1. Enthalpy (H)
The heat transferred sys ↔ surr during a chemical constant P
Can’t measure H, only ΔH
At constant P, ΔH = q = mCΔT, etc.
Literally, ΔH = Hproducts - Hreactants
ΔH = + (endothermic)
Heat goes from surr into sys
ΔH = - (exothermic)
Heat leaves sys and goes into surr

2. Reactant + Energy Product
In this example, the energy of the system (reactants and products) ↑, while the energy of the surroundings ↓
Notice that the total energy does not change
Reactant + Energy Product
Endothermic Reaction
Surroundings
Surroundings
System
Energy
System
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

3. Reactant Product + Energy
In this example, the energy of the system (reactants and products) ↓, while the energy of the surroundings ↑
Notice again that the total energy does not change
Reactant Product + Energy
Exothermic Reaction
Surroundings
Surroundings
System
System
Energy
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

4. Reaction Coordinate Diagrams: Endothermic Reaction
Activation Energy Ea
D(PE)
Products
ΔHrxn= + PE
Reactants
Progress of the Reaction

5. Reaction Coordinate Diagrams: Exothermic Reaction
Activation Energy Ea
ΔHrxn= -
Reactants
D(PE) PE
Products
Progress of the Reaction

6. Reaction Coordinate Diagrams
Draw the reaction coordinate diagram for the following rxn:
C(s) + O2(g)  CO kJ
Activation Energy
EXOTHERMIC Ea
ΔHrxn= kJ
C + O2
D(PE) PE
CO2
Progress of the Reaction

7. Enthalpies of Reaction
All reactions have some ΔH associated with it
H2(g) + ½ O2(g) → H2O(l) ΔH = kJ
How can we interpret this ΔH?
Amount of energy released or absorbed per specific reaction species
Use balanced equation to find several definitions kJ
1 mol H2 kJ
½ mol O2 kJ
1 mol H2O or or
Able to use like conversion factors in stoichiometry

8. Enthalpies of Reaction
Formation of water
H2(g) + ½ O2(g) → H2O(l) ΔH = kJ
ΔH is proportional to amount used and will change as amount changes
2H2(g) + O2(g) → 2H2O(l)
For reverse reactions, sign of ΔH changes
2H2O(l) → 2H2(g) + O2(g)
Treat ΔH like reactant or product
ΔH = kJ
ΔH = kJ
H2(g) + ½ O2(g) → H2O(l) kJ (exo)

9. Enthalpies of Reaction Practice
Consider the following rxn:
C(s) + 1/2O2(g)  CO kJ
Is the ΔH for this reaction positive or negative?
NEGATIVE (E released as a product)
What is the ΔH for 2.00 moles of carbon, if all the carbon is used?
2.00 mol C kJ
= kJ
1 mol C
What is the ΔH if 50.0g of oxygen is used?
50.0 g O2
1 mol O2 kJ
= kJ
32.0 g O2
0.5 mol O2
What is the ΔH if 50.0 g of carbon monoxide decompose, in the reverse reaction?
50.0 g CO
1 mol CO
458.1 kJ
= kJ
28.0 g CO
1 mol CO

10. Why? Because enthalpy is a state function
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps
Why? Because enthalpy is a state function
Victor Hess
To review:
1. If a reaction is reversed, ΔH is also reversed
2 CH4 + O2  2 CH3OH ΔHrxn = -328 kJ
2 CH3OH  2 CH4 + O2 ΔHrxn = +328 kJ
2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer
CH4 + 2 O2  CO2 + 2 H2O ΔHrxn = kJ
2(CH4 + 2 O2  CO2 + 2 H2O) ΔHrxn = kJ

11. Example: Methanol-Powered Cars
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g) ΔHrxn = ?
- ( )
2 CH4(g) + O2(g)  2 CH3OH(l) ΔHrxn = -328 kJ
- ( )
2 ( )
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) ΔHrxn = kJ
2 ( )
2 CH3OH(l)  2 CH4(g) + O2(g) ΔHrxn = +328 kJ 3
2 CH4(g) + 4 O2(g)  2 CO2(g) + 4 H2O(g) ΔHrxn = kJ
2 CH3OH + 3 O2  2 CO2 + 4 H2O ΔHrxn = kJ

