Download Introductory Chemistry, 2nd Edition Nivaldo Tro

Introductory Chemistry, 3rd Edition
Nivaldo Tro
Chapter 8
Quantities in
Chemical Reactions
Car an octane and oxygen
molecules
and carbon dioxide and
water
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, Maqqwertd ygoijpk[l
2009, Prentice
‗? Hall
Outline
•
•
•
•
•
8.1 Global Warming : Too much Carbon Dioxide
8.2 Making Pancakes: Relationships between Ingredients
8.3 Making Molecules: Mole to Mole Conversions
8.4 Making Molecule: Mass to Mass Conversions
8.5 More Pancakes: Limiting Reactant, Theoretical Yield,
and Percent Yield
• 8.6 Limiting Reactant, Theoretical Yield, and Percent
Yield form Initial Reactants
• 8.7 Enthalpy: Measure of the Heat Evolved or Absorbed
during a Chemical Reaction
Tro's ―Introductory Chemistry‖,
Chapter 8
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8.1 Global Warming : Too
much Carbon Dioxide
Tro's ―Introductory Chemistry‖,
Chapter 8
3
Burning of Fossil
Fuels
Car and molecules
from chapter opening
2C8H18 (l) + 25O2(g)
Internal combustion engines
produe Carbon dioxide by
burning fuels.
For example octane, C8H18,
in gasoline.
16 CO2 (g) + 18H2O(g)
Global Warming
• Average 0.6 °C rise in atmospheric temperature since
1860.
• In the same period atmospheric CO2 levels have risen 25%.
• Are the two related?
Figure 8.1 and Figure 8.2
Showing global tempertature
Rise and greenhouse effect.
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Chapter 8
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Increased CO2
• The primary source of the increased CO2
levels is combustion of fossil fuels
1860 = Industrial Revolution in the U.S. and
Europe.
Methane in natural gas
CH4(g)  2 O2(g)  CO2(g)  2 H2O(g)
Tro's ―Introductory Chemistry‖,
Chapter 8
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Quantities in Chemical Reactions
• All substances in a chemical reaction are
related.
• The numerical relationships between
compounds in a chemical reaction is called
stoichiometry.
Tro's ―Introductory Chemistry‖,
Chapter 8
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8.2 Making Pancakes:
Relationships between
Ingredients
Tro's ―Introductory Chemistry‖,
Chapter 8
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Making Pancakes
• Pancakes are made with specific amounts of
ingredients. Your book forgot buttermilk!
Pancake recipe figure on page 244
1 cup flour + 2 eggs + ½ tsp baking powder +3/4 cup buttermilk
 5 pancakes
Tro's ―Introductory Chemistry‖,
Chapter 8
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Making Pancakes, Continued
• You can scale the recipe up of down.
• Notice that for every 2 eggs  5 pancakes
• What if you have 8 eggs  ? pancakes
5 pancakes
8 eggs 
 20 pancakes
2 eggs
Graphics illustrating the above equation
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Chapter 8
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8.2 Making Pancakes:
Relationships between
Ingredients
Tro's ―Introductory Chemistry‖,
Chapter 8
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Making Molecules
Mole-to-Mole Conversions
• The balanced equation = chemical ―recipe‖
• 3 H2(g)
+ N2(g)

2 NH3(g)
3 molecules + 1 molecule 
2 molcules
H2
N2
NH3
H-H
H-H
H-H
+
N-N

H
H-N-H
H
H-N-H
Tro's ―Introductory Chemistry‖,
Chapter 8
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Molecule to Molecule and
Mole to Mole conversions
• 3 H2(g)
+ N2(g)

2 NH3(g)
• This equation may be scaled up if the ratio of H2:N2:NH3 is kept
3:1:2
• Multiply by 4
• 12 H2(g)
+ 4N2(g)

8 NH3(g)
• Multiply by 6.022 x 1023
• 3x(6.022 x1023) H2(g) + 1x(6.022 x1023)N2(g)  2x(6.022 x1023)NH3(g)
• This number of molecules can be expressed as moles.
• 3 mol H2(g) + 1 mol N2(g)  2 mol NH3(g)
Tro's ―Introductory Chemistry‖,
Chapter 8
13
The coefficients in a balanced equation give the
number of molecules as well as the number of moles
of each substance
We can use the ratios of the coefficients to convert
between moles of substances in a chemical reaction
8.3 Making Molecules: Mole to
Mole Conversions
Tro's ―Introductory Chemistry‖,
Chapter 8
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Molar Ratios
•3 H2(g)
+ N2(g)

