Here is how it affects the Hess' Law formulation: Here is what I wrote above:Īnd in writing the Hess' law formulation, I want to do this: Now, I want to rearrange the negative sign on the second value in step three above. Σ E bonds formed ⇒ always a negative value Σ E bonds broken ⇒ always a positive value I'm using E to represent the bond energy per mole of bonds (for example, E for the C≡C bond is 839 kJ/mol). In step 3 just above, I wrote the ΔH calculation in the form of Hess' Law, but with words. Let's make all the bonds of the one product:ģ) ΔH = the energies required to break bonds (positive sign) + the energies required to make bonds (negative sign): Two different types of reasons for multiplying by two.Ģ) You get energy out when a bond (any bond) forms. Note there are two C−H bonds in one molecule of C 2H 2 and there is one H−H bond in each of two H 2 molecules. Let's break all the bonds of the reactants: Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane:īond enthalpies (in kJ/mol): C−C (347) C≡C (839) C−H (413) H−H (432)ġ) You have to put energy into a bond (any bond) to break it. Just so you know!Įxample #1: Hydrogenation of double and triple bonds is an important industrial process. There is no general agreement about which average values to use.Įxample #4 has a little trick in it. If you do an Internet search, you will find that other people use different values for the carbon-carbon single bond. So, the various values that are known have been averaged and, in the case of a carbon-carbon single bond, I have decided to use the value of 347 kJ/mol. The differences from one chemical environment to the next are fairly small and it would be tedious to list each and every specific chemical environment. Why? Hydrogen and chlorine have different influences on the electron density in the carbon-carbon bond and that has an influence on how much energy it takes to break the bond (more electron density means more energy needed to break). The bond enthalpy for that situation would be different if six chlorines were instead attached to the carbons. Suppose there are six hydrogens attached. The chemical environment for that bond differs depending on what is attached to the two carbons. Here is what I mean: take a carbon-carbon single bond (C−C). Bond enthalpies actually differ slightly from substance to substance. This is because the bond enthalpy values used are averages. One final note before solving some problems: the ΔH values determined via this technique are only approximations. However, there is a difficulty: we wind up with a Hess' Law formulation that is slightly different than we use when we manipulate chemical equations with their associated enthalpies. I'm going to solve example #1 using a non-Hess' Law approach, then below it I'll go into a Hess' law discussion and then solve example #1 again. Here is where I got most (not all) of the bond enthalpy values used in the problems below. Using bond enthalpies Hess' Law: bond enthalpies - Probs 1-10 Hess' Law: two equations and their enthalpies Hess' Law: bond enthalpies - Probs 11-20 Hess' Law: three equations and their enthalpies Hess' Law: four or more equations and their enthalpies Hess' Law: standard enthalpies of formation Thermochemistry menu ChemTeam: Hess' Law - using bond enthalpies
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