## Hess' Law of Constant Heat SummationUsing bond enthalpies

I'm going to solve problem #1 using a non-Hess' Law approach, then below it I'll go into a Hess' law discussion and then solve problem #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.

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

Problem #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.

Problem #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

Problem #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

Problem #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

Problem #4a: 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.

Problem #4b: 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

Problem #5: Considering bonds broken and formed ONLY, what is the enthalpy change for the following reaction:

C40H82 ---> C16H34 + 2C12H24

Solution:

Comment: this is a bit of a trick question. Why? I'll let that evolve during the discussion.

1) Hess' Law for bond enthalpies is:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

2) Let's consider the total bonds in one molecule of the reactant, C40H82:

C−C ⇒ 39
C−H ⇒ 82

3) Let's consider the total bonds in the three product molecules:

one C16H34:

C−C ⇒ 15
C−H ⇒ 34
two C12H24:
C−C ⇒ 24
C−H ⇒ 48
total:
C−C ⇒ 15 + 24 = 39
C−H ⇒ 34 + 48 = 82
Comment: knowing the C−C bonds in the C12H24 molecule is, to the ChemTeam, the key. Note that its formula is of the form CnH2n. That indicates either (a) one double bond between two carbons or (b) a cyclic structure with only single bonds between carbons. I took the second assumption in solving this problem, because this is the assumption that makes this a trick question.

4) Calculate the ΔH:

Since there are an equal number of C−C and C−H bonds on each side, the ΔH equals zero.

Comment: suppose you were to assume that double bonds were to form. In that case, you would have this:

reactant

C−C ⇒ 39
C−H ⇒ 82
product
C−C ⇒ 15 + 20 = 35
C=C ⇒ 2
C−H ⇒ 34 + 48 = 82
Net result:
reactant ⇒ four C−C broken

I won't carry the solution any further.

Problem #6: This reaction:

BBr3(g) + BCl3(g) ---> BBr2Cl(g) + BCl2Br(g)

has a ΔH very close to zero. Explain why ΔH is so small.

Solution:

1) Let's examine the reactant bonds broken:

BBr3 ⇒ three B−Br bonds
BCl3 ⇒ three B−Cl bonds

2) Let's examine the product bonds made:

BBr2Cl ⇒ two B−Br bonds and one B−Cl bond
BCl2Br ⇒ two B−Cl bonds and one B−Br bond

Three B−Br bonds made & broken and three B−Cl bonds made & broken predicts that ΔH = zero.

3) Why then is the actual value of ΔH not zero, but very close to zero?

Remember the chemical environment argument made above. Consider the one B−Cl bond that gets made in the BBr2Cl. The enthalpy involved in that bond is different than the enthalpy involved in a B−Cl bond in the molecule BCl3. Why? In the first molecule, the bond exists in the presence of two B−Br bonds while in the second, the B−Cl bond exists in the presence of two B−Cl bonds. A slight chemical difference between Br and Cl to be sure, but a difference nonetheless.

The same type of statement can be made regarding the two chemical environments that the B−Br bond exists in.

The differences between Cl and Br are slight, but they do make for a difference that can be measured experimentally. By the way, I do not know the enthalpy for this reaction, but I suspect it is positive. I won't go into why.

Problem #7: Determine the enthalpy of reaction for the following:

H2(g) + (1/2)O2(g) ---> H2O(g)

Using the following bond enthalpies (in kJ/mol): H−H (432); O=O (496); H−O (463)

Solution:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

ΔH = [432 + (0.5)(496)] - [(2) (463)]

ΔH = 680 - 926

ΔH = -246 kJ

The 0.5 comes from the coefficient in front of the O2. We have only 1/2 mole of O2 bonds to break and the 496 value is in kJ/mol. The two used with the H−O of 463 comes from the subscript of two. There are two H−O bonds in every water molecule; there are two moles of H−O bonds in one mole of water.

Comment: note that this is not the formation reaction for water. That reaction creates water in its standard state, which is liquid. Note also a limitation of the bond enthalpy method: it would give the same answer for H2O(l) as it does for H2O(g). However, the truth is that there is a difference in the enthalpy values for the two reactions.

