CHEM1011 · Fundamentals Of Chemistry 1a
Thermodynamics
Thermodynamics in CHEM1011 is about the energy bookkeeping of a reaction: enthalpy H, the sign of ΔH (exothermic vs endothermic), and the two interchangeable routes to a reaction enthalpy — from standard enthalpies of formation and from average bond enthalpies. It is examined as short-answer calculation, so the marks come from showing the right setup, keeping every minus sign, and explaining why a hot (exothermic) reaction can still be slow because the rate is set by activation energy, not by ΔH.
What this chapter covers
- 011. Enthalpy H and the sign of ΔH: exothermic (ΔH < 0) vs endothermic (ΔH > 0)
- 022. The energy profile: reactants → transition state → products, and reading ΔH off the diagram
- 033. Hess's Law: enthalpy as a state function — reverse, scale, add reactions
- 044. Standard enthalpy of formation ΔHf°, and why elements in their standard state are 0
- 055. ΔH_rxn from formation enthalpies: Σn·products − Σn·reactants
- 066. ΔH_rxn from average bond enthalpies: Σ bonds broken − Σ bonds formed
- 077. Combustion: heat of combustion (per mole) vs calorific value (per gram)
- 088. Kinetics vs thermodynamics: activation energy Ea sets the rate, independent of ΔH
Reaction enthalpy of methane combustion from standard enthalpies of formation
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g),
given ΔHf°: CH₄(g) = −74.8, CO₂(g) = −393.5, H₂O(g) = −241.8 kJ mol⁻¹. State whether the reaction is exothermic or endothermic.
- +1Write the formula and recall that an element in its standard state has ΔHf° = 0, so O₂(g) = 0: ΔHrxn = Σn·ΔHf°(products) − Σn·ΔHf°(reactants).
- +1Sum the products, each weighted by its coefficient n: (−393.5) + 2(−241.8) = −393.5 − 483.6 = −877.1 kJ.
- +1Sum the reactants: (−74.8) + 2(0) = −74.8 kJ.
- +1Subtract reactants from products: ΔHrxn = −877.1 − (−74.8) = −802.3 kJ mol⁻¹; the negative sign means heat is released.
Key terms
- Enthalpy (H)
- The heat content of a system at constant pressure. We never measure H directly, only the change ΔH = H(products) − H(reactants) across a reaction.
- Exothermic / Endothermic
- Exothermic: ΔH < 0, heat is released and products lie lower in energy than reactants. Endothermic: ΔH > 0, heat is absorbed and products lie higher.
- Hess's Law
- Because enthalpy is a state function, ΔH depends only on the start and end states, not the path. Reactions can be reversed (flip ΔH sign), scaled (multiply ΔH), and added (add ΔH) to reach an unknown reaction enthalpy.
- Standard enthalpy of formation (ΔHf°)
- The enthalpy change when 1 mol of a compound forms from its elements in their standard states (most stable form, 1 atm, usually 25°C). ΔHf° = 0 for an element in its standard state, e.g. O₂(g), C(graphite).
- Bond enthalpy
- The energy needed to break 1 mol of a particular bond in the gas phase; always a positive (energy-in) value. ΔH_rxn ≈ Σ(bonds broken) − Σ(bonds formed), and is an estimate because tabulated values are averages.
- Activation energy (Ea)
- The energy barrier from reactants up to the transition state at the top of the energy profile. It sets the reaction rate and is independent of ΔH, so a very exothermic reaction can still be slow.
Thermodynamics FAQ
What is the difference between calculating ΔH from formation enthalpies and from bond enthalpies?
Both give a reaction enthalpy but set up oppositely. From standard enthalpies of formation you do Σn·ΔHf°(products) − Σn·ΔHf°(reactants), which gives an essentially exact value. From average bond enthalpies you do Σ(bonds broken) − Σ(bonds formed), where every bond term is a positive look-up value; because the values are averages, this route only gives an estimate. If a question supplies ΔHf° use the first method; if it supplies bond energies use the second.
How do I get the sign of ΔH right?
Decide physically first: if heat is released and products end up lower in energy, ΔH is negative (exothermic); if heat is absorbed, ΔH is positive (endothermic). In the formation-enthalpy formula always compute products MINUS reactants in that order, and keep the minus sign through every bracket — ΔHf° values are usually negative, and a single dropped sign flips an exothermic answer to endothermic.
Why can an exothermic reaction still be slow?
ΔH only tells you the energy difference between products and reactants — the destination, not the travel time. The rate is set by the activation energy Ea, the height of the barrier up to the transition state, which is independent of ΔH. The classic example is H₂ + O₂ forming water: it is hugely exothermic yet a mixture sits unchanged for years until a spark or catalyst lowers the barrier. Spontaneous does not mean fast.
Why is O₂ worth zero in a formation-enthalpy calculation?
O₂(g) is oxygen in its standard state — the most stable form of the element at 1 atm and 25°C — and by definition the standard enthalpy of formation of any element in its standard state is 0. The same applies to H₂(g), N₂(g), C(graphite) and Na(s). So those terms contribute nothing to Σn·ΔHf°, but you must still write them in to show you accounted for them.
What gets examined on thermodynamics in CHEM1011?
It comes up as short-answer calculation: state the sign of ΔH and what it means, combine reactions by Hess's Law (reverse → flip sign, scale → multiply, add → add ΔH, intermediates must cancel), compute ΔH_rxn from formation enthalpies and from bond enthalpies, convert heat of combustion to calorific value using molar mass, and explain why a large Ea makes an exothermic reaction slow. The datasheet supplies constants and the question gives the ΔHf° and bond-enthalpy values.
Exam move
Treat every thermodynamics question as a setup-then-arithmetic drill: first identify which tool the data hand you — formation enthalpies (Σn·products − Σn·reactants) or bond enthalpies (Σ broken − Σ formed) — and write the formula down verbatim before plugging numbers. Practise the two methods until the layout is automatic: build a small table with species, the supplied value, the coefficient n, and the product n·ΔH, then sum and subtract, because organised work catches the two killers (forgetting a coefficient and dropping a minus sign). Always finish the answer with a word: quote "−802.3 kJ mol⁻¹, exothermic", since the interpretation earns marks a bare number does not. Finally, rehearse the one conceptual line examiners love — that ΔH is independent of Ea, so an exothermic reaction can still be slow — and be ready to point to the energy profile to justify it.
