Hess' Law of Constant Heat Summation
Using bond enthalpies

Hess' Law: bond enthalpies - Probs 1-10      Hess' Law: two equations and their enthalpies      Thermochemistry menu
       Hess' Law: three equations and their enthalpies
Hess' Law: four or more equations and their enthalpies      Hess' Law: standard enthalpies of formation

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:

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


Example #6: Calculate the enthalpy change for the reaction of ethene and hydrogen, given the following bond energy values in kJ/mol:

H−H 436; C−H 412; C=C 612; C−C 348

Solution:

1) The chemical reaction is this:

C2H4 + H2 ---> C2H6

List the bonds broken and the bonds made:

reactant bonds broken: four C−H bonds, one C=C bond, one H−H bond
product bonds made: one C−C bond, six C−H bonds

3) You can reduce that to this:

reactant bonds broken: one C=C bond, one H−H bond
product bonds made: one C−C bond, two C−H bonds

Note that four C−H bonds were removed from each side.

4) Hess' Law:

ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken

ΔH = [(C=C) + (H−H)] − [(C−C) + (2) (C−H)]

ΔH = [612 + 436] − [348 + 824]

ΔH = −124 kJ


Example #7: Calculate the C=C bond energy in ethene:

H2C=CH2(g) + H2(g) --> H3C−CH3(g) ΔH = −138 kJ/mol

Bond enthalpies (kJ/mol): C−C = 348; H−H = 436; C−H = 412

Solution:

1) Hess' Law for bond enthalpies is:

ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken

2) Let's insert symbols, not numbers:

ΔH = Σ ([(C=C) + (4) (C−H) + (H−H)] − Σ [(6) (C−H) + (C−C)]

Do not forget that C−C bond! I initially forgot it when I solved this problem prior to formatting it for the web site.

3) Cancel 4 C−H bonds:

ΔH = Σ ([(C=C) + (H−H)] − Σ [(2)(C−H) + (C−C)]

4) Put numbers in place and solve:

−138 = Σ ([(x) + (436)] − Σ [(2)(412) + (348)]

x = 598 kJ


Example #8: The following two equations produce methane and ethane:

C + 4H ---> CH4ΔH = −1652 kJ/mol
2C + 6H ---> C2H6ΔH = −2825 kJ/mol

(a) Calculate the bond enthalpy of a C−H bond.
(b) Calculate the bond enthalpy of a C−C bond.

Solution:

In the first equation, 4 C−H bonds are formed.

−1652 kJ/mol divided by 4 = 413 kJ <--- that's the bond enthalpy of a C−H bond

Note that bond enthalpies are expressed as a positive value (energy put into the bond to break it), so I ignored the minus sign on the 1652 value.

For the second reaction, note that six C−H bonds are formed and one C−C bond is formed.

413 times 6 = 2478

2825 − 2478 = 347 kJ <--- that's the bond enthalpy of a C−C bond


Example #9: Calculate the mean bond enthalpy of the Si-F bond in SiF4(g) given:

enthalpy of formation of SiF4(g) = −1615 kJ mol¯1
enthalpy of atomization of silicion = +456 kJ mol¯1
enthalpy of atomization of fluorine = +79 kJ mol¯1

Here is the Wikipedia entry for enthalpy of atomization.

Solution:

1) Let's write out what we are given using chemical equations:

Si(s) + 2F2(g) ---> SiF4(g)ΔHf = −1615 kJ
Si(s) ---> Si(g)ΔHa = +456 kJ
12F2(g) ---> F(g)ΔHa = +79 kJ

2) Here's the reaction we want:

SiF4(g) ---> Si(g) + 4F(g)

The mean bond enthalpy is the energy required to break a bond, in this case one Si−F bond. Important point: the above reaction breaks four Si−F bonds. That will come into play below.

3) Adjust the given reactions as follows:

eq 1 ---> flip
eq 2 ---> leave unchanged
eq 3 ---> multiply by 4

4) Resulting in:

SiF4(g) ---> Si(s) + 2F2(g)ΔH = +1615kJ
Si(s) ---> Si(g)ΔH = +456kJ
2F2(g) ---> 4F(g)ΔH = +316 kJ

5) Add the three equations and their enthalpies to get:

+2387 kJ

6) That is the enthalpy to disrupt four Si-F bonds, so divide by 4 to get:

+596.75 kJ

which, to three significant figures, rounds off to +597 kJ


Example #10: The decomposition reaction of tetrahedral P4 is as follows:

P4(g) ---> 2P2(g); ΔH = +217 kJ

If the bond energy of a single P−P bond is 200 kJ mol¯1, what is the energy of the PP triple bond in P2?

Solution:

Say you break all 6 P−P bonds in P4, that is 6 x 200 = +1200.

1200 − 217 = 983 which is released when two P≡P bonds form.

So 983 / 2 = bond energy of a P≡P bond. To three sig figs, the answer is 492 kJ/mol.


Bonus Example #1: What would be the reaction enthalpy of the following reaction?

N2 + H3O+ ---> NO2¯ + NH3

Solution:

1) The equation, as given, is not ready to be examined using Hess' Law. It must first be balanced for atoms and charge. We start with the oxygen since the nitrogen is already balanced:

Balance oxygen:
N2 + 2H3O+ ---> NO2¯ + NH3

Balance hydrogens:

N2 + 2H3O+ ---> NO2¯ + NH3 + 3H+

This last step also balances the charge.

2) The equation is still not ready for treatment with Hess' Law. Notice that, chemically speaking, H3O+ and H+ stand for the same thing. Eliminate two H+ for the equation:

N2 + 2H2O ---> NO2¯ + NH3 + H+

3) One last step to get the equation ready. Nitrous acid is a weak acid, so let us represent it in an unionized way:

N2 + 2H2O ---> HNO2 + NH3

The equation is now ready.

4) Reactant bonds broken:

N≡N, four O−H

5) Product bonds broken:

N=O, N−O, O−H, three N−H

A link to the Lewis structure for HNO2

6) ΔHrxn = Σ Ereactant bonds broken − Σ Eproduct bonds broken

ΔHrxn = [945 + (3)(463)] − [607 + 201 + (3)(391)]

I eliminated one O−H from each side.

ΔHrxn = 2334 − 1981 = 353 kJ


Bonus Example #2: Given that a chlorine-oxygen bond in ClO2(g) has an enthalpy of 243 kJ/mol, an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the standard enthalpy of formation of ClO2(g) is 102.5 kJ/mol, use Hess's law to calculate the value for the enthalpy of formation per mole of ClO(g).

Solution:

1) This is our target equation:

12Cl2(g) + 12O2(g) ---> ClO(g); ΔH = ???

2) Let's see what we know:

(a) we know an enthalpy:
12Cl2(g) + O2(g) ---> ClO2(g); ΔH = 102.5 kJ

(b) and we know two bond enthalpies:

Cl−O = 243 kJ/mol
O=O = 498 kJ/mol

3) The first step is to use a bond enthalpy calculation to determine the bond enthalpy of the Cl-Cl bond in Cl2

ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken

102.5 = [12(x) + 498] − [(2) (243)]

102.5 = 12(x) + 12

x = 181 kJ

4) Now, use a second bond enthaphy calculation for the equation in step 1 above:

ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken

x = [12(181) + 12 (498)] − 243

x = 96.5 kJ

A different approach to solving this problem may be found here.


Hess' Law: bond enthalpies - Probs 1-10      Hess' Law: two equations and their enthalpies      Thermochemistry menu
       Hess' Law: three equations and their enthalpies
Hess' Law: four or more equations and their enthalpies      Hess' Law: standard enthalpies of formation