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Revision notes on thermochemistry-energetics - solving enthalpy problems using Hess's Law - for Advanced Level Theoretical-Physical Chemistry:
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Advanced Level Theoretical Physical Chemistry AS A2 Level Revision Notes Basic Thermodynamics GCE Thermodynamicsthermochemistry subindex links below Part 1 ΔH Enthalpy Changes The thermochemistry of enthalpies of reaction, formation, combustion and neutralisation Parts 1.2ab(i)(iii) Thermochemistry Hess's Law and Enthalpy Calculations This page describes Hess's Law of 'constant heat summation' and the use of thermochemical cycles to determine unknown enthalpy changes from known data e.g. involving the calculation of enthalpy of reaction, enthalpy of formation, enthalpy of combustion. As well as Hess's Law calculations other methods of solving numerical enthalpy problems without using thermochemical cycles are also described. |
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1.2 Thermochemistry Hess's Law and Enthalpy Calculations 1.2a Hess's Law and its importance Hess's Law is a version of the general law of conservation of energy i.e. the first law of thermodynamics which can be stated as energy cannot be destroyed or created but merely changed in form or distributed in different ways. Hess's Law states that the energy change from reactants A to products B is independent of the pathway taken no matter how many stages it involves. This is shown in the diagram of enthalpy cycles below.
ΔH1 = ΔH(A to B) Pathway 1 the most direct route with no intermediate stages, or ΔH1 = ΔH2 + ΔH3 pathway 2 involving one set of intermediates C, or or ΔH1 = ΔH4 + ΔH5 + ΔH6 pathway 3 involving two sets of intermediates D and E etc. etc. there is no limit to the complexity of the Hess's Law Cycle as long as A and B are constant. Many problems can be set out as a 'Hess's Law' triangle e.g.
Note: Superscript θ means a standard enthalpy value (see further down the page) and each ΔHθ could be a composite of several ΔHθ values - see enthalpy of combustion calculations below. 1.2b Using Hess's Law to perform enthalpy calculations 1.2b(i) Using known enthalpies of reaction or combustion etc. Using a Hess's Law Cycle you can calculate enthalpies of formation which you could not determine by laboratory experiment. However, using these calculated enthalpies of formation and experimentally determined enthalpies of combustion a huge variety of other enthalpies for other reactions can then be calculated. The 1st example of calculating the enthalpy of formation of methane illustrates the principles of using Hess's Law, especially as this cannot be determined by direct laboratory experiment! Given the following data below from text/data book calculate the enthalpy of formation of methane.
All of the three enthalpies above can be very accurately determined by direct experiment in a calorimeter.
2nd example of Hess's Law Cycle calculation From the following thermochemical data
Calculate the enthalpy change for the reaction
3rd example of Hess's Law Cycle calculation From the following thermochemical data to do
1.2b(ii) Solving enthalpy problems using an 'algebraicequation style' method Given the following data
Calculate the standard enthalpy of combustion of propane
What you do is rearrange, if necessary, the data equations and add up the results both equation components and delta H values, cancelling out the equation components should leave you with the correct equation whose enthalpy value you require.
Note that here you are using factors of 3 and 4 to make the cycle balance in molar terms 2nd example of 'algebraic' calculation style Given the following thermochemical data:
Calculate the enthalpy change for the reaction
Note that you double the enthalpy of formation hydrogen chloride to make the cycle balance in molar terms 1.2b(iii) A method using the summation of enthalpies of reactants and products A very simple example of enthalpy summation method
ΔHreaction = ∑Hproducts ∑Hreactants ΔHθreaction,298 = ∑ΔHθf,298(products) ∑ΔHθf,298(reactants) ΔHθesterification = {ΔHθf(ethyl ethanoate) + ΔHθf(water)} {ΔHθf(ethanoic acid) + ΔHθf(ethanol)} ΔHθesterification = {481 + 286} {487 + 278} ΔHθ(esterification reaction) = (767) (765) = 2 kJmol1 What would be ΔHθ(hydrolysis)? Answer! just reverse the sign! i.e. +2 kJmol1 2nd example of enthalpy summation method A more complicated example where you need to think more about mole ratios in the equation. Given the following data from laboratory measurements
Calculate the standard enthalpy of formation of butane gas, which cannot be determined by experiment.
To solve this you can use equation (3) and the data from equations (1) and (2) to obtain a value for equation (4)
Since oxygen is an element, ΔHθf(oxygen) = 0, therefore after rearranging we get
Enthalpy calculation problems with worked out answers based on enthalpies of reaction, formation, combustion Energetics-Thermochemistry-Thermodynamics Notes INDEX
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