Solutions for Bond Energies Practice Problems
23. Consider the following equation for the combustion of butane.
C4H10 (g) 13/2 O2 (g) → 4CO2(g) 5H2O(g)
(a) Use bond energies to estimate the enthalpy change for the reaction as written.
i)
For the reactants (+), there are:
1 mole of (3 C-C bonds)
1 mole of (10 C-H bonds)
13/2 moles of (1 O=O bond)
ii)
= 1 (3 x 347 kJ)
= 1 (10 x 338 kJ)
= 6.5 (1 x 498 kJ)
= 1041 kJ
= 3380 kJ
= 3237 kJ
7658 kJ
For the products (-), there are:
4 moles of (2 C=O bonds)
5 moles of (2 O-H bonds)
= - 5960 kJ
= -4600 kJ
- 10560 kJ
Note: Bond energy values from the data booklet were used in the above calculations.
iii)
= 4 (2 x -745 kJ)
= 5 (2 x -460 kJ)
ΔHrxn= Σ +bond energies (breaking) + Σ - bond energies (making)
= 7658 kJ + - 10560 kJ = - 2902 kJ
Thus, 2902 kJ of energy is released in this exothermic combustion reaction.
(b) Compare your answer to the accepted value using the standard molar enthalpy of combustion of butane.
Note: If we look up the value of combustion of butane gas C4H10 in Table 16.3, the ∆H value is given as
-3003.0 kJ/mol. Note: The calculated answer is not always exactly the same as that indicated in the enthalpy of
formation or of combustion tables. Bond energies are averages, so they are used only to estimate the energy
of a reaction.
24. Consider the following equation for the formation of ammonia from its elements.
N2(g) 3H2 (g) → 2NH3(g)
(a) Use bond energies to estimate the enthalpy change for the reaction as written (see Appendix E).
i)
For the reactants (+), there are:
1 mole of (1 N = N bonds)
3 moles of (1 H-H bonds)
ii)
= 945 kJ
= 1308 kJ
2253 kJ
For the products (-), there are:
2 moles of (3 N-H bonds)
iii)
= 1 (945 kJ)
= 3 (436 kJ)
= 2 (3 x -339 kJ)
= - 2034 kJ
ΔHrxn= Σ +bond energies (breaking) + Σ - bond energies (making)
= 2253 kJ + - 2034 kJ = 219 kJ
Thus, 219 kJ of energy is absorbed in this endothermic formation reaction.
(b) Compare your answer to the accepted value using the standard molar enthalpy of formation of ammonia
From the photocopied table, we see that ΔHf = –46.1 kJ/mol. This value differs considerably from the
calculated value of 219 kJ in (a). This is an indication that the bond energies are average values and give
only an estimate of the measured value for enthalpy of formation.
25. (a) Use bond energies to estimate the standard molar enthalpy of formation of gaseous water.
H2 (g) ½ O2 (g) → H2O(g)
i)
For the reactants (+), there are:
1 mole of (1 H-H bond)
½ mole of (1 O=O bond)
ii)
= 436 kJ
= 249 kJ
685 kJ
For the products (-), there are:
1 mole of (2 O-H bonds)
iii)
= 1 (436 kJ)
= ½ (498kJ)
= 1 (2 x -460 kJ)
= -920 kJ
ΔHrxn= Σ +bond energies (breaking) + Σ - bond energies (making)
= 685 kJ + - 920 kJ = - 235 kJ
Thus, 235 kJ of energy is released in this exothermic formation reaction.
(b) Compare your answer to the accepted value.
From the photocopied table, we see that ΔHf (H2O(g)) = -241.8 kJ/mol. This value is very close to the -235
kJ value calculated in (a)
26. Consider the following equation for the reaction of methane with chlorine.
CH4(g) 3Cl2(g) → CHCl3(g) 3HCl(g)
(a) Use bond energies to estimate the enthalpy change for the reaction.
i)
For the reactants (+), there are:
1 mole of (4 C-H bonds)
3 moles of (1 Cl-Cl bond)
ii)
= 1 (4 x 338 kJ)
= 3 (243 kJ)
For the products (-), there are:
=1352 kJ
= 729 kJ
2081 kJ
1 mole of (3 C-Cl bonds)
1 mole of (C-H bond)
3 moles of (H-Cl bonds)
= 1 (3 x 397 kJ)
= 1 (338 kJ)
= 3 (432 kJ)
= 1191 kJ
= 338 kJ
= 1296 kJ
-2825kJ
Note: Bond energy values from the data booklet were used in the above calculations.
iii)
ΔHrxn= Σ +bond energies (breaking) + Σ - bond energies (making)
= 2081 kJ + - 2825 kJ = - 744 kJ
Thus, 744 kJ of energy is released in this exothermic reaction.
(b) Use standard molar enthalpies to determine the enthalpy change for the reaction.
(For gaseous CHCl3, H°f 103.18 kJ/mol.) Compare your results from (a) and (b).
CH4(g) 3Cl2(g) → CHCl3(g) 3HCl(g)
Known - Data table information:
∆H f ° of CH4 = - 74.8 kJ/mol
∆H f ° of Cl2 = 0 kJ/mol (by definition)
∆H f ° of CHCl3 =103.18 kJ/mol
∆H f ° of HCl = -92.3 kJ/mol
Solution:
CH4(g) 3Cl2(g) → CHCl3(g) 3HCl(g)
1 mol (- 74.8 kJ/mol) + 3 mol (0 kJ/mol) → 1 mol (103.18 kJ/mol) + 3 mol (-92.3 kJ/mol)
- 74.8 kJ → -103.18 kJ + -276.9 kJ
- 74.8 kJ → -380.1 kJ
∆H ° = ∆H°final - ∆H f °initial = (-380.1 kJ) – (-74.8 kJ ) = -305.3 kJ