1. AP Chapter 5
Thermochemistry

2. Definitions Energy is the capacity to do work, or to transfer heat
Potential energy is the energy due to position or composition (energy of a chemical bond)
Kinetic energy is the energy due to motion
KE = ½ mv2
Temperature is a property that reflects the random motion of the particles of the substance.
Heat is the transfer of energy between 2 objects due to temperature difference. Heat is always transferred from a higher temperature object to a lower temperature object.

3. Work is equal to force over distance
Work is equal to force over distance. The part of the Universe that we will study, is called the System. Everything outside of the system is called the Surroundings.
calorie is the amount of energy required to raise the temperature of 1 g H2O through 1 oC
joule = Kg m2/s joule = 1 calorie

4. Law of Conservation of Energy states that energy can be converted from one form to another, but can be neither created nor destroyed The Energyuniverse is constant.
Open system allows heat and matter to be transferred between the system and surroundings.
Closed system allows the transfer for heat, but does NOT allow the transfer of matter.
Isolated system does NOT allow the transfer of either heat or matter. (NO SYSTEM is perfectly isolated)

5. Electrostatic Potential Energy (Eel) Eel = K Q1 Q2 / d
K = 8.99 x 109 Jm/C2
Q1 and Q2 = charges on particles
Eel = 0 infinite separation of particles
Eel = (-) charges are different + and –
Eel = (+) charges are the same
either both + or both -
See diagram on page 168

6. DE = q + w Energy of system = heat + work
H = E + PV
Enthalpy = internal energy + (Pressure volume)
W = -P DV
Work = - (pressure)(change in volume)

7. * heat of reaction = change in enthalpy H = m c DT q = m c DT
Enthalpy (H) at constant pressure, the change in enthalpy (DH) of a system is equal to the energy flow as heat
* heat of reaction = change in enthalpy
H = m c DT
q = m c DT
q = heat in joules
m = mass in grams
c = specific heat capacity is the number of joules required to raise the temperature of 1 g of substance through 1 oC.
DT = temperature change, Tfinal – Tinitial
If q is (-) negative, heat is being lost, exothermic
If q is (+) postive, heat is gained, endothermic

8. Endothermic is flow of energy into the system.
Exothermic is flow of energy out of the system.
Energy is an extensive property. It depends on the amount of material.

9. 0 = qsytem + qsurroundings 0 = (m c DT)system + (m c DT)surroundings
Remember the First Law of Thermodynamics Law of Conservation of Energy states that energy can be converted from one form to another, but can be neither created nor destroyed The Energyuniverse is constant.
Heat lost = Heat gain
0 = qsytem + qsurroundings
0 = (m c DT)system + (m c DT)surroundings
For a chemical reaction
DH = Hproducts - Hreactants

10. 0 = qsytem + qsurroundings 0 = (m c DT)system + (m c DT)surroundings
A 0.4 Kg iron horseshoe initially at 500 oC is dropped into a bucket containing 20 Kg of water at 22 oC. What is the final equilibrium temperature? Neglect any heat transfer to or from the surroundings. 0 = q Fe + q H2O

11. 0 = qsytem + qsurroundings 0 = (m c DT)system + (m c DT)surroundings
150 g of Pb is heated to 100 oC and placed in 50 g of H2O which is at 22 oC. The final temperature of the H2O and Pb is 28.8 oC. What is the specific heat of the Pb?
0 = q Pb + q H2O

12. Lead pellets, each of mass 1 g, are heated to 200 oC
Lead pellets, each of mass 1 g, are heated to 200 oC. How many pellets must be added to 500 g of H2O initially at 20 oc to make the equilibrium temperature 25 oC? Neglect any heat transfer to or from the container.

13. An unknown liquid of mass 400 g at a temperature of 80 oC is poured into 400 g of water at 40 oC. The final equilibrium temperature of the mixture is 49 oC. What is the specific heat of the unknown liquid?

