Back to homeChemistry · 6th Year HL · Change subject
Start for free
6th Year HL · Chemistry
Chemical transformations · Chemical thermodynamics

Hess's law and bond enthalpies

You want the heat of formation of ethyne (acetylene), the gas used in welding torches. Reacting graphite directly with hydrogen to make ethyne cannot be done cleanly in a school laboratory, or in any other. Hess's law lets you work out the value from reactions that are easy to measure, such as burning.

Hess's law

Burn 1 mol of carbon completely in oxygen and ΔH = −393.5 kJ/mol (heat is released), forming CO₂(g). Burn it in two stages instead, first to CO(g) and then CO to CO₂(g), and the two heats add up to −110.5 + (−283.0) = −393.5 kJ/mol, exactly the same. Why does the total not depend on the route?

This is the conservation of energy applied to a reaction. The energy profile diagram below shows an exothermic reaction (products lower than reactants, ΔH < 0) and an endothermic one (products higher, ΔH > 0); the size of ΔH is the same by any route.

Two energy profile diagrams: an exothermic reaction with products below reactants and an endothermic reaction with products above reactants. exothermic: ΔH < 0 endothermic: ΔH > 0 reactants products reactants products

If two routes gave different totals, you could go round one route and back by the other and create or destroy energy.

Energy cycle comparing the direct combustion of carbon to carbon dioxide with the two-stage route through carbon monoxide. C(graphite) + O₂(g) CO₂(g) CO(g) + ½O₂(g) ΔH₁ = −393.5 kJ/mol ΔH₂ = ?(C → CO) ΔH₃ =−283.0 kJ/mol
Practice★☆☆
At $298\,\mathrm{K}$, the conversion of sulfur dioxide into sulfur trioxide, a central step in the industrial synthesis of sulfuric acid, is to be evaluated:…

At $298\,\mathrm{K}$, the conversion of sulfur dioxide into sulfur trioxide, a central step in the industrial synthesis of sulfuric acid, is to be evaluated:

$$\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_3(\mathrm{g})$$

Experimental data are available for the following combustion and oxidation reactions at the same temperature and pressure:

(1) $\mathrm{S}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_2(\mathrm{g}) \quad \Delta H_1^\circ = -296.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

(2) $\mathrm{S}(\mathrm{s}) + \frac{3}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_3(\mathrm{g}) \quad \Delta H_2^\circ = -395.7\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

  1. a) Using Hess's law, combine equations (1) and (2) algebraically to deduce the standard enthalpy of the target reaction.
  2. b) Explain briefly which property of the state function enthalpy Hess's law is based on.
Solve
Practice★★☆
The combustion of graphite carbon can take place completely to give carbon dioxide or in stages via the intermediate carbon monoxide. The energy relationships between…

The combustion of graphite carbon can take place completely to give carbon dioxide or in stages via the intermediate carbon monoxide. The energy relationships between these transformations at $298\,\mathrm{K}$ are summarised in the following triangular diagram:

Direct route ΔH₁° = −393.5 kJ/mol Stage 1 ΔH₂° = ? Stage 2 ΔH₃° = −283.0 kJ/mol C(graphite) + O₂(g) CO₂(g) CO(g) + 1/2 O₂(g)

The thermochemical data at $298\,\mathrm{K}$ and $1\,\mathrm{bar}$ are:

(Direct route) $\mathrm{C}(\text{grafito}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \quad \Delta H_1^\circ = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

(Second stage) $\mathrm{CO}(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \quad \Delta H_3^\circ = -283.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

  1. a) Determine the value of the standard enthalpy of formation of gaseous carbon monoxide, $\Delta H_2^\circ = \Delta H_\mathrm{f}^\circ[\mathrm{CO}(\mathrm{g})]$, corresponding to the first stage.
  2. b) Justify in detail the path independence, using the cycle in the diagram.
Solve

Heats of formation

There are millions of known reactions, and nobody could measure the heat of every one. Is there a single reference value for each compound that lets you predict ΔH for any reaction?

