Applied Medical Sciences

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General Chemistry

Chapter 3
Calculations with Chemical Formulas and Equations

Halima Thwaib
Chapter 3: Calculations with Chemical
Formulas and Equations
Mass and Moles of Substance
3.1 Molecular Mass and Formula Mass
3.2 The Mole Concept

Determining Chemical Formulas

3.3 Mass Percentages from the Formula
3.4 Elemental Analysis: Percentages of Carbon,
Hydrogen, and Oxygen
3.5 Determining Formulas

Stoichiometry: Quantitative Relations in

Chemical Reactions
3.6 Molar Interpretation of a Chemical Equation
3.7 Amounts of Substances in a Chemical Reaction
3.8 Limiting Reactant; Theoretical and Percentage
Yields
Mass and Moles of Substance
3.1 Molecular Weight and Formula Weight

• Chemistry requires a method for determining the numbers of

molecules in a given mass of a substance.

– This allows the chemist to carry out “recipes” for

compounds based on the relative numbers of atoms
involved.

– The calculation involving the quantities of reactants and

products in a chemical equation is called stoichiometry.

General Chemistry 9th-Ebbing.Gammon 3

3.1 Molecular Weight and Formula Weight

• The molecular weight of a substance is the sum of the

atomic weights of all the atoms in a molecule of the
substance.

– For, example, a molecule of H2O contains 2 hydrogen

atoms (at 1.0 amu each) and 1 oxygen atom (16.0 amu),
giving a molecular weight of 18.0 amu.

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3.1 Molecular Weight and Formula Weight

• The formula weight of a substance is the sum of the

atomic weights of all the atoms in one formula unit of the
compound, whether molecular or not.

– For example, one formula unit of NaCl contains 1 sodium atom

(23.0 amu) and one chlorine atom (35.5 amu), giving a formula
weight of 58.5 amu.

– The molecular weight and the formula weight calculated from

the molecular formula of a substance are identical

General Chemistry 9th-Ebbing.Gammon 5

3.2 The Mole Concept
• A mole (symbol mol) is defined as the quantity of a given substance that
contains as many molecules or formula units as the number of atoms in
exactly 12 grams of carbon–12.

• The number of atoms in a 12-gram sample of carbon–12 is called

Avogadro’s number (to which we give the symbol NA). The value of
Avogadro’s number is 6.02 x 1023.
• For example, a mole of oxygen atoms (with the formula O) contains 6.02 x
1023 O atoms. A mole of oxygen molecules (formula O2) contains 6.02 x 1023
O2 molecules—that is, 2 x 6.02 x 1023 O atoms.

6
General Chemistry 9th-Ebbing.Gammon
Important note
1 mole of an atom, molecules or formula units
contains 6.02 x 1023 atom, molecules formula
unit

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3.2 The Mole Concept

• The molar mass of a substance is the mass of one mole of

a substance.

– For all substances, molar mass, in grams per mole, is

numerically equal to the formula weight in atomic mass
units.
– That is, one mole of any element weighs its atomic mass in grams.
– 1mole Na =23g

General Chemistry 9th-Ebbing.Gammon 8

3.2 The Mole Concept

•Determining the Formula/Molar Mass of Ammonium

Sulfate

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3.2 The Mole Concept
• Mole calculations

– Suppose we have 100.0 grams of iron (Fe). The atomic weight

of iron is 55.8 g/mol. How many moles of iron does this
represent?
– Number of moles = Mass (g)/molar.mass (g/mole)

100.0 g Fe
moles Fe 
55.8 g/mol

 1.79 moles of Fe

General Chemistry 9th-Ebbing.Gammon 10

3.2 The Mole Concept
• Mole calculations

– Conversely, suppose we have 5.75 moles of magnesium (atomic

wt. = 24.3 g/mol). What is its mass?
– # moles (n) = mass/M.wt
– 5.75 = mass / 24.3

mass Mg  (5.75 moles)  (24.3 g/mol)

 140 grams of Mg
Or 1.40 x 102 grams of Mg

General Chemistry 9th-Ebbing.Gammon 11

3.2 The Mole Concept

• Mole calculations

– This same method applies to compounds. Suppose we have

100.0 grams of H2O (molecular weight = 18.0 g/mol). How
many moles does this represent?
– # mole = mass / Mwt
100.0 g H 2O
moles H 2O 
18.0 g/mol

 5.56 moles of H 2O

General Chemistry 9th-Ebbing.Gammon 12

3.2 The Mole Concept

• Mole calculations

– Conversely, suppose we have 3.25 moles of glucose,

C6H12O6 (molecular wt. = 180.0 g/mol). What is its
mass? # mole = mass /Mwt
– 3.25 = mass / 180.0

mass C6 H12O 6  ( 3.25 moles)  (180.0 g/mol)

 585 grams of C6 H12O 6

General Chemistry 9th-Ebbing.Gammon 13

3.2 The Mole Concept
•Mass and Moles and Number of Molecules or Atoms
• The number of molecules or atoms in a sample is related to
the moles of the substance:

1 mole HCl  6.02  10 HCl molecules 23

1 mole Fe  6.02  10 23 Fe atoms

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• Suppose we have a 3.46-g sample of hydrogen chloride, HCl.
How many molecules of HCl does this represent?
23
• I mole HCl molecules = 6.02 * 10 molecules
• 0.0948 mole = ?? Molecules
23
• 6.02 * 10 * 0.0948 /1 =  5.7110 HCl molecules
22

• # moles = mass / M.wt

• = 3.46 / (1+35.5) =0.0948 mole

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3.2 The Mole Concept

•We can read formulas in terms of moles of atoms or ions.

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3.2 The Mole Concept

•How many atoms?

How many atoms are in 0.10 moles of Uranium atoms?
Convert to
Moles Atoms
23
• 1 mole of U atoms → 6.02 * 10 atoms U
• 0.10 mole of U atoms →??? atoms

23 23
• 0.10 mole * 6.02 * 10 / 1 = 6.0 * 10 atoms

General Chemistry 9th-Ebbing.Gammon 17

3.2 The Mole Concept

What is the mass of 0.10 moles of Uranium atoms?

# moles = mass / M.wt

0.1 = mass / 238.029

General Chemistry 9th-Ebbing.Gammon 18

•Practice with Propane (C3H8)

• How many molecules are in 2.00 mole of Propane?

1 mole Propane molecules → 6.02* 10^23 molecules
2.00 mole Propane molecules →???
# molecules = 1.20 * 10^ 24 molecules

• What is the mass of 1.00 mole of propane molecules?

Molar mass= 44 g/ mole
# moles = mass/ M.wt
1.00 = mass / 44
44*1.00 = mass

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3.2 The Mole Concept
 How many H atoms are in 2.00 mole of Propane? C3H8
 1 mole H atoms → 6.02* 10^23 atoms
 2* 8 mole H atoms →??? Atoms
 (2*8 * 6.02* 10^23)/1 = ??? * 1 /1

 9.63 * 10^24 atoms of H

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3.2 The Mole Concept

• How many grams of carbon are there in 2.4g C3H8?

• # moles C3H8 = mass/ M.wt = 2.4/(12*3 +8*1) =

0.054 moles

• #moles C = mass C / M.wt C

• 3 * 0.054 = mass / 12
• mass = 1.9 grams = 2 g

General Chemistry 9th-Ebbing.Gammon 21

22
3.3 Mass Percentages from the Formula

• The percent composition of a compound is the mass

percentage of each element in the compound.

– We define the mass percentage of “A” as the parts of “A” per

hundred parts of the total, by mass. That is,

mass of " A" in whole

mass % " A"   100%
mass of the whole

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• Example :
Formaldehyde, CH2O, is a toxic gas with a pungent odor. Large quantities
are consumed in the manufacture of plastics .Calculate the mass percentages
of the elements in formaldehyde?

•1mol CH2O has a mass of 30.0 g and contains 1 mol C (12.0 g), 2
mol H (2 1.01 g), and 1 mol O (16.0 g). You divide each mass of
element by the molar mass, then multiply by 100, to obtain the mass
percentage.

•You can calculate the percentage of O in the same way, but it can also
be found by subtracting the percentages of C and H from 100%:

General Chemistry 9th-Ebbing.Gammon 24

3.3 Mass Percentages from the Formula

• % composition NH3?

• %N= (mass of N / total mass of cpd) *100%

• = (14*1/(14*1+3*1)) *100% = 82.4%

• % H = (mass of H / total mass of cpd) *100%

• = ( 1*3 / 17) * 100% = 17.64 %

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3.3 Mass Percentages from the Formula

• What is the % composition of C6H12O6?

• M.wt C6H12O6 = (12.0*6)+ (12*1) + (16*6) =180 g/ mole
• % C = (mass of C / total mass of cpd) *100%
• = ( 12*6 / 180) * 100% = 40%

• % O = = (mass of O / total mass of cpd) *100%

• = ( 16*6) / 180) *100% = 53.3 %

• % H = 100% – ( 40%+53.3%) = 6.66%

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3.4 Elemental Analysis: Percentages of Carbon,
Hydrogen, and Oxygen

• Is one method of determining empirical formulas in the laboratory.

• This method is used primarily for simple organic compounds (that

contain carbon, hydrogen, oxygen).

– The organic compound is burned in oxygen.

– The products of combustion (usually CO2 and H2O) are weighed.

– The amount of each element is determined from the mass of products

General Chemistry 9th-Ebbing.Gammon 27

3.5 Determining Chemical Formulas

• Determining the formula of a compound from the percent

composition.

– The percent composition of a compound leads directly to its

empirical formula.

– An empirical formula (or simplest formula) for a compound

is the formula of the substance written with the smallest integer
(whole number) subscripts.

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3.5 Determining Chemical Formulas
• Determining the empirical formula from the percent
composition.

– Benzoic acid is a white, crystalline powder used as a food

preservative. The compound contains 68.8% C, 5.0% H, and
26.2% O by mass. What is its empirical formula?

– In other words, give the smallest whole-number ratio of the

subscripts in the formula

C x H y Oz
General Chemistry 9th-Ebbing.Gammon 29
3.5 Determining Chemical Formulas
• Benzoic acid is a white, crystalline powder used as a food preservative.
The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What
is its empirical formula?

• We assume we have 100 g of benzoic acid

• C = 68.8 g , H= 5.0 g , O= 26.2 g
# moles C= mass/M.wt = 68.8/ 12 = 5.73mole
#moles H= mass/ M.wt= 5.0/1= 5.0 mole
#moles O= mass/ M.wt= 26.2/16= 1.63 mole
C H O
5.73/1.63 5.0 /1.63 1.63 /1.63
( 3.5 3.0 1.0) 2
7 6 2

• The empirical formula is C7H6O2 .

General Chemistry 9th-Ebbing.Gammon 30
3.5 Determining Chemical Formulas

• Determining the molecular formula from the empirical

formula.

– An empirical formula gives only the smallest whole-number ratio of

atoms in a formula.

– The molecular formula should be a multiple of the empirical formula

(since both have the same percent composition).

– To determine the molecular formula, we must know the molecular

weight of the compound.

General Chemistry 9th-Ebbing.Gammon 31

3.5 Determining Chemical Formulas
 For example, suppose the empirical formula of a compound is
CH2O and its molecular weight is 60.0 g/mol.
– The molar weight of the empirical formula (the empirical
weight) is only 30.0 g/mol.
– This would imply that the molecular formula is actually the
empirical formula doubled, or

– Molecular Formula = M.wt Molecular Formula X empirical Formula

M.wt empirical Formula
= (60/ 30) X CH2O
= 2 X CH2O
C2H4O2
General Chemistry 9th-Ebbing.Gammon 32
3.5 Determining Chemical Formulas
What is the Empirical Formula if the % composition is 40.0% C,
6.7% H, and 53.3% O?
C H O

•The Empirical Formula is CH2O (MW =30.026)

•If MW of the real formula is 180.155, what is the actual ( molecular formula) formula?

