1. Thermochemistry
Part 2 – enthalpy

2. Endothermic or Exothermic?

3. Endothermic or Exothermic?
A negative ∆H indicates energy is transferred from the system to the surroundings therefore it is exothermic

4. Endothermic or Exothermic?
A positive ∆H indicates energy is transferred from the surroundings to the system therefore it is endothermic

5. q = mCDT Heat (q) absorbed or released:
The specific heat (C) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
Heat (q) absorbed or released:
q = mCDT
q: is heat, Joules or kilojoules
m: mass in grams
C: specific heat capacity
DT: change in temperature,°C

6. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ

7. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ

8. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ

9. Thermochemical Equations
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ

10. Thermochemical Equations
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
266 g P4 = 1

11. Thermochemical Equations
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
266 g P4
1 mol P4 = 1
123.9 g P4

12. Thermochemical Equations
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
266 g P4
1 mol P4
3013 kJ = 1
123.9 g P4
1 mol P4
This is like the mole-to-mole ratio,
except it uses ratio of DH to mole P4

13. Thermochemical Equations
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
266 g P4
1 mol P4
3013 kJ
= 6470 kJ 1
123.9 g P4
1 mol P4

14. Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy
, pressure, volume, temperature
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

15. Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

16. Calculating the enthalpy of reaction (DH):
aA + bB cC + dD DH
dDH(D)
cDH(C) = [ + ]
bDH(B)
aDH(A)
DH0
rxn
nDH0 (products) f = S
nDH0 (reactants) -
∑n is the sum of moles
∆H0f is the heat of formation provided in tables
Substances in their elemental state, such as metals and diatomic molecules, have a ∆H0f = 0 kJ/mol

17. Benzene (C6H6) burns in air to produce carbon dioxide and liquid water
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)

18. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
nDH0 (reactants) -

19. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
nDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
*Remember: diatomic molecules are in their “elemental state” and will have a ∆H0f = 0 kJ/mol which is why O2 is NOT included in the ∆H0rxn calculation!

20. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
nDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[12(–393.5 kJ) + 6(–187.6 kJ)] – [2(49.04 kJ)] = kJ
Substance
DH0 f (kJ/mol)
CO2
-393.5
H2O
-187.6
C6H6
49.04 O2

21. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
nDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[12(–393.5 kJ) + 6(–187.6 kJ)] – [2(49.04 kJ)] = kJ
-5946 kJ
2 mol
= kJ/mol C6H6

22. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
nDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[12(–393.5 kJ) + 6(–187.6 kJ)] – [2(49.04 kJ)] = kJ
-5946 kJ
2 mol
= kJ/mol C6H6