Energy cycle that breaks the reactants down into elements and then builds the products from those elements. elements (ΔHf = 0) −ΣΔH_f (reactants) +ΣΔH_f (products) reactants products ΔH

Heats of combustion

Many heats of formation cannot be measured directly (as with ethyne), but heats of combustion are easy to measure in a calorimeter. The heat of combustion is the heat released when 1 mol of a substance burns completely in oxygen. Build an energy cycle with the combustion products at the bottom. The same cycle works for any compound, including an alcohol: ΔHf = Σ(heats of combustion of the elements) − (heat of combustion of the compound).

Energy cycle for ethyne: forming it from its elements then burning it equals burning the elements directly, both ending at carbon dioxide and water. 2C(graphite) + H₂(g) C₂H₂(g) 2CO₂(g) + H₂O(l) ΔHf = ? burn elements burn ethyne
Practice★☆☆
At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, the thermal decomposition reaction of calcium carbonate for the industrial production of quicklime is studied:…

At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, the thermal decomposition reaction of calcium carbonate for the industrial production of quicklime is studied:

$$\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$
  1. a) Calculate the standard enthalpy of this reaction from the standard molar enthalpies of formation: $\Delta H_\mathrm{f}^\circ[\mathrm{CaCO}_3(\mathrm{s})] = -1206.9\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $\Delta H_\mathrm{f}^\circ[\mathrm{CaO}(\mathrm{s})] = -635.1\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{f}^\circ[\mathrm{CO}_2(\mathrm{g})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
  2. b) Justify the thermodynamic sign obtained for the process and state, with justification, what the enthalpy of formation is for any chemical element in its most stable reference form at that temperature and at standard pressure.
Solve
Practice★★☆
At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, samples of graphite carbon, hydrogen gas and acetylene gas (ethyne, $\mathrm{C}2\mathrm{H}2$) are completely burned in an…

At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, samples of graphite carbon, hydrogen gas and acetylene gas (ethyne, $\mathrm{C}_2\mathrm{H}_2$) are completely burned in an oxygen atmosphere, and the following standard enthalpies of combustion are measured:

(1) $\mathrm{C}(\text{grafito}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \quad \Delta H_\mathrm{c,1}^\circ = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

(2) $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l}) \quad \Delta H_\mathrm{c,2}^\circ = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

(3) $\mathrm{C}_2\mathrm{H}_2(\mathrm{g}) + \frac{5}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow 2\,\mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) \quad \Delta H_\mathrm{c,3}^\circ = -1299.6\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

  1. a) Applying Hess's law, calculate the standard molar enthalpy of formation of gaseous acetylene from its elements in their standard states: $2\,\mathrm{C}(\text{grafito}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_2\mathrm{H}_2(\mathrm{g})$.
  2. b) Calculate the heat exchanged at constant pressure, stating whether it is absorbed or released, when $52.0\,\mathrm{g}$ of acetylene are formed under these conditions. (Molar mass of $\mathrm{C}_2\mathrm{H}_2 = 26.0\,\mathrm{g}\times\mathrm{mol}^{-1}$).
Solve
Practice★★★
The direct synthesis of ethanol from its elements under standard conditions,…

The direct synthesis of ethanol from its elements under standard conditions, $2\,\mathrm{C}(\text{grafito}) + 3\,\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})$, is experimentally unfeasible directly with quantitative purity owing to the simultaneous formation of multiple hydrocarbons and oxygenated compounds.

  1. a) Design a Hess cycle that allows $\Delta H_\mathrm{f}^\circ[\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})]$ to be calculated at $298\,\mathrm{K}$ given that the standard molar enthalpies of combustion are:

$\Delta H_\mathrm{c}^\circ[\mathrm{C}(\text{grafito})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $\Delta H_\mathrm{c}^\circ[\mathrm{H}_2(\mathrm{g})] = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{c}^\circ[\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})] = -1366.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.

