1. FLOW OF ENERGY Heat, Enthalpy, & Thermochemical Equations
THERMOCHEMISTRY
FLOW OF ENERGY
Heat, Enthalpy,
& Thermochemical Equations

2. THERMOCHEMICAL EQUATIONS
Although it is NOT form of matter, in a chemical equation enthalpy change (DH) for the reaction can be written as either a reactant or a product. RECALL:

3. THERMOCHEMICAL EQUATIONS & HEATS OF REACTION
EXOTHERMIC REACTION CaO(s) +H2O(l) Ca(OH) kJ A chemical equation that includes the enthalpy change is known as a thermochemical equation Heat of Reaction is the enthalpy change for the chemical equation exactly as written

4. THERMOCHEMICAL EQUATIONS & HEATS OF REACTION
EXOTHERMIC REACTION
CaO(s) +H2O(l) Ca(OH) kJ
The above equation can be written as:
CaO(s) +H2O(l) Ca(OH) DH = kJ
Where the DH is written on the side. (Recall it is negative because it is exothermic.)

5. THERMOCHEMICAL EQUATIONS & HEATS OF REACTION
ENDOTHERMIC REACTION
2NaHCO3(s) kJ Na2CO3(s) + H2O(g) + CO2(g) OR
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
DH =+129 kJ
DH IS POSITIVE BECAUSE ENDOTHERMIC

6. Enthalpy diagrams
ENTHALPY DIAGRAMS are graphic representations of energy changes in a system

7. ENTHALPY DIAGRAMS

8. ENTHALPY DIAGRAM EXAMPLE
ENDOTHERMIC

9. ENTHALPY DIAGRAM EXAMPLE
EXOTHERMIC

10. Laws of Thermochemistry
In order to make effective use of thermochemical equations, certain basic laws can be applied:
Here come the LAWS….

11. 23.5 kcal of heat is produced when 2 moles decompose
1. ∆H is directly proportional to the amount of substance produced or reacting in a reactions
Consider hydrogen peroxide decomposing:
2H2O2 (l) H2O (g) + O2 (g) kJ
23.5 kcal of heat is produced when 2 moles decompose

12. What if 1 gram of H2O2 Decompose?
2H2O2 (l) H2O (g) + O2 (g) kJ
2 moles = 2(34 g H2O2)= 68 g
if-
68 g H2O2 produce kJ
then-
1 g produces X kJ
X kJ = .35 kJ

13. 2. ∆H for a reaction is equal in magnitude but opposite in sign from the reverse reaction
if-
2H2O2 (l) H2O (g) + O2 (g) ∆H= kJ
then-
2 H2O (g) + O2 (g) H2O2 (l) ∆H=+23.5 kJ

14. Law of Hess or Hess’s Law
3. If a reaction can be regarded as the sum of 2 or more other reactions, ∆H for the overall reaction must be the sum of the enthalpies of the component reactions

15. Consider: Sn(s) +Cl2(g) SnCl2 (s) ∆H= -83.6 kJ
SnCl2(s) +Cl2(g) SnCl4(l) ∆H = kJ
then
Sn(s) +2l2(g) SnCl4(l) ∆H =H1 + H2
∆H = kJ
Another example

16. A fairly easy one!
Calculate the heat evolved in the formation of 1 mole of PbSO4(s) from its elements, given the following: Pb(s) + S(s) → PbS(s) + 94 kJ/mol PbS(s) + 2O2(g) → PbSO4(s) kJ/mol a) ∆H = 730 kJ/mol b) ∆H = -730 kJ/mol c) ∆H = 918 kJ/mol d) ∆H = -918 kJ/mol
 -918 kJ/mol 

17. Here are three more- a tad more challenging
1. Calculate the standard enthalpy change, ΔHo, for the formation of 1 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements.
-1220 kJ

18. 2. The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is
+15.3 kJ

19. This one is REALLY a test of your knowledge!!
3. One reaction involved in the conversion of iron ore to the metal is
-11.0 kJ

20. Let’s go on-- ANOTHER METHOD of Finding ∆H
HEATS OF FORMATION 1

21. As a concise way of recording thermochemical data,
tables listing the quantities of heat accompanying many chemical changes have been compiled, in units of both joules and calories.

22. HeATs OF FORMATION
The Heat of Formation (∆Hf) is the amount of heat released when one mole of a substance is formed from its elements at 25oC and 1 atm pressure. Here are some Heats of Formation- Write the reactions they represent(1st one is given):
C(s) + ½ O2(g) –> CO(g)

23. HEATS OF FORMATION
Here’s what a real table of Standard Heats of Formation (DHºf) looks like:

24. ΔHreaction =S∆Hfproducts – S∆Hfreactants
These values allow us to calculate ΔH for a chemical reaction indirectly
If we know the ∆Hf of the reactants and the ∆Hf of the products (these will be given), we can use the following equation:
ΔHreaction =S∆Hfproducts – S∆Hfreactants
This states that the enthalpy change of a reaction is equal to the heats of formation of the products minus the heats of formation of the reactants. S MEANS SUMMATION!

