Hess' Law of Constant Heat Summation
Using Bond Enthalpies
Problems 11 - 20

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

Problem #11: 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


Problem #12: 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


Problem #13: 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


Problem #14: 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


Problem #15: 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.


Problem #16: 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.


Problem #17: Estimate the bond enthalpy for the C−C bond in ethane, given the following information:

Standard enthalpy of formation of ethane = −85 kJ/mole
Mean bond enthalpy of C−H = +413 kJ/mole
Enthalpy of atomization for carbon = +717 kJ/mole
Enthalpy of atomization for hydrogen = +218 kJ/mole

Solution:

1) Write out the information using chemical equations:

2C(s, gr) + 3H2(g) ---> C2H6(g)ΔHf = −85 kJ
C(s, gr) ---> C(g)ΔHa = +717 kJ
12H2(g) ---> H(g)ΔHa = +218 kJ

Here is a question about what the enthalpy of atomization represents.

2) This is the reaction we want:

C2H6(g) ---> 2C(g) + 6H(g)

This equation represents the breaking of one C−C and six C−H bonds. The C−C bond enthalpy is the one we want.

3) Adjust the given reactions as follows:

eq 1 ---> flip
eq 2 ---> multiply by 2
eq 3 ---> multiply by 6

4) Resulting in:

C2H6(g) ---> 2C(s, gr) + 3H2(g)ΔHf = +85 kJ
2C(s, gr) ---> 2C(g)ΔHa = +1434 kJ
3H2(g) ---> 6H(g)ΔHa = +1308 kJ

5) Add the three enthalpies:

85 + 1434 + 1308 = +2827 kJ

6) We must remove the enthalpy associated with breaking the six C−H bonds:

2827 − (6) (413) = +349 kJ

Here is a table of bond enthalpy values with the 349 value. Other tables use 348.


Problem #18: An unknown gas, X2(g), which behaves much like N2(g) (N≡N), is analysed and the following enthalpies of formation are obtained:

Gasstd. heat of formation
X2H4(g)+137 kJ/mol
X(g)+389 kJ/mol
H(g)+217 kJ/mol

We also know that the X−H bond energy is +342 kJ/mol. Use this information to estimate the X−X bond energy in the X2H4 molecule.

Solution:

1) Write out our information as chemical equations:

X2(g) + 2H2(g) ---> X2H4(g)+137 kJ
12X2(g) ---> X(g)+389 kJ
12H2(g) ---> H(g)+217 kJ

The equations with X2 and H2 are also the equations for the enthalpy of atomization.

2) The reaction we need to determine the enthalpy for is this:

X2H4(g) ---> 2X(g) + 4H(g)

The reaction shows an X−X bond being broken as well as four X−H bonds. The X−X bond enthalpy is the one we want to determine.

3) Adjust the given reactions as follows:

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

4) Resulting in:

X2H4(g) ---> X2(g) + 2H2(g)+137 kJ
X2(g) ---> 2X(g)+778 kJ
2H2(g) ---> 4H(g)+868 kJ

5) Add the three enthalpies:

137 + 778 + 868 = 1783 kJ

6) The enthalpy of breaking the four X−H bonds needs to be removed:

1783 − (4) (342) = 415 kJ

Problem #19: Without using bond dissociation energies, determine an approximate heat of reaction, ΔH in kJ, for the reaction of ethanol with acetic acid to yield ethyl acetate and water.

Solution:

CH3COOH + C2H5OH ---> CH3COOC2H5 + H2O

Lots of bonds stay exactly the same, so you can eliminate them from consideration. For example, there are a number of C−H bonds, all of which remain undisturbed. These can be eliminated from the calculation.

The only bonds broken are a O−H bond in the acetic acid and a C−O bond in the alcohol. Then, for bonds formed, you have only two: a C−O bond in the ethyl acetae and an O−H bond in the water.

Approximate heat of reaction is zero.


Problem #20: Using bond enthalpies, estimate ΔH for this reaction:

2C6H6(g) + 15O2(g) ---> 12CO2(g) + 6H2O(g)

Solution:

1) Bonds broken and formed:

bonds broken (reactants) ---> six C=C bonds, six C−C bonds, twelve C−H bonds, fifteen O=O bonds

bonds broken (products) ---> twenty-four C=O bonds, twelve O−H bonds

The six C=C and six C−C bonds comes from a consideration of the structure of benzene. You can consider the benzene ring to be made of three C=C bonds and three C−C bonds. This isn't actually the case, but it is the closest approximation.

2) Hess' Law:

ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken

ΔH = [(6)(C=C) + (6)(C−C) + (12)(C−H) + (15)(O=O] − [(24)(C=O) + (12)(O−H)]

ΔH = [(6)(612) + (6)(347) + (12)(413) + (15)(498] − [(24)(799) + (12)(464)]

ΔH = (3672 + 2082 + 4956 + 7470) − (19176 + 5568)

ΔH = 18180 − 24744

ΔH = −6564 kJ <--- note that this is the value for two moles of benzene

This Webbook entry for benzene gives its enthalpy for combustion as −3267 ± 20. kJ/mol.


Two links to Yahoo Answers bond enthalpy problems:

Calculate the bond enthalpy of the Cl−Cl bond.

Calculate the bond enthalpy for the N−H bond in NH3


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