1. Objectives
- understand that chemical reactions involve the making and breaking of bonds and the concept of bond enthalpy 
- be able to determine bond enthalpies from data 
- be able to use mean bond enthalpies to estimate ΔH for reactions

2. Bond-dissociation energy (D): the quantity of energy required to break one mole of covalent bonds in a gaseous species, usually expressed in kJ/mol.
Bond Energies
Bond Breakage:
H2(g)  2 H(g) DH = D(H-H) = kJ/mol
Bond Formation:
H2(g)  2 H(g) DH = -D(H-H) = kJ/mol 2

3. Bond Energies
Average bond energy: For example, the energy needed to break a bond in ethane (C2H6) will be different to the energy needed to break a bond in decane (C10H22).
In H2O, more energy is required to break the first bond than the second. The second bond broken is the O-H radical. The O-H bond energy in H2O is the average = kJ/mol.
Bond Breakage:
H-OH (g)  H(g) + OH(g) DH = D(H-OH) = kJ/mol
Bond Breakage:
O-H (g)  H(g) + O(g) DH = D(O-H) = kJ/mol 3

4. Breaking bonds
If we want to break a covalent bond between two atoms, we need to overcome the attractive force.
put energy in
put energy in C C
© Nueyer
Bond breaking is an endothermic process.

5. Bond making - exothermic
Break bond = Energy in

6. Bond breaking - endothermic
Energy is always required to be inputted to break a bond. Bond breaking is always endothermic.
put energy in

7. Bond making - exothermic
Energy is always released when a bond is formed. Bond making is always exothermic.
Bond FORMED
energy OUT (-ve)

8. Making bonds The opposite is true if we want to make new bonds.
Energy is released when new chemical bonds are formed.
Bond making is an exothermic process.
Breaking or making the same chemical bond will require the same energy to be put in or released. H 2H
∆H = 432 kJ 2H H
∆H = –432 kJ

9. Calculation of enthalpy of reaction from bond energies
Hypothetically, thermochemical data can be obtained from: gaseous reactants  gaseous atoms  gaseous products
Calculation of enthalpy of reaction from bond energies
Some of the BE are likely to be average bond energies rather than bond dissociation energies.
The sign changes because ALL bond energies are positive of are bond breakage. Products are not broken they are MADE, so the sign changes to reverse this process. 9

10. Fuse School (5min)

11. Example 1
What is the enthalpy change when hydrogen is added to ethyne to produce ethane?
C2H2 (g) + 2H2 (g) C2H6 (g)
To answer this we must look at what types of bonds must be broken in the reactants and formed in the products.

12. This will require energy to be put in.
C2H2 (g) + 2H2 (g)
C2H2 (g) + 2H2 (g) C2H6 (g)
C2H6 (g)
In this reaction, we must first break all the bonds inside the reactant molecules.
This will require energy to be put in.
Next, new bonds must be formed between the atoms in the product molecule.
This releases energy.
Potential energy
Reaction pathway

13. BOND DISSOCIATION ENTHALPY
Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.
Values Endothermic - Energy must be put in to break any chemical bond
Examples Cl2(g) ———> 2Cl(g) O-H(g) ———> O(g) + H(g)
Notes • ENERGY IN TO MAKE, OUT TO BREAK. Notice they are ALL positive, please.
Mean Values H-H H-F N-N 163
C-C H-Cl N=N 409
C=C H-Br NN 944
CC H-I P-P 172
C-O H-N F-F 158
C=O H-O Cl-Cl 242
C-H H-S Br-Br 193
C-N H-Si I-I 151
C-F P-H S-S 264
C-Cl O-O Si-Si 176
Average (mean) values are quoted because the actual value depends on the environment of the bond i.e. where it is in the molecule
UNITS = kJ mol-1

14. What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g) C2H6 (g) + C H
Bond Breaking
Bond Making
1 x C C
= 835 C
1 x
= 346
2 x C H
= 2 x 414
= 828 C H
6 x
= 6 x 414
= 2484
2 x H H
= 2 x 432
= 864
Total put in = 2527 kJ
Step Five:
Calculate the total energy put in breaking bonds and total energy given out making new bonds.
Total given out = –2830 kJ
Remember: Bond breaking is an endothermic process
Remember: Bond making is an exothermic process