(180.155)/(30.026) = 6 CH2O x 6 = C6H12O6

What is the Empirical Formula if the % composition is 40.0% C,
6.7% H, and 53.3% O?
 we assume 100 g from the compound
C= 40.0 g , H= 6.7 g , O=53.3 g
#mole C = mass/ m.wt = 40.0/ 12.0 = 3.33 mole
# mole H = mass/M.wt= 6.7/ 1.0= 6.7 mole
# mole O= mass/ M.wt = 53.3/ 16 = 3.33 mole

C H O
3.33/3.33 6.7/3.33 3.33/3.33
1 2 1
CH2O ( 12+ 1*2 + 16) = 30 g/mole
What is the molecular formula if its M.wt = 180?
Molecular formula = (M.wt Molecular formula/ M.wt empirical formula) empirical formula
= (180/ 30) CH2O
= 6 * CH2O
= C6H12O6

General Chemistry 9th-Ebbing.Gammon 34

Stoichiometry: Quantitative Relations in
Chemical Reactions
• Stoichiometry is the calculation of the quantities of
reactants and products involved in a chemical reaction.

– It is based on the balanced chemical equation and on the

relationship between mass and moles.

– Such calculations are fundamental to most quantitative work in

chemistry.

General Chemistry 9th-Ebbing.Gammon 35

3.6 Molar Interpretation of a chemical Equation

• The balanced chemical equation can be interpreted in

numbers of molecules, but generally chemists interpret
equations as “mole-to-mole” relationships.

– For example, the Haber process for producing ammonia

involves the reaction of hydrogen and nitrogen.

N 2 (g )  3 H 2 (g )  2 NH 3 (g )

General Chemistry 9th-Ebbing.Gammon 36

3.6 Molar Interpretation of a chemical Equation

• This balanced chemical equation shows that one mole of N 2

reacts with 3 moles of H2 to produce 2 moles of NH3.

N 2 (g)  3H 2 (g)  2 NH 3 (g )
1 molecule N2 + 3 molecules H2 2 molecules NH3

1 mol N 2  3 mol H 2  2 mol NH 3

– Because moles can be converted to mass, you can also give a
mass interpretation of a chemical equation.

General Chemistry 9th-Ebbing.Gammon 37

3.6 Molar Interpretation of a chemical Equation

• Suppose we wished to determine the number of

moles of NH3 we could obtain from 4.8 mol H2.

N 2 (g)  3H 2 (g)  2 NH 3 (g )
– Because the coefficients in the balanced equation represent mole-to-
mole ratios, the calculation is simple.

2 mol NH 3
4.8 mol H 2   3.2 mol NH 3
3 mol H 2
General Chemistry 9th-Ebbing.Gammon 38
3.6 Molar Interpretation of a chemical Equation

N 2 (g)  3H 2 (g)  2 NH 3 (g )
3 mole 2 mole
4.8 mole x mole

2 *4.8 = 3 * x
2*4.8 /3 = 3.2 mole NH3

General Chemistry 9th-Ebbing.Gammon 39

3.7 Amounts of Substances in a Chemical
Reaction
• Amounts of substances in a chemical reaction by mass.

– How many grams of HCl are required to react with 5.00 grams
manganese (IV) oxide according to this equation?

4 HCl(aq)  MnO 2 (s )  2 H 2O(l)  MnCl 2 (aq)  Cl 2 (g )

•Steps in a stoichiometric calculation : You convert the mass of substance A in a reaction to
moles of substance A, then to moles of another substance B, and finally to mass of substance
B.

General Chemistry 9th-Ebbing.Gammon 40

3.7 Amounts of Substances in a Chemical Reaction
4 HCl(aq)  MnO 2 (s )  2 H 2O(l)  MnCl 2 (aq)  Cl 2 (g )
4 mole 1 mole
x 0.0575 mole

# n MnO2 = mass/ M.wt = 5.00/86.9 = 0.0575 mole MnO2

0.0575* 4 = 1* x
# moles HCl = 0.230 mole

# mole HCl = mass/ M.wt

0.230 = mass/(1+35.5)
Mass = 8.395 = 8.40

General Chemistry 9th-Ebbing.Gammon 41

3.8 Limiting Reactant; Theoretical and
Percentage Yields
• The limiting reactant (or limiting reagent) is the reactant that is
entirely consumed when the reaction goes to completion. The
limiting reagent ultimately determines how much product can be
obtained.
• For example, bicycles require one frame and two wheels. If you have 20
wheels but only 5 frames, it is clear that the number of frames will determine
how many bicycles can be made.

General Chemistry 9th-Ebbing.Gammon 42

3.8 Limiting Reactant; Theoretical and
Percentage Yields

• Zinc metal reacts with hydrochloric acid by the following

reaction.

Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )

– If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl,
how many moles of H2 are produced?

General Chemistry 9th-Ebbing.Gammon 43

3.8 Limiting Reactant; Theoretical and Percentage
Yields
Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )
1 mole 1 mole
0.30 mole ??
1*0.30 /1 = 0.30 mole H2

Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )

2 mole 1 mole
0.52 mole ??

0.52*1/2 = 0.26 mole H2

HCl is the limiting reagent
General Chemistry 9th-Ebbing.Gammon 44
3.8 Limiting Reactant; Theoretical and Percentage
Yields
Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )
1 mole 2 mole
0.30 mole 0.52 mole

0.3/ 1 0.52/ 2
0.3 0.26 the smallest value so HCl is the limiting reagent

Zn(s)  2 HCl(aq)  ZnCl 2 (aq)  H 2 (g )

2 mole 1 mole
0.52 mole ???

0.52 mole X 1 = 2 X ??
# mole H2 = 0.26 mole General Chemistry 9th-Ebbing.Gammon 45
3.8 Limiting Reactant; Theoretical and
Percentage Yields

• The theoretical yield of product is the maximum amount of

product that can be obtained from given amounts of
reactants.

– The percentage yield is the actual yield (experimentally determined) expressed as a

percentage of the theoretical yield (calculated).

actual yield
%Yield   100%
theoretical yield
General Chemistry 9th-Ebbing.Gammon 46
3.8 Limiting Reactant; Theoretical and
Percentage Yields

• To illustrate the calculation of percentage yield, recall that the

theoretical yield of H2 in the previous example was 0.26 mol (or
0.52 g) H2.
• Theoretical yield: Mass = # n * M.wt
Mass= 0.26 * (2*1.00) = 0.52 g
• If the actual yield of the reaction had been 0.22 g H2, then
0.22 g H 2
%Yield   100%  42%
0.52 g H 2

General Chemistry 9th-Ebbing.Gammon 47

3.8 Limiting Reactant; Theoretical and Percentage Yields
 Example : How many grams of HI can be formed from 2.00 g H2 and
2.00 g of I2?

H2 + I2  2 HI
1 mole 1 mole
1 mole 0.0079 mole
1/1 = 1 0.0079/1 = 0.0079 the smallest value

I2 is the limiting reagent

# n H2 = mass/ m.wt = 2 / (1*2) = 1 mole

#n I2= mass/ m.wt = 2 /253.8 = 0.0079 mole

H2 + I2  2 HI
1mole 2 mole
0.0079 mole ????

2*0.0079 = 1 * ??? General Chemistry 9th-Ebbing.Gammon 48

0.0158 mole HI
3.8 Limiting Reactant; Theoretical and Percentage Yields

#n HI = mass/ m.wt
0.0158 = mass / 127.9
Mass HI= 2.016 g

Suppose the actual yeid of HI is 1.05 g calculate the percentage yeild?

% yeild = actual/ therotical * 100%
= 1.05 / 2.016 * 100%
= 52.08%

General Chemistry 9th-Ebbing.Gammon 49

• Example :

How many grams of CO2 can be formed from 1.44 g C3H8

and 2.65 g of O2?

C3H8 + 5 O2  3CO2 + 4 H2O

General Chemistry 9th-Ebbing.Gammon 50

 Applied Medical Sciences
Medical Imaging & Physiotherapy

General Chemistry

Chapter 4
Chemical Reactions

Course No: 510162

Halima Thwaib

Second Semester
General Chemistry 9th-Ebbing.Gammon 2
Chapter 4: Chemical Reactions
Ions in Aqueous Solution
4.1 Ionic Theory of Solutions and Solubility Rules
4.2 Molecular and Ionic Equations

Types of Chemical Reactions

4.3 Precipitation Reactions
4.4 Acid–Base Reactions
4.5 Oxidation–Reduction Reactions
4.6 Balancing Simple Oxidation–Reduction
Equations

Working with Solutions

4.7 Molar Concentration
4.8 Diluting Solutions

Quantitative Analysis
4.9 Gravimetric Analysis
4.10 Volumetric Analysis
 Ions in Aqueous Solution
4.1 Ionic Theory of Solutions and Solubility Rules
• In 1884, the young Swedish chemist Svante Arrhenius proposed the ionic
theory of solutions to account for this conductivity.
• He said that certain substances produce freely moving ions when they
dissolve in water, and these ions conduct an electric current in an aqueous
solution.

• Suppose you dissolve sodium chloride, NaCl, in water. You may remember
that sodium chloride is an ionic solid consisting of Na + and Cl -, held in a
regular, fixed array. When you dissolve solid NaCl in water, the Na + and Cl -
ions go into solution as freely moving ions.

General Chemistry 9th-Ebbing.Gammon 4

4.1 Ionic Theory of Solutions and Solubility Rules

• Now consider pure water. Water consists of molecules, each of which is

electrically neutral. Since each molecule carries no net electric charge, it
carries no overall electric charge when it moves. Thus, pure water is a
nonconductor of electricity.

• In summary, although water is itself nonconducting, it has the ability to

dissolve various substances, some of which go into solution as freely
moving ions. An aqueous solution of ions is electrically conducting.

General Chemistry 9th-Ebbing.Gammon 5

4.1 Ionic Theory of Solutions and Solubility Rules
Electrolytes and Nonelectrolytes

• Many ionic compounds dissociate into independent ions when

dissolved in water

H 2O

NaCl( s )  Na  ( aq )  Cl  ( aq )

• These compounds that “freely” dissociate into independent ions in aqueous

solution are called electrolytes. Their aqueous solutions are capable of
conducting an electric current.

Electrolytes are substances that dissolve in water to give an electrically

conducting solution.
•Thus, in general, ionic solids that dissolve in water are electrolytes.
General Chemistry 9th-Ebbing.Gammon 6
4.1 Ionic Theory of Solutions and Solubility Rules
• Not all electrolytes are ionic compounds. Some molecular
compounds such as acids, dissociate into ions.
 
HCl ( aq )  H ( aq )  Cl ( aq )
• The resulting solution is electrically conducting, and so we say that the
molecular substance is an electrolyte.

• Some molecular compounds dissolve but do not dissociate into

ions.
C 6 H 12 O 6 ( s ) ( glucose )  C 6 H 12 O 6 ( aq )
H 2O

• These compounds are referred to as nonelectrolytes. They dissolve in water

to give a nonconducting solution.
General Chemistry 9th-Ebbing.Gammon 7
Observing the Electrical Conductivity of a Solution

Figure 4.3 shows a simple apparatus that allows you to observe the
conductivity of a solution.

• If the solution is conducting, the circuit is complete and the bulb lights.
• If the solution is nonconducting, the circuit is incomplete and the bulb
does not light.

General Chemistry 9th-Ebbing.Gammon 8

Strong and Weak Electrolytes
• A strong electrolyte is an electrolyte that exists in solution almost
entirely as ions.

• An example is sodium chloride. We can represent the dissolution of

sodium chloride in water by the following equation:

 
NaCl( s )  Na ( aq )  Cl ( aq )
H 2O

•Most ionic solids that dissolve in water do so almost completely as ions, so

they are strong electrolytes.

General Chemistry 9th-Ebbing.Gammon 9

Strong and Weak Electrolytes
• A weak electrolyte is an electrolyte that dissolves in water to give a
relatively small percentage of ions.
• These are generally molecular substances. Ammonia, NH3, is an example.


