GENERAL CHEMISTRY - · PDF file1 GENERAL CHEMISTRY 1. Scientific Notation, Significant...

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GENERAL CHEMISTRY

1. Scientific Notation, Significant Figures, and the Factor-Label Method of Solving

Problems

2. Naming Inorganic Compounds Naming Inorganic Compounds

3. Mole Definitions

4. Empirical and Molecular Formulas

5. Chemical Stoichiometry Problems

6. Types of Chemical Reactions; Writing Balanced Ionic Equations

7. Oxidation-Reduction Reactions

8. Working with Solutions

9. pH and Titrations

10. Ideal Gases

11. Ideal Gas Mixtures

12. Calorimetry Exercises

13. The Enthalpy of Chemical Change: Calculations using Hess's Law and Heats of

Formation

14. Electromagnetic Radiation and the Spectrum of Atomic Hydrogen

15. Quantum Numbers, Orbitals, and Electron Configurations

16. Periodic Trends

17. Hybridization of Carbon Diagram

18. Molecular Geometry Summary Chart

19. Predicting Molecular Geometry and Hybridization

20. Molecular Orbital Diagrams for the First and Second Rows

21. Intermolecular Forces

22. Chemical Kinetics: Introductory Concepts

23. How to Determine the Rate Law from Experimental Data

24. How to Work from a Mechanism to a Rate Law

25. How to Solve Equilibrium Problems

26. Le Chatelier's Principle

27. Table of Relative Strengths of Conjugate Acid-Base Pairs

28. How to Calculate the pH of an Acidic or Basic Solution

29. Strong and Weak Acids and Bases

30. Predicting the pH of Salt Solutions

31. Buffers

32. Predicting Solubility

33. Thermodynamics and Equilibrium: Important Equations

34. How to Balance Equations for Oxidation-Reduction Reactions

35. Voltaic Cells

36. Factors Affecting Corrosion

37. Electrolysis

http://spiepho.sbc.edu/worksheets/Gen_Chem_1/Chp6,Calorimetry.doc

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Scientific Notation, Significant Figures, and the

Factor-Label Method of Solving Problems

Scientific Notation

Scientific notation is a type of exponential notation in which only one digit is kept to the left

of the decimal point. Example: 8.4050 x 10-8

.

Significant Figures

It is reasonable that a calculated result can be no more precise than the least precise piece of

information that went into the calculation. Thus it is common practice to write numbers in

scientific notation with only the last place containing any uncertainty. When we do this we are

keeping only the “significant figures”.

To determine the number of significant figures in a number, you read the number from left to

right and count all digits starting with the first non-zero digit. Do not count the exponential part.

Thus the number 0.002050 contains 4 significant figures and is written in scientific notation as

2.050 x 10-3

. The trailing zeros in a non-decimal number such as 1200 may or may not be

significant: the number may be written as 1.2 x 102, 1.20 x 10

2 or 1.200 x 10

2 depending on

whether it has 2, 3, or 4 significant figures.

Significant Figures in Derived Quantities

When doing calculations, you should use all the digits allowed by your calculator in all

intermediate steps. Then in the final step, round off your answer to the appropriate number of

“significant figures” such that only the last decimal place contains any uncertainty. You do this

by following the rules:

• When adding or subtracting, first express all numbers with the same exponent. Then the

number of decimal places in the answer should be equal to the number of decimal places in

the number with the fewest decimal places.

• In multiplication or division, the number of significant figures in the answer should be the

same as that in the factor with the fewest significant figures.

When using these rules, assume that exact numbers have an infinite number of significant figures

(or decimal places). For example, there are exactly 12 inches in one foot.

Solving Problems Using Dimensional Analysis: The Factor-Label Method

Units may be used as a guide in solving problems. First decide what units you need for your

answer. Then determine what units you are given in the problem, and what conversion factors

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will take you from the given units to the desired units. If the units cancel out properly, chances

are that you are doing the right thing! The basic set up is

? units desired= units given x new units

y units given

.. ..

Conversion factors are added until the new units are the same as the units desired. Each

conversion factor has a denominator equivalent to the numerator but in different units.

Examples

1. Carry out the following mathematical operations and express your answer in scientific notation

using the proper number of significant figures.

(a) (4.28 x 10-4

) + (3.564 x 10-2

)

(b) (0.00950) x (8.501 x 107)

3.1425 x 10-11

_____________________________________________________________________

2. Use the factor-label method to solve the following problem. Show your work, and give your

answer in scientific notation using the proper number of significant figures.

The calorie (1 cal = 4.184 J) is a unit of energy. The burning of a sample of gasoline produces

4.0 x 102 kJ of heat. Convert this energy to calories. (10

3 J = 1 kJ.)

____________________________________________________________________

3. Use the factor-label method to solve the following problem. Show your work, and give your

answer in scientific notation using the proper number of significant figures.

The distance from the sun to the earth is 93 million miles. How many minutes does it take light

from the sun to reach earth?

Useful information: 1 km = 0.6214 mile, c = speed of light = 3.00 x 108 m/s)

Answers:

1. (a) 3.607 x 10-2

; (b) 2.57 x 1016

2. 9.6 x 104 calories

3. 8.3 minutes

____________________________________________________________________

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Naming Inorganic Compounds

To name a compound you must first decide whether the substance is an ionic or molecular

compound. Ionic compounds are easily recognized since they usually contain both metallic and

non-metallic elements. The most common exception to this rule are ionic compounds containing

the ammonium ion, NH4+, such as (NH4)2CO3 or NH4Br which contain no metal ions. Molecular

compounds typically contain only non-metallic atoms (and metalloids).

Note that when naming these molecular compounds, the number of atoms of a given type is

commonly indicated with a prefix (di-, tri-, tetra, etc.).

______________________________________________________________________________

Exercises

1. Complete the following chart of corresponding ion names and formulas.

Cation Name Formula Anion Name Formula

(1) potassium ion (11) nitrate ion

(2) Fe3+

(12) H2PO4-

(3) ammonium ion (13) hydrogen carbonate

(or bicarbonate) ion

(4) Ba2+

(14) MnO4-

(5) silver ion (15) perchlorate ion

(6) Cu2+

(16) S2-

(7) zinc ion (17) acetate ion

(8) Co2+

(18) dichromate ion

(9) hydrogen ion (19) CO32-

(10) chromium(III) ion (20) sulfite ion

______________________________________________________________________________

2. Complete the following chart of corresponding compound names and formulas. Circle the

names of all non-ionic (i.e., molecular) compounds.

Compound Name Formula Compound Name Formula

(1) silver nitrate (11) sodium hydrogen

phosphate

(2) Ni(CH3CO2)2 (12) SO3

(3) ammonium sulfate (13) potassium permanganate

(4) P2O5 (14) Al2S3

(5) sodium oxide (15) cobalt(III) sulfate

(6) NH4NO3 (16) Ag2CrO4

(7) nitrogen trichloride (17) SrF2

(8) NaHCO3 (18) sulfur hexafluoride

(9) iron(II) acetate (19) NH3

(10) carbon tetrachloride (20) LiClO3

Answers

1. Complete the following chart of corresponding ion names and formulas.

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Cation Name Formula Anion Name Formula

(1) potassium ion K+ (11) nitrate ion NO3

-

(2) iron(III) ion

(or ferric ion)

Fe3+

(12) dihydrogen

phosphate ion

H2PO4-

(3) ammonium ion NH4+

(13) hydrogen carbonate

(or bicarbonate) ion

HCO3-

(4) barium ion Ba2+

(14) permanganate ion MnO4-

(5) silver ion Ag+ (15) perchlorate ion ClO4

-

(6) copper(II) ion

(or cupric ion)

Cu2+

(16) sulfide ion S2-

(7) zinc ion Zn2+

(17) acetate ion CH3CO2-

(or C2H3O2-)

(8) cobalt(II) ion Co2+

(18) dichromate ion Cr2O72-

(9) hydrogen ion H+ (19) carbonate ion CO3

2-

(10) chromium(III) ion Cr3+

(20) sulfite ion SO32-

______________________________________________________________________________

2. Complete the following chart of corresponding compound names and formulas. Circle the

names of all non-ionic (i.e., molecular) compounds. (NOTE: In these answers I have used

shading (instead of circles) to indicate the molecular compounds.)

Compound Name Formula Compound Name Formula

(1) silver nitrate AgNO3 (11) sodium hydrogen

phosphate

Na2HPO4

(2) nickel(II) acetate Ni(CH3CO2)2 (12) sulfur trioxide SO3

(3) ammonium sulfate (NH4)2SO4 (13) potassium permanganate KMnO4

(4) diphosphorus

pentoxide

(or diphosphorus

pentaoxide)

P2O5 (14) aluminum sulfide Al2S3

(5) sodium oxide Na2O (15) cobalt(III) sulfate Co2(SO4)3

(6) ammonium nitrate NH4NO3 (16) silver chromate Ag2CrO4

(7) nitrogen trichloride NCl3 (17) strontium fluoride SrF2

(8) sodium hydrogen

carbonate (or sodium

bicarbonate)

NaHCO3 (18) sulfur hexafluoride

SF6

(9) iron(II) acetate Fe(CH3CO2)2

or

Fe(C2H3O2)2

(19) ammonia

NH3

(10) carbon tetrachloride CCl4 (20) lithium chlorate LiClO3

Atomic Mass, Moles, and the Periodic Table

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Atomic Mass and Molar Mass

Isotopic masses cannot be obtained by summing the masses of the elementary particles

(neutrons, protons, and electrons) from which the isotope is formed. This process would give

masses slightly too large, since mass is lost when the neutrons and protons come together to form

the nucleus.

