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 brokenΔH = [(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 ---> CH4 ΔH = −1652 kJ/mol 2C + 6H ---> C2H6 ΔH = −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 1⁄2F2(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 kJwhich, 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:
1⁄2Cl2(g) + 1⁄2O2(g) ---> ClO(g); ΔH = ???
2) Let's see what we know:
(a) we know an enthalpy:1⁄2Cl2(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 broken102.5 = [1⁄2(x) + 498] − [(2) (243)]
102.5 = 1⁄2(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 brokenx = [1⁄2(181) + 1⁄2 (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 1⁄2H2(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 kJHere 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:
Gas std. 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 1⁄2X2(g) ---> X(g) +389 kJ 1⁄2H2(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 + H2OLots 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 bondsbonds 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 brokenΔH = [(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. For one mole of benzene it is −3282 kJ
This Webbook entry for benzene gives its enthalpy for combustion as −3267 ± 20. kJ/mol.