Hess' Law of Constant Heat Summation
Using Bond Enthalpies
Problems 1 - 10

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

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Problem #1: 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
product ⇒ two C=C made

I won't carry the solution any further.


Problem #2: 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 #3: Determine the enthalpy of reaction for the following:

H2(g) + 12O2(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 12 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 #4: Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows:

4NH3(g) + 7O2(g) ---> 4NO2(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):

BondBond Energy
O−H464
N−H389
O=O498

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 #5: 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:

BondBond Energy
C−C347
C=C611
C−H414
C=O736
O=O498
O−H464

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 #6: 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.


Problem #7: Calculate the bond dissociation energy for one mole of O-F bonds, given the following data. (Hint: oxygen is the central atom of OF2)

F-F bond dissociation energy = 159 kJ
O=O bond dissociation energy = 498 kJ

F2(g) + 12O2(g) ---> OF2 (g) ; ΔH = 28 kJ

Solution:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

reactant bonds broken ---> one F-F; one-half O=O <--- note: 12, not 1

product bonds broken ---> two O-F

28 = [(1) (159) + (12) (498)] - 2x

28 = 408 - 2x

2x = 380

x = 190 kJ


Problem #8: Using the following bond enthalpy (in kJ mol-1) values, determine the heat of formation of methane:

H2 = 436 and C-H = 414

as well as the sublimation energy of C(s, gr) = 713 kJ/mol-1

Solution:

Note the approach to the solution. The bond enthalpy values are each associated with a specific chemical equation. The three equations are then added together to yield the target equation. This is not the usual way in which bond enthalpy values are used.

1) This is the equation needed:

C(s, gr) + 2H2(g) ---> CH4(g)

2) Adding the following equations will yield the equation needed:

Equation (1) is the sublimation energy. It is endothermic.
Equation (2) is the energy needed to break 2 moles of H2 into 2H(g) . Bond breaking is always an endothermic process.
Equation (3) is the formation of four C-H bonds. Bond formation is always exothermic, hence the negative sign.

(1) C(s, gr) ---> C(g)ΔH = +713 kJ
(2) 2H2(g) ---> 4H(g)ΔH = +872 kJ (from 436 x 2)
(3) C(g) + 4H(g) ---> CH4(g)ΔH = -1664 kJ (from 414 x 4)

3) Add the three equations and their enthalpies:

C(s, gr) + 2H2(g) ---> CH4(g)ΔH = -79 kJ

The heat of formation of CH4(g) is -79 kJ/mole

The Chemistry Webbook gives the value as being close to -75 kJ/mol

Source for sublimation of C


Problem #9: An unknown gas, X2, which behaves much like nitrogen gas (N≡N), is analyzed and the following enthalpies of formation are obtained:

X(g) = 412 kJ/mol
H(g) = 217 kJ/mol
X2H4(g) = 3 kJ/mol

The X-H bond energy is known to be 383 kJ/mol. Use this information to estimate the X-X single-bond energy in the X2H4 molecule.

Solution:

1) Write the full chemical equations for each enthalpy of formation:

12X2(g) ---> X(g)ΔH = 412 kJ/mol
12H2(g) ---> H(g)ΔH = 217 kJ/mol
X2(g) + 2H2(g) ---> X2H4(g)ΔH = 3 kJ/mol

2) Rewrite the first two equations from the step just above:

X2(g) ---> 2X(g)ΔH = 824 kJ/mol
H2(g) ---> 2H(g)ΔH = 434 kJ/mol
X2(g) + 2H2(g) ---> X2H4(g)ΔH = 3 kJ/mol

This was done so as to generate an equation where one mole of the reactant is split apart (as opposed to one mole of product being produced, which is what an enthalpy of formation calls for). This gives us the value for the bond enthalpy, when one mole of X2 bonds are broken and when one mole of H2 bonds are broken.

3) We now use the Hess' Law formulation used when bond enthalpies are involved:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

3 = [824 + (2) (434)] - [x + (4) (383)]

3 = [1692] - [x + 1532]

3 = -x + 160

x = 157 kJ/mol


Problem #10: Calculate enthalpy of this reaction:

C(s) + 2H2(g) ---> CH4(g)

given the following two bond dissociation energies:

H-H = 436 kJ/mol
C-H = 415 kJ/mol

and the following enthalpy of formation:

C(g) = +715 kJ/mol

Solution:

1) Think of this reaction in two steps:

C(s) ---> C(g) ΔHf = +715 kJ

then:

C(g) + 2H2(g) ---> CH4(g)

2) Calculate the ΔH of the second reaction in step one of the solution using bond enthalpies:

2(436) - 4(415) = -788 kJ

3) The sum of the two reactions in step one of the solution is the reaction we wish. Add the two enthalpies for the answer:

+715 + (-788) = -73 kJ

C(s) + 2H2(g) ---> CH4(g) ΔH = -73 kJ


Bonus Problem: Calculate ΔH for this reaction:

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

using the following data:

BondBond Energy
C−C347
C=C611
C−H414
C−O360
O−H464

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:

SubstanceΔH°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."

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

Go to Hess' Law - using bond enthalpies

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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