NH 4 OH ( aq ) 

NH 4 ( aq )  OH ( aq ) 

• Most soluble molecular compounds are either nonelectrolytes or weak

electrolytes.

General Chemistry 9th-Ebbing.Gammon 10

Strong and Weak Electrolytes

•Figure 4.4 illustrates the conductivity of weak versus strong electrolytes.

•Solutions of weak electrolytes contain only a small percentage of ions. We

denote this situation by writing the equation with a double arrow.

General Chemistry 9th-Ebbing.Gammon 11

Strong and Weak Electrolytes

• In summary, substances that dissolve in water are either

electrolytes or nonelectrolytes.

– Nonelectrolytes form nonconducting solutions because they

dissolve as molecules.

– Electrolytes form conducting solutions because they dissolve as

ions.
• Electrolytes can be strong or weak.

– Almost all ionic substances that dissolve are strong electrolytes.

– Molecular substances that dissolve are either nonelectrolytes or

weak electrolytes.
General Chemistry 9th-Ebbing.Gammon 12
Solubility Rules

• Substances vary widely in their solubility, or ability to dissolve, in

water.

•Some compounds, such as sodium chloride and ethyl alcohol

(CH3CH2OH), dissolve readily and are said to be soluble. Others,
such as calcium carbonate and benzene (C6H6), have quite limited
solubilities and are thus said to be insoluble

•Table 4.1 lists eight solubility rules for ionic compounds.

General Chemistry 9th-Ebbing.Gammon 13

Table 4.1 lists eight solubility rules for ionic compounds.

General Chemistry 9th-Ebbing.Gammon 14

•Let’s Practice Determining Solubility

•Which of the following are soluble in water?

Na2CO3 yes Cu(OH)2 no

CaCl2 yes Ba(OH)2 yes

AgCl no Ca3(PO4)2 no

BaSO4 no Pb(NO3)2 yes

(NH4)2S yes PbCl2 no

General Chemistry 9th-Ebbing.Gammon 15

4.2 Molecular and Ionic Equations
Molecular Equations

•Molecular equation, which is a chemical equation in which the reactants

and products are written as if they were molecular substances, even though they
may actually exist in solution as ions.

•The molecular equation is useful because it is explicit about what the reactant solutions
are and what products you obtain.

•The equation says that you add aqueous solutions of calcium hydroxide and sodium
carbonate to the reaction vessel. As a result of the reaction, the insoluble, white calcium
carbonate solid forms in the solution; that is, calcium carbonate precipitates. After you
remove the precipitate, you are left with a solution of sodium hydroxide.
General Chemistry 9th-Ebbing.Gammon 16
Complete Ionic Equations

•Although a molecular equation is useful in describing the actual reactant and

product substances, it does not tell you what is happening at the level of ions.
That is, it does not give you an ionic-theory interpretation of the reaction.

• Consider the reaction of calcium hydroxide, Ca(OH)2, and sodium carbonate,

Na2CO3. Both are soluble ionic substances and therefore strong electrolytes;
when they dissolve in water, they go into solution as ions. The complete
equation is:

• Complete ionic equation is a chemical equation in which strong electrolytes

(such as soluble ionic compounds) are written as separate ions in the solution.

General Chemistry 9th-Ebbing.Gammon 17

Net Ionic Equations

• In the complete ionic equation representing the reaction of calcium hydroxide and
sodium carbonate, some ions (OH and Na) appear on both sides of the equation.
This means that nothing happens to these ions as the reaction occurs. They are called
spectator ions.
• A spectator ion is an ion in an ionic equation that does not take part in the
reaction.
• You can cancel such ions from both sides to express the essential reaction that
occurs

• The resulting equation is

• This is the net ionic equation, an ionic equation from which spectator ions have
been canceled. It shows that the reaction that actually occurs at the ionic level is
between calcium ions and carbonate ions to form solid calcium carbonate.
General Chemistry 9th-Ebbing.Gammon 18
Net Ionic Equations

AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)

Molecular Equation

1. Divorce Ag++NO3- + Na++ Cl- 

2. Change Partners Ag++NO3- + Na++ Cl-  AgCl + NaNO3

3. Soluble? Ag++NO3- + Na++ Cl-  AgCl(s) + Na+ NO3-

Total Ionic Equation

4. Cross out Ag++ Cl-  AgCl(s)

Spectator Ions

5. Balance Ag++ Cl-  AgCl(s)

Net Ionic Equation

General Chemistry 9th-Ebbing.Gammon 19

Net Ionic Equations

Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq)

Molecular Equation

1. Divorce Pb+2+2NO3- + 2Na++2 l- 

2. Change Partners Pb+2+2NO3- + 2Na++2 l-  Pbl2 + 2NaNO3

3. Soluble? Pb+2+2NO3- + 2Na++2 l -  Pbl2(s) + 2Na+ +2NO3-

Total Ionic Equation

4. Cross out Pb+2+ 2 l-  Pbl2(s)

Spectator Ions

5. Balance Pb+2+ 2 l-  Pbl2(s)

Net Ionic Equation

General Chemistry 9th-Ebbing.Gammon 20

Net Ionic Equations

BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)

Molecular Equation

1. Divorce Ba+2 + 2Cl- + 2Na++ SO4-2 

2. Change Partners Ba+2 +2 Cl- + 2Na++ SO4-2  BaSO4 + 2NaCl

3. Soluble? Ba+2 +2 Cl- +2 Na++ SO4-2  BaSO4(s)+ 2Na+ + 2Cl-

Total Ionic Equation

4. Cross out Ba+2+ SO4-2  BaSO4(s)

Spectator Ions

5. Balance Ba+2+ SO4-2  BaSO4(s)

Net Ionic Equation

General Chemistry 9th-Ebbing.Gammon 21

 Types of Chemical Reactions

Most of the reactions we will study belong to one of three types:

1. Precipitation reactions. In these reactions, you mix solutions of two

ionic substances, and a solid ionic substance (a precipitate) forms.

2. Acid–base reactions. An acid substance reacts with a substance

called a base. Such reactions involve the transfer of a proton between
reactants.

3. Oxidation–reduction reactions. These involve the transfer of

electrons between reactants.

General Chemistry 9th-Ebbing.Gammon 22

 Types of Chemical Reactions
4.3 Precipitation Reactions

• A precipitation reaction occurs in aqueous solution because one

product is insoluble.

– A precipitate is an insoluble solid compound formed during a

chemical reaction in solution.

– For example, the reaction of sodium chloride with silver nitrate forms
AgCl(s), an insoluble precipitate

NaCl ( aq )  AgNO 3 ( aq )  AgCl ( s )   NaNO 3 ( aq )

General Chemistry 9th-Ebbing.Gammon 23

Predicting Precipitation Reactions
•Suppose you mix together solutions of magnesium chloride, MgCl2, and silver
nitrate, AgNO3. You can write the potential reactants as follows:

MgCl2+ AgNO3

•When you write a precipitation reaction as a molecular equation, the reaction has
the form of an exchange reaction. An exchange (or metathesis) reaction is a
reaction between compounds that, when written as a molecular equation, appears
to involve the exchange of parts between the two reactants.

•In a precipitation reaction, the anions exchange between the two cations (or vice
versa).

MgCl2(aq) +2 AgNO3(aq)  Mg(NO3)2(aq) + 2AgCl(s)

General Chemistry 9th-Ebbing.Gammon 24

 Types of Chemical Reactions
4.4 Acid–Base Reactions
– Acids and bases are some of the most important electrolytes. (See Table 4.2)

– Acids have a sour taste. Solutions of bases, on the other hand, have a bitter
taste and a soapy feel. They can cause color changes in certain dyes called
acid-base indicators.

– An acid–base indicator is a dye used to distinguish between acidic and

basic solutions by means of the color changes it undergoes in these solutions.

– Household acids and bases

25
General Chemistry 9th-Ebbing.Gammon
Table 4.2 Common acids and bases
Definitions of Acid and Base

• The Arrhenius Concept

– The Arrhenius concept defines acids as substances that produce hydrogen

ions, H+, when dissolved in water.

– An example is nitric acid, HNO3, a molecular substance that dissolves in

water to give H+ and NO3-.

 
HNO 3 ( aq )  H ( aq )  NO 3 ( aq )
H 2O

General Chemistry 9th-Ebbing.Gammon 27

Definitions of Acid and Base

• The Arrhenius Concept

– The Arrhenius concept defines bases as substances that produce hydroxide

ions, OH-, when dissolved in water.

– An example is sodium hydroxide, NaOH, an ionic substance that dissolves

in water to give sodium ions and hydroxide ions.
 
NaOH ( s )  Na ( aq )  OH ( aq )
H 2O

– The molecular substance ammonia, NH3, is a base in the Arrhenius view,


 
NH 3 ( aq )  H 2 O ( l )  NH 4 ( aq )  OH  ( aq )
– because it yields hydroxide ions when it reacts with water
General Chemistry 9th-Ebbing.Gammon 28
 Definitions of Acid and Base
• The Brønsted-Lowry Concept

– The Brønsted-Lowry concept defines an acid as the species (molecule or ion)

that donates a proton (H+) to another species in a proton-transfer reaction.

– A base is defined as the species (molecule or ion) that accepts the proton (H+)
in a proton-transfer reaction.

In the reaction of ammonia with water,

  
NH 3 ( aq )  H 2 O ( l )  NH 4 ( aq )  OH ( aq )

H+
•The H2O molecule is the acid because it donates a proton. The NH3 molecule is a
base, because it accepts a proton.
29
General Chemistry 9th-Ebbing.Gammon
Definitions of Acid and Base

• The Brønsted-Lowry Concept

The H+(aq) ion associates itself with water to form H3O+(aq).

 
H ( aq )  H 2 O ( l )  H 3 O ( aq )
This “mode of transportation” for the H+ ion is called the hydronium ion.

General Chemistry 9th-Ebbing.Gammon 30

Definitions of Acid and Base

• The Brønsted-Lowry Concept

The dissolution of nitric acid, HNO3, in water is therefore a proton-

transfer reaction


HNO 3 ( aq )  H 2 O ( l )  NO 3 ( aq )  H 3 O  ( aq )

H+
where HNO3 is an acid (proton donor) and H2O is a base (proton
acceptor).

General Chemistry 9th-Ebbing.Gammon 31

Definitions of Acid and Base

• In summary, the Arrhenius concept and the Brønsted-Lowry

concept are essentially the same in aqueous solution.

– The Arrhenius concept

acid: proton (H+) donor
base: hydroxide ion (OH-) donor

– The Brønsted-Lowry concept

acid: proton (H+) donor
base: proton (H+) acceptor

General Chemistry 9th-Ebbing.Gammon 32

Strong and Weak Acids and Bases

• A strong acid is an acid that ionizes completely in water; it is a strong

electrolyte.


HNO 3 ( aq )  H 2 O ( l )  NO 3 ( aq )  H 3 O  ( aq )
 
HCl ( aq )  H 2 O ( l )  Cl ( aq )  H 3 O ( aq )

33
General Chemistry 9th-Ebbing.Gammon
Strong and Weak Acids and Bases

• A weak acid is an acid that only partially ionizes in water; it is a weak

electrolyte.

• The hydrogen cyanide molecule, HCN, reacts with water to produce a small
percentage of ions in solution.

  
HCN ( aq )  H 2 O ( l )  CN ( aq )  H 3 O ( aq )

General Chemistry 9th-Ebbing.Gammon 34

Strong and Weak Acids and Bases

• A strong base is a base that is present entirely as ions, one of which is

OH-; it is a strong electrolyte.

H O
 
NaOH ( s )  Na ( aq )  OH ( aq )
2

•The hydroxides of Group IA and IIA elements, except for beryllium

hydroxide, are strong bases.

General
Presentation
Chemistry
of Lecture
9th-Ebbing.Gammon
Outlines, 4–35
Strong and Weak Acids and Bases

– A weak base is a base that is only partially ionized in water; it

is a weak electrolyte.

– Ammonia, NH3, is an example.

