1. Thermochemistry
Enthalpy:
Heat of Reaction

2. Enthalpy Heat of Reaction
So we have discussed the energy changes when materials are warming and cooling simply because the materials were simply at different temperatures, but what about the energy changes caused by a chemical reaction?

3. The Molecular View
Where does the energy come from during chemical reactions?
It does not come simply from the kinetic energy of the atoms or molecules of the system.
It comes from the chemical potential energy that arises from the electrostatic forces of the protons and electrons within the system.
During the reaction, bonds break, atoms rearrange, & bonds form resulting in a different arrangement.
Arrangement resulting in lower potential energy, exothermic, E is negative = more stable, more favorable.
Arrangement resulting in higher potential energy, endothermic, E is positive = less stable, less favorable.

4. The Energy of Bonds Bond breaking Bond forming
Endothermic
Bond forming
exothermic
C6H12O6 + 6O2 → 6CO2 + 6H2O + E
For this reaction the net result of the energy in for bonds broken combined with the energy out from bonds formed, give off heat energy.

5. The Energy of Bonds Bond breaking Bond forming
Endothermic
Bond forming
Exothermic
ATP + water → ADP + Phosphate-water + E
For this reaction the net result of the energy in for bonds broken combined with the energy out from bonds formed, give off heat energy.

6. Measuring thermal heat of a reaction.
Symbolized by ∆HRxn Called Enthalpy
The amount of heat released or absorbed during a chemical reaction.
∆HRx positive (+), means energy is absorbed (temp goes down)
∆HRx negative (−), means energy is released (temp goes up)
CH4 + 2O2 → CO2 + 2H2O + Energy
∆Hcombustion = −802 kJ/mol of CH4
Thus q = ∆Hrxn x amount
∆H is intensive (notice, joules per mole)
q is extensive (notice, just joules)

7. Relationships involving ∆HRxn
If a chemical equation is multiplied by some factor, the ∆HRxn is also multiplied by the same factor.
CH4 + 2O2 → CO2 + 2H2O kJ
∆Hcombustion = −802 kJ/mol of CH4
2CH4 + 4O2 → 2CO2 + 4H2O kJ
∆Hcombustion = −1604 kJ/ 2 mol of CH4

8. Relationships involving ∆HRxn
If a chemical equation is reversed, then the ∆HRxn changes sign.
CH4 + 2O2 → CO2 + 2H2O kJ
∆Hcombustion = −802 kJ/mol of CH4
CO2 + 2H2O kJ → CH4 + 2O2
∆Hreaction = +802 kJ/mol of CH4

9. ∆Hformation values Check out the Thermodynamic Table
Look at the first column ∆Hºf (kJ/mol)
What does ºf mean, and per mole of what?
∆H = Enthalpy,
f = "of formation”; forms 1 mole of the substance listed, from its elements in their standard condition
º = at standard conditions (1 M and 1 atm)
measured at 25ºC

10. ∆Hformation Reactions
Look up gaseous butane
∆Hºf = − kJ/mol
∆ tells us change, but change for what??
For a reaction….but what reaction?
For the formation of gaseous butane from its elements
4C + 5H2 → C4H10(g) ∆H = − kJ/mol

11. ∆Hformation Reactions
Look up calcium hydroxide
∆Hºf = −986.2 kJ/mol
For what reaction?
For the formation of solid calcium hydroxide from its elements
Ca + O2 + H2 → Ca(OH)2(s) ∆H = −986.2 kJ/mol

12. ∆Hºf By Definition
Chemists have compiled long lists of energy values for formation reactions.
A formation reaction is defined as: The formation of one mole of a substance from the substances’ elements in their standard state.
Formation of liquid water?
H2(g) + ½ O2(g) → H2O(L) ∆Hfº = −286 kJ/mol
Formation of solid sugar (glucose)?
6Cs(graphite) + 6H2(g) + 3O2(g) → C6H12O6(s) ∆Hfº = −1,273 kJ/mol
These values can be used to calculate the energy for many reactions.