Problem #8: Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows:

4 NH3(g) + 7 O2(g) ---> 4 NO2(g) + 6H2O(g)

Use the following data for bond energies to determine the bond energy of the N−O bond of NO2, using the follow values (given in kJ/mol):

 Bond Bond Energy O−H 464 N−H 389 O=O 498

Solution:

1) Let's see how many bonds are involved:

reactant

N−H ⇒ 12
O=O ⇒ 7

product

N−O ⇒ 8 (this is the unknown)
O−H ⇒ 12

2) State Hess' Law and substitute values:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

-1135 kJ = [(12)(389) + (7)(498)] - [(8)(x) + (12)(464)]

x = 465 kJ/mol (to three sig figs)

3) A problem with the answer:

NO2 actually has resonance structures that result in a bond order of 1.5, the equivalent of one and one-half bonds, if you will. The N−O bond energy is 222 kJ/mol and that for N=O is 590 kJ/mol. An average of those two values is 406 kJ/mol, which is, more or less, what we should expect from 1.5 bonds between N and O.

So, why is the answer to this problem 465? The ChemTeam does not know for sure. It might have something to do with a more technical analysis finding a bond order of 1.7, but the ChemTeam just does not know if this is the case.

Problem #9: Determine the enthalpy of the following reaction:

CH3CH=CH + 4.5 O=O ---> 3 O=C=O + 3 H-O-H

using the following bond enthalpy values:

 Bond Bond Energy C−C 347 C=C 611 C−H 414 C=O 736 O=O 498 O−H 464

Solution:

1) Let's see how many bonds are involved:

reactant

C=C ⇒ 1
C−C ⇒ 1
C−H ⇒ 5
O=O ⇒ 4.5

product

C=O ⇒ 6
O−H ⇒ 6

2) State Hess' Law and substitute values:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

ΔH = [(1)(611) + (1)(347) + (5)(414) + (4.5)(498)] - [(6)(736) + (6)(464)]

ΔH = 5269 - 7200 = -1931 kJ

Problem #10: Determine the enthalpy for the following reaction:

C(s) + CO2(g) ---> 2CO(g)

Note: The enthalpy of sublimation of graphite, C(s) is 719 kJ/mol

Solution:

Hess' Law for bond enthalpies is:

ΔHrxn = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

The enthalpy of sublimation can be considered to be the bond dissociation energy for solid carbon (that is, for this reaction: C(s) ---> C(g)).

x = [(1) (719) + (2) (736)] - [(2) 1073)]

x = 45 kJ

The bond enthalpy for CO was located here.

Bonus Problem: Calculate ΔH for this reaction:

H2C=CH2(g) + H2O(l) ---> CH3-CH2-OH(l)

using the following data:

 Bond Bond Energy C−C 347 C=C 611 C−H 414 C−O 360 O−H 464

Solution:

1) Let's see how many bonds are involved:

reactant

C=C ⇒ 1
C−H ⇒ 4
O−H ⇒ 2

product

C−H ⇒ 5
C−C ⇒ 1
C−O ⇒ 1
O−H ⇒ 1

Comment: be careful as you examine the structure. You might miss that there is a C-C bond or that there is a C-O bond. It has been known to happen!

2) This time, I'd like to eliminate like bonds on each side before using Hess' Law:

reactant

C=C ⇒ 1
O−H ⇒ 1

product

C−H ⇒ 1
C−C ⇒ 1
C−O ⇒ 1

3) State Hess' Law and substitute values:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

ΔH = (611 + 464) - (414 + 347 + 360)

ΔH = 1075 - 1121 = -46 kJ

Comment: as a comparison, I'd like to calculate ΔH using the following data:

 Substance ΔH°f C2H4(g) +52.7 kJ/mol H2O(l) -285.40 kJ/mol CH3CH2OH(l) -277.63 kJ/mol

Solution:

ΔH = Σ ΔH°f, products minus Σ ΔH°f, reactants

ΔH = (-277.63) - [+52.7 + (-285.40)] = -44.93 kJ/mol

Altough it is not great, why is there a difference between the two answers (-46 kJ and -44.93)?

Bond enthalpy values are averages of many different bond enthalpies. For example, the C-H bonds in C2H4 are in a different chemical environment than the C-H bonds in CH3CH2OH. Consequently, their exact bond enthalpies are going to be different. Since it is too difficult to tabulate all the different C-H bond enthalpies of every different chemical environment, an average C-H bond enthalpy is used (and different sources use different averages!). Consequently, enthalpy calculations using bond enthalpies are only a rough guide to the enthalpy of a given reaction.

In addition, the chemical environment changes as you remove a given bond. For example, I have a link several problems above to a table of bond enthalpy values. At that link is this comment:

"Average bond energies are the averages of bond dissociation energies. For example the average bond energy of O-H in H2O is 464 kj/mol. This is due to the fact that the H-OH bond requires 498.7 kj/mol in order to dissociate, while the O-H bond needs 428 kj/mol."