  
NH 3 ( aq )  H 2 O ( l )  NH 4 ( aq )  OH ( aq )

General Chemistry 9th-Ebbing.Gammon 36

4.4 Acid–Base Reactions
Strong and Weak Acids and Bases

– You will find it important to be able to identify an acid or base as strong

or weak.

– When you write an ionic equation, strong acids and bases are
represented as separate ions.

– Weak acids and bases are represented as undissociated “molecules” in

ionic equations.

General Chemistry 9th-Ebbing.Gammon 37

Neutralization Reactions
– One of the chemical properties of acids and bases is that they neutralize one
another.

– A neutralization reaction is a reaction of an acid and a base that results in an

ionic compound and water.

– The ionic compound that is the product of a neutralization reaction is called a

salt.
H CN ( aq )  K O H ( aq )  K CN ( aq )  H 2 O ( l )
acid base salt
– The net ionic equation for each acid-base neutralization reaction involves a
transfer of a proton.
– Consider the reaction of the strong acid , HCl(aq) and a strong base, KOH(aq).

HCl ( aq )  KOH ( aq )  KCl ( aq )  H 2 O ( l )

General Chemistry 9th-Ebbing.Gammon 38
Neutralization Reactions

– Writing the strong electrolytes in the form of ions gives the complete ionic
equation.
H  ( aq )  C l  ( aq )  K  ( aq )  O H  ( aq ) 
K  ( aq )  C l  ( aq )  H 2 O ( l )

– Canceling the spectator ions results in the net ionic equation. Note the proton
transfer.
   
H ( aq )  Cl ( aq )  K ( aq )  OH ( aq ) 
K  ( aq )  Cl  ( aq )  H 2 O ( l )
H  ( aq )  O H  ( aq )  H 2 O ( l )

General Chemistry 9th-Ebbing.Gammon 39

H+
Neutralization Reactions

– In a reaction involving HCN(aq), a weak acid, and KOH(aq), a strong base,

the product is KCN, a strong electrolyte.

– The net ionic equation for this reaction is

HCN ( aq )  OH  ( aq )  CN  ( aq )  H 2 O ( l )

Note the proton transfer.

H+

General Chemistry 9th-Ebbing.Gammon 40

Acid-Base Reactions with Gas Formation
•Certain salts, notably carbonates, sulfites, and sulfides, react with acids to form
a gaseous product.

•Consider the reaction of sodium carbonate with hydrochloric acid. Carbonates

react with acids to form CO2, carbon dioxide gas. The molecular equation for
the reaction is: H2CO3

Na 2 CO 3  2 HCl  2 NaCl  H 2 O  CO 2 
– Sulfites react with acids to form SO2, sulfur dioxide gas.

Na 2 SO 3  2 HCl  2 NaCl  H 2 O  SO 2 

General Chemistry 9th-Ebbing.Gammon 41

Acid-Base Reactions with Gas Formation

– Sulfides react with acids to form H2S, hydrogen sulfide gas.

Na 2 S  2 HCl  2 NaCl  H 2 S 

General Chemistry 9th-Ebbing.Gammon 42

 Types of Chemical Reactions

4.5 Oxidation–Reduction Reactions

– Oxidation-reduction reactions involve the transfer of electrons from one
species to another.
– Oxidation is defined as the loss of electrons.( ‫زيادة في رقم التأكسد‬
– Reduction is defined as the gain of electrons ‫ نقصان في رقم التأكسد‬.
– Oxidation and reduction always occur simultaneously.

– The reaction of an iron nail with a solution of copper(II) sulfate, CuSO4, is an

oxidation- reduction reaction (See Figure 4.11).
– The molecular equation for this reaction is:

Fe ( s )  CuSO 4 ( aq )  FeSO 4 ( aq )  Cu ( s )
General Chemistry 9th-Ebbing.Gammon 43
 FIGURE 4.11 : Reaction of iron with Cu+2(aq)
Left: Iron nail and copper(II) sulfate solution, which has a blue color.
Center: Fe reacts with Cu+2 (aq) to yield Fe+2(aq) and Cu(s). In the molecular view,
water and the sulfate anion have been omitted.
Right: The copper metal plates out on the nail
General Chemistry 9th-Ebbing.Gammon 44
4.5 Oxidation–Reduction Reactions

– Complete ionic equation :

– Fe (s) + Cu +2 + SO4 -2 Fe +2 + SO4 -2 + Cu (s)

– The net ionic equation shows the reaction of iron metal with Cu2+(aq) to
produce iron(II) ion and copper metal.

Loss of 2 e-1 oxidation

2 2
Fe ( s )  Cu ( aq )  Fe ( aq )  Cu ( s )
Gain of 2 e-1 reduction

General Chemistry 9th-Ebbing.Gammon 45

4.5 Oxidation–Reduction Reactions

Oxidation Numbers

– The concept of oxidation numbers is a simple way of keeping track of

electrons in a reaction.

– The oxidation number (or oxidation state) of an atom in a substance is the

actual charge of the atom if it exists as a monatomic ion.

– Alternatively, it is hypothetical charge assigned to the atom in the substance

by simple rules.

– The reaction of calcium metal with chlorine gas The chemical equation is

General Chemistry 9th-Ebbing.Gammon 46

Oxidation Numbers Rules

Rule Applies to Statement

1 Elements The oxidation number of an atom in an element is
zero.

2 Monatomic ions The oxidation number of an atom in a monatomic

ion equals the charge of the ion.

3 Oxygen The oxidation number of oxygen is –2 in most of its

compounds. (An exception is O in H2O2 and other
peroxides, where the oxidation number is –1.)

General Chemistry 9th-Ebbing.Gammon 47

Oxidation Numbers Rules

Rule Applies to Statement

4 Hydrogen The oxidation number of hydrogen is +1 in most of
its compounds.

5 Halogens Fluorine is –1 in all its compounds. The other

halogens are –1 unless the other element is another
halogen or oxygen.

6 Compounds and The sum of the oxidation numbers of the atoms in a

ions compound is zero. The sum in a polyatomic ion
equals the charge on the ion.

General Chemistry 9th-Ebbing.Gammon 48

•What are the Oxidation Numbers for each element in the
following?

H2O +1 for H, -2 for O

N2 Zero for N, elemental state

1(+1 K)+ 4(-2 O) + 1(Mn) = 0
KMnO4 +1 for K, -2 for O, +7 for Mn 1+-8 + Mn = 0
Mn = +7
CO2 -2 for O, +4 for C

+1 for H, -4 for C
CH4 1(+1 H) + 3(-1 Cl) +1 (C) = 0

CHCl3 +1 for H, -1 for Cl, +2 for C 1 + -3 + C =0

C = +2
He Zero for He, elemental state

Cu Zero for Cu, elemental state

2(+1 Na)+ 7(-2 O) + 2( Cr) = 0
Na2Cr2O7 +1 for Na, -2 for O, +6 for Cr 2+ -14 + 2 (Cr) = 0
-12 = -2 (Cr)
General Chemistry 9th-Ebbing.Gammon 49
+ 6 = Cr
•Are the Following Oxidation-Reduction Reactions? What is Oxidized and What is
Reduced?

NaCl + LiBr  NaBr + LiCl No

CH4 + 2O2  CO2 + 2H2O Yes C ox, O red

4Fe + 3O2  2Fe2O3 Yes Fe ox, O red

2NaBr + MgO  MgBr2 + Na2O No

P4 + 6 Br2  4 PBr3 Yes P ox, Br red

Zn + Cu2+  Zn2+ + Cu Yes Zn ox, Cu red

General Chemistry 9th-Ebbing.Gammon 50

4.5 Oxidation–Reduction Reactions
Describing Oxidation-Reduction Reactions

– Look again at the reaction of iron with copper(II) sulfate.

2 2
Fe ( s )  Cu ( aq )  Fe ( aq )  Cu ( s )
– We can write this reaction in terms of two half-reactions.

– A half-reaction is one of the two parts of an oxidation-reduction reaction.

– One involves the loss of electrons (oxidation) and the other involves the
gain of electrons (reduction)

2 
Fe ( s )  Fe ( aq )  2 e oxidation half-reaction

2 
Cu ( aq )  2 e  Cu ( s ) reduction half-reaction

51
General Chemistry 9th-Ebbing.Gammon
Describing Oxidation-Reduction Reactions

– An oxidizing agent is a species ‫ نوع‬that oxidizes another species; it is

itself reduced.

– A reducing agent is a species that reduces another species; it is itself

oxidized.

Loss of 2 e- oxidation
reducing agent
2 2
Fe ( s )  Cu ( aq )  Fe ( aq )  Cu ( s )
oxidizing agent
Gain of 2 e- reduction
52
General Chemistry 9th-Ebbing.Gammon
Some Common Oxidation-Reduction Reactions

– Most of the oxidation-reduction reactions fall into one of the following

simple categories:

– Combination Reaction

– Decomposition Reactions

– Displacement Reactions

– Combustion Reactions

General Chemistry 9th-Ebbing.Gammon 53

Some Common Oxidation-Reduction Reactions

• Combination Reactions
– A combination reaction is a reaction in which two substances combine
to form a third substance.

2 N a (s )  C l 2 (g )  2 N a C l( s )

54
General Chemistry 9th-Ebbing.Gammon
Some Common Oxidation-Reduction Reactions

• Combination Reactions

Other combination reactions involve compounds as reactants and are

not oxidation– reduction reactions. For example,

CaO ( s )  SO 2 ( g )  CaSO 3 ( s )

General Chemistry 9th-Ebbing.Gammon 55

Some Common Oxidation-Reduction Reactions

• Decomposition Reactions

– A decomposition reaction is a reaction in which a single

compound reacts to give two or more substances.

2 H g O ( s )  2 H g ( l)  O 2 ( g )
•Decomposition reaction of mercury(II) oxide

General Chemistry 9th-Ebbing.Gammon 56

Some Common Oxidation-Reduction Reactions

• Displacement Reactions

–A displacement reaction (also called a single-

replacement reaction) is a reaction in which an
element reacts with a compound, displacing an
element from it.

Zn ( s )  2 HCl ( aq )  ZnCl 2 ( aq )  H 2 ( g )

• Displacement reaction of zinc and hydrochloric acid

General Chemistry 9th-Ebbing.Gammon 57

• Displacement Reactions

• Whether a reaction occurs between a given

element and a monatomic ion depends on the
relative ease with which the two species gain or
lose electrons.

• Table 4.6 shows the activity series of the

elements, a listing of the elements in decreasing
order of their ease of losing electrons during
reactions in aqueous solution.

• The metals listed at the top are the strongest

reducing agents (they lose electrons easily); those
at the bottom, the weakest.
General Chemistry 9th-Ebbing.Gammon 58
Some Common Oxidation-Reduction Reactions

 Combustion Reactions

– A combustion reaction is a reaction in which

a substance reacts with oxygen, usually with
the rapid release of heat to produce a flame.

4 F e (s ) + 3 O 2 (g )  2 F e 2O 3 (s )
• Combustion reaction of iron wool

General Chemistry 9th-Ebbing.Gammon 59

4.6 Balancing Simple Oxidation–Reduction Equations

– At first glance, the equation representing the reaction of zinc metal

with silver(I) ions might appear to be balanced.

 2
Zn ( s )  Ag ( aq )  Zn ( aq )  Ag ( s )

– However, a balanced equation must have a charge balance as well as a

mass balance.

General Chemistry 9th-Ebbing.Gammon 60

4.6 Balancing Simple Oxidation–Reduction Equations

– Since the number of electrons lost in the oxidation half-reaction must equal
the number gained in the reduction half-reaction,

2 
Zn ( s )  Zn ( aq )  2 e oxidation half-reaction

 
2 Ag ( aq )  2 e  2Ag ( s ) reduction half-reaction

we must double the reaction involving the reduction of the silver.

General Chemistry 9th-Ebbing.Gammon 61

4.6 Balancing Simple Oxidation–Reduction Equations

– Adding the two half-reactions together, the electrons cancel,

Zn ( s )  Zn 2  ( aq )  2 e  oxidation half-reaction

2 Ag  ( aq )  2 e   2 Ag ( s ) reduction half-reaction

 2+
Z n ( s )  2 A g ( a q )  Z n (a q )  2 A g ( s )
which yields the balanced oxidation-reduction reaction.