13. Using Thermochem values to find ∆HRxn
Let’s say you were asked to find the ∆HRxn for the combustion of ethanol gas.
*Take out your reference sheets.
We can calculate ∆HRxn using the formula:
∆Hrxn = Σn∆Hºf products − Σn∆Hºf reactants
= sigma = sum
n = coefficient from balanced equation

14. ∆Hrxn = Σn∆Hºf products − Σn∆Hºf reactants
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g)
ΔH°f : kJ/mol kJ/mol kJ/mol kJ/mol
∆HRxn = [ 2(-394) + 3(-242) ] − [ (-235) + 3(0) ] = −1279
∆HRxn = −1279 kJ/mol

15. An example problem
A student was burning methanol to heat mL of water at 18.5°C. If 4.25 grams of methanol heated the water up to 95.5°C, calculate the experimental ΔHRxn in kJ/mole.
q = - ΔHRxn x amount (mol)
m c ΔT = - ΔHRxn x mol
- m c ΔT = ΔHRxn
mol

16. 1a.) A student was burning methanol to heat 255 mL of water at 18.5°C. If 4.25 grams of methanol heated the water up to 95.5°C, calculate the experimental ΔHRxn in kJ/mole.
4.25 g CH3OH
__________
1 mol
x =
.133 mol
32 g
(255)(4.18)(77)
ΔHRxn
= = J/mol
.133
= kJ/mol

17. 2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
1a.) A student was burning methanol to heat 255 mL of water at 18.5°C. If 4.25 grams of methanol heated the water up to 95.5°C, calculate the experimental ΔHRxn in kJ/mole.
b.) Write a balanced equation for the combustion of methanol, CH3OH.
2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)

18. 1a.) A student was burning methanol to heat 255 mL of water at 18.5°C. If 4.25 grams of methanol heated the water up to 95.5°C, calculate the experimental ΔHRxn in kJ/mole.
b.) Write a balanced equation for the combustion of methanol, CH3OH.
c.) Using the balanced equation & the heat of formation values on your reference sheet, calculate the theoretical value for the heat of combustion of propane.

19. 2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
c.) Using the balanced equation & the heat of formation values on your reference sheet, calculate the theoretical value for the heat of combustion of propane.
2 CH3OH (g) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
kJ/mol kJ/mol kJ/mol kJ/mol
∆Hrxn = Σn∆Hºf products − Σn∆Hºf reactants
∆Hrxn = [2(-393.5) + 4( )] – [2(-201.2) + 3(0)]
∆Hrxn = kJ/mol

20. Part a: - 726 kJ/mol (experimental value) Part b: - 1351.88 kJ/mol
d.) Using your values from the calculations above to calculate the percent yield (in this case it can also be a % efficiency) of the heat transferred from the flame to the water.
Part a: kJ/mol (experimental value)
Part b: kJ/mol
- 726 kJ/mol
kJ/mol
x 100 = 53.7%

21. Clicker Questions

22. In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is defined as
activation energy
ionization energy
heat of vaporization
heat of reaction
kinetic energy
lattice energy

23. In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is defined as
activation energy
ionization energy
heat of vaporization
heat of reaction, ∆H
kinetic energy
lattice energy

24. or time, or progress of reaction
Which number in the potential energy diagram shown below, represents the heat of the reaction, ∆H? 1 2 3 4 6 5
Products
Reactants
or time, or progress of reaction

25. Which letter in the potential energy diagram shown below, represents the heat of the reaction, ∆H?1 2 3 4 6 5
∆H represents the “net” energy difference between the reactants and the products
Products
Reactants

26. The reaction represented by this graph1 2 3 4 6 5
is exothermic
is endothermic
There is not enough information to determine the energetics of this reaction.
Products
Reactants

