1. Energy Part III: Calculation of ΔH from a) Thermochemical Equations b) Heat of Formation Chapter 6 Sec 6 – Sec 8 of Jespersen 7TH ed
Dr. C. Yau
Spring 2015 1

2. Standard Heat of Reaction
ΔH = “enthalpy of reaction”
or “heat of reaction”
= heat transferred in a rxn
(usually in kJ, not kJ/mol)
ΔHo= standard heat of reaction
= ΔH at standard conditions
o means standard conditions
(1 atm, 1M if aq soln, usually 25oC)
Remember these conditions! 2

3. Thermochemical Equations
N2 (g) + 3H2 (g) NH3 (g) ΔHo = kJ
This tells us that 1 mole of N2 would produce kJ of heat,
that 3 moles of H2 would produce kJ of heat,
that production of 2 moles of NH3 would be accompanied by a release of kJ of heat.
NOTE that ΔHo is in kJ and not kJ/mol.
The amt of heat transferred is directly proportional to the # moles in thermochemical eqn shown. 3

4. Thermochemical Equations
Example: Magnesium burns in air to produce a bright light and is often used in fireworks displays.
2 Mg (s) + O2 (g) MgO (s) Ho= kJ
How many grams of Mg is needed to produce 400. kJ of heat?
p.295 #6.75
How much heat (in kJ) is liberated by the combustion of 6.54 g of Mg?
Set these problems up in dimensional analysis.

5. Thermochemical Equations
N2 (g) + 3H2 (g) NH3 (g) ΔHo = kJ
The value of ΔHo depends on the coefficients in the equation.
If coefficients are doubled, ΔHo would be doubled:
2N2 (g) + 6H2 (g) NH3 (g) ΔHo = x2 kJ
Note also that ΔHo is dependent on the physical states as stated in the equation.
CH4 (g) + 2O2 (g) CO2(g) + 2H2O(l) ΔHo = kJ
CH4 (g) + 2O2 (g) CO2(g) + 2H2O(g) ΔHo = kJ 5

6. Example 6.6 p.276
The following thermochemical equation is for the exothermic reaction of hydrogen and oxygen that produces water.
2H2 (g) + O2 (g) H2O (l) ΔHo = kJ
What is the thermochemical equation for this rxn when it is conducted to produce mol H2O?
Do Pract Exer 15 & 16 p.277

7. Thermochemical Equations
If we reverse a reaction, the magnitude of ΔHo is the same but the sign is changed:
C (s) + O2 (g)  CO2 (g) H° =  kJ
CO2 (g)  C (s) + O2 (g) H° = kJ

8. Determination of ΔH by manipulation of Eqns.
Use the two equations below to determine the standard enthalpy change for the reaction:
H2O2 (l)  H2O (l) + ½ O2 (g)
(1) H2 (g) + O2 (g)  H2O2 (l) H° = 188 kJ
(2) H2 (g) + ½ O2 (g)  H2O (l) H° = 286 kJ
Strategy:
We need H2O2 on the left side, so Eqn 1 must be reversed.
Eqn 2 probably can stay as is. WE MUST CHECK. How?
On my exams you are expected to “show your work” as we are doing here in class. Take notes!

9. Determination of ΔH by manipulation of Eqns.
Ethylene glycol, HOCH2CH2OH, is used as antifreeze. It is produced from ethylene oxide, C2H4O, by the reaction
(1) C2H4O (g) + H2O (l)  HOCH2CH2OH (l)
What is the heat of reaction of this reaction...
Given:
(2) 2C2H4O (g) + 5O2 (g)  4CO2 (g) + 4H2O (l)
H°=  kJ
(3) HOCH2CH2OH(l) + 5/2 O2(g) 2CO2(g) +3H2O(l)
H°=  Kj
What is the strategy?
Do Pract Exer 19 & 20 p.282

10. Determination of ΔH by manipulation of Eqns.
2Cu (s) + O2 (g)  2CuO (s) H° = 310 kJ
2Cu (s) + ½ O2 (g)  Cu2O (s) H° = 169 kJ
Use the two equations above to determine the H° of this reaction:
Cu2O (s) + ½ O2 (g)  2CuO (s)
Is this exothermic or endothermic?
Solve the problem by manipulating the given eqns.
ANS -141 kJ

11. Determination of ΔH by manipulation of Eqns.
Example 6.8 p.281
Fe2O3 (s) + 3CO (g)  2Fe(s) + 3CO2(g) H° =  26.7 kJ
CO(g) + ½ O2 (g)  CO2 (g) H° =  kJ
Calculate the value of H° for the following reaction:
2 Fe (s) + O2 (g)  Fe2O3 (s)

12. Enthalpy Diagrams C (s) + ½ O2 (g)  CO (g) H° = 110.5 kJ
Construct an enthalpy diagram for the reaction.
Learn the terminology, "enthalpy diagram."
Know what is asked for on an exam.