General Chemistry 9th-Ebbing.Gammon 62

Summary Types of Reactions

Combination Reaction: a reaction in which two substances

chemically combine to form a third.
4 Fe(s) + 3O2(g) 2 Fe2O3(s)

Decomposition Reaction: a reaction in which a single compound

breaks up into two or more substances.
2 AgCl(s) 2 Ag(s) + Cl2(g)

Single-Replacement Reactions: one single reactant replaces another.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Combustion Reactions: organic plus oxygen gives carbon dioxide

and water.
C 6 H 1 2 O 6 (s ) + 6 O 2 (g )  6 C O 2 (g )  6 H 2 O (g )
63
General Chemistry 9th-Ebbing.Gammon
 Working with Solutions
4.7 Molar Concentration
• The majority of chemical reactions discussed here occur in
aqueous solution.
– When you run reactions in liquid solutions, it is convenient to
dispense the amounts of reactants by measuring out volumes of
reactant solutions.

• When we dissolve a substance in a liquid, we call the substance the

solute and the liquid the solvent.

– The general term concentration refers to the quantity of solute in

a standard quantity of solution.

General Chemistry 9th-Ebbing.Gammon 64

4.7 Molar Concentration

• Molar concentration, or molarity (M), is defined as the moles of

solute dissolved in one liter (cubic decimeter) of solution.

moles of solute
Molarity (M) 
liters of solution

General Chemistry 9th-Ebbing.Gammon 65

4.7 Molar Concentration

• Let’s try an example.

– A sample of 0.0341 mol iron(III) chloride, FeCl3, was dissolved in water

to give 25.0 mL of solution. What is the molarity of the solution?

– 1 Liter = 1000 ml
– ??? = 25 ml
moles of FeCl 3
– Since
molarity 
liters of solution

0.0341 mole of FeCl 3

then
M  1.36 M FeCl 3
0.0250 liter of solution
General Chemistry 9th-Ebbing.Gammon 66
• Example :Sample of NaNO3 weighing 0.38 g is placed in 50.0 ml
flask, then it was filled with water. What is the molarity of the
resulting solution?

• Molareity = moles of solute / Liter of solution

= 0.00447 / (50/1000)
= 0.089 M

#n = mass/ m.wt
= 0.38/ (23+14+(16*3)
= 0.00447 mole
General Chemistry 9th-Ebbing.Gammon 67
4.8 Diluting Solutions

• The molarity of a solution and its volume are inversely

proportional. Therefore, adding water makes the solution less
concentrated.
• When the solution is diluted by adding more water, the concentration and
volume change to Mf (the final molar concentration) and Vf (the final
volume), and the moles of solute equals :
Moles of solute = Mf X Vf

• Because the moles of solute has not changed during the dilution
– This inverse relationship takes the form of:

M i  Vi  M f  V f
– So, as water is added, increasing the final volume, Vf, the final molarity,
General Chemistry 9th-Ebbing.Gammon 68
Mf, decreases.
4.8 Diluting Solutions

•Let’s Practice!

 What is the volume of 6.0 M HCl that can be made from 5.0 mls of
12.0 M HCl?

M1= 12.0 M, V1 = 5.0 mls, M2 = 6.0 M

M1 V 1 = M2 V 2
(12.0 M)(5.0 mls) = (6.0 M)V2

Solving for V2
V2 = (12.0 M)(5.0 mls)/(6.0 M) = 10 mls

General Chemistry 9th-Ebbing.Gammon 69

 Quantitative Analysis

• Analytical chemistry deals with the determination of

composition of materials-that is, the analysis of materials

– Quantitative analysis involves the determination of the amount

of a substance or species present in a material.

General Chemistry 9th-Ebbing.Gammon 70

4.9 Gravimetric Analysis

• Gravimetric analysis is a type of quantitative analysis in which

the amount of a species in a material is determined by converting
the species into a product that can be isolated and weighed.

– Precipitation reactions are often used in gravimetric analysis.

– The precipitate from these reactions is then filtered, dried,
and weighed.

General Chemistry 9th-Ebbing.Gammon 71

4.9 Gravimetric Analysis

• Consider the problem of determining the amount of lead in a

sample of drinking water.

– Adding sodium sulfate (Na2SO4) to the sample will precipitate lead(II)

sulfate.

2 
Na 2 SO 4 ( aq )  Pb ( aq )  2 Na ( aq )  PbSO 4 ( s )
– The PbSO4 can then be filtered, dried, and weighed.

General Chemistry 9th-Ebbing.Gammon 72

4.9 Gravimetric Analysis

• Suppose a 1.00 L sample of polluted water was analyzed for lead(II) ion,
Pb2+, by adding an excess of sodium sulfate to it. The mass of lead(II)
sulfate that precipitated was 229.8 mg. What is the mass of lead in a liter of
the water? Express the answer as mg of lead per liter of solution.

2 
Na 2 SO 4 ( aq )  Pb ( aq )  2 Na ( aq )  PbSO 4 ( s )

• First we must obtain the mass percentage of lead in lead(II) sulfate, by

dividing the molar mass of lead by the molar mass of PbSO4, then
multiplying by 100.

General Chemistry 9th-Ebbing.Gammon 73

4.9 Gravimetric Analysis

207.2 g/mol
% Pb   100  68.32%
303.3 g/mol

– Then, calculate the amount of lead in the PbSO4 precipitated.

Am ount Pb in sam ple  229.8 m g PbSO 4  0.6832  157.0 m g Pb

General Chemistry 9th-Ebbing.Gammon 74

4.10 Volumetric Analysis

• An important method for determining the amount of a particular

substance is based on measuring the volume of the reactant
solution.

– Titration is a procedure for determining the amount of substance A by

adding a carefully measured volume of a solution with known
concentration of B until the reaction of A and B is just complete (See
Figure 4.22).

– Volumetric analysis is a method of analysis based on titration.

General Chemistry 9th-Ebbing.Gammon 75

FIGURE 4.22 : Titration of an unknown amount
of HCl with NaOH

General Chemistry 9th-Ebbing.Gammon 76

4.10 Volumetric Analysis

• Consider the reaction of sulfuric acid, H2SO4, with sodium

hydroxide, NaOH:

H 2 SO 4 ( aq )  2 NaO H ( aq )  2 H 2 O ( l )  Na 2 SO 4 ( aq )

– Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many

milliliters of 0.250 M NaOH must be added to completely react with the
sulfuric acid?

General Chemistry 9th-Ebbing.Gammon 77

4.10 Volumetric Analysis

– First we must convert the 0.0350 L (35.0 mL) to moles of

H2SO4 (using the molarity of the H2SO4).

– Then, convert to moles of NaOH (from the balanced chemical

equation).
– Finally, convert to volume of NaOH solution (using the molarity of
NaOH).

0.175 mole H 2SO 4 2 mol NaOH 1 L NaOH soln.

( 0.0350L )    
1 L H 2SO 4 solution 1 mol H 2SO 4 0.250 mol NaOH
0 . 0490 L N aO H solu tion ( or 49.0 m L of N aO H solution)
General Chemistry 9th-Ebbing.Gammon 78
 Applied Medical Sciences
Medical Imaging & Physiotherapy

General Chemistry

Chapter 5

The Gaseous State

Course No: 510162

Halima Thwaib

Second Semester
Chapter 5 : The Gaseous State
Gas Laws
1. Gas Pressure and Measurement
2. Empirical Gas Laws
3. The Ideal Gas Law
4. Stoichiometry and Gas Volumes
5. Gas Mixtures: Law of Partial Pressures

Kinetic-Molecular Theory
6. Kinetic Theory of an Ideal Gas
7. Molecular Speeds; Diffusion and Effusion
8. Real Gases
2
 Gas Laws
•Most substances composed of small molecules are gases under normal
conditions or else are easily vaporized liquids. Table 5.1 lists selected
gaseous substances and some of their properties.

3
5.1 Gas Pressure and Its Measurement
• Pressure is defined as the force exerted( ‫) المبذولة‬per unit area
of surface by molecules in motion

P = Force/unit area
•The SI unit of pressure, kg/(m.s2), is given the name
pascal (Pa)

•A barometer is a device for measuring the pressure of

the atmosphere

4
5.2 Empirical Gas Laws
❑Boyle’s Law: Relating Volume and Pressure

• Boyle’s Law: The volume of a sample of gas at a given

temperature varies inversely with the applied pressure.

• V  1/P
(constant moles and T)

•Boyle’s Law:
PV = constant

Pf  Vf = Pi  Vi
5
5.2 Empirical Gas Laws

• A Problem to Consider :
• A sample of chlorine gas has a volume of 1.8 L at 1.0 atm. If the
pressure increases to 4.0 atm (at constant temperature), what
would be the new volume?

using Pf  Vf = Pi  Vi
Pi  Vi (1.0 atm) (1.8 L)
Vf = =
Pf (4.0 atm)
Vf = 0.45 L
6
5.2 Empirical Gas Laws
❑ Charles’s Law: Relating Volume and Temperature

• Charles’s Law: The volume occupied by any sample of gas at

constant pressure is directly proportional to its absolute temperature.

V  Tabs (constant moles and P)

or

Vf
Tf
= Vi
Ti
7
5.2 Empirical Gas Laws

•A Problem to Consider
• A sample of methane gas that has a volume of 3.8 L at 5.0°C is
heated to 86.0°C at constant pressure. Calculate its new volume.

Vf = Vi
using
Tf Ti
Vi Tf
Vf = Ti
= (3.8L)(359K )
(278K )
Vf = 4.9 L
8
5.2 Empirical Gas Laws

• Gay-Lussac’s Law: The pressure exerted by a gas at constant

volume is directly proportional to its absolute temperature.

P  Tabs (constant moles and V)

or

Pf
Tf
= Pi
Ti

9
5.2 Empirical Gas Laws

•A Problem to Consider
• An aerosol can has a pressure of 1.4 atm at 25°C. What pressure
would it attain at 1200°C, assuming the volume remained
constant?

Pf = Pi
using
Tf Ti

Pf = PT
i
Ti
f
= (1.4atm)(1473K )
(298K )

Pf = 6.9atm
10
5.2 Empirical Gas Laws
❑Combined Gas Law: Relating Volume, Temperature, and
Pressure

• Combined Gas Law: In the event that all three parameters, P, V,

and T, are changing, their combined relationship is defined as
follows:

Pi Vi
Ti
= Pf Vf
Tf
11
5.2 Empirical Gas Laws

•A Problem to Consider

• A sample of carbon dioxide occupies 4.5 L at 30°C and 650 mm

Hg. What volume would it occupy at 800 mm Hg and 200°C?

PiVi Pf Vf
using =
Ti Tf
PV Tf (650 mm Hg)(4.5 L)(473 K)
Vf = P T =
i i
f i (800 mm Hg)(303 K)
Vf = 5.7L
12
5.2 The Empirical Gas Laws
❑ Avogadro’s Law: Relating Volume and Amount

• Avogadro’s Law: Equal volumes

of any two gases at the same
temperature and pressure contain 22.4 L/mol

the same number of molecules.

• The volume of one mole of gas is called the molar gas

volume, Vm.
• Volumes of gases are often compared at standard
temperature and pressure (STP), chosen to be 0 oC and 1
atm pressure.
•At STP, the molar gas volume is found to be 22.4 L/mol
13
5.2 The Empirical Gas Laws
• Avogadro’s Law

– At STP, the molar volume, Vm, that is, the volume occupied by one mole
of any gas, is 22.4 L/mol

– So, the volume of a sample of gas is directly proportional to the number

of moles of gas, n.

V  n
14
5.2 The Empirical Gas Laws
•A Problem to Consider

• A sample of fluorine gas has a volume of 5.80 L at 150.0oC and 10.5

atm of pressure. How many moles of fluorine gas are present?

