temperature rise means ΔH negative.
- $\Delta H \approx \sum E(\text{broken}) - \sum E(\text{formed})$: gases only.
- $\Delta H^\circ = \sum n\Delta_f H^\circ(\text{products}) - \sum n\Delta_f H^\circ(\text{reactants})$
- $\Delta S^\circ = \sum nS^\circ(\text{products}) - \sum nS^\circ(\text{reactants})$: units J K⁻¹ mol⁻¹.
- $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$: T in K; ΔS in kJ K⁻¹ mol⁻¹. Crossover $T = \Delta H/\Delta S$.
Putting it together
Here is one reaction followed through the whole module: ethanol burning, C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g).
- Measured (4C): a bomb calorimeter gives ΔcH = −1367 kJ mol⁻¹ (to liquid water); a spirit burner gives about half that because of heat loss.
- Profile (4D): highly exothermic, but it needs a flame to supply Eₐ before it sustains itself.
- Formation data (4G): ΔH° = [2(−393.5) + 3(−241.8)] − (−277.6) = −1234.8 kJ mol⁻¹ with steam as product. The 132 kJ difference from −1367 is the heat released by condensing 3 mol of steam.
- Entropy (4H): 3 mol gas → 5 mol gas, so ΔS° is positive: [2(213.7) + 3(188.8)] − [160.7 + 3(205.1)] = +217.8 J K⁻¹ mol⁻¹.
- Gibbs (4I): ΔH negative and ΔS positive, so ΔG is negative at every temperature: at 298 K, ΔG° = −1234.8 − 298(0.2178) = −1299.7 kJ mol⁻¹. It is spontaneous, yet ethanol sits safely in a bottle because Eₐ is high.
Test yourself
0.40 g of propane (M = 44.09 g mol⁻¹) heats 150 g of water by 12.4 °C. What is the experimental ΔcH?
q = 150 × 4.18 × 12.4 = 7775 J = 7.775 kJ; n = 0.40 ÷ 44.09 = 0.00907 mol; ΔcH = −7.775 ÷ 0.00907 = −857 kJ mol⁻¹ (much less than the data value because of heat loss).
Which statement about a catalyst is correct?
A catalyst changes the rate only; ΔH, ΔS and ΔG are unchanged.
Given 2NO(g) + O₂(g) → 2NO₂(g), ΔH = −114.2 kJ, and N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ, ΔfH° of NO₂(g) is
Adding gives N₂ + 2O₂ → 2NO₂, ΔH = +66.4 kJ; per mole of NO₂ that is +33.2 kJ mol⁻¹.
A reaction has ΔH° = −92 kJ mol⁻¹ and ΔS° = −199 J K⁻¹ mol⁻¹. At 600 K the reaction is
ΔG = −92 − 600(−0.199) = −92 + 119.4 = +27 kJ mol⁻¹. The crossover is 462 K.
Which change has a negative ΔH and a positive ΔS?
Decomposing hydrogen peroxide is exothermic and makes a gas from liquids.
Bond-energy estimates of ΔH are least reliable when
Bond energies ignore intermolecular forces in liquids, and a small error in large sums gives a large percentage error in a small ΔH.
Label the energy profile with and without a catalyst.
Which quantity does each statement describe?
An exothermic reaction has a ΔH. The enthalpy change of a reaction is independent of the route, which is law. A catalyst lowers the but does not change ΔH. A reaction is spontaneous when ΔG is negative; for a reaction with positive ΔH and positive ΔS this happens at temperatures. Entropy is a measure of the of particles and energy.
Hydrogen is promoted as a clean fuel. It burns according to 2H₂(g) + O₂(g) → 2H₂O(l).
Data: ΔfH°(H₂O(l)) = −285.8 kJ mol⁻¹. S° (J K⁻¹ mol⁻¹): H₂ 130.7, O₂ 205.1, H₂O(l) 69.9.
(a) Calculate ΔH° and ΔS° for the reaction as written. (3 marks)
(b) Calculate ΔG° at 298 K and use it to explain why hydrogen and oxygen can be stored mixed in a balloon without reacting, yet explode when a spark is applied. (4 marks)
(a) ΔH° = 2(−285.8) − [2(0) + 0] = −571.6 kJ mol⁻¹. ΔS° = 2(69.9) − [2(130.7) + 205.1] = 139.8 − 466.5 = −326.7 J K⁻¹ mol⁻¹.
(b) ΔG° = −571.6 − 298 × (−0.3267) = −571.6 + 97.4 = −474.2 kJ mol⁻¹. Entropy decreases (3 mol of gas become a liquid), which opposes the reaction, but the very large negative ΔH outweighs the TΔS term at 298 K, so ΔG is negative and the reaction is spontaneous. Spontaneous only means the reaction is thermodynamically favoured; it says nothing about its rate. The H–H and O=O bonds must be broken first, so the activation energy is high and at room temperature almost no collisions have enough energy, so the mixture is stable. A spark provides the activation energy for a few molecules; the heat they release then supplies Eₐ to their neighbours, so the reaction spreads explosively.