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.
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What this dot point is asking
You must use Hess's law with enthalpy cycles and standard enthalpies, and 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.
This means we can construct an energy cycle: if a direct reaction is hard to measure, we go via intermediate steps whose enthalpies are known, and add them up. Reversing a step changes the sign of its ; multiplying a step multiplies its .
Using enthalpies of formation
The standard enthalpy of formation is the enthalpy change when one mole of a compound forms from its elements in their standard states. Elements in their standard state have . For any reaction:
Using bond enthalpies
A reaction breaks bonds in the reactants (requiring energy in) and makes bonds in the products (releasing energy). The estimated enthalpy change is:
Bond enthalpies are averages taken over many molecules, so this method gives an estimate rather than an exact value, and it only applies to gaseous species.
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.
2022 TASC3 marksThe enthalpy of formation of benzene (Equation 1: 3H2(g) + 6C(s) -> C6H6(l)) cannot be measured directly. Use Hess's law and the combustion reactions below to calculate delta H for Equation 1. Equation 2: C6H6(l) + 15/2 O2(g) -> 6CO2(g) + 3H2O(l), delta H = -3267 kJ mol-1; Equation 3: H2(g) + 1/2 O2(g) -> H2O(l), delta H = -286 kJ mol-1; Equation 4: C(s) + O2(g) -> CO2(g), delta H = -394 kJ mol-1.Show worked answer →
Use Hess's law: target = (6 x Equation 4) + (3 x Equation 3) - (Equation 2). Reversing Equation 2 changes the sign of its delta H.
delta H = 6(-394) + 3(-286) - (-3267)
= -2364 - 858 + 3267
= +45 kJ mol-1.
So the enthalpy of formation of benzene is about +45 kJ mol-1 (endothermic, consistent with benzene being less stable than its elements). (3 marks: correct combination, sign change on reversal, and value.)
2021 TASC3 marksFor the reaction N2(g) + 3H2(g) reversible 2NH3(g), use the average bond energies below to determine the theoretical delta H for the production of ammonia. H-H = 436 kJ mol-1; N-H = 391 kJ mol-1; N(triple bond)N = 941 kJ mol-1.Show worked answer →
delta H = (sum of bonds broken in reactants) - (sum of bonds made in products).
Bonds broken: one N(triple)N = 941, plus three H-H = 3 x 436 = 1308. Total = 941 + 1308 = 2249 kJ.
Bonds made: each NH3 has three N-H bonds, and there are 2 NH3, so 6 x 391 = 2346 kJ.
delta H = 2249 - 2346 = -97 kJ. The reaction is exothermic (close to the accepted value of about -92 kJ). (3 marks: bonds broken, bonds made, value with sign.)
2023 TASC3 marksPropanol combusts: C3H8O(l) + 9/2 O2(g) -> 3CO2(g) + 4H2O(g). Using C(s) + O2(g) -> CO2(g) delta H = -393.5 kJ mol-1; 3C(s) + 4H2(g) + 1/2 O2(g) -> C3H8O(l) delta H = -303.0 kJ mol-1; H2(g) + 1/2 O2(g) -> H2O(g) delta H = -241.8 kJ mol-1, show that the heat of combustion of propanol is approximately -1850 kJ mol-1.Show worked answer →
Combine the given equations to build the combustion of propanol: target = (3 x equation for CO2) + (4 x equation for H2O) - (equation forming C3H8O). Reverse the formation of propanol so its sign flips.
delta H = 3(-393.5) + 4(-241.8) - (-303.0)
= -1180.5 - 967.2 + 303.0
= -1844.7 kJ mol-1.
This rounds to approximately -1850 kJ mol-1, as required. (3 marks: correct combination, sign change, value.)