12. Tips for applying Hess’s Law…
Look at the final equation that you are
trying to create first…
Find a molecule from that eq. that is only in one of the given equations (i.e. CH3OH, CO2)
Make whatever alterations are necessary to those
Once you alter a given equation, you will not alter it again
Continue to do this until there are no other options
Next, alter remaining equations to get things to cancel that do not appear in the final equation

13. * * 1. Given the following data:
S(s) + 3/2O2(g) → SO3(g) ΔH = kJ
2SO2(g) + O2(g) → 2SO3(g) ΔH = kJ .
Calculate ΔH for the following reaction:
S(s) + O2(g) → SO2(g)
- ½ ( )
- ½ ( ) *
S(s) + 3/2O2(g) → SO3(g) ΔH = kJ
2SO3(g) → O2(g) + 2SO2(g) ΔH = kJ *
SO3(g) → ½ O2(g) + SO2(g) ΔH = kJ
S(s) + O2(g) → SO2(g)
ΔH = kJ

14. * * * 2. Given the following data:
C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = kJ
C(s) + O2(g) → CO2(g) ΔH = -394 kJ
H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ
Calculate ΔH for the following reaction:
2C(s) + H2(g) → C2H2(g)
-( )
2( )
2( ) *
2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ *
2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g) ΔH = kJ *
H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ
2C(s) + H2(g) → C2H2(g) ΔH = +226 kJ

15. * * * 3. Given the following data: 2O3(g) → 3O2(g) ΔH = - 427 kJ -½( )
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = kJ
Calculate ΔH for the following reaction:
NO(g) + O(g) → NO2(g)
-½( )
-½( )
-½( )
-½( ) *
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = kJ *
O(g) → ½ O2(g) ΔH = kJ *
3/2 O2(g) → O3(g) ΔH = kJ
NO(g) + O(g) → NO2(g) ΔH = kJ

16. 4. Given the following data:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -23 kJ
3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g) ΔH = -39 kJ
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ΔH = 18 kJ
Calculate ΔH for the following reaction:
FeO(s) + CO(g) → Fe(s) + CO2(g)

17. Hess’s Law HW Questions
1. A  B ∆H = kJ
C  B ∆H = kJ
Calculate ΔH for the following reaction: A  C
2. Suppose you are given the following reactions:
4X  2Y DH = kJ
X  ½ Z DH = kJ
Calculate ΔH for the following reaction: Y  Z

18. 3. From the following heats of reaction:
2 H2 (g) + O2 (g)  2 H2O (g) DH = kJ
3 O2 (g)  2 O3 (g) DH = kJ
Calculate the heat of the reaction (∆H):
3 H2 (g) + O3 (g)  3 H2O (g)

19. 4. From the following enthalpies of reaction:
H2 (g) + F2 (g)  2 HF (g) DH = kJ
C (s) F2 (g)  CF4 (g) DH = kJ
2 C (s) H2 (g)  C2H4 (g) DH = kJ
Calculate the DH for the reaction of ethylene with F2.
C2H4 (g) + 6F2 (g)  2 CF4 (g) HF(g)

20. 5. Given the following data:
N2 (g) + O2 (g)  2 NO (g) DH = kJ
2 NO (g) + O2 (g)  2 NO2 (g) DH = kJ
2 N2O (g)  2 N2 (g) + O2 (g) DH = kJ
Calculate DH for the reaction below:
N2O (g) + NO2 (g)  3 NO (g) DH = ?

