chapter 8 molecules and carbon dioxide and quantities in ...whitener/courses/sp2011/... · yield...

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, Maqqwertd ygoijpk[l

‗?

Introductory Chemistry, 3rd Edition

Nivaldo Tro

Chapter 8

Quantities in

Chemical Reactions

2009, Prentice Hall

Car an octane and oxygen

molecules

and carbon dioxide and

water

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

2Tro's ―Introductory Chemistry‖,

Chapter 8

8.1 Global Warming : Too

much Carbon Dioxide


Chapter 8

Burning of Fossil

Fuels

Internal combustion engines

produe Carbon dioxide by

burning fuels.

For example octane, C8H18,

in gasoline.

2C8H18 (l) + 25O2(g) 16 CO2 (g) + 18H2O(g)

Car and molecules

from chapter opening


Chapter 8

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.


Chapter 8

Increased CO2

• The primary source of the increased CO2

levels is combustion of fossil fuels

1860 = Industrial Revolution in the U.S. and Europe.

(g)(g)(g)(g) OH 2 CO O 2 CH 2224

Methane in natural gas


Chapter 8

Quantities in Chemical Reactions

• All substances in a chemical reaction are related.

• The numerical relationships between compounds in a chemical reaction is called stoichiometry.

8.2 Making Pancakes:

Relationships between

Ingredients


Chapter 8


Chapter 8

Making Pancakes

• Pancakes are made with specific amounts of

ingredients. Your book forgot buttermilk!

1 cup flour + 2 eggs + ½ tsp baking powder +3/4 cup buttermilk

5 pancakes

Pancake recipe figure on page 244


Chapter 8

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

pancakes 20 eggs 2

pancakes 5eggs 8

Graphics illustrating the above equation

8.2 Making Pancakes:

Relationships between

Ingredients


Chapter 8


Chapter 8

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-N-H H-N-H

H-H

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)


Chapter 8

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


Chapter 8

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


Chapter 8

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


Chapter 8

8.4 Making Molecule: Mass to

Mass Conversions


Chapter 8


Chapter 8

21

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 mol of B g of B

Molar

mass of AMolar

mass of B

Coefficients

of the Balanced

Equation


Chapter 8

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)

Molar masses needed (sum masses of all atoms)

CO2 = 44.01 g/mol

Glucose = 180.16 g/mol

g of CO2 mol of CO2 mol of C6H12O6 g of C6H12O6

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 = 0.222 mol C6H12O6

6 mol CO2

0.222 mol g CO2 180.16 g C6H12O6

x = 40.0 g C6H12O6

1 mol C6H12O6

Another stoichiometry exampleHow 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 mol of mol of g of

Ba(NO3)2 Ba(NO3)2 Al2(SO4)3 Al2(SO4)3

molar masses needed

Ba(NO3)2 = 261.34g/mol

Al2(SO4)3 = 342.15 g/mol

Grams moles moles grams

Al2(SO4)3(aq) + 3Ba(NO3)2(aq) 3BaSO4(s) +Al(NO3)3(aq)

24.7 g Ba(NO3)2 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


Chapter 8

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.27Tro's ―Introductory Chemistry‖,

Chapter 8

% 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) HgO(s)

Molar masses:

Hg = 200.59 g/mol

HgO = 216.59 g/mol28

% 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


Chapter 8

75.0 g HgO

% Yield = x 100

108 g HgO

= 69.4 % Yield of HgO

.


Chapter 8

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

32

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.


Chapter 8

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.

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 O234

Example 8.7—How Much Heat Is Associated with the

Complete Combustion of 11.8 x 103 g of C3H8(g)?

1 mol C3H8 = -2044 kJ, Molar mass = 44.11 g/mol

The sign is correct and the value is reasonable.

11.8 x 103 g C3H8,

heat, kJ

Check:

Solution:

Solution Map:

Relationships:

Given:

Find:

g 09.44

HC mol 1 83

kJ 1047.5HC mol 1

kJ 2044-

HC g 44.11

HC mol 1 HC g 1011.8 5

8383

8383

3

83HC mol 1

kJ 2044-

mol C3H8 kJg C3H8


Chapter 8

Practice—How Much Heat Is Evolved When a 0.483 g

Diamond Is Burned?

(DHcombustion = −395.4 kJ/mol C)

chapter 8 molecules and carbon dioxide and quantities in ...whitener/courses/sp2011/... · yield...

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