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Concept 05 / 06Thermodynamics

Thermochemistry: Standard Enthalpies and Hess's Law

Thermochemical equations, standard states, standard enthalpies of formation, combustion, atomisation, neutralisation, solution and phase transitions, Hess's law of constant heat summation, calculating reaction enthalpies from formation data and bond enthalpies, and lattice enthalpy via Born-Haber cycles.

Figure 1
Thermochemical equations. A balanced equation with physical states and the enthalpy change, for example C(graphite)+O2(g)→CO2(g)\mathrm{C(graphite) + O_{2}(g) \rightarrow CO_{2}(g)}, ΔrH∘=−393.5 kJ mol−1\Delta_{r}H^{\circ} = -393.5\,\mathrm{kJ\,mol^{-1}}. The value refers to the equation as written: reversing it changes the sign; multiplying the equation multiplies ΔH\Delta H. Standard state: the pure substance in its most stable form at 1 bar1\,\mathrm{bar} and the specified temperature (usually 298 K298\,\mathrm{K}); marked with ∘^{\circ}. Types of enthalpy change. Standard enthalpy of formation ΔfH∘\Delta_{f}H^{\circ}: for forming one mole of a compound from its elements in their standard states. By convention ΔfH∘=0\Delta_{f}H^{\circ} = 0 for elements in their reference forms (graphite, O2(g)\mathrm{O_{2}(g)}, Br2(l)\mathrm{Br_{2}(l)}, rhombic sulfur, white phosphorus). Diamond (+1.9+1.9) and O3\mathrm{O_{3}} (+142.7+142.7) are not zero. Standard enthalpy of combustion ΔcH∘\Delta_{c}H^{\circ}: complete burning of one mole in oxygen; always negative. Calorific value is heat released per gram. Enthalpy of atomisation: breaking one mole of a substance completely into gaseous atoms. Enthalpy of neutralisation: about −57.1 kJ mol−1-57.1\,\mathrm{kJ\,mol^{-1}} for strong acid with strong base (just H++OH−→H2O\mathrm{H^{+} + OH^{-} \rightarrow H_{2}O}); less negative for weak acids or bases, because some energy goes into ionising them. Enthalpy of solution, of hydration, and of phase transitions (ΔfusH\Delta_{fus}H, ΔvapH\Delta_{vap}H, ΔsubH=ΔfusH+ΔvapH\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H). Hess's law of constant heat summation. Because HH is a state function, the enthalpy change of a reaction is the same whether it occurs in one step or several. Thermochemical equations can be added and subtracted like algebraic equations (see figure). Consequences: ΔrH∘=∑ν ΔfH∘(products)−∑ν ΔfH∘(reactants)\Delta_{r}H^{\circ} = \sum \nu\,\Delta_{f}H^{\circ}(\text{products}) - \sum \nu\,\Delta_{f}H^{\circ}(\text{reactants}). ΔrH∘=∑ν ΔcH∘(reactants)−∑ν ΔcH∘(products)\Delta_{r}H^{\circ} = \sum \nu\,\Delta_{c}H^{\circ}(\text{reactants}) - \sum \nu\,\Delta_{c}H^{\circ}(\text{products}). Bond enthalpies. For gas-phase reactions, ΔrH≈∑\Delta_{r}H \approx \sum (bond enthalpies of bonds broken) −∑- \sum (bond enthalpies of bonds formed). Mean bond enthalpies are averages over many compounds, so this gives estimates. Lattice enthalpy cannot be measured directly; it is obtained from a Born-Haber cycle (see Chemical Bonding). Key Takeaways: - ΔrH∘=∑ΔfH∘(products)−∑ΔfH∘(reactants)\Delta_{r}H^{\circ} = \sum\Delta_{f}H^{\circ}(\text{products}) - \sum\Delta_{f}H^{\circ}(\text{reactants}). - ΔrH=∑BE(broken)−∑BE(formed)\Delta_{r}H = \sum \text{BE(broken)} - \sum \text{BE(formed)}. - ΔfH∘=0\Delta_{f}H^{\circ} = 0 for elements in their reference states only. - Strong acid plus strong base: −57.1 kJ mol−1-57.1\,\mathrm{kJ\,mol^{-1}}. - Pitfall: the formation and combustion formulas have opposite orders (products minus reactants versus reactants minus products). Brain-Teaser 1: The enthalpies of combustion of graphite, H2\mathrm{H_{2}} and CH4\mathrm{CH_{4}} are −393.5-393.5, −285.8-285.8 and −890.3 kJ mol−1-890.3\,\mathrm{kJ\,mol^{-1}}. Find ΔfH∘\Delta_{f}H^{\circ} of methane. Solution: Target: C+2H2→CH4\mathrm{C + 2H_{2} \rightarrow CH_{4}}. By Hess's law, ΔfH=ΔcH(C)+2ΔcH(H2)−ΔcH(CH4)=−393.5−571.6+890.3=−74.8 kJ mol−1\Delta_{f}H = \Delta_{c}H(\mathrm{C}) + 2\Delta_{c}H(\mathrm{H_{2}}) - \Delta_{c}H(\mathrm{CH_{4}}) = -393.5 - 571.6 + 890.3 = -74.8\,\mathrm{kJ\,mol^{-1}}. Brain-Teaser 2: Estimate ΔH\Delta H for C2H4(g)+H2(g)→C2H6(g)\mathrm{C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)} using bond enthalpies (kJ mol−1\mathrm{kJ\,mol^{-1}}): C=C 614614, C-C 347347, C-H 414414, H-H 436436. Solution: Broken: one C=C and one H-H, 614+436=1050614 + 436 = 1050. Formed: one C-C and two new C-H, 347+2×414=1175347 + 2 \times 414 = 1175. ΔH=1050−1175=−125 kJ mol−1\Delta H = 1050 - 1175 = -125\,\mathrm{kJ\,mol^{-1}} (the measured value is −137-137; the difference reflects the use of average bond enthalpies). Brain-Teaser 3: The enthalpy of neutralisation of HCl\mathrm{HCl} with NaOH\mathrm{NaOH} is −57.1 kJ mol−1-57.1\,\mathrm{kJ\,mol^{-1}} and that of HCN\mathrm{HCN} with NaOH\mathrm{NaOH} is −12.1 kJ mol−1-12.1\,\mathrm{kJ\,mol^{-1}}. Find the enthalpy of ionisation of HCN\mathrm{HCN}. Solution: Neutralising HCN\mathrm{HCN} = ionising HCN\mathrm{HCN} + (H++OH−→H2O\mathrm{H^{+} + OH^{-} \rightarrow H_{2}O}). So −12.1=ΔionH+(−57.1)-12.1 = \Delta_{\text{ion}}H + (-57.1), giving ΔionH=+45.0 kJ mol−1\Delta_{\text{ion}}H = +45.0\,\mathrm{kJ\,mol^{-1}}.

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