Hess' Law of Constant Heat Summation
Using bond enthalpies

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Here is where I got most (not all) of the bond enthalpy values used in the problems below.

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.

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.

One final note before solving some problems: the ΔH values determined via this technique are only approximations. This is because the bond enthalpy values used are averages. Bond enthalpies actually differ slightly from substance to substance.

Here is what I mean: take a carbon-carbon single bond (C−C). The chemical environment for that bond differs depending on what is attached to the two carbons. Suppose there are six hydrogens attached. The bond enthalpy for that situation would be different if six chlorines were instead attached to the carbons. 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 differences from one chemical environment to the next are fairly small and it would be tedious to list each and every specific chemical environment. 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.

If you do an Internet search, you will find that other people use different values for the carbon-carbon single bond. There is no general agreement about which average values to use.

Example #4 has a little trick in it. Just so you know!


Example #1: Hydrogenation of double and triple bonds is an important industrial process. Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane:

H−C≡C−H(g) + 2H2(g) ---> H3C−CH3(g)

Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432)

Solution:

1) You have to put energy into a bond (any bond) to break it. Bond breaking is endothermic. Let's break all the bonds of the reactants:

one C≡C ⇒ +839 kJ
two C−H ⇒ 413 x 2 = +826 kJ
two H−H ⇒ 432 x 2 = +864 kJ

The sum is +2529 kJ

Note there are two C−H bonds in one molecule of C2H2 and there is one H−H bond in each of two H2 molecules. Two different types of reasons for multiplying by two.

2) You get energy out when a bond (any bond) forms. Bond making is exothermic. Let's make all the bonds of the one product:

one C−C ⇒ −347 kJ
six C−H ⇒ −413 x 6 = −2478

The sum is −2826 kJ

3) ΔH = the energies required to break bonds (positive sign) plus the energies required to make bonds (negative sign):

+2529 + (−2825) = −296 kJ/mol

In step 3 just above, I wrote the ΔH calculation in the form of Hess' Law, but with words. Let's try some symbols:

ΔH = Σ Ebonds broken plus Σ Ebonds formed

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). Also, a reminder:

Σ Ebonds broken ⇒ always a positive value
Σ Ebonds formed ⇒ always a negative value

Now, I want to rearrange the negative sign on the second value in step three above. Here is what I wrote above:

+2529 + (−2825) = −296 kJ/mol

and in writing the Hess' law formulation, I want to do this:

+2529 − (+2825) = −296 kJ/mol

Here is how it affects the Hess' Law formulation:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

Notice how I changed the subscript on the second Σ E. Since it is now a positive value, it is the Σ E for the bonds of the product being broken.

The formulation of Hess' Law just above is the one usually used in textbooks.

There is an important point to be made if you decide to use the Hess' Law formulation:

use all bond enthalpies as positive numbers

One final point. We're using Hess' Law and bond enthalpies. Notice how it is reactant values minus product values. In the other Hess' Law tutorials, it was the product values minus the reactant values. Be aware of the difference.


Example #1 (again): Hydrogenation of double and triple bonds is an important industrial process. Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane:

H−C≡C−H(g) + 2H2(g) ---> H3C−CH3(g)

Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432)

Solution:

1) Hess' Law for bond enthalpies is:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

2) On the reactant side, we have these bonds broken:

Σ [two C−H bonds + one C≡C bond + two H−H bonds]

Σ [(2 x 413) + 839 + (2 x 432)] = 2529 kJ

3) On the product side, we have these bonds broken:

Σ [one C−C bond + six C−H bonds]

Σ [347 + (6 x 413)] = 2825 kJ

4) Using Hess' Law, we have:

ΔH = 2529 minus 2825 = −296 kJ

Example #2: Using bond enthalpies, calculate the reaction enthalpy (ΔH) for:

CH4(g) + Cl2(g) ---> CH3Cl(g) + HCl(g)

Bond enthalpies (in kJ/mol): C−H (413); Cl−Cl (239); C−Cl (339); H−Cl (427)

Solution:

1) Hess' Law for bond enthalpies is:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

2) On the reactant side, we have these bonds broken:

Σ [four C−H bonds + one Cl−Cl bond]

Σ [(4 x 413) + 239] = 1891 kJ

3) On the product side, we have these bonds broken:

Σ [three C−H bonds + one C−Cl bond + one H−Cl bond]

Σ [(3 x 413) + 339 + 427] = 2005 kJ

4) Using Hess' Law, we have:

ΔH = 1891 minus 2005 = −114 kJ

Comment: you may have noticed that four C−H bonds were involved on the reactant side and three C−H bonds were involved on the product side. You might be wondering about elimiminating three C−H bonds to make a problem seem a bit simpler. You may do that, if you wish. You'd get this:

reactant side: [one C−H bond + one Cl−Cl bond] = 413 + 239 = 652
product side: [one C−Cl bond + one H−Cl bond] = 339 + 427 = 766
ΔH = 652 minus 766 = −114 kJ

Example #3: What is the enthalpy of reaction for the following equation:

2 CH3OH(l) + 3 O2(g) ---> 2 CO2(g) + 4 H2O(g)

Given the following bond enthalpies (in kJ/mol): C−H (414); C−O (360); C=O (799); O=O (498); O−H (464)

Solution:

1) Bonds broken:

reactants ⇒ 3 O=O bonds; 6 C−H bonds; 2 C−O bonds; 2 O−H bonds

2) Bonds formed:

products ⇒ 4 C=O bonds; 8 O−H bonds

3) Using Hess' Law:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

ΔH = [(3)(498) + (6)(414) + (2)(360)] - [(4)(799) + (6)(464)]

Notice how I eliminated two O−H bonds from each side.

ΔH = 4698 - 5980

ΔH = -1282 kJ


Example #4: Calculate the bond energy of the Cl-F bond using the following data:

Cl2 + F2 ---> 2ClF ΔH = −108 kJ

Bond enthalpies (in kJ/mol): Cl−Cl (239); F−F (159)

Solution:

Hess' Law for bond enthalpies is:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

−108 = [239 + 159] − 2x

−2x = −506

x = 253 kJ/mol

Note the use of 2x because there are two ClF molecules.


Example #5: The reaction of H2 with F2 produces HF with ΔH = -269 kJ/mol of HF. If the H-H and H-F bond energies are 432 and 565 kJ/mol, respectively, what is the F-F bond energy?

H2(g) + F2(g) ----> 2HF(g)

Solution:

Hess' Law for bond enthalpies is:

ΔHrxn = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

The ΔH is given per mole of HF, so we need to use -269 x 2 = -538 kJ for the enthalpy of the reaction.

-538 = [432 + x] - [(2) (565)]

x = 160 kJ


Go to Hess' Law - using bond enthalpies - Problems 1 - 10

Go to Hess' Law - using bond enthalpies - Problems 11 - 15

Go to Hess' Law - using two equations and their enthalpies

Go to Hess' Law - using three equations and their enthalpies

Go to Hess' Law - using four or more equations and their enthalpies

Go to Hess' Law - using standard enthalpies of formation

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