2 NH3(g)
•This equation has 6 molar ratios relating the
reactant and products.
•For example H2 and N2 are related by
1 mole N2
3 mol H2
or
3 mol H2
1 mole N2
Mole to Mole Conversions
• 3 H2(g)
+ N2(g)

2 NH3(g)
• 12 moles of H2 requires how many mole of
N2
• The ratio must be H2:N2 is 3:1, so so 12:4 is
the same so 4 mole of N2 is needed.
• How many moles of N2 are needed for 1.74
mol of H2
Tro's ―Introductory Chemistry‖,
Chapter 8
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Mole to Mole Conversion
•
•
•
•
3 H2(g)
+ N2(g)

2 NH3(g)
How many moles of N2 are needed for 1.74 mol of H2?
We need to use the molar ratio.
mol H2  mole N2
1 mole N2
1.74 mol H2 x
= 0.580 mole N2
3 mol H2
Tro's ―Introductory Chemistry‖,
Chapter 8
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8.4 Making Molecule: Mass to
Mass Conversions
Tro's ―Introductory Chemistry‖,
Chapter 8
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Tro's ―Introductory Chemistry‖,
Chapter 8
20
Making Molecules
Mass-to-Mass Conversions
• We learned previously to convert between moles and grams
using the molar mass.
• Combining this with moles to moles conversions allows us to
related grams of one substrance in an equation to grams of
another substance.
g of A
mol of A
Molar
mass of A
mol of B
Coefficients
of the Balanced
Equation
g of B
Molar
mass of B
21
Example 8.2—How Many Grams of Glucose Can
Be Synthesized from 58.5 g of CO2 in
Photosynthesis?
• Photosynthesis:
6 CO2(g) + 6 H2O(g)  C6H12O6(s) + 6 O2(g)
g of CO2
mol of CO2
mol of C6H12O6
g of C6H12O6
Molar masses needed (sum masses of all atoms)
CO2 = 44.01 g/mol
Glucose = 180.16 g/mol
Tro's ―Introductory Chemistry‖,
Chapter 8
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Grams  moles  moles  grams
6 CO2(g) + 6 H2O(g)  C6H12O6(s) + 6 O2(g)
58.5 g CO2
1 mol CO2
x
= 1.33 mol CO2
44.01 g CO2
1.33 mol g CO2 1 mol C6H12O6
x
6 mol CO2
= 0.222 mol C6H12O6
0.222 mol g CO2 180.16 g C6H12O6
x
=
1 mol C6H12O6
40.0 g C6H12O6
Another stoichiometry example
How many g of Al2(SO4)3 are required to completely
react with 24.7 g of Ba(NO3)2?
Al2(SO4)3(aq) + 3Ba(NO3)2(aq)
3BaSO4(s) + 2Al(NO3)3(aq)
g of
Ba(NO3)2
mol of
Ba(NO3)2
mol of
Al2(SO4)3
molar masses needed
Ba(NO3)2 = 261.34g/mol
Al2(SO4)3 = 342.15 g/mol
g of
Al2(SO4)3
Grams  moles  moles  grams
Al2(SO4)3(aq) + 3Ba(NO3)2(aq)
24.7 g Ba(NO3)2
3BaSO4(s) +Al(NO3)3(aq)
1 mol Ba(NO3)2
x
= 0.0945 mol Ba(NO3)2
261.34 g Ba(NO3)2
0.0945 mol Ba(NO3)2
1 mol Al2(SO4)3
x
= 0.0315 mol Al2(SO4)3
3 mol Ba(NO3)2
0.0315 mol g Al2(SO4)3
342.15 g Al2(SO4)3
x
= 10.8 g Al2(SO4)3
1 mol Al2(SO4)3
8.5 More Pancakes: Limiting
Reactant, Theoretical Yield, and
Percent Yield
8.6 Limiting Reactant,
Theoretical Yield, and Percent
Yield form Initial Reactants
Tro's ―Introductory Chemistry‖,
Chapter 8
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Note in these two sections, we are only going to
cover percent yield
Actual; yield
% Yield =
x 100
Theoretical yield
Actual yield – amount actually obtained
Theoretical yield – amount obtained if