Atomic masses (also called atomic weights) are thus assigned relative to the mass of a

particular carbon isotope, 6

12C , which is assigned the mass of 12 amu exactly. Likewise 1 mole of

6

12C has a mass of exactly 12 g. Atomic masses and molar masses of other isotopes are calculated

based on their mass relative to that of Carbon-12.

Masses of “average” atoms are found by summing isotopic masses, weighting each isotopic

mass by its abundance. Thus one “average” C atom has a mass of 12.01 amu, and the mass of 1

mole of “average” carbon atoms has a mass of 12.01 g. These average masses are what are given

on the periodic chart.

What is a Mole?

Since atoms and molecules are so tiny, it is convenient to talk about a large number of them

at a time. The chemical counting unit is known as the mole. A mole is defined as the amount of

substance that contains as many elementary entities (atoms, molecules, or other particles) as there

are atoms in exactly 12 g of the 6

12C isotope. It has been found experimentally that

1 mole of particles= 6.022 x 1023

particles

This value is known as Avogadro’s number. Just like 1 dozen of anything always contains 12

items, 1 mole of anything always contains 6.022 x 1023 items.

Molecular Masses and Compound Masses

Molecular masses are found by summing atomic masses. They are often called molecular

weights. Thus the mass of 1 mole of water, H2O, would be 2 x (molar mass of H) plus 1x (molar

mass of O) or [(2 x 1.008 g) + (1 x 16.00 g)] = 18.02 g.

Ionic compounds such as NaCl do not contain molecules. Their formulas give the relative

numbers of each kind of atom in the sample. What we mean by the molar mass (or the molecular

weight) of an ionic compound is really the formula weight. The formula weight is the sum of the

atomic masses in the formula.

Percent Composition of Compounds

The percent composition by mass is the percent by mass of each element in a compound. If

there are n moles of an element per mole of compound, the percent by mass of the element is

calculated using the equation,

% Composition of Element=n molar mass of element

molar mass of compound

100%

The sum of the % compositions of all elements in a compound is 100%.

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______________________________________________________________________________

Exercises

1. The atomic mass scale gives masses in atomic mass units (amu) relative to the mass of carbon-

12.

(a) What is the mass of one 12

C atom in atomic mass units (amu)? ____________

(b) What is the mass of an average C atom in atomic mass units (amu)? ____________

(c) What is the mass of an average Cl atom in amu? ____________

(d) What is the mass of an average Br atom in amu? ____________

______________________________________________________________________________

2. The molar mass scale gives masses in grams (g) relative to the mass of 12

C.

(a) What is the mass in grams of 1 mole (mol) of 12

C? ___________

(b) What is the mass in grams of 1 mole (mol) of carbon? ___________

(c) What is the mass in grams of 1 mole (mol) of Cl? ___________

(d) What is the mass in grams of 1 mole (mol) of Na? ___________

______________________________________________________________________________

3. How many 12

C atoms are present in a mole of 12

C ?

______________________________________________________________________________

4. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its

molecular formula is C9H10O.

(a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol.

(b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g?

Answers:

1. (a) 12 amu exactly; (b) 12.01 amu; (c) 35.45 amu; (d) 79.90 amu.

2. (a) 12 g exactly; (b) 12.01 g; (c) 35.45 g; (d) 22.99 g.

3. 6.022 x 1023

atoms of 12

C.

4. (a) 80.56% C; 7.51% H; 11.93% O; (b) 2.11 x 1021

molecules of C9H10O.

______________________________________________________________________________

Empirical and Molecular Formulas

The empirical formula of a compound gives the simplest whole number ratio of different types of

atoms in the compound. All salt formulas are empirical formulas. On the other hand, the

molecular formula of a compound may or may not be the same as its empirical formula. For

example, the molecular formula of butane is C4H10 while its empirical formula is C2H5. The

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molecular formula gives the true number of each kind of atom in a molecule.

Empirical formulas may be easily determined from experimental data.

Usually you must first determine how many grams of each type of atom are in the compound. If

percent composition data is given, assume that you have 100.0 g of the compound; then the

number of grams of each element is equal to the percentage for that element.

The next task is to convert the grams of each element to moles of the element. Be sure to keep at

least three significant figures in your answers.

The final step is to write the molar amounts of each element as subscripts in the formula. Then

divide all molar subscripts by the smallest value in the set. At this point, the subscripts may all

be very close to whole numbers; if so, you are finished. If one (or more) of the subscripts is not

close to a whole number, multiply all molar subscripts by the simple factor which makes all

subscripts whole numbers.

Once the empirical formula is determined, the molecular formula is easily found if the molar

mass (molecular weight) of the molecule is also known. You first calculate the molar mass of the

empirical formula. Then you divide the molar mass of the molecule by the molar mass of the

empirical formula. The division should give a simple whole number. That number is the factor

by which all subscripts in the empirical formula must be multiplied to obtain the molecular

formula.

______________________________________________________________________________

Exercises

1. The molecular formula of the antifreeze ethylene glycol is C2H6O2. What is the empirical

formula?

______________________________________________________________________________

2. A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula

C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound?

______________________________________________________________________________

3. Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One

of them is a blue solid that comes apart, reversibly, in the gas phase. It contains 36.84% N.

What is the empirical formula of this oxide?

______________________________________________________________________________

4. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine

What is the empirical formula of the indium compound?

9

______________________________________________________________________________

Answers:

1. CH3O

2. Molar mass of empirical formula is 58.06 g/mol. Thus molecular formula is C4H8N2O2.

3. The ratios are N1.00O1.50 . Since 1.50 is not close to a whole number, we multiply both

subscripts by 2. The empirical formula is thus N2O3. (The name is dinitrogen trioxide.)

4. InCl3.

Chemical Stoichiometry Problems

Calculating the yield of a chemical reaction is a process at the heart of chemistry. While

there are many ways a problem can be phrased, in all cases the stoichiometric coefficients in the

balanced reaction are used to determine the mole ratios between reactants and products. Thus

the first step is usually calculating the moles of each species available. If an amount is given in

grams, the molar mass is used as a conversion factor to change grams to moles.

Limiting Reagent Problems

In some problems, amounts of more than one species are given. In that case your first task is

to determine which species is the limiting reagent. Just as you can make only 1 bicycle from 2

wheels and 4 handlebars (with 3 handlebars left over), and only 2 bicycles from 8 wheels and 2

handlebars (with 4 wheels left over), in chemical reactions some species are limiting while others

may be present in excess.

In the case of a bicycle, we need 2 wheels

1 handlebar

. We obtain analogous information about the

relative amounts of species that react from the stoichiometric coefficients in a balanced chemical

equation. For example, in Exercise (2) below the equation

CO(g) + 2 H2(g 3OH (l)

tells us we need 2 mol H2

1 mol CO

. If we have more than 2 moles of H2 for each mole of CO, CO will

be the limiting reagent and the excess H2 will not react. Conversely, if we have more than 1

mole of CO for every 2 moles of H2, H2 will be the limiting reagent and the excess CO(g) will be

left over. In each case, the yield of CH3OH is determined by the moles of limiting reagent

available.

Calculating the Theoretical Yield

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The theoretical (maximum possible) yield is based on the amount of limiting reagent

available. The yield is calculated in steps:

• Calculate moles of all reactants available. If amounts are given in grams, convert grams to

moles using the molar mass of each reactant as your conversion factor: 1 mole reactant

# g reactant

.

• NOTE: Skip this step if you have already identified the limiting reagent. To determine

which reagent is limiting, use the mole ratio obtained from the balanced equation for the

reaction to find the moles of reactant B needed to react with the available moles of reactant A.

If the moles of B available are less than the moles of B needed, reactant B is the limiting

reagent and reactant A is in excess. Conversely, if the moles of B available are more than

the moles of B needed, A is the limiting reagent and B is in excess.

• Calculate the moles of product based on the moles of limiting reagent available; use the

stoichiometric ratio of # moles product

# moles limiting reagent

as the conversion factor.

• If you are asked for the yield in grams, convert the yield in moles to a yield in grams using

the molar mass as your conversion factor: # g product

1 mole product

Percent Yield

Most reactions do not go to completion, and so the actual yield is less than the percent yield.

The percent yield is calculated as

Percent yield=actual yield

theoretical yield

100%

_____________________________________________________________________

Exercises: Use your own paper if you want more space.

1. Ammonia is produced by the reaction

3 H2(g) + N2(g 3(g)

(a) If N2(g) is present in excess and 55.6 g of H2(g) reacts, what is the theoretical yield of

NH3(g)?

(b) What is the percent yield if the actual yield of the reaction is 159 g of NH3(g)?

Answers: #1(a) 313 g NH3(g); (b) 50.8% yield.

______________________________________________________________________

2. Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction

CO(g) + 2 H2(g 3OH (l)

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A mixture of 1.20 g H2(g) and 7.45 g CO(g) are allowed to react.

(a) Which reagent is the limiting reagent?

(b) What is the yield of CH3OH? [Assume theoretical yield in g is what is wanted here.]

(c) How much of the reagent present in excess is left over?

(d) Suppose the actual yield is 7.52 g of CH3OH. What is the % yield?