First, use the combined empirical gas law to determine the

volume at STP.

PiViTstd (10.5atm)(5.80L)(273K)
VSTP = =
PstdTi (1.0atm)(423K)

VSTP = 39.3L
15
5.2 The Empirical Gas Laws
•A Problem to Consider

• Since Avogadro’s law states that at STP the molar volume is 22.4
L/mol, then

VSTP
moles of gas =
22.4 L/mol
39.3 L
moles of gas =
22.4 L/mol
moles of gas = 1.75 mol
16
5.3 The Ideal Gas Law ‫قانون الغاز المثالي‬

• From the empirical gas laws, we See that volume varies in

proportion to pressure, absolute temperature, and moles.

V  1/P Boyle's Law

V  Tabs Charles' Law
Vn Avogadro's Law

17
5.3The Ideal Gas Law

• This implies that there must exist a proportionality constant

governing these relationships.

– Combining the three proportionalities, we can obtain the

following relationship.

V ="R" ( nTabs
P
)
• where “R” is the proportionality constant referred to as the ideal gas
constant.

18
5.3 The Ideal Gas Law
• The numerical value of R can be derived using Avogadro’s law, which states
that one mole of any gas at STP will occupy 22.4 liters.

R = VP
nT

R= (22.4 L)(1.00 atm)

(1.00 mol)(273 K)

= Latm
0.0821 molK
19
5.3 The Ideal Gas Law

• Thus, the ideal gas equation, is usually expressed in the following

form:

PV = nRT
P is pressure (in atm)
V is volume (in liters)
n is number of atoms (in moles)
R is universal gas constant 0.0821 L.atm/K.mol
T is temperature (in Kelvin)

20
5.3 The Ideal Gas Law
•A Problem to Consider

• An experiment calls for 3.50 moles of chlorine, Cl2. What

volume would this be if the gas volume is measured at 34°C and
2.45 atm?

since V = nRT
P

then V = (3.50 mol)(0.0821 mol K )(307 K)

Latm

2.45 atm
then V = 36.0 L
21
• An experiment calls for 3.50 moles of chlorine, Cl2. What
volume would this be if the gas volume is measured at 34°C
and 2.45 atm?

n=3.50 mole
V=??
T= 34 C+273 = 307 K
P= 2.45 atm =

PV= nRT
2.45 * V = 3.5* 0.0821 * 307
V= 3.5 * .0821* 307 / 2.45
V= 36.0 L

22
5.3 The Ideal Gas Law

❑ Gas Density; Molecular-Mass Determination

• In Chapter 3 we showed the relationship between moles and
mass.

moles = mass
molecular mass

or

n=m
M m
23
5.3 The Ideal Gas Law
❑ Gas Density; Molecular-Mass Determination
• If we substitute this in the ideal gas equation, we obtain

PV = ( Mm )RT m

•If we solve this equation for the molecular mass, we obtain

mRT
M m
=
PV
24
5.3 The Ideal Gas Law
❑ Gas Density; Molecular-Mass Determination

• A 15.5 gram sample of an unknown gas occupied a volume

of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its
molecular mass.
mRT
Since M m =
PV
Latm )(298 K)
(15.5 g)(0.0821molK
then Mm =
(1.08 atm)(5.75 L)
M m = 61.1 g/mol
25
• A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25°C and a
pressure of 1.08 atm. Calculate its molecular mass.

• Mass= 15.5 gram

• V= 5.75 L

• T= 25 +273 = 298 K

• P = 1.08 atm

• M.wt ???

• PV = nRT

• 1.08 * 5.75 = (15.5/ M.wt ) *0.0821 * 298

• M.wt = 15.5 * .0821*298 / 1.08*5.75

• M.wt = 61.1 g/ mole

26
5.3 The Ideal Gas Law
❑ Density Determination

• If we look again at our derivation of the molecular mass

equation,

PV = ( Mm )RTm

•we can solve for m/V, which represents density.

m PMm
=D=
V RT
27
5.3 The Ideal Gas Law
❑ Density Determination
•A Problem to Consider

• Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50°C and

1.75 atm of pressure.

PM m
Since D =
RT
(1.75 atm)(48.0 g/mol)
then D=
Latm )(323 K)
(0.0821 molK
D = 3.17 g/L
28
5.3 The Ideal Gas Law
❑ Density Determination
•A Problem to Consider

• Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50°C and

1.75 atm of pressure.
Mwt = 48.0 g/mole
T = 50 +273 = 323 K
P= 1.75 atm
D=??

PV = nRT
PV = (mass/M.wt) RT
PVM.wt = Mass * R*T
P*M.wt/RT = mass/V
P*M.wt / Rt = D
1.75 * 48/ 0.0821 * 323 = D
D = 3.7 g/L
29
5.4 Stoichiometry Problems Involving Gas
Volumes
• Consider the following reaction, which is often used to generate
small quantities of oxygen.

2 KClO 3 (s) → 2 KCl(s) + 3 O 2 (g)

• Suppose you heat 0.0100 mol of potassium chlorate, KClO3,

in a test tube. How many liters of oxygen can you produce
at 298 K and 1.02 atm?

30
5.4 Stoichiometry Problems Involving Gas
Volumes

• First we must determine the number of moles of oxygen produced

by the reaction.

3 mol O 2
0.0100 mol KClO 3 
2 mol KClO 3

= 0.0150 mol O 2

31
5.4 Stoichiometry Problems Involving Gas
Volumes

• Now we can use the ideal gas equation to calculate the volume of
oxygen under the conditions given.

nRT
V=
P
V= (0.0150 mol O 2 )(0.0821 mol K )(298 K)
Latm

1.02 atm

V = 0.360 L
32
5.4 Stoichiometry Problems Involving Gas
Volumes
In the reaction of
6NaN3 (s) + Fe2O3(s) → 3Na2O(s) +2Fe(s) + 9N2 (g)

How many gram of sodium azide (NaN3) would be required to provide

75.0 L of nitrogen gas (N2)at 250C and 748 mmHg?

33
5.5 Partial Pressures of Gas
Mixtures‫الضغوط الجزئية لمخاليط الغاز‬
• Dalton’s Law of Partial Pressures: the sum of all the pressures of
all the different gases in a mixture equals the total pressure of the
mixture.
Ptot = Pa + Pb + Pc + ....

34
5.5 Partial Pressures of Gas Mixtures
❑ Collecting Gases “Over Water”

• A useful application of partial pressures arises when you collect

gases over water. (See Figure 5.20)

– As gas bubbles through the water, the gas becomes saturated with water
vapor.
– The partial pressure of the water in this “mixture” depends only on the
temperature. (See Table 5.6)

35
 Figure 5.20: Collection of gas over water.

Hydrogen, prepared by the reaction of zinc with HCl, is led to an inverted tube initially filled
with water. When the gas-collection tube is adjusted so that the water level in the tube is at the
same height as the level in the beaker, the gas pressure in the tube equals the barometric pressure
(769 mmHg). The total gas pressure equals the sum of the partial pressure of the hydrogen (752
mmHg) and the vapor pressure of water (17 mmHg). For clarity, HCl is shown in light brown and
36
hydrogen gas in light blue.
9.8 30 31.8

37
5.5 Partial Pressures of Gas Mixtures

•A Problem to Consider
• Suppose a 156 mL sample of H2 gas was collected over water at
19oC and 769 mm Hg. What is the mass of H2 collected?
– First, we must find the partial pressure of the dry H2.

PH = Ptot − PH 0
2 2
–(See Table 5.6)
– Table 5.6 lists the vapor pressure of water at 19oC as 16.5 mm Hg.

PH = 769 mm Hg - 16.5 mm Hg
2

PH = 752 mm Hg
2 38
•A Problem to Consider

• Now we can use the ideal gas equation, along with the partial
pressure of the hydrogen, to determine its mass.

PH = 752 mm Hg  7601mm
2
atm
Hg
= 0.989 atm

V = 156 mL = 0.156 L
T = (19 + 273) = 292 K
n=?
39
•A Problem to Consider
• From the ideal gas law, PV = nRT, you have

PV (0.989 atm)(0.156 L)
n= =
Latm )(292 K )
RT (0.0821 molK
n = 0.00644mol
– Next,convert moles of H2 to grams of H2.

2.02 g H 2
0.00644 mol H 2  = 0.0130 g H 2
1 mol H 2
40
 Kinetic-Molecular Theory
6. Kinetic Theory of an Ideal Gas
❑ Postulates of Kinetic Theory

• Volume of particles is negligible

• Particles are in constant motion
• No inherent attractive or repulsive forces
• When molecules collide with one another, the collisions are
elastic
• The average kinetic energy of a collection of particles is
proportional to the temperature (K)

41
5.7 Molecular Speeds; Diffusion and Effusion

• The root-mean-square (rms) molecular speed, u, is a type of

average molecular speed, equal to the speed of a molecule having
the average molecular kinetic energy. It is given by the following
formula:

3RT
u=
Mm
R= 8.314
M m = in Kg

42
5.7 Molecular Speeds; Diffusion and Effusion

• Diffusion is the transfer of a gas through space or another gas over

time.
• Effusion is the transfer of a gas through a membrane or orifice

• – The equation for the rms velocity of gases shows the following
relationship between rate of effusion and molecular mass.

1
Rate of effusion 
Mm
43
5.7 Molecular Speeds; Diffusion and Effusion

• According to Graham’s law, the rate of effusion or diffusion is

inversely proportional to the square root of its molecular mass.
(See Figures 5.28 and 5.29)

Rate of effusion of gas "A" Mm of Gas B

=
Rate of effusion of gas "B" Mm of gas A

44
Figure 5.28: Figure 5.29: Hydrogen
Gaseous Effusion Fountain

45
5.7 Molecular Speeds; Diffusion and Effusion
❑ A Problem to Consider
• How much faster would H2 gas effuse through an opening than
methane, CH4?

Rate of H 2 M m (CH 4 )
=
Rate of CH 4 M m (H 2 )
Rate of H 2 16.0 g/mol
= = 2.8
Rate of CH 4 2.0 g/mol
•So hydrogen effuses 2.8 times faster than CH4
46
5.8 Real Gases

• Real gases do not follow PV = nRT perfectly. The van der Waals
equation corrects for the non ideal nature of real gases.

(P + n 2a )(V - nb)
V2
= nRT

a corrects for interaction between atoms.

b corrects for volume occupied by atoms.

47
5.8 Real Gases

• In the van der Waals equation,

V becomes (V - nb)

where “nb” represents the volume

occupied by “n” moles of molecules.

48
5.8 Real Gases
• Also, in the van der Waals equation,

P becomes (P + n 2a )
V2
•where “n2a/V2” represents the effect on pressure to intermolecular attractions
or repulsions.

Table 5.7 gives values of van der Waals constants for

various gases.

49
50
5.8 Real Gases
❑ A Problem to Consider

• If sulfur dioxide were an “ideal” gas, the pressure at 0°C exerted by

1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der
Waals equation to estimate the “real” pressure.

Table 5.7 lists the following values for SO2

a = 6.865 L2.atm/mol2
b = 0.05679 L/mol

51
❑ A Problem to Consider

• First, let’s rearrange the van der Waals equation to solve for
pressure.
2
nRT n a
P= - 2
V - nb V
R= 0.0821 L. atm/mol. K
T = 273.2 K a = 6.865 L2.atm/mol2
V = 22.41 L b = 0.05679 L/mol
52
❑ A Problem to Consider

nRT n2a
P= -
V - nb V 2

Latm )(273.2K)
(1.000 mol)(0.08206 molK atm )
(1.000 mol)2 (6.865 Lmol
2

P= -
2

22.41 L - (1.000 mol)(0.05679 L/mol) (22.41 L)2

P = 0.989 atm
• The “real” pressure exerted by 1.00 mol of
SO2 at STP is slightly less than the “ideal”
pressure.
53
 Applied Medical Sciences
Medical Imaging & Physiotherapy

General Chemistry

Chapter 6
Thermochemistry

Course No: 510162

Halima Thwaib

Second Semester
Chapter 6 : Thermochemistry

Understanding Heats of Reaction

6.1 Energy and Its Units
6.2 Heat of Reaction
6.3 Enthalpy and Enthalpy Change
6.4 Thermochemical Equations
6.5 Applying Stoichiometry to Heats of
Reaction
6.6 Measuring Heats of Reaction

Using Heats of Reaction

6.7 Hess’s Law
6.8 Standard Enthalpies of Formation
6.9 Fuels—Foods, Commercial Fuels, and
Rocket Fuels

2
Thermochemistry

•Nearly all chemical reactions involve either the release or the absorption of
heat, a form of energy. The burning of coal and gasoline are dramatic examples
of chemical reactions in which a great deal of heat is released. Such reactions
are important sources of warmth and energy. Chemical reactions that absorb
heat are usually less dramatic.