reaction goes to 100% completion.
Tro's ―Introductory Chemistry‖,
Chapter 8
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% yield example
A is ran with 100 g of Hg reacting with
oxygen. 75.3 grams of HgO is isoloted.
What is the % yield?
Hg(s) + O2(g)
Molar masses:
Hg = 200.59 g/mol
HgO = 216.59 g/mol
HgO(s)
28
% yield example (cont)
• First calculate the theoretical yield.
• g Hg  mol Hg  mol HgO  g HgO
• 100 g Hg
1 mol Hg
x
= 0.499 mol Hg
200.59 g Hg
• 0.499 mol Hg 1 mol HgO
x
= 0.499 mol HgO
1 mol Hg
• 0.499 mol Hg 216.59 g HgO
x
= 108 g HgO (theoretical
1 mol HgO
yield)
29
% yield example
75.0 g HgO
% Yield =
x 100
108 g HgO
= 69.4 % Yield of HgO
.
Tro's ―Introductory Chemistry‖,
Chapter 8
30
Enthalpy Change
• We previously described processes as
exothermic if they released heat, or
endothermic if they absorbed heat.
• The enthalpy of reaction is the amount of
thermal energy that flows through a process.
At constant pressure.
DHrxn
Tro's ―Introductory Chemistry‖,
Chapter 8
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Sign of Enthalpy Change
• For exothermic reactions, the sign of the enthalpy
change is negative when:
 Thermal energy is produced by the reaction.
 The surroundings get hotter.
 DH = ─
 For the reaction CH4(s) + 2 O2(g)  CO2(g) + 2 H2O(l), the
DHrxn = −802.3 kJ per mol of CH4.
• For endothermic reactions, the sign of the enthalpy
change is positive when:
 Thermal energy is absorbed by the reaction.
 The surroundings get colder.
 DH = +
 For the reaction N2(s) + O2(g)  2 NO(g), the
DHrxn = +182.6 kJ per mol of N2.
32
Enthalpy and Stoichiometry
• The amount of energy change in a reaction depends
on the amount of reactants.
 You get twice as much heat out when you burn twice as
much CH4.
• Writing a reaction implies that amount of energy
changes for the stoichiometric amount given in the
equation.
For the reaction C3H8(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
DHrxn = −2044 kJ
So 1 mol C3H8  5 mol O2  3 mol CO2  4 mol H2O 
−2044 kJ.
Tro's ―Introductory Chemistry‖,
Chapter 8
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Enthalpy Calculation
• C3H8(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
DHrxn = −2044 kJ
• How much heat is released when 53.8 g of O2
completely reacts with propane?
• g O2  mol O2  heat (J)
• 53.8 g of O2 1 mol O2
x
= 1.68 mol O2
32.00 g O2
• 1.68 mol of O2 -2044 J
x
= 687 J of heat
5 mol O2
34
Example 8.7—How Much Heat Is Associated with the
Complete Combustion of 11.8 x 103 g of C3H8(g)?
Given:
Find:
Solution Map:
11.8 x 103 g C3H8,
heat, kJ
g C3H8
mol C3H8
1 mol C3H8
44.09 g
Relationships:
kJ
- 2044 kJ
1 mol C3H 8
1 mol C3H8 = -2044 kJ, Molar mass = 44.11 g/mol
Solution:
11.8  103 g C3H8 
1 mol C3H8
- 2044 kJ

 5.47  105 kJ
44.11 g C3H8 1 mol C3H8
Check: The sign is correct and the value is reasonable.
Practice—How Much Heat Is Evolved When a 0.483 g
Diamond Is Burned?
(DHcombustion = −395.4 kJ/mol C)
Tro's ―Introductory Chemistry‖,
Chapter 8
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