CHEM1011 · Chemistry 1a
Thermochemistry
Week 8 develops energy bookkeeping: the first law relating heat q, work w and internal energy ΔU, calorimetry, enthalpy ΔH from standard formation data and Hess's law, entropy ΔS as energy dispersal, and their combination into Gibbs energy ΔG to predict spontaneity. The ΔG = ΔH − TΔS spontaneity analysis and the ΔG° = −RT ln K link (Mastery) are recurring final-exam calculations.
What this chapter covers
- 01System / surroundings / universe; isolated, closed and open systems; state functions vs path functions
- 02First law ΔU = q + w; sign convention (q > 0 heat in, w > 0 work on system); pressure–volume work w = −pΔV
- 03Calorimetry: q = m·c·ΔT or n·C·ΔT; heat released by reaction from the calorimeter temperature change
- 04Enthalpy H = U + pV; at constant pressure ΔH = q_p; standard enthalpy of formation ΔfH° (elements = 0)
- 05Reaction enthalpy ΔrH° = Σn·ΔfH°(products) − Σn·ΔfH°(reactants); Hess's law and Born–Haber cycles (Mastery)
- 06Entropy S as energy dispersal; ΔrS° from standard molar entropies; the second and third laws
- 07Gibbs energy ΔG = ΔH − TΔS; ΔG < 0 spontaneous; the four sign combinations and their temperature dependence
- 08ΔG° = −RT ln K and ΔG = ΔG° + RT ln Q; the van 't Hoff temperature dependence of K (Mastery)
Spontaneity and the crossover temperature from ΔH° and ΔS°
- +1Use ΔG° = ΔH° − TΔS°. First put ΔS° in the same energy units as ΔH°: 161 J K⁻¹ mol⁻¹ = 0.161 kJ K⁻¹ mol⁻¹.
- +1At 298 K: ΔG° = 178 − (298)(0.161) = 178 − 48.0 = +130 kJ mol⁻¹. ΔG° > 0, so the reaction is non-spontaneous at 298 K.
- +1With ΔH° > 0 and ΔS° > 0 the reaction is spontaneous only at high temperature. The crossover is where ΔG° = 0, i.e. T = ΔH°/ΔS°.
- +1T = 178000 J mol⁻¹ / 161 J K⁻¹ mol⁻¹ = 1106 K (about 833 °C). Above ~1106 K, TΔS° exceeds ΔH°, ΔG° < 0, and the reaction becomes spontaneous.
Key terms
- First law of thermodynamics
- Energy is conserved: ΔU = q + w, where q is heat and w is work. The sign convention is q > 0 for heat into the system (endothermic) and w > 0 for work done on the system.
- Enthalpy (ΔH)
- H = U + pV; at constant pressure the heat of reaction ΔH = q_p. Standard reaction enthalpy ΔrH° = Σn·ΔfH°(products) − Σn·ΔfH°(reactants), using standard enthalpies of formation (elements in their standard state = 0).
- Hess's law
- Reaction enthalpy is path-independent, so the enthalpy of a target reaction equals the sum of the enthalpies of any set of steps that add to it (reverse a step → change sign, scale a step → scale ΔH). Includes Born–Haber cycles (Mastery).
- Entropy (ΔS)
- A measure of energy dispersal/disorder (J K⁻¹ mol⁻¹). ΔrS° = Σn·S°(products) − Σn·S°(reactants); it increases on melting, vaporising and forming more gas particles. The third law sets S = 0 for a perfect crystal at 0 K.
- Gibbs energy (ΔG)
- ΔG = ΔH − TΔS decides spontaneity at constant T and pressure: ΔG < 0 spontaneous, ΔG > 0 non-spontaneous, ΔG = 0 at equilibrium. The sign can depend on temperature through the TΔS term.
- ΔG° and K
- ΔG° = −RT ln K links the standard free-energy change to the equilibrium constant; a negative ΔG° gives K > 1 (product-favoured). For non-standard conditions ΔG = ΔG° + RT ln Q (Mastery).
Thermochemistry FAQ
How does the sign of ΔG tell me if a reaction is spontaneous?
At constant temperature and pressure, ΔG < 0 means the reaction is spontaneous in the forward direction, ΔG > 0 means it is non-spontaneous (spontaneous in reverse), and ΔG = 0 means the system is at equilibrium. Because ΔG = ΔH − TΔS, temperature can flip the sign: the four ΔH/ΔS combinations give 'always', 'never', 'low-T only' and 'high-T only' spontaneity.
What is Hess's law and when do I use it?
Hess's law says reaction enthalpy is path-independent — a state-function property — so you can build a target reaction from other thermochemical equations and sum their enthalpies. Reverse a step and you flip the sign of its ΔH; scale a step and you scale its ΔH. Use it whenever you can't measure a reaction directly but can add up steps (including formation reactions or a Born–Haber cycle) that combine to give it.
Why do I have to watch units when combining ΔH and ΔS?
Because ΔH is almost always tabulated in kJ mol⁻¹ while ΔS is in J K⁻¹ mol⁻¹ — a factor of 1000 apart. In ΔG = ΔH − TΔS you must convert them to the same energy unit before subtracting; forgetting to do so is the single most common way to get ΔG wrong by three orders of magnitude. Convert ΔS to kJ K⁻¹ mol⁻¹ (or ΔH to J) first.
Can Sia help me with thermochemistry calculations?
Yes. Sia can run a Hess's-law combination, compute ΔrH° and ΔrS° from formation and entropy tables, evaluate ΔG = ΔH − TΔS with the units handled correctly, find a crossover temperature, and convert between ΔG° and K. It explains the method and checks your working; it does not do graded assessment, and UNSW academic-integrity rules apply.
Exam move
Keep the energy bookkeeping tidy: fix the first-law sign convention (q > 0 heat in, w > 0 work on the system) and the state-function versus path-function distinction, then practise calorimetry with q = mcΔT and per-mole scaling. Make the formation-data routine automatic for both enthalpy and entropy — Δr° = Σn·(products) − Σn·(reactants) — and rehearse Hess's law as reverse-and-scale steps that sum to the target. The highest-value skill is the ΔG = ΔH − TΔS analysis: always match units first, evaluate ΔG at the stated temperature, classify the ΔH/ΔS quadrant, and find the crossover T = ΔH°/ΔS° when both terms share a sign. For the Mastery layer, drill the ΔG° = −RT ln K link in both directions and the van 't Hoff estimate of K at a new temperature. This is where cross-topic exam questions live (thermodynamics ↔ equilibrium ↔ electrochemistry), so practise moving between ΔG°, K and E°cell. When a units or sign step slips, ask Sia to redo it carefully.
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