15. What is the enthalpy change when hydrogen is added to ethyne, producing ethane?
C2H2 (g) + 2H2 (g) C2H6 (g) + C H
Bond Breaking
Bond Making
1 x C C
= 835 C
1 x
= 346
2 x C H
= 2 x 414
= 828 C H
6 x
= 6 x 414
= 2484
2 x H H
= 2 x 432
= 864
Total put in = 2527 kJ
Total given out = –2830 kJ ∆H =
2527 +
(–2380)
Step Six:
Calculate the enthalpy change for the reaction. =
2527 –
2380 =
– 303 kJ mol-1

16. Calculating ΔH
CH4(g) + 2O2(g) 2H2O(l) + CO2(g)

17. Calculating ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bonds broken = 4 x (C-H) + 2 x (O=O)
= 4 x x 498
= = 2658 KJ/mol

18. Calculating ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bonds broken = 4 x (C-H) + 2 x (O=O)
= 4 x x 498
= = 2658 KJ/mol
Bonds made = 4 x (O-H) + 2 x (C=O)
= 4 x x -805
= = KJ/mol

19. Calculating ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bonds broken = 4 x (C-H) + 2 x (O=O)
= 4 x x 498
= = 2658 KJ/mol
Bonds made = 4 x (O-H) + 2 x (C=O)
= 4 x x -805
= = KJ/mol
Overall Energy change = = -808 KJ/mol
(Exothermic)

20. Enthalpy Cycles and Hess’s Law
Think of a hike to your nice wood cabin in the French Alps.
The wood cabin is your destination – but you can either walk through the forest or over the mountains. Either way you will still end up at the cabin!

21. Calculate the potential energy of each climber taking route 1 and route 2
+87KJ
+102KJ
-163KJ
-193KJ
+52KJ
+125KJ
-147KJ
-269KJ
+7KJ
-137KJ
Regardless of the route the climber and the miner took they ended up having the same amount of potential energy!!

22. Hess’ Law
The enthalpy change of a reaction is independent of the pathway of that reaction
i.e. All that matters is the start and finish points
Note: add when going ‘with’ an arrow, subtract when going against an arrow.
The first law of thermodynamics relates to the conservation of energy. It is sometimes expressed in the following form: Energy cannot be created or destroyed, it can only change form.
A + B
C + D
E + F
∆H1
∆H2
∆H3
∆H1 = ∆H2 + ∆H3
∆H1
A + B
C + D
∆H2
∆H4
∆H3
E + F
G + H
∆H1 = ∆H2 - ∆H3 + ∆H4

23. Energy Diagrams https://www.youtube.com/watch?v=YV0vgbs7vUk
This clip is 20 minutes long

24. HESS’S LAW HR = H1 + H2 THE TOTAL ENTHALPY CHANGE OF A REACTION
CYCLE 1
THE TOTAL ENTHALPY CHANGE OF A REACTION
IS INDEPENDENT OF THE REACTION ROUTE.
HR
REACTANTS
PRODUCTS
Alternative route
Direct measurement of ΔHR may not be possible because :
H1
H2
INTERMEDIATES
1. Reaction incomplete
2. Other reactions occur
HR =
H1 + H2
By Hess’s Law:
3. Reaction too slow
Intermediates =
elements if ΔHf known,
oxides if ΔHC known,
or gaseous atoms if E[X-Y] (Bond Enthalpy) known

25. The total enthalpy change for route 1 is the same as for route 2
HESS’S LAW
EXAMPLE OF CYCLE 1
Hess’s Law states that the total enthalpy change is independent of the route taken
ΔHr = -66.4kJmol-1
2NO2(g)
N2(g) +2O2(g)
Route 1
This is a thermo-chemical cycle
+114.4kJ
-180.8kJ
Route 2
2NO(g) + O2(g)
Route 2
ΔHr = (-180.8) = -66.4kJmol-1
The total enthalpy change for route 1 is the same as for route 2

26. HESS’S LAW HR = -H1 + H2 HR REACTANTS PRODUCTS H1 H2
CYCLE 2
We can consider situations where the conversions are the other way round. In this case the intermediate will convert to both reactants and products. (Look at the direction of the arrows!)
HR
REACTANTS
PRODUCTS
Alternative route
Pathways:
I  R = H1
I  P = H2
H1
H2
INTERMEDIATES
HR =
-H1 + H2
Therefore By Hess’s Law:
Intermediates =
elements if ΔHf known,
oxides if ΔHC known,
or gaseous atoms if E[X-Y] (Bond Enthalpy) known