  1. b) Determine the standard enthalpy of formation of ethanol in the gas phase given that the enthalpy of vaporisation of liquid ethanol at $298\,\mathrm{K}$ is $\Delta H_\mathrm{vap}^\circ[\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})] = +42.3\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, and interpret the effect of intermolecular condensation on the enthalpy difference between liquid and gas (do not deduce spontaneity from that datum alone).
Solve

Bond enthalpies

Hydrogen and chlorine gas mixed in sunlight react explosively to give hydrogen chloride. If only covalent bonds are involved, where does all the energy released come from?

Energy level diagram showing the enthalpy change as the energy absorbed breaking bonds in reactants minus the energy released making bonds in products. break: +E make: −E reactants free atoms products ΔH ≈ Σ(bonds broken) − Σ(bonds made)

Why do the two methods disagree?

Some of the exercises below ask you to compare a bond-enthalpy value with a value from heats of formation, or to include a heat of vaporisation. Always note the state symbols.

Heats of combustion of alcohols

Burn a series of primary alcohols and the standard heat of combustion becomes more negative as the chain gets longer. The balanced equation for complete combustion of a primary alcohol, for example ethanol, is: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). For propan-1-ol it is: C₃H₇OH(l) + 9/2 O₂(g) → 3CO₂(g) + 4H₂O(l). In general, for a primary alcohol with n carbon atoms, each extra CH₂ adds one CO₂ and one H₂O to the products.

Alcohol Formula Heat of combustion (kJ/mol)
Methanol CH₃OH −726
Ethanol C₂H₅OH −1,367
Propan-1-ol C₃H₇OH −2,021
Butan-1-ol C₄H₉OH −2,676

Each extra CH₂ adds roughly the same amount (about −650 kJ/mol per CH₂). Burning a longer chain means breaking and making more C–H and C–C bonds, and forming one more CO₂ and one more H₂O, so each CH₂ unit contributes a similar amount to the overall ΔH.

In the spirit-burner experiment you find the experimental value from the water heated: q = mcΔT, where m is the mass of water (g), c = 4.18 J g⁻¹ K⁻¹ and ΔT is the temperature rise. Divide q by the moles of alcohol burned (mass lost ÷ molar mass) to get kJ/mol, and give it a negative sign because the reaction is exothermic. In a spirit-burner experiment the value you measure is usually less exothermic than the data-book value, because heat is lost to the surroundings and combustion may be incomplete.

Practice★☆☆
The catalytic hydrogenation of ethene in the gas phase leads to the formation of ethane according to the following thermochemical equation at $298\,\mathrm{K}$:…

The catalytic hydrogenation of ethene in the gas phase leads to the formation of ethane according to the following thermochemical equation at $298\,\mathrm{K}$:

$$\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3(\mathrm{g})$$
  1. a) Estimate the standard enthalpy change of the process using only the mean bond energy values: $E(\mathrm{C}=\mathrm{C}) = 611.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}-\mathrm{C}) = 347.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}-\mathrm{H}) = 414.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $E(\mathrm{H}-\mathrm{H}) = 436.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
  2. b) Explain why mean bond energies allow an estimate in the gas phase, not an exact value, and what corrections would have to be added if any species were condensed.
Solve
Practice★★☆
The aim is to estimate the standard enthalpy of formation of liquid water, $\Delta H_\mathrm{f}^\circ[\mathrm{H}2\mathrm{O}(\mathrm{l})]$, at $298\,\mathrm{K}$ using…

The aim is to estimate the standard enthalpy of formation of liquid water, $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})]$, at $298\,\mathrm{K}$ using mean bond energies, for which the following are available:

$E(\mathrm{H}-\mathrm{H}) = 436.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{O}=\mathrm{O}) = 498.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{O}-\mathrm{H}) = 463.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, and the enthalpy of vaporisation of water at $298\,\mathrm{K}$: $\Delta H_\mathrm{vap}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = +44.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.