25. Heats of formation EXAMPLE
CH4 + 2O2  2H2O + CO2
ΔHf of CH4 = -75 kJ/mol
ΔHf of O2 = 0 kJ/mol
ΔHf of H2O = -242 kJ/mol
ΔHf of CO2 = -394 kJ/mol
ΔHreaction = [product energy – reactant energy]
= [2 (-242) + (-394)] – [ (0)] = -803 kJ
GIVEN in TABLE

26. Calculate ΔH for the following reaction:
Your turn
Calculate ΔH for the following reaction:
8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s)
ΔH = Σ ΔHf products - Σ ΔHf reactants
ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s)
ΔH = 4( kJ) - 3( kJ)
ΔH = kJ
Heats of formation Table

27. A Question from 2000 AP Chem Exam

28. ANOTHER METHOD of Finding ∆H
Bond Energies 1

29. The strength of a chemical bond is measured by the energy required to break the bond, i.e. to separate the atoms and leave them as distinct isolated gaseous atoms. Bond Energies are ALWAYS +

30. Bond Energies for commonly occurring diatomic molecules range from:
N kJ to 150 kJ I2
HF 569 kJ to 295 kJ HI

31. Molecules with 3 or more atoms have 2 or more bonds
The ∆Hf for such molecules are equal to the sum of the Bond Energies

32. ∆Hf for NH3
Equals 3 B.E. N-H

33. ∆Hf for CH3OH
Equals 3B.E.’s C-H
+ BE C-0 +
BE 0-H

34. B.E.’s can be used to calculate ∆H values for reactions – IF reaction is viewed as the forming and breaking of bonds

35. NH3 + H2O ---> NH4OH H [H-N-H+ ] +[ O-H]- H ∆H = BE H-O + (-BE N-H)
H-N-H + H-O-H --> H
H [H-N-H+ ] +[ O-H]- H
∆H = BE H-O + (-BE N-H)
Note: BE’s are + when bonds broken
BE’s are - when bonds formed
Bond Formed
Bond Broken

36. C(s) + 2H2(g) --> CH4(g) ∆H = 2 (+BE H-H) + 4(-BE C-H)
2nd Example:
C(s) H2(g) --> CH4(g)
∆H = 2 (+BE H-H) + 4(-BE C-H)
= 2(432 kJ) + 4(-413k j)
∆H = KJ

37. BREAK EVERYTHING IN THE REACTANTS--- FORM EVERYTHING IN THE PRODUCTS!!
Generally-
For those who don’t want to think about what bonds are broken or formed-
BREAK EVERYTHING IN THE REACTANTS---
FORM EVERYTHING IN THE PRODUCTS!!

38. Use the formula-
∆H=
S BE’s bonds broken - S BE’s
bonds formed

39. A Table of Bond Energies looks like:
H C N O F S Cl Br I H C N O F
REMEMBER:
∆H= S BE’s bonds broken - S BE’sbonds formed

40. We could use this table to find ∆H for H2 + Cl2 --> 2HCl
∆H = BE H-H + + BE Cl-Cl - 2BE H-Cl
From Table
∆H = kj kj - 2(432 kJ)
∆H = 186 kj/mol

41. SUMMARY

42. ∆H= S BE’s bonds broken - S BE’sbonds formed
Source
The net energy change during a reaction is the sum of the energy required to break the reactant bonds and the energy released in forming the product bonds. The net energy change may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released.
Any bond that can be formed can be broken. These processes are in opposition. (their enthalpy changes are equal in magnitude, opposite sign)
ΔH bonds breaking  ENDOTHERMIC (+)
ΔH bonds forming  EXOTHERMIC (-)
To find ΔHrxn, apply Hess’s Law:
ΔHrxn = ΣΔH bonds breaking (+) + Σ ΔH bonds forming (-)
To calculate or estimate ΔHrxn from Bond Energy:
Draw the Lewis Structure. Don’t forget about double and triple bonds!
Add up ΔH bonds breaking. It’s + (kJ)
Add up ΔH bonds forming. It’s - (kJ).
Add the two terms. Units are kJ/mol rxn. .
Table of Bond Energies
H C N O F S Cl Br I H C N O F
∆H= S BE’s bonds broken - S BE’sbonds formed

43. Video REVIEW

44. ∆H= S BE’s bonds broken - S BE’sbonds formed
Source
The net energy change during a reaction is the sum of the energy required to break the reactant bonds and the energy released in forming the product bonds. The net energy change may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released.
Any bond that can be formed can be broken. These processes are in opposition. (their enthalpy changes are equal in magnitude, opposite sign)
ΔH bonds breaking  ENDOTHERMIC (+)
ΔH bonds forming  EXOTHERMIC (-)
To find ΔHrxn, apply Hess’s Law:
ΔHrxn = ΣΔH bonds breaking (+) + Σ ΔH bonds forming (-)
To calculate or estimate ΔHrxn from Bond Energy:
Draw the Lewis Structure. Don’t forget about double and triple bonds!
Add up ΔH bonds breaking. It’s + (kJ)
Add up ΔH bonds forming. It’s - (kJ).
Add the two terms. Units are kJ/mol rxn. .
Table of Bond Energies
H C N O F S Cl Br I H C N O F
∆H= S BE’s bonds broken - S BE’sbonds formed
Video

46. ΔH= Bonds Broken – Bonds Formed = [H-H + Cl-Cl]-[2(H-Cl)]
= [436 kJ/mol+242 kJ/mol] – [2(431)]
= -184 kJ/mol
= [4 H-C + C=C + F-F] – [C-C + 4 C-H + 2 C-F]
= [4(413) ]-[ (413) + 2(485)]
= 2421 – 2970 kJ/mol
= -549 kJ/mol
= [6 H-C + 2 C-O + 2 O-H] – [6 H-C + 2 C-O + 2 O-H]
= [6(413) + 2(358) + 2(463)]-[ 6(413) + 2(358) + 2(463)]
= 4120 – 4120 kJ/mol
= 0 kJ/mol
= [H-C + 3 C-Cl] – [2 C-Cl + C=O + H-Cl]
= [ (328)]- [2(328) ]
= 1397 – 1886 kJ/mol
= -489 kJ/mol
= [4 H-C + 2 O=O] – [2 C=O + 4 H-O]
= [4(413) + 2(495)]- [2(799) + 4(463)]
= 2642 – 3450 kJ/mol
= -808 kJ/mol