Answers: # 2(a) CO is the limiting reagent; (b) 8.52 g CH3OH; (c) 0.13 g H2; (d) 88.3%

Types of Chemical Reactions

One skill that chemists learn over time is that of writing and balancing equations. The first

task is deciding what type of reaction is taking place. In this chapter we study three types:

Precipitation Reactions: In these reactions two soluble salts usually react to form to an

insoluble salt (the precipitate!) and a soluble salt. The cations of the reacting salts exchange

anions.

Acid-Base Reactions: Most commonly an acid of the type HX or H2X reacts with a basic

hydroxide to form a salt plus water. Alternatively, the acid may react with ammonia (NH3) to

form an ammonium salt (but no water). These are proton transfer reactions in which H+ (the

proton) is transferred from the acid to the base.

Oxidation-Reduction Reactions: These are reactions in which one type of atom increases in

oxidation number (is oxidized) and another type of atom decreases in oxidation number (is

reduced). A large number of oxidation-reduction (redox) reactions contain one or more

reactants or products, which are pure elements.

Note that hydroxides can react with acids in acid-base reactions, and also with other salts in

precipitation reactions.

Writing Balanced Ionic Equations

The first step in writing a balanced equation is predicting the products of the reaction as

discussed above. Then the steps below are completed in sequence:

Balance the Molecular Equation: In the “molecular” equation, nothing is broken up into

ions. Salt formulas are written so that the cation charges exactly balance out the anion

charges so that the salt is neutral. Then the equation is balanced for atoms.

Balance the Total Ionic Equation: The first step in writing an ionic equation is to decide

what species should be broken up into ions. The rules below should help!

Break up into Ions Do NOT break up! Leave “as is”!

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Strong Acids. HCl, HBr, HI, HNO3, HClO4,

and H2SO4 are the most common examples;

assume other acids are weak.

Strong Bases. NaOH, KOH, or Ba(OH)2 are the

most common examples; assume other bases are

weak.

Soluble Salts. Salts of the alkali metals, salts

containing the NH4+ ion, the NO3

- ion, and other

salts as specified in Chang, Table 4.2, p. 119.

Weak Acids. Nearly all acids are

weak.

Weak Bases. Nearly all bases are

weak.

Insoluble Salts. Most salts are

insoluble.

Non-electrolytes or Weak

Electrolytes. Examples include H2O,

gases, pure elements, hydrocarbons,

and alcohols.

Balance the Net Ionic Equation: Identify all spectator ions: these are ions that are identical

on both sides of the balanced total ionic equation. Remove the spectator ions from the

equation. What remains is the net ionic equation. Finally, simplify the stoichiometric

coefficients if all of them are divisible by a common factor.

If all the ions are spectator ions so that nothing is left for your net ionic equation, no reaction

has taken place!

______________________________________________________________________________

Exercises

For each of the following reactions, complete the chart. Be sure to balance all of your

equations.

1. Mg(OH)2(s) + HCl(aq)

(a) Reaction type:

Formulas of Products Formed:

(b) Molecular Equation

(c) Total Ionic Equation

(d) Net Ionic Equation

Answer to 1.(d): Mg(OH)2(s) + 2 H+(aq

2+(aq) + 2 H2O(l)

______________________________________________________________________________

2. AgNO3(aq) + K2Cr2O7(aq)

(a) Reaction type:





Answer to 2.(d): 2 Ag+(aq) + Cr2O7

2-(aq 2Cr2O7(s)

______________________________________________________________________________

3. NH3(aq) + HC2H3O2(aq)

(or CH3COOH)

13

(a) Reaction type:





Answer to 3.(d): NH3 + HC2H3O2 4+(aq) + C2H3O2

-(aq)

______________________________________________________________________________

4. NaOH(aq) + H2SO4(aq)

(a) Reaction type:





Answer to 4.(d): OH-(aq) + H

+(aq 2O(l) (obtain this after all coefficients have been

divided by 2)

______________________________________________________________________________

5. H2S(aq) + Ba(OH)2(aq)

(a) Reaction type:





Answer to 5.(d): H2S(aq) + Ba2+

(aq) + 2 OH-(aq s) + 2 H2O(l)

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change.

Oxidation numbers are either real charges or formal charges which help chemists keep track of

electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device.

Oxidation cannot occur without reduction.

In a redox reaction, the substance oxidized contains atoms which increase in oxidation

number. Oxidation is associated with electron loss (helpful mnemonic: LEO = Loss of

Electrons, Oxidation).

The substance reduced contains atoms which decrease in oxidation number during the

reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of

Electrons, Reduction).

An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a

reducing agent is a substance that reduces another reactant: it itself is oxidized.

A disproportionation reaction is a reaction in which the same element is both oxidized and

14

reduced.

How to Assign Oxidation Numbers: The Fundamental Rules

The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is

zero.

The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number

of Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is

-2.

The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if

an ion.

The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in

compounds is +2. The oxidation number of F is -1 in all its compounds.

The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and

the ionic hydrides, such as NaH (where H = -1).

The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O =

0) and peroxides, such as H2O2 or Na2O2, where O = -1.

For other elements, you can usually use the sum rule above to solve for the unknown

oxidation number.

Examples:

NO(g) has O = -2, so N = +2.

NO2 (g) has two oxygen atoms and each has O = -2. Thus N + 2(-2) = 0, so N = +4.

SO42-

has O = -2. Thus S + 4(-2) = -2. Solving the equation gives S = -2 + 8 = +6.

K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(-2) = 0; 2 Cr = 12; Cr = +6.

Recognizing Oxidation-Reduction Reactions

Oxidation-reduction reactions are reactions in which one type of atom increases in

oxidation number (is oxidized) and another type of atom decreases in oxidation number (is

reduced). Thus to show that a reaction is a redox reaction, you need to calculate oxidation

numbers for the atoms in the reactants and products, and document that changes are taking place.

There are, however, a few useful generalizations.

A large number (but not all!) of oxidation-reduction reactions contain one or more reactants

or products which are pure elements. Why is this true? Also, all electrochemical reactions

are redox reactions.

Most acid-base reactions and most precipitation reactions are not redox reactions. Why?

Give some examples!

Working with Solutions

15

Solutions are uniform mixtures on the molecular level of two or more substances. The

substance present in largest amount is called the solvent (usually water) and any substance

dissolved in the solvent is called a solute.

Molarity, M

The molar concentration or molarity, M, of a solution is used to indicate the number of moles

of solute per liter of solution:

Molarity M mol

L

(no.of moles solute)

(no. of liters of solution)

The molarity of a solution is often used as a conversion factor between moles of solute and

volume of solution: it is a “molar density”.

Dilution

One common type of “lab assistant” problem is the preparation of a dilute solution from a

more concentrated solution. For example, we might want to prepare 250 mL of a 0.500 M

NaOH solution from a 6.00 M NaOH solution as in exercise 5 below. There is a shortcut way to

work dilution problems which is based on the knowledge that the # of moles of solute you need

for the dilute solution all come from the concentrated solution. Thus

#moles concentrated #moles dilute and since M(mol L-1

) x V(L) = n (mol), it follows that

Mconcentrated Vconcentrated Mdilute Vdilute

or, in the notation of Chang,

Minitial Vinitial Mfinal Vfinal

In the laboratory this equation is often used to determine the Vconcentrated that needs to be diluted

to give the desired volume of a more dilute solution.

Stoichiometry of Reactions in Solution

Problems involving solutions are very similar to the chemical stoichiometry problems we

have discussed earlier. The only difference is that the moles of reactant or product may need to

be calculated from a solution volume using the molarity (M = mol/L) as a conversion factor

between volume and moles.

Lab Assistant Problems

The problems below will introduce you to calculations involving molarity. I call these “lab

assistant problems” since we do this kind of calculation all the time when setting up labs! When

working these problems, it is useful to recall that 1 L = 1000 mL.

16

______________________________________________________________________

Exercises:

1. What is the molarity of a solution containing 21.0 g NaCl in 200 mL of solution?

Answer: 1.80 M NaCl

_____________________________________________________________________

2. What ions exist in a 0.245 M Na2CO3 solution? Give the molar concentration of each ion.

Answer: 0.490 M Na+ and 0.245 M CO32-

______________________________________________________________________

3. How many mL of a 0.420 M NaCl solution should be measured out to provide 1.5 g NaCl?

Answer: 61 mL of 0.420 M NaCl

_____________________________________________________________________

4. How many grams of CaCl2 are needed to make 200 mL of a 0.500 M Cl- solution? (Note:

CaCl2 is a soluble salt. The molar mass of CaCl2 is 110.98 g/mol.)

Answer: 5.55 g CaCl2

_____________________________________________________________________

5. How do you prepare 250 mL of a 0.500 M NaOH solution from a 6.00 M NaOH solution?

Answer: Take 20.8 mL of 6.00 M NaOH and dilute to a total volume of 250 mL.