  1. a) Calculate the standard enthalpy of the gas-phase reaction $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{g})$ from the bond energies, and from it deduce $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})]$ by including the change of state.
  2. b) Compare the result with the experimental literature value of $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and state two factors that cause the mean bond energy method to introduce deviations.
Solve
Practice★★★
For the combustion of propane gas at $298\,\mathrm{K}$: $$\mathrm{CH}3\mathrm{CH}2\mathrm{CH}3(\mathrm{g}) + 5\,\mathrm{O}2(\mathrm{g}) \rightarrow…

For the combustion of propane gas at $298\,\mathrm{K}$:

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_3(\mathrm{g}) + 5\,\mathrm{O}_2(\mathrm{g}) \rightarrow 3\,\mathrm{CO}_2(\mathrm{g}) + 4\,\mathrm{H}_2\mathrm{O}(\mathrm{g})$$
  1. a) Calculate the standard enthalpy change of this reaction using only the mean bond energy data: $E(\mathrm{C}-\mathrm{C}) = 347.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}-\mathrm{H}) = 414.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{O}=\mathrm{O}) = 498.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}=\mathrm{O}) = 799.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ (in $\mathrm{CO}_2$) and $E(\mathrm{O}-\mathrm{H}) = 463.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
  2. b) Calculate the enthalpy of combustion from the standard molar enthalpies of formation: $\Delta H_\mathrm{f}^\circ[\mathrm{C}_3\mathrm{H}_8(\mathrm{g})] = -103.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $\Delta H_\mathrm{f}^\circ[\mathrm{CO}_2(\mathrm{g})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{g})] = -241.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. Discuss the physical origin of the quantitative discrepancy observed between the two analytical methods.
Solve
Practice★★★
In a calorimetric laboratory, the standard enthalpy of combustion of the solvent cyclohexane, $\mathrm{C}6\mathrm{H}{12}(\mathrm{l})$, is evaluated. The following…

In a calorimetric laboratory, the standard enthalpy of combustion of the solvent cyclohexane, $\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l})$, is evaluated. The following reaction data at $298\,\mathrm{K}$ are obtained by combustion at constant volume and correction to constant pressure:

(1) $\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l}) + 9\,\mathrm{O}_2(\mathrm{g}) \rightarrow 6\,\mathrm{CO}_2(\mathrm{g}) + 6\,\mathrm{H}_2\mathrm{O}(\mathrm{l}) \quad \Delta H_1^\circ = -3919.6\,\mathrm{kJ}\times\mathrm{mol}^{-1}$

The standard enthalpies of formation of $\mathrm{CO}_2(\mathrm{g})$ and of liquid water are also available: $\Delta H_\mathrm{f}^\circ[\mathrm{CO}_2(\mathrm{g})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, as well as the enthalpy of vaporisation of cyclohexane: $\Delta H_\mathrm{vap}^\circ[\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l})] = +33.1\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.

  1. a) Determine the standard enthalpy of formation of liquid cyclohexane, $\Delta H_\mathrm{f}^\circ[\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l})]$, and of gaseous cyclohexane, $\Delta H_\mathrm{f}^\circ[\mathrm{C}_6\mathrm{H}_{12}(\mathrm{g})]$.
  2. b) If the experimental combustion in the calorimeter had been incomplete, forming a molar $1.0\,\%$ of $\mathrm{CO}(\mathrm{g})$ (whose $\Delta H_\mathrm{f}^\circ[\mathrm{CO}(\mathrm{g})] = -110.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$) instead of $\mathrm{CO}_2(\mathrm{g})$, explain whether the heat measured experimentally would be an overestimate or an underestimate relative to the theoretical stoichiometric value.
Solve

Back to the start: To find ΔHf of ethyne we combined the heats of combustion of graphite (−393.5 kJ/mol), hydrogen (−285.8 kJ/mol) and ethyne (−1,299.5 kJ/mol): ΔHf = 2(−393.5) + (−285.8) − (−1,299.5) = +226.7 kJ/mol. We never needed to make ethyne directly.

Summary

See all the exercises in this section