14. An aluminum calorimeter of mass 100 g contains 250 g of H2O
An aluminum calorimeter of mass 100 g contains 250 g of H2O. They are in thermal equilibrium at 10 oC. Two metallic blocks are placed in the H2O. One is a 50 g piece of Cu at 80oC. The other sample has a mass of 70 g and is originally at a temperature of 100 oC. The entire system stabilizes at a final temperature of 20 oC. Determine the specific heat of the unknown second sample.

15. Calorimetry is the measurement of heat flow
Calorimeter is a device used to determine the heat associated with a chemical reaction.
Heat Capacity (m c) is the heat energy required to raise through some temperature increase.
q = m c DT

16. Heat Capacity of a Bomb Calorimeter
Bomb Calorimeter is a highly accurate device used to measure heat associated with chemical reactions.
Heat Capacity of a Bomb Calorimeter
= mcalorimeter . ccalorimeter
*Pressure remains constant; therefore,
Denthalpy = heat change

17. Calculate the heat of combustion of Octane (C8H18). A 0
Calculate the heat of combustion of Octane (C8H18). A g sample of octane is placed in a bomb calorimeter with heat capacity of 11.3 Kj/oC. DT = 2.25 oC
Molar heat capacity is the heat capacity per mole of substance. (kilojoule/ mol or kilojoule/ mole) Calculate the molar heat capacity for the octane.

18. A g sample of Naphthaline, C10H8 (s), is completely burned in a bomb calorimeter. Naphthaline has a heat of combustion (Molar heat capacity) of Kj/mol. If the temperature of the bomb calorimeter increases from oC, to oC. What is the heat capacity of the bomb calorimeter?

19. The combustion of g Sucrose, C12H22O11, causes the temperature to raise from oC to oC. The heat capacity of the bomb calorimeter is 4.89 Kj/oC. What is the heat of combustion (Molar heat capacity) of sucrose in Kj/mol ?

20. DH = heat of reaction is the enthalpy change in a chemical reaction
DH = heat of reaction is the enthalpy change in a chemical reaction. It is an extensive property, dependent on the amount of substance.
Remember:
If a reaction is exothermic, a negative (-) sign is placed in front of the numerical value (-DH) and the heat term goes on the right side of the chemical reaction.
2 H2 (g) + O2 (g)  2 H2O (g) DH = -242 Kj/mol H2O
2 H2 (g) + O2 (g)  2 H2O (g) Kj
If a reaction is endothermic, a positive (+) sign is placed in front of the numerical value (+DH) and the heat term goes on the left side of the chemical reaction.
N2 (g) O2 (g)  2 NO2 (g) DH = +34 Kj/mol NO2
N2 (g) O2 (g) Kj  2 NO2 (g)

21. C(s) + O2 (g)  CO (g) DH = -110.5 KJ/mol CO C (s) + H2 (g)  C3H8 (g)
DH = KJ/mol C3H8
S (s) O2 (g)  SO3 (g)
DH = KJ/mol SO3
N2 (g) O2 (g)  NO (g)
DH = KJ/mol NO

22. Ca+2 (aq) + OH-1 (aq)  Ca(OH)2 (s)
N2 (g) O2 (g)  NO2 (g)
DH = KJ/mol NO2
N2 (g) H2 (g)  N2H4 (g)
DH = KJ/mol N2H4
Ca+2 (aq) OH-1 (aq)  Ca(OH)2 (s)
DH = KJ/molCa(OH)2
C (s) H2 (g)  C6H6 (s)
DH = KJ/mol C6H6

23. Hess’s Law
DH is an extensive property – dependent on the amount of chemical
DH changes sign when a reaction is reversed
Enthalpy changes (DH) and equations can be added to determine an unknown enthalpy

24. To use Hess’s Law of summation, the following manipulations may be used:
Equations may be multiplied by multiplying all coefficients and heat term by the same number.
Multiple the following equation by 2
C(s) + ½ O2(g)  CO(g) DH = KJ/mole CO
Multiple the following equation by 3
½ H2(g) + ½ I2(s)  HI(g) DH = KJ/mole HI

