1. THERMOCHEMISTRY
CP Unit 9
Chapter 17

2. Thermochemistry 17.1
Thermochemistry is the study of energy changes (HEAT) that occur during chemical reactions and changes in state.

3. ENTHALPY 17.2 q = ΔH > 0 (positive values)
Enthalpy = a type of chemical energy, sometimes referred to as “heat content”, ΔH (the heat of reaction for a chemical reaction)
endothermic reactions (feels cold):
q = ΔH > 0 (positive values)
exothermic reactions (feels hot):
q (heat) = ΔH (enthaply, heat of rxn) < 0 (negative values)

4. Endo vs. Exo-
Endothermic reactions: absorbs heat from surroundings (+).
If you touch an endothermic reaction it feels COLD
Exothermic reactions: release heat to the surroundings (-)
If you touch an exothermic reaction it feels HOT

5. Magnitude of Heat Flow Units of heat energy:
1 kcal = 1,000 cal = 1 Cal (nutritional)
1 kJ = 1,000 J
1 calorie = J
1 kcal = kJ

6. Thermochemical Equations
A chemical equation that shows the enthalpy (H) is a thermochemical equation.

7. Rule #1
The magnitude (value) of H is directly proportional to the amount of reactant or product.
H2 + Cl2  2HCl H = kJ
* meaning there are 185 kJ of energy RELEASED for every:
1 mol H2
1 mol Cl2
2 moles HCl

8. Rules of Thermochemistry
Example 1:
H2 + Cl2  2HCl H = kJ
Calculate H when 2.00 moles of Cl2 reacts.

9. Rules of Thermochemistry
Example 2: Methanol burns to produce carbon dioxide and water:
2CH3OH + 3O2  2CO2 + 4H2O H = kJ
What mass of methanol is needed to produce 1820 kJ?

10. Rule #2
H for a reaction is equal in the magnitude but opposite in sign to H for the reverse reaction.
(If 6.00 kJ of heat absorbed when a mole of ice melts, then 6.00 kJ of heat is given off when 1.00 mol of liquid water freezes)

11. Rules of Thermochemistry
Example 1:
Given:
H2 + ½O2  H2O H = kJ
Reverse:
H2O  H2 + ½O2 H = kJ

12. Example 2 CaCO3 (s)  CaO (s) + CO2 (g) H = 178 kJ
What is the H for the REVERSE RXN?
CaO (s) + CO2 (g)  CaCO3 (s) H = ?

13. Alternate form of thermochem. eq.
Putting the heat content of a reaction INTO the actual thermochemical eq.
H2 + ½O2  H2O H = kJ
Exothermic
(H is negative)
Heat is RELEASED as a PRODUCT
The alternate form is this: H2 + ½O2  H2O kJ

14. EX: 2 NaHCO3 + 129 kJ  Na2CO3 + H2O + CO2 Put in the alternate form
The alternate form is this:
2 NaHCO3  Na2CO3 + H2O + CO2 H = kJ

15. Put the following in alternate form
H2 + Cl2  2 HCl H = -185 kJ
2 Mg + O2  2 MgO kJ
2 HgO  2 Hg + O2 H = kJ

16. Enthalpies of Formation
standard conditions
enthalpy
change (delta)
formation

17. Enthalpies of Formation
usually exothermic
see table for Hf value (Table A3)
enthalpy of formation of an element in its stable state = 0
these can be used to calculate H for a reaction

18. Standard Enthalpy Change
Standard enthalpy change, H, for a given thermochemical equation is = to the sum of the standard enthalpies of formation of the product – the standard enthalpies of formation of the reactants.
sum of (sigma)

19. Standard Enthalpy Change
elements in their standard states can be omitted:
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
ΔHrxn = (ΔHfproducts) - (ΔHfreactants)
ΔHrxn = ΔHfAl2O3 - ΔHfFe2O3
ΔHrxn = ( kJ) – ( kJ)
ΔHrxn = kJ

20. Standard Enthalpy Change
the coefficient of the products and reactants in the thermochemical equation must be taken into account:
O2(g) + 2SO2 (g)  2SO3 (g)
ΔHrxn = (ΔHfproducts) - (ΔHfreactants)
ΔHrxn = 2ΔHfSO3 - 2ΔHfSO2
ΔHrxn = 2( kJ) – 2( kJ)
ΔHrxn = kJ

21. Don’t forget the coefficents!
Standard Enthalpy Change - Calculate the standard heat for formation of benzene, C6H6, given the following thermochemical equation:
Don’t forget the coefficents!
C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) H = kJ
-393.5kJ
-285.8kJ
Total X
kJ = [6(-393.5kJ)+3(-285.8kJ)]–X
kJ = –X
-49.0kJ = – X
X = kJ