1. Formation of Ammonia

2. Proportional Relationships
Stoichiometry
mass relationships between substances in a chemical reaction
based on the mole ratio
Mole Ratio
indicated by coefficients in a balanced equation
2 Mg + O2  2 MgO
Courtesy Christy Johannesson

3. Stoichiometry Island Diagram
Known Unknown
Substance A Substance B
Mass
Mass
1 mole = molar mass (g)
1 mole = molar mass (g)
Use coefficients
from balanced
chemical equation
Mole
Mole
1 mole = x 1023 particles
(atoms or molecules)
1 mole = x 1023 particles
(atoms or molecules)
Particles
Particles
Stoichiometry Island Diagram

4. Stoichiometry Steps Core step in all stoichiometry problems!!
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
Mole ratio = moles  moles
Molar mass = moles  grams
Avogadro’s number = particles  moles
Mole ratio = moles  moles
Core step in all stoichiometry problems!!
4. Check answer.
Courtesy Christy Johannesson

5. Stoichiometry Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol KClO3
Courtesy Christy Johannesson

6. Stoichiometry Problems
How many grams of silver will be formed from 12.0 g copper?
Cu AgNO3  2Ag Cu(NO3)2
12.0 g
? g
12.0
g Cu
1 mol Cu
63.55
g Cu
2 mol Ag
1 mol Cu
107.87
g Ag
1 mol Ag
= 40.7 g Ag
Courtesy Christy Johannesson

7. Rocket Fuel B2H6 + O2 B2O3 + H2O B2H6 + O2 B2O3 + H2O 3 3
The compound diborane (B2H6) was at one time
considered for use as a rocket fuel. How many grams
of liquid oxygen would a rocket have to carry to burn
10 kg of diborane completely?
(The products are B2O3 and H2O).
B2H6 + O2
B2O3 + H2O
Chemical equation
Balanced chemical equation
B2H O B2O H2O 3 3
10 kg x g
1000 g B2H6
1 mol B2H6
3 mol O2
32 g O2
x g O2 = 10 kg B2H6
1 kg B2H6
28 g B2H6
1 mol B2H6
1 mol O2
X = 34,286 g O2

8. Limiting Reactants Limiting Reactant Excess Reactant
used up in a reaction
determines the amount of product
Excess Reactant
added to ensure that the other reactant is completely used up
cheaper & easier to recycle
Courtesy Christy Johannesson

9. Percent Yield actual yield % yield = x 100 theoretical yield
measured in lab
actual yield
% yield =
x 100
theoretical yield
calculated on paper
Courtesy Christy Johannesson

10. When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
actual yield
46.3 g
K2CO  2KCl + H2O + CO2
2HCl
45.8 g
excess
? g
theoretical yield
Theoretical yield
1 mol K2CO3
2 mol KCl
74.5 g KCl
x g KCl = 45.8 g K2CO3
= 49.4 g KCl
49.4 g
49.4 g KCl
138 g K2CO3
1 mol K2CO3
1 mol KCl
Example courtesy Christy Johannesson
% Yield =
Actual Yield
Theoretical Yield
46.3 g KCl
% Yield =
x 100
% Yield = 93.7% efficient

11. Tro's "Introductory Chemistry", Chapter 3
Thermodynamics
Temperature: the measurement of the average kinetic energy of the particles in an object.
Energy: the ability (or capacity) of a system to do work or supply (or produce) heat.
Heat: heat is the transfer of energy between two objects due to temperature differences.
Tro's "Introductory Chemistry", Chapter 3

12. Tro's "Introductory Chemistry", Chapter 3
Temperature
K = 0C + 273
Water boils at 100 0C or 212 0F or 373 K
Water freezes at 0 0C or 32 0F or 273 K
Absolute Zero (0 K = -2730C): theoretical temperature at which all atoms cease motion
Tro's "Introductory Chemistry", Chapter 3

13. Law of Conservation of Energy
“Energy can neither be created nor destroyed.”
The total amount of energy in the universe is constant. There is no process that can increase or decrease that amount.
However, we can transfer energy from one place (a system) in the universe to another (surroundings), and we can change its form.
Energy is stored in chemical bonds of matter.

14. Units of Energy
Calorie (cal) is the amount of energy needed to raise one gram of water by 1 °C.
kcal = energy needed to raise 1000 g of water 1 °C.
food calories = kcals.
Energy Conversion Factors
1 calorie (cal) =
4.184 joules (J)
1 Calorie (Cal)
1000 calories (cal)

15. Burning of a Match (chemical change)
System
Surroundings
D(PE)
(Reactants)
Potential energy
Energy released to the surrounding as heat
(Products)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

16. An Energy Diagram (chemical change)
activated
complex
activation
energy Ea
reactants
products
Potential energy
course of reaction

17. Exothermic Reaction (chemical change)
Reactants  Products + Thermal Energy
(heat)
Energy of reactants
Energy of products
Potential Energy
Reactants
-DHrxt Exothermic
Products
Reaction Progress

18. An Exothermic Reaction
Surroundings
reaction
Potential energy
Reactants
Products
Amount of energy released

19. The Zeppelin LZ 129 Hindenburg catching fire on May 6, 1937 at Lakehurst Naval Air Station in New Jersey.

20. Endothermic Reaction (chemical change)
Thermal Energy + Reactants  Products
(heat)
Activation
Energy
Energy of reactants
Energy of products
Products
Potential Energy
+DHrxt Endothermic
Reactants
Reaction progress

