1. Lesson 4
Bond Enthalpies

2. Refresh Using the equations below:
C(s) + O2(g) → CO2(g) ∆Ho = –394 kJ mol-1
Mn(s) + O2(g) → MnO2(s) ∆Ho = –520 kJ mol-1
What is ∆H, in kJ, for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
914
126
–126
–914

3. We Are Here

4. Lesson 4: Bond Enthalpies
Objectives:
Understand bond enthalpies and what they mean
Use bond enthalpies to calculate enthalpy changes
Use bond enthalpies to help you predict the outcome of an experiment

5. Playing with Magnets Take two magnets (preferably round ones)
Place them near each other and allow them to attract each other.
Is this an endothermic or an exothermic process? How do you know?
Pull your two magnets apart.
Experiment with magnets of different strengths.
How do your energy changes relate to the strength of the magnets?
What has all this got to do with chemical bonds?
Are there any limits to this analogy (places where it breaks down)?

6. Average Bond Enthalpies
Bond enthalpies are defined in terms of breaking bonds
What are some of your favourite bonds?
Check their values in Table 10 of the Data Booklet
Try to come up with a suitable definition for bond enthalpies
Include all relevant detail: amounts, states, conditions and so on
What sign do bond enthalpies have to be and why?

7. The bond breaking reaction:
For O=O: O2(g)  2O(g) H = 498 kJ mol-1
For H-F: HF(g)  H(g) + F(g) H = 568 kJ mol-1
For O-H: ½H2O(g)  H(g) + ½O(g) H = 464 kJ mol-1
NOT: H2O(l)  2H (g) + O (g)
Note: The decomposition of water involves breaking 2 O-H bonds. Bond enthalpies given in tables are averaged values.
From gaseous covalent bonds to gaseous atoms

8. Average Bond Enthalpy
The average energy required to break one mole of a chemical bond to produce two moles of gaseous atoms at 298K.
Bond enthalpies are always positive
These are average values, they do not take into account:
Variations in strengths of the same bond in different compounds
For example: the strength of a C-H bond will vary slightly depending on whether it is in methane, ethene, glucose and so on
Energy changes due to state changes (from intermolecular forces)
Energy changes due to dissolution in water
This means enthalpy changes calculated from bond enthalpies will generally be less accurate than those calculated by other means

9. Bond Enthalpy and Enthalpy Change
As the enthalpy increases, bonds are being broken, as it decreases, bonds are being made. H
Explain in terms of bond enthalpies why some reactions are endothermic and others are exothermic…
∆H is negative
Weaker bonds broken
Stronger bonds made
∆H is positive
Stronger bonds broken
Weaker bonds made

10. Bond Enthalpies in Calculations
Determine the enthalpy change for the following reaction
H2(g) Cl2(g)  2 HCl(g)
H2(g) Cl2(g)  HCl(g)
2 H(g) Cl(g)
∆H = 679 – 864 = -185 kJ mol-1
Note: the experimentally measured value is also -185 kJ mol-1
1 x H-H + 1 x Cl-Cl
= = 679
2 x H-Cl
= 2 x 432 = 864
Delta H = Bond energy (bonds broken) – Bond energy (bonds formed)

11. Bond Enthalpies in Calculations
Determine the enthalpy change for the following reaction using bond enthalpies.
CH4(g) O2(g)  CO2(g) + 2H2O(l)
∆H = BONDS BROKEN BONDS FORMED
= (4 x C-H + 2 x O=O) (2 x C=O + 4 x O-H)
= (4 x x 498) (2 x x 464)
= 2648 – 3348
= -700 kJ mol-1
Note: the experimentally measured value is -890 kJ mol-1
Why the big difference?

12. Bond Enthalpies in Calculations – Method 2
Determine the enthalpy change for the following reaction using the enthalpies of formation.
CH4(g) O2(g)  CO2(g) + 2H2O(l)
∆H = PRODUCTS REACTANTS
= ( x -286) - ( x 0) =
= -891 kJ mol-1
Note: the experimentally measured value is -890 kJ mol-1
Why the big difference?

13. Bond Enthalpies in Practice
Hydrogen peroxide decomposes to form water and oxygen.
In this experiment, you will design a method to measure the enthalpy change and compare it to one calculated from bond enthalpies.
Follow the instructions here

14. Key Points
Bond enthalpies tell you the amount of energy required to break a bond in gaseous molecules
Bond enthalpies are always positive (endothermic)
Enthalpies of reaction calculated from bond enthalpies are less accurate because the value of the bond enthalpies are averages of the bonds being broken.

15. More Practice
Consider the reaction of chlorine with methane to produce dichloromethane and hydrogen chloride:
CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g)
Determine the enthalpy change for the following reaction using bond enthalpies.

16. Answer
In this reaction, two Cl–Cl bonds and two C–H bonds are broken, and two new C–Cl and H–Cl bonds are formed. The net change associated with the reaction is
2(C–H) + 2(Cl–Cl) – 2(C–Cl) – 2(H–Cl) = ( –660 – 864) kJ
which comes to –208 kJ per mole of methane; this agrees quite well with the observed heat of reaction, which is – 202 kJ/mol.