_____________________________________________________________________

pH and Titration

17

Relation between pH and Hydrogen Ion Concentration

The pH scale is widely used to report the molar concentration of hydrogen ion, H+(aq), in

aqueous solution. The pH of a solution is defined as

pH log10[H] (1)

where [H+] = the molar concentration of H+(aq) in the solution. (In chemistry, square brackets

around a chemical symbol mean "the molar concentration of" whatever they enclose.) Equation

(1) above may be solved for [H+] to give

[H]10pH (2)

(Here we use the well known rule that if y log10 x , then x 10y

.) In practice, the pH scale is

only used when [H+(aq)] is less than 1.0 M. Acidic, basic, and neutral solutions can be

distinguished as shown below:

Type of Solution pH [H+] Color of litmus

Acidic < 7.00 > 1.0 107 pink

Neutral = 7.00 = 1.0 107 in between

Basic > 7.00 < 1.0 107 blue

Titration

A titration is a procedure in which a solution of known concentration is used to determine the

concentration of another solution with which it reacts. The reaction must be rapid and should go

to completion. It may be an acid-base reaction, an oxidation-reduction reaction, or a

precipitation reaction.

Typically a titration is conducted by filling a buret with one solution and transferring an exact

amount of the second solution to an Erlenmeyer (conical) flask with a pipet. Indicator is added

to the flask, and the first solution is added drop wise from the buret until the indicator changes

color. The point of color change is called the endpoint, the equivalence point, or the

stoichiometric point of the titration: all of these terms are synonymous. The indicator is chosen

so the color change occurs when stoichiometric amounts of the reactants have been added to the

flask.

The concentration of the unknown solution is calculated as illustrated in the exercise 3 below.

___________________________________________________________________________

Exercises:

18

1. Use Eq (1) and the log10 button on your calculator to determine the pH of solutions with the

specified hydrogen ion concentrations [H+]:

(a) (b) (c) (d) (e) (f)

[H+] 0.10 M 0.0010 M 10

-7 M 5.0 x 10-10 M 6.0 M 1.0 M

pH

Acidic,

basic

or

neutral

?

Answers: pH values are

(a) 1.00 (b) 3.00 (c) 7 (d) 9.30 (e) - 0.78 (f) 0.00

Ideal Gas Problems

Gases at low pressures obey the ideal gas law,

p V = n R T (1)

where R is a constant (known as the gas constant) that has the value

R = 0.08206 L atm K-1 mol-1 (2)

Appropriate units to use for p, V, n, and T in the ideal gas equation are those used for R above.

Thus the pressure (p) should be in atm, the volume (V) in L, the temperature (T) in degrees K,

and the amount of gas (n) should be in moles. Useful conversion factors are

Pressure:1 atm = 760 Torr = 760 mmHg = 101.3 kPa = 1.013 bar

Temperature:K = 273 + oC

Volume:1 L = 1000 mL = 1000 cm3

Since pV

nT R , and R is a constant, it follows that

p1V1

n1T1

p2V2

n2T2

(3)

19

where the subscript “1” represents one set of conditions, and the subscript “2” represents another

set of conditions. More specialized equations may be derived from Eq(3) when one or more of

the variables is held constant. For example, you can easily derive the familiar equations given

below in this manner (convince yourself that this works!):

Boyle‟s law: p1V1 p2V2 (obtained when n1 = n2 and T1 = T2)

Charles‟s law:

V1

T1

V2

T2

(obtained when n1 = n2 and p1 = p2)

Avogadro‟s Principle:

V1

n1

V2

n2

(obtained when T1 = T2 and p1 = p2)

STP

Often you will see gas volumes reported at STP (standard temperature and pressure). STP is

defined as T = 273 K (0oC) and p = 1 atm. Substitution of these values into Eq(1) shows that

the volume of 1 mol of any gas is approximately 22.4 L at STP. (You should verify this for

yourself using Eq(1)!).

Gas Density (d) and Molar Mass (M)

Rearranging the ideal gas equation and using the definitions of density d and molar mass M

gives

n(mol)

V(L)

p

RT and d

g

L

n(mol)

V(L) M

g

mol

pM

RT (4)

Note: M (in italics) is molar mass in g/mol, while M (no italics) is molarity in g/L

___________________________________________________________________________

1. What is the volume occupied by 35.4 g of nitrogen gas at 35oC and 735 Torr?

Answer: 33.0 L

___________________________________________________________________________

2. A scuba diver inhales a lung-full (350 mL) of air at a depth of 33 ft where the pressure is

approximately 2.0 atm and the water temperature is 18oC. If the diver holds her breath (not a

good idea!!), what volume will the same amount of air occupy at sea level where the pressure is

approximately 1.0 atm and the air temperature is 35oC?

20

Answer: 741 mL (Note: You can either use Eq(3) or you can use Eq(1) twice to solve this

problem.)

___________________________________________________________________________

3. (a) Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide gas. If 35.3 g

of calcium carbonate reacts with 100 mL of 6.00 M HCl, how many liters of carbon dioxide gas

will be produced at 745 mmHg and 23.0oC?

Hints:

• Begin by writing the balanced equation for the reaction.

• This is a limiting reagent problem (why?), so you will next need to determine whether

calcium carbonate or hydrochloric acid is the limiting reagent.

• Once you have determined the identity of the limiting reagent, you can calculate the moles of

carbon dioxide produced.

• The last step is to find the volume of carbon dioxide using the ideal gas law.

(b) What volume of carbon dioxide gas would be obtained at STP?

Answers: (a) 7.43 L of carbon dioxide; (b) 6.72 L of carbon dioxide

Ideal Gas Mixtures

Dalton’s Law of Partial Pressures

The partial pressure of a gas in a mixture is the pressure it would exert if alone in the

container. Dalton’s law of partial pressures says that the total pressure of a mixture of gases is

the sum of the partial pressures. Thus, for a mixture of n gases,

pTotal p1 p2 p3 ... pi

i1

n

(1)

where

pi ni

RT

V

, pTotal nTotal

RT

V

and nTotal n1 n2 n3 ... nii1

n

(2)

It follows from Eq(2) that

21

pi

pTotal

ni(RT V )

nTotal(RT V )

ni

nTotal

Xi (3)

where Xi is the mole fraction of component i.

Vapor Pressure, Relative Humidity, and the Dew Point

A gas commonly present in gas mixtures is water vapor, H2O(g), which exerts a partial

pressure, pH 2O . The maximum value of pH 2O at a given temperature is the vapor pressure of

water. The vapor pressure of a gas is the pressure exerted by a vapor in equilibrium with its

liquid in a closed container. In experiments in which a gas is collected over water, pH 2O

contributes to the total gas pressure in the container (see exercise (1) below).

Weather reports often give the relative humidity which is the percent of the equilibrium vapor

pressure at the reported temperature which is actually present in the atmosphere:

Relative Humidity = pH2O

(vapor pressure)100% (4)

Since the vapor pressure increases as the temperature increases, a relative humidity of 90%

indicates a much higher pH 2O on a hot day in summer than on a cold day in winter! As the

temperature drops for a given pH 2O , the relative humidity increases (since the vapor pressure

decreases as T decreases); when the relative humidity reaches 100% the dew point has been

reached, and water vapor begins to condense (as “dew” or “frost”).

Average Molecular Speed (Root-Mean-Square Speed)

At a given temperature a molecule of any gas has the same average kinetic energy,

KE 1

2mu

2. This implies that gas molecules with low molar masses must have higher average

speeds u2

than those with high molar masses (why?). The average (root-mean-square) speed of a

gas molecule with molar mass M may be calculated at a temperature T :

average speed = root-mean-square speed = urms u2

3RT

M (5)

Units: We want to calculate the average speed in units of meters per second (m s-1

). To do this

the gas constant R is expressed in energy units (R = 8.314 J K-1

mol-1

where 1 J = 1 kg m2 s

-2),

and the molar mass M is expressed in units of kilograms per mole (kg mol-1

).

It follows from Eq (5) that gases with low molar masses have higher average speeds and

hence higher rates of effusion and diffusion than gases with higher molar masses. (The

spreading of one substance through another substance is called diffusion. The escape of one

substance through a small hole is called effusion.) Gases with low molar masses leak easily out

of containers and, if they have an odor, the smell will quickly permeate a room.

22

_____________________________________________________________________________

Exercises.

1. A mixture of NO(g) and NO2(g) was collected over water at 28.0oC and 745 mmHg (i.e.,

pTotal = 745 mmHg). At 28.0oC the vapor pressure of water is 28.3 mmHg.

(a) If pNO = 368 mmHg, what is the partial pressure of NO2(g) in the mixture?

(b) Calculate the mole fraction of NO2(g) in the mixture.

Answer: (a) partial pressure of NO2 XNO 2 = 0.468

_____________________________________________________________________________

2. Two flasks at the same temperature are joined by a glass tube with a stopcock. Flask A is a

4.0 L flask containing N2(g) at 2.0 atm, while flask B is a 10.0 L flask containing CO(g) at 1.4

atm. What is the final pressure in the flasks after the stopcock is opened?

Hints:

• Determine the final volume for the gases (easy!).

• Use Boyle's law (why?) to find the final partial pressure for both N2 and CO individually.

• Finally, use Dalton's law of partial pressures to find the total final pressure.

Answer: After the stopcock is opened, p(N2)= 0.57 atm, p(CO)= 1.0 atm, and pTotal = 1.57 atm

tm

______________________________________________________________________________

3. Calculate the root-mean-square speeds of He and benzene (C6H6) at 25oC. Give two reasons

why He will leak more readily than benzene out of a minute opening in a container.

Answer: He: u2

= 1.36 x 103 m s

-1; Benzene: u

2= 3.08 x 10

2 m s

-1; He atoms are much smaller

than benzene molecules and will leak out of smaller openings, and u2

is much larger for He than

for benzene.