27. Enthalpy of reaction from enthalpies of formation
EXAMPLE OF CYCLE 2
Enthalpies of formation:water-286,NO +33 and nitric acid -173 kJ mol-1.
2H2O(l) + 4NO2(g) +O2(g)  HNO3(l)
4 x (-173)
or -692kJ
2 x (-286)+4x(+33)+0
Which equals -440kJ
2H2 + 6O2 + 2N2
DH°r = kJ kJ or -692kJ kJ
ANSWER = kJ
DH =  DHf of products –  DHf of reactants

28. HESS’S LAW HR = H1 + -H2 HR REACTANTS PRODUCTS H1 H2
CYCLE 3
We can consider situations where the conversions are the other way round. In this case the reactants and products will both convert to intermediates. (Look at the direction of the arrows!)
HR
REACTANTS
PRODUCTS
Alternative route
Pathways:
R  I = H1
P  I = H2
H1
H2
INTERMEDIATES
HR =
H H2
Therefore By Hess’s Law:
Intermediates =
elements if ΔHf known,
oxides if ΔHC known,
or gaseous atoms if E[X-Y] (Bond Enthalpy) known

29. Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
EXAMPLE OF CYCLE 3
Calculate the enthalpy change for the hydrogenation of ethene
Alternative route
DH2 1 x C=C bond @ = kJ
4 x C-H bonds @ = kJ
1 x H-H bond @ = kJ
Total energy to break bonds of reactants = kJ
DH3 1 x C-C bond @ = kJ
6 x C-H bonds @ = kJ
Total energy to break bonds of products = kJ
Applying Hess’s Law DH1 = DH DH3 = (2699 – 2824) = – 125 kJ

30. Enthalpy of reaction from bond enthalpies
energy released making bonds (OUT) EXOTHERMIC
energy used to break bonds (IN) ENDOTHERMIC
Step 1 Energy is put in to break bonds
Step 2 The gaseous atoms then combine to form bonds and energy is released
Applying Hess’s Law DHr = Step Step 2
Or DH =  bond enthalpies of reactants  bond enthalpies of products

31. Using Hess’s Law
Although carbon and hydrogen do not combine directly to form propane, C3H8, the enthalpy change for the reaction is:
Alternative route

32. Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
DH2 1 x C=C bond @ = kJ
4 x C-H bonds @ = kJ
1 x H-H bond @ = kJ
Total energy to break bonds of reactants = kJ
DH3 1 x C-C bond @ = kJ
6 x C-H bonds @ = kJ
Total energy to break bonds of products = kJ
Applying Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ

33. Enthalpy of reaction from bond enthalpies
energy released making bonds (OUT) EXOTHERMIC
energy used to break bonds (IN) ENDOTHERMIC
Step 1 Energy is put in to break bonds
Step 2 The gaseous atoms then combine to form bonds and energy is released
Applying Hess’s Law DHr = Step Step 2
Or DH =  bond enthalpies –  bond enthalpies
of reactants of products

34. Hess’s Law and Enthalpy of Formation and Combustion
Learning Objectives:
Define Hess’s Law
Using calculations show that enthalpies for formation of products are the same regardless of the route taken
Calculate change in enthalpy of reactions using enthalpy of formation or combustion data

35. Calculations linked to Enthalpy Cycles (Hess’s Law)
Enthalpy of Formation
Enthalpy of Combustion
Bond Enthalpy
You must know the difference between these!

36. Hess’ Law
The enthalpy change of a reaction is independent of the pathway of that reaction
i.e. All that matters is the start and finish points
Note: add when going ‘with’ an arrow, subtract when going against an arrow.
The first law of thermodynamics relates to the conservation of energy. It is sometimes expressed in the following form: Energy cannot be created or destroyed, it can only change form.
A + B
C + D
E + F
∆H1
∆H2
∆H3
∆H1 = ∆H2 + ∆H3
∆H1
A + B
C + D
∆H2
∆H4
∆H3
E + F
G + H
∆H1 = ∆H2 - ∆H3 + ∆H4