Questions: Thermochemistry and Standard Formation Properties

Question 1

C(graphite) has ΔH°_f = 0, but C(diamond) has ΔH°_f = +1.9 kJ/mol. What is the correct explanation for this difference?

• A. Graphite has no chemical energy because it is the most abundant form of carbon
• B. Diamond is a higher-energy allotrope, so it has a positive formation enthalpy, while graphite is the reference state assigned zero by convention
• C. Diamond cannot be formed from pure elements, so its formation enthalpy must be measured indirectly
• D. The zero assignment for graphite reflects that graphite is perfectly stable and cannot release energy under any conditions

Answer: (B)

The zero assignment is a convention: the most stable elemental form at 25°C and 1 atm is the reference state and is assigned ΔH°_f = 0. For carbon, the most stable form is graphite — not because graphite has no energy, but because it is the chosen baseline. Diamond is less stable than graphite (it would convert to graphite given enough activation energy), so forming diamond from graphite requires energy input: ΔH°_f(diamond) = +1.9 kJ/mol. This is NOT because graphite has 'no energy' — it has plenty of chemical energy — but simply because the convention places graphite at zero. Option D is wrong: graphite absolutely can release energy (burning graphite in O₂ produces CO₂ with a large negative ΔH).

Question 2

You need to calculate ΔH°_rxn for the combustion of ethanol, but you cannot measure it directly. You have standard formation enthalpies for ethanol, CO₂, H₂O, and O₂. What is the correct calculation?

• A. Add all ΔH°_f values for products and reactants together regardless of sign
• B. ΔH°_rxn = Σ(ν_products × ΔH°_f,products) − Σ(ν_reactants × ΔH°_f,reactants)
• C. ΔH°_rxn equals the formation enthalpy of the products minus the formation enthalpy of the fuel only
• D. Use Kirchhoff's law directly since the reaction occurs at constant pressure

Answer: (B)

Hess's law in numerical form: you conceptually 'un-form' all reactants back to their elements (subtracting their formation enthalpies, multiplied by stoichiometry) and then 'form' the products from those elements (adding their formation enthalpies). Because enthalpy is a state function, the path does not matter — only the initial and final states. Note that ΔH°_f(O₂) = 0 since O₂ is the reference form for oxygen, so it drops out of the calculation. Option D is wrong because Kirchhoff's law corrects for temperature, not for calculating ΔH°_rxn at 25°C from formation data.

Question 3

Elements in their standard reference state have ΔH°_f = 0 because they contain no chemical energy at 25°C and 1 atm.

• T. True
• F. False

Answer: False

The zero assignment is purely conventional, not a physical fact about energy content. Elements in their reference state contain substantial chemical energy — burning carbon (graphite) in oxygen releases ~394 kJ/mol as CO₂, and hydrogen gas burns with ~286 kJ/mol released as water. These energies are real, but they cannot be measured on an absolute scale; we can only measure differences. By setting all reference elements to zero, we establish a consistent baseline so that ΔH°_f values for all compounds are comparable and can be combined via Hess's law. The convention is a bookkeeping device, not a claim about absolute energy.

Question 4

Kirchhoff's law is needed when applying standard formation enthalpies to reactions occurring at temperatures significantly different from 25°C.

• T. True
• F. False

Answer: True

Standard formation enthalpies are tabulated at 25°C (298.15 K). For reactions at other temperatures, Kirchhoff's law provides the correction: ΔH_rxn(T) = ΔH°_rxn + ∫₂₉₈^T ΔCₚ dT, where ΔCₚ is the stoichiometry-weighted difference in heat capacities of products minus reactants. For moderate temperature deviations and small ΔCₚ, the correction may be negligible. But for high-temperature industrial processes (furnaces, turbines, combustion chambers operating at 1000°C+), ignoring the Kirchhoff correction can introduce significant error in energy balances and equilibrium predictions.

Question 5

Explain why the standard enthalpy of formation of O₂(g) is defined as zero, while the standard enthalpy of formation of O(g) is +249 kJ/mol.

Model answer: The zero for O₂(g) is a convention: O₂ is the most stable form of elemental oxygen at 25°C and 1 atm, so it is chosen as the reference state and assigned ΔH°_f = 0. This does not mean O₂ has no energy — it means all formation enthalpies for oxygen-containing compounds are measured relative to O₂. Atomic oxygen O(g) is not the reference form; forming it requires breaking an O=O bond: ½O₂(g) → O(g). This process requires +249 kJ/mol of energy input. Therefore O(g) has ΔH°_f = +249 kJ/mol — it is 249 kJ/mol higher in enthalpy than the reference form. The positive value correctly captures that O(g) is a high-energy, reactive species relative to the stable diatomic reference.

This distinction — reference form (zero by convention) vs. non-reference form (nonzero because it costs energy to form from the reference) — is a common source of error. Students sometimes use O(g) when they mean O₂(g) or vice versa, changing the calculated ΔH°_rxn by hundreds of kJ/mol. The same issue arises for H (atomic, +218 kJ/mol) vs. H₂ (reference, 0) and for S(g) vs. S(rhombic, reference).