Calorimetry Exercises ____________________________________________________________________________

1. When 12.29 g of finely divided brass (60% Cu, 40% Zn) at 95.0oC is quickly stirred into

40.00 g of water at 22.0oC in a calorimeter, the water temperature rises to 24.0

oC. Find the

specific heat of brass.

Hints:

• The heat lost by the brass is gained by the surroundings (the water plus the calorimeter).

What relation can you therefore write between qbrass and qsurr?

• Since no information is given about the heat capacity of the calorimeter, you should assume it

23

is negligible.

• The final temperature of the brass is the same as the final temperature of the water. The

specific heat of water, s(H2O), is 4.184 J g-1

oC

-1.

Answers: qsurr = q(H2O) = 334.7 J; qbrass = - 334.7 J; sbrass = 0.38 J g-1

oC

-1.

______________________________________________________________________________

2. In an experiment, 400. mL of 0.600 M HNO3(aq) is mixed with 400. mL of 0.300 M

Ba(OH)2(aq) in a constant-pressure calorimeter having a heat capacity of 387 J/oC. The initial

temperature of both solutions is the same at 18.88oC, and the final temperature of the mixed

solution is 22.49oC. Calculate the heat of neutralization in kJ per mole of HNO3.

Hints:

The heat evolved in the neutralization reaction is gained by the surroundings (the mixed

solution plus the calorimeter). What relation can you therefore write between qrxn and qsurr?

There are two contributions to qsurr. What are they? What assumptions (if any) need to be

made in calculating these contributions?

Is this a limiting reagent problem, or are reactants supplied in the stoichiometric ratio given

by the equation? (Why do we care about this?)

We want our answer in kJ per mole of HNO3. How do we calculate that?

Answers: ∆T = 3.61oC; qsolution = 12083.4 J; qcalorimeter = 1397.1 J; qsurr = 13481 J; qrxn = - 13481 J;

neut(kJ/mol HNO3) = - 56.2 kJ/ mol HNO3.

The Enthalpy of Chemical Change:

Calculations using Hess's Law and Heats of Formation

Enthalpy of reaction values have been determined experimentally for numerous reactions,

and these ∆H values may be used to calculate ∆H values for other reactions involving the same

chemical species. The reason this is possible is that enthalpy H is a state property so ∆H is

independent of path. (Similarly, the height of a mountain above sea level is independent of the

path you follow to climb the mountain.) Because ∆H is independent of path, we can determine

the enthalpy of foods by burning them in a bomb calorimeter in the laboratory to produce the

same products that are obtained by the complicated metabolic pathways in our body!

There are two principle methods used to calculate ∆H values for a reaction, both of which are

based on the idea that ∆H for a reaction is independent of the path used to go from reactants to

products. The first makes use of Hess's Law while the second employs tabulated heats of

formation H fo

(kJ/mol).

Use of Hess's Law to Calculate ∆H

Hess's Law states that ∆H for a reaction can be found indirectly by summing ∆H values for

any set of reactions which sum to the desired reaction. Usually before reactions are added

together, some of them must be reversed and/or multiplied by a factor n in order that they sum to

the desired reaction. In this process the rules are:

24

• Whenever you multiply a reaction by n, ∆H for the reaction is also multiplied by n.

• If you reverse a reaction, ∆H changes sign.

Problem (1) below is an example of how this procedure is used.

Use of Tabulated Heats of Formation to Calculate ∆H

The Standard Heat of Formation H fo

(kJ/mol) for a compound is the heat absorbed (or

released) in forming one mole of the compound from its elements in their standard states at 1 bar

(≈ 1 atm) pressure and the specified temperature (usually 25oC). Thus H fo

(kJ/mol) for acetone

CH3COCH3 is the heat of the reaction

3 C(s) + 3 H2(g) + 1/2 O2(g) CH3COCH3(l) ∆H = – 246.8 kJ (1)

and so H fo

(CH3COCH3(l)) = – 246.8 kJ/mol). By definition, H fo

(kJ/mol) = 0 for any

element in its standard state at 25oC and 1 bar.

Tabulated heats of formation H fo

(kJ/mol) are given in Table 6.4, p. 238, and Appendix 3, pp.

A-8 to A-12, in Chang. These may be used to calculate the standard enthalpy change, Hrxn

o, for

any reaction for which the heats of formation of all reactants and products are known:

Hrxn

o nprodH f

o( prod)

products

nreactH f

o(react)

reactants

(2)

This equation tells us to sum the enthalpies of formation of each product multiplied by its

stoichiometric coefficient in the reaction equation and then to subtract the enthalpy of formation

of each reactant multiplied by its stoichiometric coefficient. We use this equation to work

problem (2).

Hrxn

o applies for a balanced equation with specific stoichiometric amounts. If a different

number of moles reacts, the heat absorbed or evolved will change proportionately (problem 3).

______________________________________________________________________________

1. From the following heats of reaction

2 SO2(g) + O2(g) 2 SO3(g) ∆H = – 196 kJ (a)

2 S(s) + 3 O2(g) 2 SO3 (g) ∆H = – 790 kJ (b)

calculate the heat of reaction for

S(s) + O2(g) SO2(g) ∆H = ? kJ (c)

Method: Use Hess‟s Law to solve this problem:

• Identify a species in the target equation (c) which is on the correct side in only one of the

listed equations, (a) or (b). Multiply the entire equation, and its ∆H value by the factor n

necessary to make the stoichiometric coefficient for the species identical to that in equation

(c).

• Reverse a listed equation, (a) or (b), and change the sign of its ∆H value if it contains a

species which is on the wrong side of the target equation (c); next multiply the entire reversed

25

equation by the factor n necessary to make the stoichiometric coefficient for the species

identical to that in equation (c). The ∆H value for the rewritten equation is ( - n) times that of

the original equation. (Ignore any species present in both equations (a) and (b).)

• Test to see if your rewritten equations now sum to the desired equation (c). If they do, the ∆H

value for equation (c) is the sum of the values of the rewritten equations.

Answer: (1/2) Eq(b) + ( - 1/2) Eq(a) = Eq(c); Thus, (1/2) ∆H (b) + ( – 1/2) ∆H (a) = ∆H (c) so

∆H(c) =(1/2)( – 790 kJ) + ( – 1/2)( – 196 kJ) = kJ

______________________________________________________________________________

2. Calculate the standard reaction enthalpy for the photosynthesis reaction,

6 CO2(g) + 6 H2O(l 6H12O6(s) + 6 O2(g) Hrxn

o = ? kJ

Note: The heat of formation of glucose, C6H12O6(s), is given in Appendix 3 on pg A-12.

Answer: Use Eq (2) with H fo

(CO2(g)) = – 393.5 kJ/mol, H fo

(H2O(l)) = – 285.8 kJ/mol, H fo

(C6H12O6(s)) = – 1274.5 kJ/mol, and H fo

(O2(g)) = 0 kJ/mol. Thus ∆Ho = 2801.3 kJ ≈ 2801

kJ for the photosynthesis reaction.

______________________________________________________________________________

3. Is the photosynthesis reaction above endothermic or exothermic? How much heat is absorbed

or evolved if 11.0 g of CO2(g) reacts completely with excess water to form glucose and oxygen

gas?

Answers: Endothermic. 117 kJ of heat is absorbed.

Electromagnetic Radiation and the Spectrum of Atomic Hydrogen

To make sense of the chemistry of the elements we need to understand the electronic

structure of atoms. It is the atom‟s electronic structure which governs everything from molecular

geometry to chemical reactivity. Electromagnetic radiation is the probe we use to obtain

knowledge of electronic structure, so we begin by looking at some of its properties.

Light has a dual nature – it is both wave-like and particle-like. Thus light and all other forms

of electromagnetic radiation obey two equations, one of which shows the inverse relation

between wavelength and frequency (both properties of waves) and the other which relates the

energy of light photons (“particles”) to their frequency :

Wave model (m) (s1

) c(m s) (1)

Particle model Ephoton(J) h(J s) (s1

) (2)

The constants and variables in these equations are

26

Speed of light = c = 3.00 x 108 m/s = 3.00 x 108 m s-1

Planck's constant = h = 6.626 x 10-34 J s

Wavelength = = Greek lower case lambda

Frequency = = Greek lower case nu

Photon Energy = Ephoto n

Common units include

Wavelength : 1 nm = 10-9 m = 10 Å

Frequency : 1 s-1 = 1 Hz

The H Atom Emission Spectrum

Our modern theory of the electronic structure of atoms began with attempts to understand the

emission spectrum of the simplest atom – hydrogen. By the end of the nineteenth century it was

established that the frequencies of the spectral lines in the H atom emission spectrum all fit a

very simple formula (see Eq (6) ahead). The famous Bohr model of the H atom proposed by Bohr

in 1914 was an attempt to understand where this formula came from. Bohr recognized that the

frequencies would be predicted correctly if the H atom energy levels obeyed the equation

En RH

1

n2

for n = 1, 2, 3, ... (Applies only to H atom!!) (3)

where

RH = Rydberg constant = 2.18 x 10-18 J

While the Bohr model was incorrect in assuming that the electrons orbit the nucleus like planets

orbit the sun, it was correct in predicting the energies of the H atom energy levels.