13. Enthalpy Diagrams N2 (g) + O2 (g)  2 NO (g) H° = +181 kJ
Draw the enthalpy diagram for this reaction.
We are skipping Example 6.7, Pract Exer 17 & 18. You do not need to know how to draw enthalpy diagrams of that sort. However, you should know how to add equations together to determine the enthlapy change for the overall reaction.
You should know how H diagrams differ between an endo- and exothermic reactions.

14. Hess' Law
The value of H° for any reaction that can be written in steps equals the sum of the values of H° of each of the individual steps.
This is based on the fact that enthalpy is a state function.
The implication is that regardless of how many steps are taken the overall enthalpy change is the same.

15. Enthalpy as a State Function
Enthalpy (H) depends only on its current state and not the path taken to get there.
It is not affected by how many steps are used to get there.
A  B
C F
D E
H1
H2
H3
H4
H5
H6
H1 = H2 + H3 + H4+H5 + H6

16. Hess’ Law is extremely useful because if H1 cannot be measured, it can be calculated from H2 and H3.
For example:
A B C
D + E
H1=?
H2
H3
H1 = H2 + H3

17. Hess' Law Watch out for the directions of the arrows.
What is H1 in terms of H2 and H3?
F G J
K + L
H1
H2
H3
H1 = H2 - H3

18. Standard Heat of Combustion
ΔHoc = standard heat of combustion
It is the amount of heat released when one mole of a fuel substance is completely burned in pure oxygen gas with all reactants and products brought to 25oC and 1 bar pressure (1 atm).
Combustion reactions are always exothermic. Therefore, ΔHoc is always negative.

19. Example 6.9 p.283
How many moles of carbon dioxide gas are produced by a gas-fired power plant for every 1.00 MJ (megajoule) of energy it produces? The plant burns methane, CH4 (g), for which ΔHoc is -890 kJ/mol
Do Pract Exer 21 & 22 p.284

20. Standard Enthalpy of Formation
LEARN THIS DEFINITION!
ΔHfo = standard enthalpy of formation
It is the amount of heat absorbed or evolved when specifically one mole of substance is formed at 25oC at 1 atm from its elements in their standard states.

21. ΔHfo for solid potassium sulfate is -1433. 7 kJ
ΔHfo for solid potassium sulfate is kJ. Write the thermochemical equation corresponding to this value.
ΔHfo for solid ammonium chloride is – kJ. Write the thermochemical equation corresponding to this value.
ΔHfo for solid calcium hydroxide is kJ. Write the thermochemical equation corresponding to this value.
Do Example 6.10 and Pract Exer 23 & 24 p.286

22. ΔHfo of Elements
Write the thermochemical equation corresponding to the ΔHfo of chlorine gas.
What do you think the value of ΔHfo would be for chlorine gas?
What about the value of ΔHfo of solid silver? of liquid mercury?
Remember this! You will not be provided ΔHfo elements in their standard states. They are always ZERO kJ.

23. Applying Hess' Law to Heats of Formation
Hrxn = ?
A B C D
Hf (A)
Hf (B)
Hf (C)
Hf (D)
All elements in their standard states.
What is heat of reaction of A+B C+D?
Hrxn = -Hf (A) -Hf (B) Hf (C) +Hf (D)
Hrxn = - {Hf (A) +Hf (B) } + {Hf (C) +Hf (D)}
Hrxn = +{Hf (C) +Hf (D)} - {Hf (A) +Hf (B) }
Hrxn = SumH (products) - Sum H(reactants)

24. Hess' Law of Summation
Remember Hf is per mole.
You must multiply it with n (the number of moles from the coefficients of the balanced equation).
ΔHfo values are given in Table 7.2 p. 285 and will be provided on exams.

25. Example 6.11 p.287
Some chefs keep baking soda, NaHCO3, handy to put out grease fires. When thrown on the fire, baking soda partly smothers the fire and the heat decomposes it to give CO2, which further smothers the flame. The eqn is
2NaHCO3 (s)  Na2CO3 (s) + H2O (l) + CO2 (g)
Use the data in Table 7.2 (p.285) to calc the ΔHo for this reaction in kilojoules.
Do not put out a kitchen fire with water!

26. Calculating ΔH For Reactions Using ΔHf°
2Fe(s) + 6H2O(l) → 2Fe(OH)3(s) + 3H2(g)
ΔH°f =
in kJ mol-1
CO2(g) + 2H2O(l) → 2O2(g) + CH4(g)
ΔH°f =
Do Pract Exer 25, 26 & 27 p. 288

27. SUMMARY
What are the different ways you can determine the ΔH of a reaction?
Measure q from calorimetry expt. In open containers, q = ΔH
Calculate by manipulating given thermochemical equations.
Calculate from ΔHfo
Calculate from bond energies (handout)
Also, remember how to calculate amt of heat from stoichiometry & thermochemical eqn.