25. Equations may be reversed by writing everything on other side of arrow.
Reverse the following equation
½ N2(g) /2 H2(g)  NH3(g) DH = KJ/mole NH3

26. Reverse the following equation
Equations may be divided by dividing all coefficients by the same number
Divide the following equation by 4
4 H2(g) O2(g)  4 H2O (L) DH = KJ/mole H2O
Reverse the following equation
3 C(s) H2(g)  C3H8(g) DH = KJ/mole C3H8

27. Equations may be added by adding the coefficients of like terms
Add the following 2 equations
2 C(s) H2(g)  C2H6(g) DH = KJ/mole C2H6
3 C(s) H2(g)  C3H8(g) DH = KJ/mole C3H8

28. Add the following 2 equations C(s) + ½ O2(g)  CO(g). DH = -110
Add the following 2 equations C(s) ½ O2(g)  CO(g) DH = KJ/mole CO C(s) O2(g)  CO2(g) DH = KJ/mole CO2

29. When adding equations, cancel any like terms on opposite sides of the equations
NO(g)  ½ N2(g) + ½ O2(g)
DH = KJ/mole NO
½ N2(g) + O2(g)  NO2(g) DH = KJ/mole NO2

30. Add the following equations,
cancel any like terms
SO2(g)  S(s) O2(g) DH = KJ/mole SO2
S(s) /2 O2(g)  SO3(g) DH = KJ/mole SO3

31. Add the following equations, cancel any like terms
H2O(L)  H2(g) + ½ O2(g)
DH = KJ/mole H2O
SO2(g)  S(s) + O2(g)
DH = KJ/mole SO2
H2(g) + S(s) + 2O2(g)  H2SO4(L)
DH = KJ/mole H2SO4

32. Determine DH for the following chemical reaction:
Determine DH for the following chemical reaction: C2H4 (g) + Cl2 (g)  C2H4Cl2 (L)
Using the following knowns:
Cl2 (g)+ H2O(L) 2 HCl (g)+ ½ O2 (g) DH = KJ
C2H4Cl2 (L) + H2O (L) ½ O2(g) + 2 HCl (g) + C2H4 (g)
DH = KJ

33. Determine DH for the following chemical reaction:
Determine DH for the following chemical reaction: 2 B (s) H2 (g)  B2H6 (g)
Using the following knowns:
2 B(s) + 3/2 O2 (g)  B2O3(s) DH = KJ
B2H6 (g) O2 (g)  B2O3 (s) H2O (g) DH = KJ
H2 (g) + ½ O2 (g)  H2O (L) DH = -286 KJ
H2O(L)  H2O (g) DH = 44 KJ

34. Standard Enthalpies of Formation (DHof) is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their natural (standard state)
The standard states of a compound is at 25 oC (room temp.)
1. for gaseous substance is at 1 atm
2. for liquid or solid condensed is as a liquid or solid
3. substance present in solution is 1 M solution
The standard state for an element is at 25 oC and 1 atm, is zero. The state shown on the periodic table (s, l, or g), or for the diatomic elements is X2.

35. n = moles DHoRXN = Sn DHof (products) – Sn DHof (reactants)
DHo = S DHof products - S DHof reactants
n = moles
S = (sigma) the sum of
DHof = enthalpy of formation
DHo or DHoRXN = enthalpy of reaction
Shown in text
Shown on formula sheet

36. Find the DHo for C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (L)

37. Mg (s) ½ O2 (g)  MgO (s)
100 cm Mg = x cm Mg
g .25 g
Mg mass 2016
Mass of H2O used: (______ M HCl) (_____ L HCl) = _____mole HCl (______mol HCl) ( _______m) = _____ g HCl _______ g H2O = (______ g HCl solution) – (________ g HCl)

38. Foods and Fuels
1 gram + O2(g)  fuel value Carbohydrate 17 KJ/g 4 Calorie/g Protein 17 KJ/g 4 Calorie/g Fat 38 KJ/g 9 Calorie/g

39. Present Energy Sources

40. Heat of vaporization
H = Hv m
Heat of fusion
H = Hf m