27. The reaction represented by this graph1 2 3 4 6 5
is exothermic
is endothermic
There is not enough information to determine the energetics of this reaction.
Products
Reactants

28. 2H2O(L) + 570 kJ → 2H2(g) + O2(g) The ∆Hºf for liquid water is
Consider the following thermochemical equation:
2H2O(L) kJ → 2H2(g) + O2(g) The ∆Hºf for liquid water is
(+570)2
(−570)2
−570
(+570)½
+570
(-570)½

29. The ∆Hºf for liquid water is
Consider the following thermochemical equation:
2H2O(L) kJ → 2H2(s) + O2(g)
The ∆Hºf for liquid water is
(+570)2
+570
(−570)2
(+570)½
(-570)½
−570
When you take the thermochemical value out of the equation above, because it is exothermic, you need to represent it as negative.
In addition, ∆Hºf values are always per mole of whatever is being formed from its elements.

30. Which energy diagram below best represents the formation of liquid water?
None of the choices are valid.

31. Which energy diagram below best represents the formation of liquid water?
None of the choices are valid.
since we learned in the last slide that the formation of liquid water is exothermic.

32. Use your Thermodynamic Tables to calculate the ∆H for the process of vaporizing solid iodine, I2
Write your numeric answer on a whiteboard.

33. Use your Thermodynamic Tables to calculate the ∆H for the process of vaporizing solid iodine, I2
I2(s) → I2(g)
∆Hvap = Σn∆Hºf products - Σn∆Hºf reactants
kJ/mole - 0 kJ/mole
thus ∆Hvap = /mole I2(s)

34. ∆Hcombustion = −2061 kJ/mole alcohol
Use your Thermo Tables, and the enthalpy for the reaction given below to calculate the ∆Hºf for gaseous propanol.
C3H7OH(g) + ⁹/2O2(g) → 3CO2(g) + 4H2O(g)
∆Hcombustion = −2061 kJ/mole alcohol
Enter a value rounded to the nearest whole number.

35. ∆Hcombustion = −2061 kJ/mole
Use your Thermo Tables, and the enthalpy for the reaction given below to calculate the ∆Hºf for gaseous propanol.
C3H7OH(g) + ⁹/2O2(g) → 3CO2(g) + 4H2O(g)
∆Hcombustion = −2061 kJ/mole
∆Hrx = Σn∆Hºf products − Σn∆Hºf reactants
[3(−393.5) + 4(−241.82)] − [(x) + (0)] = −2061 kJ
x = Hºf for C3H7OH(g) = −87 kJ

36. Which reaction best represents the ∆Hºf reaction for solid sodium nitrate?
Na+(aq) + NO3−(aq) → NaNO3(aq)
Na+(aq) + NO3−(aq) → NaNO3(s)
Na+(g) + NO3(g) → NaNO3(s)
Na(s) + NO3(s) → NaNO3(s)
2Na(s) + N2(g) + 3O2(g) → 2NaNO3(s)
Na(s) + ½N2(g) + ³/₂O2(g) → NaNO3(s)

37. Which reaction best represents the ∆Hºf reaction for solid sodium nitrate?
Na+(aq) + NO3−(aq) → NaNO3(aq)
Na+(aq) + NO3−(aq) → NaNO3(s)
Na+(g) + NO3(g) → NaNO3(s)
Na(s) + NO3(s) → NaNO3(s)
2Na(s) + N2(g) + 3O2(g) → 2NaNO3(s)
Na(s) + ½N2(g) + ³/₂O2(g) → NaNO3(s)
Heat of formation reactions are always per mole of the “formed” chemical.

38. Select the unit label(s) that would be appropriate for ∆HºfkJ
kJ/mole
kJ/moleºC
J/g
J/ºC
J/gºC
calories/mg

39. Select the unit label(s) that would be appropriate for ∆HºfkJ
kJ/mole
kJ/moleºC
J/g
J/ºC
J/gºC
calories/mg