1.2.3 (HL)—Standard enthalpy changes

Quantity	Equation constraint
ΔHf°	Form one mole of compound from elements in standard states
ΔHc°	Completely burn one mole of substance in oxygen under standard conditions

Use correct standard states and coefficients; do not mix formation and combustion definitions in one equation.

A formation equation must produce exactly one mole of the compound from elements in their standard states, so fractional coefficients may be necessary. A combustion equation must burn exactly one mole completely in O₂; keep the stated standard state of water because changing H₂O(l) to H₂O(g) changes ΔH°.

Worked scaling example: $\Delta H_c^\circ(\ce{C2H6})=-1560\,\mathrm{kJ\,mol^{-1}}$ refers to complete combustion of one mole of ethane. For the corresponding balanced combustion of two moles, multiply the complete equation and its enthalpy by two: $\Delta H^\circ=2(-1560)=-3120\,\mathrm{kJ}$ . Do not report $\mathrm{kJ\,mol^{-1}}$ for the two-mole equation unless the result is normalized back to one mole.

Representative question

[Maximum number: 3]

Calculate the enthalpy of reaction, in $\mathrm{kJ} \mathrm{mol}^{-1}$ , when 1 mol of potassium reacts with water. Use section 12 of the data booklet. $\Delta H_{\mathrm{f}}$ of KOH(aq) is $-481.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Alternative 1:

« $\Delta H_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right)=$ » -285.8 «kJ mol ${ }^{-1}$ »

$\ll \Delta H_{\text {reaction }}=\Sigma \Delta H_{\mathrm{f}}($ products $)-\Sigma \Delta H_{\mathrm{f}}($ reactants ) 》

$« \Delta H_{\text {reaction }}=(2(-481.8)+0)-(0+2(-285.8)) »$

« $\Delta H_{\text {reaction }}=$ » -392.0 «kJ»

$<\Delta H=>-196.0<\mathrm{kJ} \mathrm{mol}^{-1} \gg$

Alternative 2:

« $\Delta H_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right)=$ » -285.8 «kJ mol ${ }^{-1}$ »

$\mathrm{K}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{KOH}(\mathrm{aq})+1 / 2 \mathrm{H}_{2}(\mathrm{~g})$

OR

$\ll \Delta H_{\text {reaction }}=\Sigma \Delta H_{\mathrm{f}}($ products $)-\Sigma \Delta H_{\mathrm{f}}($ reactants ) »

$« \Delta H_{\text {reaction }}=»(-481.8)-(-285.8)$

« $\Delta H_{\text {reaction }}=$ » -196.0 «kJ»

Marking guidance:

Award [3] for correct final answer.

M1 may be awarded from working.

Retrieve the route: count bond breaking/forming, manipulate Hess equations, define standard formation/combustion values, apply product–reactant sums, and track every Born–Haber energy term.

Check equation direction, coefficients, signs, standard states, lattice enthalpy convention, and one-versus-two-electron steps.

1.2.3 (HL)—Standard enthalpy changes

Standard Formation and Combustion Enthalpies

Writing Standard-Enthalpy Equations

Assessment in practice

Energy Cycles Summary

Question 1