Absorption and Emission

Absorption of electromagnetic radiation occurs when electrons make transitions from lower

to higher energy levels. Photons provide the energy required for the jumps. Similarly, when

electrons move from higher to lower energy levels, photons carry off the excess energy:

electromagnetic radiation is emitted and an emission spectrum is obtained. In absorption and

emission, photon energies and frequencies obey the equations

Energy Gained in Absorption Ephoton h Eupper Elower (4)

Energy Lost in Emission Ephoton h Eupper Elower (5)

These equations apply to all atoms and molecules.

In the case of the H atom Eq (3) gives the energies of Eupper and Elower once the quantum

numbers, nupper and nlower are specified. Using these values in Eq (5) allowed Bohr to predict the

correct frequencies for the emission lines of the H atom:

E photon h RH 1

nlower

2

1

nupper

2

(Applies only to the H atom!!) (6)

27

______________________________________________________________________________

1. A violet photon has a frequency of 7.100 x 1014 Hz. (a) What is the wavelength (in nm) of

the photon? (b) What is the wavelength in Å ? (c) What is the energy of the photon? (d) What is

the energy of 1 mole of these violet photons?

Answer: (a) 422.5 nm ; (b) 4225 Å ; (c) 4.704 x 10-19 J ; (d) 283.3 kJ

______________________________________________________________________________

2. The lines in the visible region of the emission spectrum of the H atom are called the Balmer

lines. All of them involve transitions in which nlower = 2. Calculate (a) the frequency and (b) the

wavelength of the lowest energy Balmer emission line. (c) What color is this line?

Hint: Use Eq (6) with nupper = 3 (why?); RH = Rydberg constant = 2.18 x 10-18 J.

Answer: (a) 4.57 x 1014 s-1; (b) 656 nm; (c) It‟s the Balmer red line! (see Chang, pg 267).

______________________________________________________________________________

3. Derive Eq (6) from Eq (5) for the H atom.

Hint: First use Eq (3) to calculate Eupper and Elower in terms of nupper and nlower

respectively. Then substitute these expressions into Eq (5) and rearrange.

Quantum Numbers, Orbitals, and Electron Configurations

Quantum Mechanics and the H Atom

Bohr theory (1914) ran into problems when it was applied to atoms other than H, and was

soon replaced by quantum mechanics (1926). When quantum mechanics was applied to the H

atom, the same energies were calculated for the H atom energy levels as in the Bohr model,

En RH

1

n2

for n = 1, 2, 3, ... (Applies only to H atom!!) (1)

but the method by which they were calculated was entirely different.

Quantum mechanics assumes the electron is wave-like, and the planetary orbits of Bohr

theory (which views the electron as a particle) are discarded. Orbitals, which give us a picture of

the most probable locations for the electron in a particular energy state, replace the Bohr orbits.

Since the electron is viewed as a wave, it is impossible to describe its precise location: an

averaged picture is the best we can do! Another change is that while the quantum number n still

plays the prominent role, it is augmented by the quantum numbers l, ml, and ms. The allowed

quantum numbers for the H atom are given in the tables below.

28

Table 1.

Quantum

Number

Allowed Values Name and Meaning

n n = 1, 2, 3, ...... Principal quantum number: orbital energy

and size.

l l = (n-1), (n-2), ...., 0 Azimuthal (or orbital) quantum number:

orbital shape (and energy in a multi-electron

atom), letter name for subshell (s, p, d, f)

ml ml = l, (l-1), ..., 0, ..., (-l+1), -l Magnetic quantum number: orbital

orientation

ms ms = 1/2, -1/2 Electron spin quantum number: spin up ( )

or spin down ( ).

Table 2

l Value

Letter Equivalent

to l Value

No. of

Orbitals

in Set

Approximate Shape of Orbitals

with Specific l Values

0 s 1 spherical

1 p 3 px, py, pz are dumbbells along x, y, and z

axes

2 d 5 mostly cloverleaf shapes

3 f 7 very complicated shapes!

Table 3

Shell

(n)

Subshell

(l)

Orbital

Name (nl)

Orientations

(ml)

No. of

Orbitals

Maximum

Occupancy

n = 1 l = 0 1s ml = 0 1 2 e-

n = 2 l = 0 2s ml = 0 1 2 e-

l = 1 2p ml = 1, 0 -1

(or px, py, pz)

3 6 e-

n = 3 l = 0 3s ml = 0 1 2 e-

l = 1 3p ml = 1, 0, -1

(or px, py, pz)

3 6 e-

l = 2 3d ml = 2, 1, 0, -1, -2

(or dxy, dyz, dxz, dx2 y 2

, dz2

)

5 10 e-

Orbital Energies and Electron Configurations of Multi-Electron Atoms

For the H atom the orbital energy depends only on n, so all orbitals with the same value of n

have the same energy. This is not true, however, for any other atom!

The H atom orbitals may be used to approximate the orbitals for multi-electron atoms. But

since these atoms have more than one electron, electrons in the outer orbitals are shielded

somewhat from the nucleus: they do not feel the full nuclear charge. Orbitals with a lower l

value penetrate closer to the nucleus and are less shielded and have a lower energy than those

with a higher l value. The result is that for a given value of n the energy order is s < p < d < f.

29

Orbitals are filled from lowest energy to highest energy. Each orbital holds 2 electrons (Pauli

Exclusion Principle), one with spin up ( ) and one with spin down ( ). If more than one

orbital has the same energy (e.g., px, py, pz ), electrons first enter different orbitals with spins

parallel (Hund’s Rule); once each orbital in the set contains one electron, additional electrons

form pairs.

The order of orbital filling is easily remembered if correlated with the periodic chart. The

order is 1s (first row of chart); 2s, 2p (second row); 3s, 3p (third row); 4s, 3d, 4p (fourth row); 5s,

4d, 5p (fifth row); 6s, 4f, 5d, 6p (sixth row); 7s, 5f, 6d. There are a few exceptions to these rules.

______________________________________________________________________________

1. (a) List all the orbitals for which n = 4. (b) How many orbitals are there in all?

Answer: (a) When n = 4, allowed values of l are: l = 0=s, l = 1=p, l = 2=d, and l = 3=f. Thus we

have one 4s, three 4p, five 4d, and seven 4f orbitals. See Table 1 on p. 1 for the allowed ml

values for each value of l. For example, when l = 3, ml = 3, 2, 1, 0, -1, -2, -3. (b) 1 + 3 + 5 + 7

= 16 orbitals (each holds 2 e-).

______________________________________________________________________________

2. Which of the following subshells cannot exist: (a) 1p; (b) 4f; (c) 2d; (d) 5p; (e) 3f? Why not?

Answer: (a), (c), (e) [See Table 1 for the relation between the allowed values of l and value of n]

______________________________________________________________________________

3. List all possible values of ml for each of the indicated subshells. What role does the principal

quantum number n play in determining your answer?

Subshell Values of ml

(a) 4s

(b) 2p

(c) 3d

(d) 5f

Answer: (a) 0; (b) 1, 0, -1; (c) 2, 1, 0, -1, -2; (d) 3, 2, 1, 0, -1, -2, -3; Principal quantum number

plays no role: answers do not depend on n

______________________________________________________________________________

4. Give the formula that relates the number of possible values of ml to the value of l.

Answer: 2l +1 (why?)

30

Periodic Trends

Here we summarize trends for the main group elements (Columns 1A - 8A). Trends for the

transition metals, the lanthanides, and the actinides may differ.

Sizes of Atoms and Ions

Neutral Atoms (or Ions with the Same Charge).

• Size increases as you go down a column. Why? As you go down a column, electrons are

filling orbitals farther and farther out from the nucleus. Each row adds a new shell. Outer

electrons are shielded from the nucleus by electrons in inner shells; thus they are less tightly

held (in spite of the much increased nuclear charge).

• Size decreases as you go across a row. In this case electrons are being added to the same

shell. Thus they experience little additional shielding. On the other hand, the nuclear charge

of the atom increases with the atomic number. Thus as you go across a row, the electrons are

held more tightly and the size decreases.

Isoelectronic Series. These are series of atoms and ions in which the number of electrons stays

constant, but the number of protons increases with the atomic number. In this type of series, the

size of the atom decreases as the number of protons increases. The reason for the size decrease

is that more protons are pulling in the same number of electrons. Examples include the series

below in which the largest member of the series is listed first:

10 electron series: Ne > Na+ > Mg2+ > Al3+

18 electron series: P3- > S2- > Cl – > Ar

Cation Size as Compared to Parent Atom. The size decreases when cations form. The effect is

particularly pronounced when all the valence electrons are lost and only the noble gas core of

electrons remains. For example, the Mg2+ ion (65 pm radius) is considerably smaller than the

Mg atom (160 pm radius).

Anion Size as Compared to Parent Atom. The size increases when anions form. The added

electrons are going into the same shell. They repel each other and so the size increases. Thus the

Cl– ion (181 pm radius) is considerably larger than the Cl atom (99 pm radius).

Ionization Energies

The ionization energy I is the minimum energy needed to remove an electron from the

ground state of a gaseous atom, A(g).

A(g) A+(g) + e–(g) ∆E = I = I1

More precisely, this is the first ionization energy I1. Additional electrons may be removed with

ionization energies I2, I3, etc., for the removal of the second, third, etc., electrons. Ionization is

always an endothermic process: it requires energy to remove an electron from an atom or ion.

The overall trends in ionization energy are opposite to those for atomic and ionic radii. The

more tightly electrons are held, the higher the ionization energy, and the smaller the atom or ion

size. Some generalities are as follows:

• Noble gases have the highest ionization energies of the atoms in each row.