A reaction that absorbs heat

Two crystalline substances, barium hydroxide octahydrate and an ammonium salt, are mixed thoroughly in a
flask. Then the flask, which feels quite cold to the touch, is set in a puddle of water on a board. In a couple of
minutes, the flask and board are frozen solidly together. The board can then be inverted with the flask frozen
to it. 3
Understanding Heats of Reaction

•Thermodynamics is the science of the relationship between heat and other

forms of energy.

•Thermochemistry is the study of the quantity of heat absorbed or evolved by

chemical reactions.
•An example of a heat-evolving reaction is the burning of fuel.

• There may be practical reasons why you want to know the quantity of heat
evolved during the burning of a fuel: you could calculate the cost of the fuel per
unit of heat energy produced; you could calculate the quantity of heat obtained
per unit mass of rocket fuel; and so forth.

General Chemistry 9th-Ebbing.Gammon

4
6.1 Energy and Its Units

Types of Energy
1. Kinetic energy
2. Potential energy
3. Chemical energy
4. Heat energy
5. Electric energy
6. Radiant energy

General Chemistry 9th-Ebbing.Gammon

5
6.1 Energy and Its Units
•Energy is the potential or capacity to move matter.

•According to this definition, energy is not

a material thing but rather a property of
matter. Energy exists in different forms that
can be interconverted.

• In this chapter, we will be especially concerned with the energy of substances,

or chemical energy, and its transformation during chemical reaction into heat
energy. To prepare for this, we will first explore the quantitative meaning of the
energy of motion (kinetic energy). Then we will look at the concepts of
potential energy and of the internal energy of substances, which is defined in
terms of the kinetic and potential energies of the particles making up the
substance. General Chemistry 9th-Ebbing.Gammon
6
6.1 Energy and Its Units
➢Kinetic Energy; Units of Energy

•Kinetic energy is the energy associated with an object by virtue of its

motion. An object of mass m and speed or velocity v has kinetic energy Ek
equal to

Ek = 1 mv 2
2
– This shows that the kinetic energy of an object depends on both its mass
and its speed.
– A heavy object can move more slowly than a light object and still
have the same kinetic energy.

General Chemistry 9th-Ebbing.Gammon

7
➢Kinetic Energy; Units of Energy

• Problem: Consider the kinetic energy of a person whose mass is

130 lb (59.0 kg) traveling in a car at 60 mph (26.8 m/s).

Ek = 1 mv 2
2
Ek = 1 ( 59.0 kg )  ( 26.8 m / s ) 2
2
E k = 2.12  10 kg  m / s
4 2 2

E k = 2.12  10 J 4

– The SI unit of energy, kg.m2/s2, is given the name Joule.

General Chemistry 9th-Ebbing.Gammon, 6–

.
8
➢Kinetic Energy; Units of Energy

•The calorie (cal) is a non-SI unit of energy commonly used by chemists,

originally defined as the amount of energy required to raise the
temperature of one gram of water by one degree Celsius.

•This is only an approximate definition, however, because we now know that

the energy needed to heat water depends slightly on the temperature of the
water. In 1925 the calorie was defined in terms of the joule:

General Chemistry 9th-Ebbing.Gammon

9
➢Kinetic Energy; Units of Energy

Example 6.1: A good pitcher can throw a baseball so that it travels between 60
and 90 miles per hour (although speeds in excess of 100 miles per hour have
been recorded). A regulation baseball weighing 143 g (0.143 kg) travels
75 miles per hour (33.5 m/s). What is the kinetic energy of this baseball in
joules? in calories?

E k = 12 mv 2

General Chemistry 9th-Ebbing.Gammon

10
➢Potential Energy

• Potential energy is the energy an object has by virtue of its position in a

field of force.

• Potential Energy: This energy depends on

the “position” (such as height) in a “field of
force” (such as gravity).

• For example, water of a given mass m at the

top of a dam is at a relatively high “position”
h in the “gravitational field” g of the earth.

E p = mgh
General Chemistry 9th-Ebbing.Gammon, 6–
.
11
➢Potential Energy

•The potential energy of the water at the top of the dam is converted to kinetic
energy when the water falls to a lower level. As the water falls, it moves more
quickly. The potential energy decreases and the kinetic energy increases. Figure
shows the potential energy of water being converted to kinetic energy as the
water falls over a dam.

General Chemistry 9th-Ebbing.Gammon

12
➢Potential Energy

• Problem :Consider the potential energy of 1000 lb of water (453.6

kg) at the top of a 300 foot dam (91.44 m).

E p = mgh
E p = ( 453 .6 kg )  ( 9.80 m / s )  ( 91.44 m ) 2

E p = 4.06  10 kg  m / s
5 2 2

E p = 4.06  10 J 5

General Chemistry 9th-Ebbing.Gammon

.
13
➢Internal Energy

• Internal Energy is the energy of the particles making up a

substance.
• The sum of the kinetic and potential energies of the particles
making up a substance is referred to as the internal energy, U, of
the substance.
• The total energy of a system is the sum of its kinetic energy,
potential energy, and internal energy, U.

E tot = E k + E p + U

. General Chemistry 9th-Ebbing.Gammon, 14

➢Law of Conservation of Energy

• We have discussed situations in which one form of energy can be converted

into another form of energy.
• For example, when water falls over a dam, potential energy is converted
into kinetic energy. Some of the kinetic energy of the water may also be
converted into random molecular motion—that is, into internal energy of the
water. The total energy, E tot, of the water, however, remains constant, equal
to the sum of the kinetic energy, Ek, the potential energy, Ep, and the
internal energy, U, of the water.

• This result can be stated more generally as the law of conservation of

energy:
Energy may be converted from one form to another, but the total
quantities of energy remain constant.

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6.2 Heat of Reaction
• In chemical reactions, heat is often transferred from the “system” to its
“surroundings,” or vice versa.
• To understand heat of reaction concept, you need to know what is meant by
a thermodynamic system and its surroundings and to have a precise
definition of the term heat.
• The substance or mixture of substances under study in which a change
occurs is called the thermodynamic system (or simply system.)
• The surroundings are everything in the vicinity ‫ قرب‬of the thermodynamic
system

Figure: Illustration of a
thermodynamic system

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6.2 Heat of Reaction
➢Definition of Heat

• Heat is defined as the energy that flows into or out of a system because of a
difference in temperature between the system and its surroundings.

• As long as a system and its surroundings are in thermal contact (that is, they
are not thermally insulated from one another), energy (heat) flows between
them to establish temperature equality, or thermal equilibrium.

• Heat flows from a region of higher temperature to one of lower

temperature; once the temperatures become equal, heat flow stops.

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➢Definition of Heat
•We can explain this flow of energy between two regions of different
temperatures in terms of kinetic-molecular theory. Imagine two vessels in
contact, each containing oxygen gas and the one on the left being hotter (Figure
below).

• According to kinetic theory, the average speed of molecules in the hotter gas
is greater than that of molecules in the colder gas. But as the molecules in their
random motions collide with the vessel walls, they lose energy to or gain
energy from the walls.
•The faster molecules tend to slow down, while the slower molecules tend to
speed up. Eventually, the average speeds of the molecules in the two vessels
(and therefore the temperatures of the two gases) become equal. The net result
is that energy is transferred through the vessel walls from the hot gas to the cold
gas; that is, heat has flowed from the hotter vessel to the cooler one.
➢Definition of Heat

• Heat is denoted by the symbol q.

– The sign of q is positive if heat is absorbed by the
system.
– The sign of q is negative if heat is evolved by the
system.

– Heat of Reaction is the value of q required to

return a system to the given temperature at the
completion of the reaction.

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➢Heat of Reaction
•The heat of reaction (at a given temperature) is the value of q required to
return a system to the given temperature at the completion of the reaction.

• Chemical reactions or physical changes are classified as exothermic or

endothermic.
• An exothermic process is a chemical reaction or physical change in which
heat is evolved (q is negative).
• An endothermic process is a chemical reaction or physical change in which
heat is absorbed (q is positive).

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➢Heat of Reaction
6.3 Enthalpy and Enthalpy Change

• The heat absorbed or evolved by a reaction depends on the conditions

under which it occurs.

• Usually, a reaction takes place in an open vessel, and therefore at the

constant pressure of the atmosphere.
• The heat of this type of reaction is denoted qp, the heat at constant
pressure.

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6.3 Enthalpy and Enthalpy Change
• Enthalpy, denoted H, is an extensive property of a substance that can be
used to obtain the heat absorbed or evolved in a chemical reaction.
– An extensive property is one that depends on the quantity of
substance.
• Enthalpy is a state function, a property of a system that depends only on
its present state, which is determined by variables such as temperature
and pressure, and is independent of any previous history of the system or
independent of path.
• This means that a change in enthalpy does not
depend on how the change was made, but only on
the initial state and final state of the system.

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6.3 Enthalpy and Enthalpy Change
▪Enthalpy of Reaction

• Consider a chemical reaction system. At first, the enthalpy of the system

is that of the reactants. But as the reaction proceeds, the enthalpy changes
and finally becomes equal to that of the products
• The change in enthalpy for a reaction at a given temperature and
pressure (called the enthalpy of reaction) is obtained by subtracting the
enthalpy of the reactants from the enthalpy of the products.

 H = H (products) − H ( reactants )
Because H is a state function, the value of H is independent of the details of the
reaction. It depends only on the initial state (the reactants) and the final state
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6.3 Enthalpy and Enthalpy Change
▪Enthalpy of Reaction

• The change in enthalpy is equal to the heat of reaction at constant

pressure. This represents the entire change in internal energy (U)
minus any expansion “work” done by the system.

H = q p

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6.3 Enthalpy and Enthalpy Change

• Enthalpy and Internal Energy

– The internal energy of a system, U, is precisely defined as the heat at

constant pressure plus the work done by the system:

U = qp + w
– In chemical systems, work is defined as a change in volume at a given
pressure, that is:

w = − PV
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6.3 Enthalpy and Enthalpy Change

• Enthalpy and Internal Energy

• Since the heat at constant pressure, qp, represents Δ H, then

U = H − PV
– So ΔH is essentially the heat obtained or absorbed by a reaction in an
open vessel where the work portion of ΔU is unmeasured.

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6.4Thermochemical Equations

• A thermochemical equation is the chemical equation for a reaction

(including phase labels) in which the equation is given a molar
interpretation, and the enthalpy of reaction for these molar amounts is
written directly after the equation.

N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g );  H = -91.8 kJ
•This equation says that 1 mol of N2 reacts with 3 mol of H2 to produce 2 mol
of ammonia, and -91.8 kJ of heat evolves.

Note that the thermochemical equation includes phase labels. This is because
the enthalpy change, H, depends on the phase of the substances.

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6.4Thermochemical Equations

• In a thermochemical equation it is important to note phase labels

because the enthalpy change, ∆H, depends on the phase of the
substances.
•Consider the reaction of hydrogen and oxygen to produce water. If the product
is water vapor, 2 mol of H2 burn to release 483.7 kJ of heat.

2 H 2 ( g ) + O 2 ( g ) → 2 H 2O ( g ) ;  H = - 483.7 kJ o

On the other hand, if the product is liquid water, the heat released is 571.7 kJ.