21. An Endothermic Reaction
Surroundings
reaction
Potential energy
Products
Reactants
Amount of energy absorbed

22. Effect of Catalyst on Reaction Rate
What is a catalyst? What does it do during a chemical reaction?
Catalyst lowers the activation energy for the reaction.
No catalyst
activation energy
for catalyzed reaction
reactants
Potential Energy
A catalyst lowers the activation energy for the reaction. This allows the reaction to occur at a much faster rate. The catalyst is not a reactant or product. It is not consumed during the chemical reaction.
products
Reaction Progress

23. Barbecue
An LP gas tank in a home barbecue contains 11.8 X 103g of propane (C3H8). Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. The heat of reaction is kJ/mol C3H8.
__C3H8 + __O2(g)  __CO2(g) + __H2O(g)
1 mol C3H8
-2044 kJ
kJ = X g C3H8
44 g C3H8
1 mol C3H8
kJ = X 105 kJ

24. Water Molecules in Hot and Cold Water
Hot water Cold Water
90 oC oC
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

25. Heat Transfer (ex: physical change)
Surroundings
System
ENDOthermic
EXOthermic
System
H2O(s) + heat  H2O(l)
melting
H2O(l)  H2O(s) + heat
freezing
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207

26. Heating Curve of Water Gas - KE  Boiling - PE  Liquid - KE 
140
Gas - KE 
120
100
Boiling - PE  80 60 40
Liquid - KE 
Temperature (oC) 20
Melting - PE 
-20
-40
Solid - KE 
-60
-80
-100
Thermal Energy added from Surroundings

27. Heat Gain or Loss by an Object
The amount of heat energy gained or lost by an object depends on 3 factors: 1. how much material there is 2. what the material is 3. how much the temperature changed. Amount of Heat = Mass x Specific Heat x Temperature Change
q = m x C x DT
Tro's "Introductory Chemistry", Chapter 3

28. Thermometer
Styrofoam
cover
cups
Stirrer
A Coffee Cup Calorimeter Calorimetry: experimental technique for investigating heat transfer
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

29. Heat Capacity
Heat capacity is the amount of heat a substance must absorb to raise its temperature by 1 °C.
cal/°C or J/°C.
Metals have low heat capacities; insulators have high heat capacities.
Specific heat = heat capacity of 1 gram of the substance.
cal/g°C or J/g°C.
Water’s specific heat = J/g°C for liquid.
Or cal/g°C.
It is less for ice and steam.

30. Specific Heat Capacities

31. Calorimetry q = m . C . DT ΔT = TF - TI m . C . DT = - [m . C . DT]
Object gaining heat = Object losing heat
+q = q
m . C . DT = [m . C . DT]
Material whose temperature increases gains heat while
materials whose temperature decreases lose heat.
Metals generally lose heat, while liquids (such as water)
generally gain heat in a system. We must account for all of the
heat energy in the system due to conservation of energy.

32. Enthalpy Changes H2(g) + ½ O2(g) H2O(g) H2O(l) DH = +242 kJ
Endothermic
-242 kJ
Exothermic
-286 kJ
Endothermic
DH = -286 kJ
Exothermic
H2O(g)
Energy
44 kJ
Exothermic
+44 kJ
Endothermic
H2O(l)
H2(g) + 1/2O2(g)  H2O(g) kJ DH = kJ
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

33. Enthalpy Changes DH3 = DH1 + DH2 (Hess’s Law)
Change in enthalpy does not depend on path of reaction
a) H2(g) + 1/2O2(g)  H2O(g) DH1 = kJ
b) H2O(g)  H2O(l) DH2 = kJ
c) H2(g) + 1/2O2(g)  H2O(l) DH3 = kJ

34. Hess’s Law a) N2(g) + O2(g)  2NO(g) DH1 = +180 kJ
b) 2NO(g) + O2(g)  2NO2(g) DH2 = kJ
c) N2(g) + 2O2(g)  2NO2(g) DH3 = kJ

35. Hess’s Law
Calculate the enthalpy of formation of carbon dioxide from its elements.
C(g) + 2O(g)  CO2(g)
Use the following data:
2O(g)  O2(g) DH = kJ
C(s)  C(g) DH = kJ
CO2(g)  C(s) + O2(g) DH = kJ
2O(g)  O2(g) DH = kJ
C(g)  C(s) DH = kJ
C(s) + O2(g)  CO2(g) DH = kJ
C(g) + 2O(g)  CO2(g) DH = kJ

36. Hess’s Law
Calculate DH for the synthesis of diborane from its elements.
2B(s) + 3H2(g)  B2H6(g) DH = ?
a) 2B(s) + 3/2O2(g)  B2O3(s) DH = kJ
b) B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) DH = kJ
c) H2(g) + 1/2O2(g)  H2O(l) DH = kJ
d) H2O(l)  H2O(g) DH = kJ

37. Visualizing a Chemical Reaction2
Na Cl NaCl 2
___ mole Na 10 10
___ mole Cl2 5 5
___ mole NaCl 10 10 ?