• Alkali metals have the lowest ionization energies of the atoms in each row.

31

• In general, ionization energies increase as you go across a row, but there are a few local ups

and downs. Dips occur with the loss of the first and the fourth p electron: Thus in the second

row, there are dips for boron and for oxygen.

• The ionization energy decreases for atoms as you go down a column.

• Higher ionization energies are always larger than lower ionization energies: I1< I2< I3 , etc.

A huge jump in ionization energy occurs when you first pull an electron out of the noble gas core.

Electron Affinity

The electron affinity EA is the energy released when an electron is added to a gas-phase

atom (or ion) of the element. The sign convention is opposite to that for ∆H. If the process is

exothermic, ∆H is (-) and EA is (+); if it is endothermic, ∆H is (+) and EA is (-).

A(g) + e–(g) A–(g) ∆H = - EA

While ionization energies are always positive numbers, electron affinities can be either positive

or negative. A high positive EA (and thus a (-) value of ∆H) indicates that gaining an electron is

a very favorable process. The halogens have the most positive electron affinities of all the

elements.

Electronegativity

The electronegativity (Greek letter chi) is a measure of the ability of an atom to attract and

hold electrons. Elements that readily form negative ions have high electronegativities, while a

low electronegativity correlates with the tendency to lose electrons and form positive ions.

Values of range from a high of = 4.0 for F to a low of = 0.7 for Cs. In general

electronegativities increase diagonally from the lower left (Cs) to the upper right (F) of the

periodic chart.

In practice, chemists use electronegativities far more than ionization energies or electron

affinities.

______________________________________________________________________________

Exercises:

1. In each of the following pairs, circle the species with the higher first ionization energy:

(a) Li or Cs (b) Cl- or Ar (c) Ca or Br (d) Na+ or Ne (e) B or Be

______________________________________________________________________________

2. In each of the following pairs, circle the species with the larger atomic radius:

(a) Mg or Ba (b) S or S2- (c) Cu+2 or Cu (d) He or H- (e) Na or Cl

______________________________________________________________________________

3. Circle the best choice in each list:

(a) highest first ionization energy: C, N, Si

(b) largest radius: S2–, Cl–, Cl

(c) highest electronegativity: As, Sn, S

32

(d) smallest atom: Na, Li, Be

(e) most paramagnetic: Fe, Co, Ni

(f) lowest first ionization energy: K, Na, Ca

(g) highest second ionization energy: Na, Mg, Al

(h) lowest second ionization energy: Ar, K, Ca

______________________________________________________________________________

Answers (be sure you can explain the reason for each answer!):

1. (a) Li; (b) Ar (isoelectronic pair); (c) Br; (d) Na+ (isoelectronic pair); (e) Be (common

exception: what is the rule here?).

2. (a) Ba; (b) S2-; (c) Cu; (d) H- (isoelectronic pair); (e) Na.

3. (a) N; (b) S2- (S2- and Cl- are isoelectronic); (c) S; (d) Be; (e) Fe (hint: determine no. of

unpaired spins for each element); (f) K; (g) Na; (h) Ca.

33

sp

C Atom

H

C

HH

H

2s 2p

Promotion Step

C Atom

2s 2p

Hybridization step( leads to energy release after bonding, since more bonds can be formed)

sp3

For 4 groups: tetrahedral

electron pair geometry

C C

Hybridization of Carbon

For 3 groups: trigonal planar


H

H

H

H

For 2 groups: linear


sp2 pz

C

py

leftover

H

H

sp py, pzTwo hybridized

AO'sleftover

sp

Example: CH 4

Use to form

3 single ()

bonds

Use pz to

form 1

pi () bond

H C

Example: C 2H4

bond

C

Each C forms three bonds and

one bond. Bond angles are 120 o.

H

xy plane

Use to form

2 single ()

bonds

Use py

and pz to

form 2 pi()

bonds

Example: C 2H2

Each C forms two bonds

and two bonds (which are

perpendicular to each other).

Bond angles are 180 o.

( requires energy)

Three hybridized AO's

Four hybridized AO's

pzxy plane

C

H

pz

H

C CH H

Use to form 4 single

(sigma = ) bonds

sp2

sp2sp2

sp3 sp3

sp3

sp3

C forms four bonds.

Bond angles are 109.5 o.

34

Total #

of

Groups of

e-

Electron Pair

Geometry

(Hybridization)

Approximate

Bond Angle

# of

Bonding

Directions

(# of X)

# of

Lone Pairs

(# of E)

Geometry Name

(VSEPR class)

Shape

Examples

2 linear

(sp)

180o 2 0 linear

(AX2)

BeH2,

CO2

3

trigonal planar

(sp2)

120o

3

0

trigonal planar

(AX3)

BF3, NO3–

2 1 bent

(AX2E)

SO2

4

tetrahedral

(sp3)

109.5o

4

0

tetrahedral

(AX4)

CH4

3 1 trigonal

pyramidal

(AX3E)

NH3

2 2 bent

(AX2E2)

H2O

5

trigonal

bipyramidal

(sp3d)

120o (in plane) &

90o (above &

below)

5

0

trigonal

bipyramidal

(AX5)

PCl5

4 1

seesaw

(AX4E)

SF4

3 2

T-shaped

(AX3E2)

ClF3

2 3

linear

(AX2E3)

XeF2

6

octahedral

(sp3d2)

90o

6

0

octahedral

(AX6)

SF6

5 1 square

pyramidal

(AX5E)

BrF5

35

4 2 square planar

(AX4E2)

XeF4

3 3 T-shaped

(AX3E3)

2 4 linear

(AX2E4)

Predicting Molecular Geometry and Hybridization

1. In each case, predict (a) the approximate bond angle(s), (b) the hybridization around the

underlined atom. (Note: It is helpful to first sketch the Lewis structure!)

Molecule or Ion (1) OF2 (2) H2CO (3) NO2+ (4) BF3 (5) SbF5

(a) No. of valence e -

„s

(b) Lewis structure

(c) Approximate

bond angle(s)

(d) Hybridization

(e) Polar or non-

polar molecule?

Ion: Not

applicable

(f) Geometry name

______________________________________________________________________________

2. For each of the molecules below fill in the indicated items in the chart. The central atoms are

underlined.

Molecule (1) SO2 (2) HBF2 (3) XeF4 (4) CH2Cl2 (5) NF3

(a) No. of valence e

- „s

(b) Lewis structure

(c) Approximate

bond angle(s)

36

(d) Hybridization

(e) Polar or non-

polar molecule?

(f) Geometry name

____________________________________________________________________________

3. Predict (a) the approximate bond angle, (b) the hybridization around the indicated atoms (the

atoms to which the arrows are drawn in the structures below). Write your answers near the

corresponding labels (1 to 5) in the drawings. (Note: the lone pairs on the F atoms are omitted.)

N

C

N

C

C

C

N

C

N

N

O

H

H H

H

H

3

4

5F S

F F

F

1

F Br

F F

F

F 2

37

Answers:

1. In each case, predict (a) the approximate bond angle(s), (b) the hybridization around the

underlined atom. (Note: It is helpful to first sketch the Lewis structure!)

Molecule or Ion (1) OF2 (2) H2CO (3) NO2+ (4) BF3 (5) SbF5

(a) No. of valence e -

„s

20 12 16 24 40

(b) Lewis structure

(c) Approximate

bond angle(s)

109.5o 120

o 180

o 120

o 90

o, 120

o

(d) Hybridization

sp3 sp

2 sp sp

2 sp

3d

(e) Polar or non-

polar molecule?

polar polar Ion: Not

applicable

non-polar non-polar

(f) Geometry name

bent trigonal planar linear trigonal

planar

trigonal

bypyramidal

______________________________________________________________________________

2. For each of the molecules below fill in the indicated items in the chart. The central atoms are

underlined.

Molecule (1) SO2 (2) HBF2 (3) XeF4 (4) CH2Cl2 (5) NF3

(a) No. of valence e

- „s

18 18 36 20 26

(b) Lewis structure

(c) Approximate

bond angle(s)

120o 120

o 90

o 109.5

o 109.5

o

(d) Hybridization

sp2 sp

2 sp

3d

2 sp

3 sp

3

(e) Polar or non-

polar molecule?

polar polar non-polar polar polar

(f) Geometry name

bent trigonal

planar

square

planar

tetrahedral trigonal

pyramidal

____________________________________________________________________________

3. Predict (a) the approximate bond angle, (b) the hybridization around the indicated atoms (the

atoms to which the arrows are drawn in the structures below). Write your answers near the

corresponding labels (1 to 5) in the drawings. (Note: the lone pairs on the F atoms are omitted.)

38

(1) 90

o, 120

o; sp

3d (2) 90

o; sp

3d

2 (3) 109.5

o; sp

3 (4) 120

o; sp

2 (5) 109.5

o; sp

3

N

C

N

C

C

C

N

C

N

N

O

H

H H

H

H

3

4

5F S

F F

F

1

F Br

F F

F

F 2

39

Note: The MO order for 2nd

row diatomics illustrated above is correct for Li2 through N2;

however, for O2 and F2 2p 2p MO so the order of these

MOs in the diagram above should be switched. This does not affect the bond order calculation.