2 H 2 ( g ) + O 2 ( g ) → 2 H 2O ( l ) ;  H = - 571.7 kJ
o

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6.4Thermochemical Equations
Example 6.2 : Aqueous sodium hydrogen carbonate solution (baking soda
solution) reacts with hydrochloric acid to produce aqueous sodium chloride,
water, and carbon dioxide gas. The reaction absorbs 12.7 kJ of heat at constant
pressure for each mole of sodium hydrogen carbonate. Write the
thermochemical equation for the reaction
6.4Thermochemical Equations

• The following are two important rules for manipulating

thermochemical equations:

– When a thermochemical equation is multiplied by any factor, the value

of ΔH for the new equation is obtained by multiplying the ΔH in the
original equation by that same factor.
– When a chemical equation is reversed, the value of ΔH is reversed in
sign.

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6.4Thermochemical Equations
Example 6.3 : When 2 mol H2(g) and 1 mol O2(g) react to give liquid water, 572
kJ of heat evolves.
2H2(g) + O2(g) → 2H2O(l ); ∆H = -572 kJ
Write this equation for 1 mol of liquid water. Give the reverse equation, in
which 1 mol of liquid water dissociates into hydrogen and oxygen.
 Thermochemistry
6.5 Applying Stoichiometry to Heats of Reactions
6.6 Measuring Heats of Reactions
6.7 Hess’s Law
6.8 Standard Enthalpies of Formation
6.9 Fuels—Foods, Commercial and Rocket Fuels
6.5 Applying Stoichiometry and Heats of Reactions

• Consider the reaction of methane, CH4, burning in the presence of

oxygen at constant pressure. Given the following equation, how much
heat could be obtained by the combustion of 10.0 grams CH4?

CH 4 ( g ) + 2O 2 ( g ) → CO 2 ( g ) + 2 H 2O ( l ); H o = -890.3 kJ
890.3 kJ = − 556 kJ
10.0 g CH 4  16.0 g
1 mol CH 4
 1−mol CH 4

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6.5 Applying Stoichiometry and Heats of Reactions

Example 6.4
6.6 Measuring Heats of Reaction

• To See how heats of reactions are measured, we must look at the heat
required to raise the temperature of a substance, because a
thermochemical measurement is based on the relationship between heat
and temperature change.
• The heat required to raise the temperature of a substance is its heat capacity.

❑Heat Capacity and Specific Heat

– The heat capacity, C, of a sample of substance is the quantity of heat
required to raise the temperature of the sample of substance one degree
Celsius.
– Changing the temperature of the sample requires heat equal to:

q = C T
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6.6 Measuring Heats of Reaction

• Example 6.5 : Suppose a piece of iron requires 6.70 J of heat to raise its
temperature by one degree Celsius. The quantity of heat required to raise the
temperature of the piece of iron from 25.0oC to 35.0oC is:

q = C T = ( 6.70 J / C )  ( 35.0 C − 25.0 C )

o o o

q = 67.0 J

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6.6 Measuring Heats of Reaction

• Heat capacities are also compared for one gram amounts of substances.
The specific heat capacity (or “specific heat”) is the heat required to
raise the temperature of one gram of a substance by one degree
Celsius.

– To find the heat required you must multiply the specific heat, s, of the
substance times its mass in grams, m, and the temperature change,
ΔT.

q = s  m  T

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 Specific Heats and Heat Capacities

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6.6 Measuring Heats of Reaction

• Example 6.6 : Calculate the heat absorbed when the temperature of 15.0
grams of water is raised from 20.0oC to 50.0oC. (The specific heat of
water is 4.184 J/g.oC.)

q = s  m  T
q= ( 4.184 gJo C )  (15.0g )  ( 50.0 − o
20.0 C )
q = 1.88  10 J 3

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 Heats of Reaction:
Calorimetry
• A calorimeter is a device
used to measure the heat
absorbed or evolved during
a physical or chemical
change.

– The heat absorbed by the

calorimeter and its contents
is the negative of the heat of
reaction.
q calorimete r = − q rxn
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 A Problem to Consider

• When 23.6 grams of calcium chloride, CaCl2, was

dissolved in water in a calorimeter, the temperature
rose from 25.0oC to 38.7oC.

If the heat capacity of the solution and the calorimeter is

1258 J/oC, what is the enthalpy change per mole of
calcium chloride?

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 Heats of Reaction: Calorimetry

• First, let us calculate the heat absorbed

by the calorimeter.
q cal = C T = (1258 o C )  ( 38.7 C −
J o o
25.0 C )

q cal = 1.72  10 J 4

• Now we must calculate the heat per mole of

calcium chloride.
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 Heats of Reaction: Calorimetry

• Calcium chloride has a molecular mass of 111.1

g, so
(1 mol CaCl 2 )
( 23.6 g CaCl 2 )  = 0.212 mol CaCl 2
111.1 g
• Now we can calculate the heat per mole of
calcium chloride.
q rxn − 17.2 kJ
H = = = −81.1 kJ / mol
mol CaCl 2 0.212 mol
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 Hess’s Law

• Hess’s law of heat summation states that

for a chemical equation that can be written
as the sum of two or more steps, the
enthalpy change for the overall equation is
the sum of the enthalpy changes for the
individual steps. (See Animation: Hess’s Law)

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 Hess’s Law

• For example, suppose you are given the following

data:

S (s ) + O 2 ( g ) → SO 2 ( g );  H = -297 kJ o

2SO 3 ( g ) → 2SO 2 ( g ) + O 2 ( g );  H = 198 kJ o

• Could you use these data to obtain the enthalpy

change for the following reaction?

2S (s ) + 3O 2 ( g ) → 2SO 3 ( g );  H = ? o

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 Hess’s Law

• If we multiply the first equation by 2 and

reverse the second equation, they will sum
together to become the third.
2S (s ) + 2O 2 ( g ) → 2SO 2 ( g );  H = ( -297 kJ)  (2) o

2SO 2 ( g ) + O 2 ( g ) → 2SO 3 ( g );  H o = (198 kJ)  (-1)

2S (s ) + 3O 2 ( g ) → 2SO 3 ( g );  H = -792 kJ o

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 Standard Enthalpies of Formation ΔHo

• The term standard state refers to the

standard thermodynamic conditions chosen
for substances when listing or comparing
thermodynamic data: 1 atmosphere pressure
and the specified temperature (usually
25oC).
– The enthalpy change for a reaction in which reactants are in
their standard states is denoted ΔHo (“delta H zero” or “delta H
naught”).

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Standard Enthalpies of Formation Hfo,

• The standard enthalpy of formation of a

substance, denoted Hfo, is the enthalpy
change for the formation of one mole of a
substance in its standard state from its
component elements in their standard state.

– Note that the standard enthalpy of formation for a pure element in

its standard state is zero.

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 Standard Enthalpies of Formation

• The law of summation of heats of formation

states that the enthalpy of a reaction is equal
to the total formation energy of the products
minus that of the reactants.

H = 
o
n  H f (products ) −
o
 m  H f (reactants )
o

− S is the mathematical symbol meaning “the sum of”, and m and n

are the coefficients of the substances in the chemical equation.

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 A Problem to Consider

• Large quantities of ammonia are used to

prepare nitric acid according to the following
equation:

4 NH 3 ( g ) + 5O 2 ( g ) → 4 NO ( g ) + 6 H 2O ( g )
– What is the standard enthalpy change for this reaction? Use Table
6.2 for data.

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 A Problem to Consider

• You record the values of Hfo under the formulas in

the equation, multiplying them by the coefficients in
the equation.

4 NH 3 ( g ) + 5O 2 ( g ) → 4 NO ( g ) + 6 H 2O ( g )
Table 6.2 Table 6.2 Table 6.2 Table 6.2

• You can calculate Ho by subtracting the values

for the reactants from the values for the
products.
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 A Problem to Consider

• Using the summation law:

H = o
n  H f (products ) −
o
 m  H f (reactants )
o

 H = [4( 90.3 ) + 6( − 241.8 )] kJ − [4( − 45.9 ) + 5( 0 )] kJ

o

 H = − 906 kJ
o

– Be careful of arithmetic signs as they are a likely source of

mistakes.

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 Fuels

• A fuel is any substance that is burned to

provide heat or other forms of energy.
• In this section we will look at: Syngas ‫الغاز المتزايد‬
– Foods as fuels
– Fossil fuels
– Coal gasification
and liquefaction

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 Fuels

• Food fills three needs of the body:

– It supplies substances for the growth and repair of tissue.
– It supplies substances for the synthesis of compounds used in the
regulation of body processes.
– It supplies energy. About 80% of the energy we need is for heat.
The rest is used for muscular action and other body processes

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 Fuels
• A typical carbohydrate food, glucose (C6H12O6)
undergoes combustion according to the following
equation.

– One gram of glucose yields 15.6 kJ (3.73 kcal) when burned.

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 Fuels
• A representative fat is glyceryl trimyristate,
C45H86O6. The equation for its combustion is:

C 45 H 86O 6 ( s ) + 127
2 O 2 (g ) →

45CO 2 ( g ) + 43 H 2O ( l );  H = -27,820 kJ o

– One gram of fat yields 38.5 kJ (9.20 kcal) when burned.

Note that fat contains more than twice the fuel per gram
than carbohydrates contain.

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 Metabolic Energy

The complex of biochemical reactions that make food energy

available for use by living organisms is called metabolism.
A kilocalorie (4.2kj) is the amount of heat needed to change the
temperature of 1 kg of water by 1° C; it is equal to one dietary
"calorie."
Food kcal
1 raw onion 5
1 dill pickle 15
1 gum drop 35
1 poached egg 75
1 banana 120
1 cupcake 130
1 broiled hamburger patty 150
1 glass of milk 165
1 cup bean soup 190
1 cup tuna salad 220
1 ice cream soda 325
½ broiled chicken 350
1 lamb chop 420

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 Energy of a Snickers Bar

There are 311 kCal in a Snickers Bar. If you

added this much heat to a gallon of water at
22oC what would be the final temperature?

311 Cal = 311,000 cal 1 gallon of water weighs 3960.4 g

311,000 cal = 1.000 cal/goC x 3960.4g x ?oC

Solving for the change of temperature gives us 78.5oC

Add this to the 22oC to get 100.5oC which is above the boiling point of water
(100oC)

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 Fuels
• Fossil fuels account for nearly 90% of the
energy usage in the United States.
– Anthracite, or hard coal, the oldest variety of coal, contains about
80% carbon.
– Bituminous coal, a younger variety of coal, contains 45% to 65%
carbon.
– Fuel values of coal are measured in BTUs (British Thermal
Units).
– A typical value for coal is 13,200 BTU/lb.
– 1 BTU = 1054 kJ

.
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 Fuels
• Natural gas and petroleum account for nearly
three-quarters of the fossil fuels consumed per year.

– Purified natural gas is primarily methane, CH4, but also contains small
quantities of ethane, C2H6, propane, C3H8, and butane, C4H10.
– We would expect the fuel value of natural gas to be close to that for the
combustion of methane.

CH 4 ( g ) + 2O 2 ( g ) → CO 2 ( g ) + 2 H 2O ( g ) ;  H o = − 802 kJ

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 Fuels
Petroleum is a very complicated mixture of
compounds.
Gasoline, obtained from
petroleum, contains many
different hydrocarbons, one of
which is octane, C8H18.

C 8 H 18 ( l ) + 252 O 2 ( g ) →
8CO 2 ( g ) + 9 H 2O ( g );  H o = -5,074 kJ
– This value of Ho is equivalent to 44.4 kJ/gram.
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 Fuels
• With supplies of petroleum estimated to be 80%
depleted by the year 2030, the gasification of coal has
become a possible alternative.
– First, coal is converted to carbon monoxide using steam.

C(s ) + H 2O ( g ) → CO ( g ) + H 2 ( g )
– The carbon monoxide can then be used to produce a variety of other
fuels, such as methane.

CO ( g ) + 3 H 2 ( g ) → CH 4 ( g ) + H 2O ( g )
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