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How do we find an enthalpy change we cannot measure directly?

Apply Hess's law and bond enthalpy data to calculate enthalpy changes for reactions.

Hess's law, energy cycles, calculating enthalpy changes from enthalpies of formation, and estimating enthalpy from average bond enthalpies, with fully worked TASC-style examples.

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What this dot point is asking

TASC expects you to use Hess's law with energy cycles and standard enthalpies of formation, and to estimate enthalpy changes from average bond enthalpies.

Hess's law

Because enthalpy is a state function, the energy change depends only on the start and end points, not the path between them.

You can therefore construct an energy cycle: if a direct reaction is hard to measure, go via intermediate steps whose enthalpies are known and add them up. Reversing a step changes the sign of its ΔH\Delta H; multiplying a step multiplies its ΔH\Delta H.

Using enthalpies of formation

The standard enthalpy of formation ΔHf\Delta H_f^\circ is the enthalpy change when one mole of a compound forms from its elements in their standard states. Elements in their standard state have ΔHf=0\Delta H_f^\circ = 0.

Using bond enthalpies

A reaction breaks bonds in the reactants (requiring energy) and makes bonds in the products (releasing energy):

ΔH(bonds broken)(bonds made).\Delta H \approx \sum(\text{bonds broken}) - \sum(\text{bonds made}).

Bond enthalpies are averages over many molecules, so this method gives an estimate, not an exact value, and it strictly applies only to gaseous species.

Why bond enthalpies give only an estimate

Bond enthalpy data are average values measured across many different molecules, so the energy of, say, a CH\text{C}-\text{H} bond in methane differs slightly from one in ethanol. When you use average bond enthalpies you are therefore approximating, and the result usually differs by a few percent from a value calculated using exact enthalpies of formation. Bond enthalpies also apply strictly to species in the gas phase, because they describe breaking bonds in isolated gaseous molecules; if a reactant or product is a liquid or solid, the additional enthalpy of vaporisation or sublimation is ignored, adding further error. For an exact answer, formation data or Hess cycles built from measured combustion enthalpies are preferred.

Building a Hess cycle

When formation data are not directly available, you can construct an energy cycle from any set of reactions that share common intermediates, typically combustion reactions or formation reactions. Write the target reaction, then arrange the known equations (reversing or scaling as needed) so that adding them reproduces the target exactly. The sum of the adjusted enthalpies is the target ΔH\Delta H. This is the method used in the benzene and propan-1-ol questions above, and it works because enthalpy is a state function: the route does not matter, only the start and end states.

In the exam, write the balanced equation first, label which method you are using, multiply every term by its coefficient, and check the sign against whether you expect the reaction to be exothermic or endothermic.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20223 marksThe enthalpy of formation of benzene, 3H2(g)+6C(s)C6H6(l)3\text{H}_2(g) + 6\text{C}(s) \rightarrow \text{C}_6\text{H}_6(l) (Equation 1), cannot be measured directly. Use Hess's law with: Equation 2, C6H6(l)+152O2(g)6CO2(g)+3H2O(l)\text{C}_6\text{H}_6(l) + \tfrac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l), ΔH=3267 kJ mol1\Delta H = -3267\ \text{kJ mol}^{-1}; Equation 3, H2(g)+12O2(g)H2O(l)\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l), ΔH=286 kJ mol1\Delta H = -286\ \text{kJ mol}^{-1}; Equation 4, C(s)+O2(g)CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH=394 kJ mol1\Delta H = -394\ \text{kJ mol}^{-1}. Calculate ΔH\Delta H for Equation 1.
Show worked answer →

Combine: target =(6×Eq 4)+(3×Eq 3)Eq 2= (6 \times \text{Eq 4}) + (3 \times \text{Eq 3}) - \text{Eq 2}. Reversing Equation 2 flips the sign of its ΔH\Delta H. (1 mark)

ΔH=6(394)+3(286)(3267)=2364858+3267=+45 kJ mol1.\Delta H = 6(-394) + 3(-286) - (-3267) = -2364 - 858 + 3267 = +45\ \text{kJ mol}^{-1}.

(2 marks)

So the enthalpy of formation of benzene is about +45 kJ mol1+45\ \text{kJ mol}^{-1} (endothermic, consistent with benzene being less stable than its elements).

TCE 20213 marksFor N2(g)+3H2(g)2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), use the average bond energies to determine the theoretical ΔH\Delta H. HH=436\text{H}-\text{H} = 436; NH=391\text{N}-\text{H} = 391; NN=941 kJ mol1\text{N}\equiv\text{N} = 941\ \text{kJ mol}^{-1}.
Show worked answer →

ΔH=(bonds broken)(bonds made)\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made}).

Bonds broken: one NN=941\text{N}\equiv\text{N} = 941, plus three HH=3×436=1308\text{H}-\text{H} = 3 \times 436 = 1308. Total =2249 kJ= 2249\ \text{kJ}. (1 mark)

Bonds made: each NH3\text{NH}_3 has three NH\text{N}-\text{H} bonds and there are two molecules, so 6×391=2346 kJ6 \times 391 = 2346\ \text{kJ}. (1 mark)

ΔH=22492346=97 kJ\Delta H = 2249 - 2346 = -97\ \text{kJ} per mole of reaction. The reaction is exothermic, close to the accepted 92 kJ mol1-92\ \text{kJ mol}^{-1}. (1 mark)

TCE 20233 marksPropan-1-ol combusts: C3H8O(l)+92O2(g)3CO2(g)+4H2O(g)\text{C}_3\text{H}_8\text{O}(l) + \tfrac{9}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g). Using C(s)+O2(g)CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH=393.5\Delta H = -393.5; 3C(s)+4H2(g)+12O2(g)C3H8O(l)3\text{C}(s) + 4\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{C}_3\text{H}_8\text{O}(l), ΔH=303.0\Delta H = -303.0; H2(g)+12O2(g)H2O(g)\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g), ΔH=241.8 kJ mol1\Delta H = -241.8\ \text{kJ mol}^{-1}, show that the heat of combustion is about 1850 kJ mol1-1850\ \text{kJ mol}^{-1}.
Show worked answer →

Combine: target =(3×CO2 eq)+(4×H2O eq)(formation of propan-1-ol)= (3 \times \text{CO}_2 \text{ eq}) + (4 \times \text{H}_2\text{O eq}) - (\text{formation of propan-1-ol}). Reverse the formation so its sign flips. (1 mark)

ΔH=3(393.5)+4(241.8)(303.0)=1180.5967.2+303.0=1844.7 kJ mol1.\Delta H = 3(-393.5) + 4(-241.8) - (-303.0) = -1180.5 - 967.2 + 303.0 = -1844.7\ \text{kJ mol}^{-1}.

(1 mark)

This rounds to approximately 1850 kJ mol1-1850\ \text{kJ mol}^{-1}, as required. (1 mark)

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