21. Heats of Formation, ΔH°f
The enthalpy change when one mole of a compound is formed from the elements in their standard states
° = standard conditions
Gases at 1 atm pressure
All solutes at 1 M concentration (remember M = mol/L)
Pure solids and pure liquids
f = a formation reaction
1 mole of product formed
From the elements in their standard states (1 atm, 25°C)
For all elements in their standard states, ΔH°f = 0
What’s the formation reaction for adrenaline, C9H12NO3(s)? 9
Cgr + 6
H2(g) +
1/2
N2(g) +
3/2
O2(g)
 C9H12NO3(s)

22. Thermite Reaction Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l) ΔHrxn = ?
Welding railroad tracks

23. ΔHrxn = nΔH°f(products) - nΔH°f(reactants)
Thermite Reaction
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
Reactants
Elements
(standard states)
Products
Fe2O3(s)
2 Fe(s)
2 Fe(l)
2 Al(s)
3/2 O2(g)
Al2O3(s)
2 Al(s)
ΔHrxn = 2ΔH°f(Fe(l)) + ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s))
ΔHrxn = 2(15 kJ) + (-1676 kJ) - (-822 kJ) – 2(0)
ΔHrxn = kJ
ΔHrxn = nΔH°f(products) - nΔH°f(reactants)

24. ∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
1. CH4(g) + 2 Cl2(g)  CCl4(g) + 2 H2(g)
ΔHrxn = ?
(- 74.8) 2
(0)
( ) 2
(0)
∆H = [(-106.7) + 0] – [(-74.8)+0] =
= kJ
2. 2 KCl(s) + 3 O2(g)  2KClO3(s)
ΔHrxn = ? 2
( ) 3
(0) 2
( )
∆H = [(2)( )] – [(2)( ) + (3)(0)] =
= kJ

25. 3. AgNO3(s) + NaCl (aq)  AgCl(s) + NaNO3(aq)
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
3. AgNO3(s) + NaCl (aq)  AgCl(s) + NaNO3(aq)
ΔHrxn = ?
(-124.4)
(-127.0)
(-446.2)
(-407.1)
∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)] =
= kJ
4. C2H5OH(l) + 7/2 O2(g)  2CO2(g) + 3H2O(g)
ΔHrxn = ?
(-277.7)
(7/2)
(0)
(2)
(-393.5)
(3)
(-241.8)
∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)] =
= kJ

26. Enthalpy Review
#2. Calculate H for the following reaction: N2H4 (g) + O2 (g) → N2 (g) + 2H2O (g) Given: H (kJ/mol) 2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (g) N2O (g) + 3 H2 (g) → N2H4 (g) + H2O (g) NH3 (g) + ½ O2 (g) → N2H4 (g) + H2O (g) -143 H2 (g) + ½ O2 (g) → H2O (g) -286

27. Bond Energies Chemical reaction ⇔ Bond breakage & bond formation
Bond energy = energy required to break a bond
Bond breaking is endothermic (raises potential energy)
Bond formation is exothermic (lowers PE)
Average energy for one type of bond in different molecules
Common Bond Energies
C-H : 413 kJ/mol C=O : 799 kJ/mol
O=O : 495 kJ/mol O-H : 467 kJ/mol
ΔHrxn = (bonds broken) – (bonds formed)
Energy required
Energy released

28. Bond Energies ex. CH4 + 2 O2  CO2 + 2 H2O
ΔHrxn = (bonds broken) – (bonds formed)
ex. CH O2  CO2 + 2 H2O
C-H : 413 kJ/mol C=O : 799 kJ/mol
O=O : 495 kJ/mol O-H : 467 kJ/mol
ΔHrxn = [4(C-H) + 2(O=O)] – [2(C=O) + 4(O-H)]
ΔHrxn = [4(413 kJ) + 2(495 kJ)] – [2(799 kJ) + 4(467 kJ)]
ΔHrxn = -824 kJ
Compare to ΔHrxn = kJ

30. Enthalpy Summary Enthalpy (ΔH) rxn = heat = q
All rxns have some ΔH
ΔH = +…endo
ΔH = - …exo
If given ΔH, can use stoich to quantify
Three ways to estimate ΔH (if not given)
Hess’s Law
known eq’s manipulated into desired eq.  ΔHrxn
Heats of Formation (ΔHf): values from appendix
ΔHrxn = nΔH°f(products) - nΔH°f(reactants)
Bond energies
ΔHrxn = bonds broken – bonds formed