MO Energy Level Diagrams

1st Row Homonuclear Diatomics

Energy

1sA 1sB

1s

*1s

Atom A Atom B

Molecule A-B

90% Contours of Orbitals

+

A B

1sA 1sB

node

*1s

anti-bonding MO

+

1sA 1sB

A B A B

1s bonding MO

2nd Row Homonuclear Diatomics

Energy

2pA2pB

*2p

*2p

2p

2p

2sA 2sB

*2s

2satom B

+

A B A B

*2p

anti-bonding MO

A

node

A B B

BA

2p

A B

*2p

bonding

anti-bonding

ABA

B

2p 2p bonding

Also: *2px, 2px formed from 2px + 2px

coordinate system

Bond Order = 1/2 (# of bonding electrons) ? 1/2 (# of anti-bonding electrons)

ZA ZB

A B

YA YB

XA XB 2p+z zA B

2p 2pz zA B

–

2pyA

2p yB

2py

2py

BA

z

+

+

–

+

+

A

molecule A-B

atom A

Electrons filled in for oxygen

+

–

node

xy plane

z

B

40

Intermolecular Forces

Types of Solids* Intermolecular Force(s) Between Particles

1. Metallic Crystals (Metals)

Examples: Na, Cu, Fe, Mn

Metallic bonding: Valence electrons form mobile sea of

electrons which comprise the metallic bond.

2. Ionic Crystals (Ionic Solids)

Examples: NaCl, MgCl2, MgO

Ionic Bonding: Attraction of charged ions for one another.

Lattice energy is a measure of ionic bond strength.

3. Covalent Crystals (Network

Solids)

Examples (small class!): C(diamond),

SiC(s), SiO2 (quartz)

Network covalent bonding. Network solids are extremely

hard compounds with very high melting and boiling points

due to their endless 3-dimensional network of covalent

bonds.

4. Molecular Crystals

Examples:

One or more of the following:

(a) Need H bonded to O, N or F: H2O,

HF, NH3.

(a) Hydrogen bonding: Hydrogen bonds are weaker than

covalent bonds, but stronger than (b) or (c) below.

(b) C6H6 (benzene), polyethylene, I2, F2,

and all the compounds from (a) above.

(b) Dispersion forces (induced dipole – induced dipole or

London dispersion forces): universal force of attraction

between instantaneous dipoles. These forces are weak for

small, low-molecular weight molecules, but large for

heavy, long, and/or highly polarizable molecules. They

usually dominate over (c) below.

(c) CHF3, CH3COCH3 (acetone) and

H2O, HF, NH3.

(c) Dipole-dipole forces: these forces act between polar

molecules. They are much weaker than hydrogen bonding.

Note: Van der Waals Forces is a category which includes both categories (b) and (c) above.

5. Atomic Crystals

Examples: He, Ne, Ar, Kr, Xe

Dispersion forces: See Section 4(b) above.

*Note: Many of the compounds given as examples are not solids at room temperature. But if

you cool them down to a low enough temperature, eventually they will become solids.

Physical properties depend on these forces. The stronger the forces between the particles,

(a) the higher the melting point.

(b) the higher the boiling point.

(c) the lower the vapor pressure (partial pressure of vapor in equilibrium with liquid or solid in a

closed container at a fixed temperature).

(d) the higher the viscosity (resistance to flow).

(e) the greater the surface tension (resistance to an increase in surface area).

41

______________________________________________________________________________

1. What type of crystal will each of the following substances form in its solid state? Choices to

consider are metallic, ionic, covalent, or molecular crystals.

(a) C2H6 __________ (b) Na2O ____________ (c) SiO2 ______________

(d) CO2 ______________ (e) N2O5 __________ (f) NaNO3 ______________

(g) Al ________________ (h) C(diamond) ______ (I) SO2 ________________

______________________________________________________________________________

2. Circle all the compounds in the following list which would be expected to form intermolecular

hydrogen bonds in the liquid state:

(a) CH3OCH3 (b) CH4 (c) HF (d) CH3CO2H (e) Br2 (f) CH3OH

(dimethyl ether) (acetic acid) (methanol)

______________________________________________________________________________

3. Specify the predominant intermolecular force involved for each substance in the space

immediately following the substance. Then in the last column, indicate which member of the

pair you would expect to have the higher boiling point.

Substance #1

Predominant

Intermolecular

Force

Substance #2

Predominant

Intermolecular

Force

Substance with

Higher Boiling

Point

(a) HCl(g) I2

(b) CH3F CH3OH

(c) H2O H2S

(d) SiO2 SO2

(e) Fe Kr

(f) CH3OH CuO

(g) NH3 CH4

(h) HCl(g) NaCl

(i) SiC Cu

Answers:

1. (a) molecular; (b) ionic; (c) covalent (network solid); (d) molecular; (e) molecular; (f) ionic;

(g) metallic; (h) covalent (network solid); (i) molecular.

2. Hint: Molecule must contain H bonded to O, N, or F, since only H bonded to O, N, or F can

form a hydrogen bond with an O, N, or F on another molecule. Thus (c), (d), and (f) should be

circled.

3. Hint: Choices for the predominant intermolecular force are metallic bonding, ionic bonding,

network covalent bonding, hydrogen bonding, and dispersion forces (induced dipole – induced

dipole forces). Dipole-dipole forces are generally dominated by dispersion forces and are rarely

predominant.

(a) dispersion forces; dispersion forces; I2.

(b) dispersion forces; hydrogen bonding; CH3OH.

42

(c) hydrogen bonding; dispersion forces; H2O.

(d) network covalent bonding; dispersion forces; SiO2.

(e) metallic bonding; dispersion forces; Fe.

(f) hydrogen bonding; ionic bonding; CuO.

(g) hydrogen bonding dispersion forces; NH3.

(h) dispersion forces; ionic bonding; NaCl.

(i) network covalent bonding; metallic bonding; SiC.

Chemical Kinetics: Introductory Concepts

Chemical kinetics is the study of the rates of chemical reactions and the mechanisms by

which reactions occur. A rate is the change of a property (in this case, concentration) per unit

time. The rate of a chemical reaction is found by following the rate of disappearance (or

decomposition) of one of the reactants or the rate of appearance (or formation) of one of the

products.

Suppose we consider the reaction

N2(g) + 3 H2(g) 2 NH3(g) (1)

Since three H2 molecules react with one N2 molecule to produce two NH3 molecules, the rate of

disappearance of H2 will be three times the rate of disappearance of N2, and the rate of

appearance of NH3 will be twice the rate of disappearance of H2. Thus,

rate

=

rate of

disappearance of N2

=

1/3 the rate of

disappearance of H2

=

1/2 times the rate of

appearance of NH3

rate

= [N2 ]

t

=

1

3

[H 2]

t

=

1

2 [NH3]

t

rate

=

vN 2

=

1

3 vH2

=

1

2 vNH 3

(2)

Note that rates are positive numbers (time doesn‟t go backwards!). That is the reason for the

negative sign in the expressions for the rate of disappearance of N2 and H2 in the Eq (2); for

example, since [N2 ] is a negative number, we need to multiply [N2 ] t by (-1) to obtain a

positive rate.

In general, for the chemical equation

a A + b B c C + d D

the rate is given by

43

rate 1

a

A

t

1

b

B

t

1

c C

t

1

d D

t

rate v 1

avA

1

bvB

1

cvC

1

dvD

(3)

Since the rates of appearance and disappearance of all reactants and products are related by

the equation stoichiometry, it doesn‟t matter which rate we actually measure – experimental

convenience governs our choice. However, since the rates differ by stoichiometric ratios, we

must specify the substance for which our rate is defined.

Rate Laws

The rate of most reactions changes with time. Initially, when concentrations of reagents are

highest, the rate is fastest. As reactants are consumed and products form, the forward reaction

slows down and the backward reaction speeds up. Eventually, either the reaction reaches

equilibrium, or it goes to completion and the rate goes to zero. At equilibrium, the forward

reaction rate equals the backwards reaction rate, and there is no further net change in

concentration.

In many cases the rate of reaction obeys a simple equation known as the rate law. A rate law

is an equation expressing the reaction rate as the product of a rate constant k and the

concentrations of species involved in the reaction raised to various powers:

rate = v = k [A]m [B]n .... (4)

A fast reaction is characterized by a large value for k. As a reaction proceeds, its rate changes

with concentration according to the rate law, but k remains constant. The rate constant k depends

on temperature, but is independent of concentration.

The powers to which the concentrations are raised (m, n, ...) are typically positive integers

(but may be fractions or negative numbers). They may also be zero, but in that case the factor is

usually omitted from the rate law, since, for example, [A]o = 1.

Reaction Order

The overall order of a reaction is the sum of the exponents in the rate law (Eq (4)). The

order of reaction in A is the power to which [A] is raised in the rate law, and so on for B, C, etc.

If the concentration of a substance does not appear in the rate law, the reaction is zeroth order in

that species. For example, if the reaction

2 A + B + C D + 2E

has the rate law

rate = v = k [A] [B]2

44

the reaction is first order in [A], second order in [B], zeroth order in [C], and third order overall.

The rate law must be determined experimentally. The reaction orders have no necessary

relation to the stoichiometric coefficients in the equation for the reaction.

Units

The units of a rate are always M s-1

= mol L-1

s-1

. Since in the rate law equation (Eq (4)), the

units on the left must equal the units on the right, the units of k depend on the overall order. For

example, if a reaction is first-order overall, k will have units of s-1

; if it is second-order overall, k

will have units of L mol-1

s-1

; and so on.

______________________________________________________________________________

1. For the reaction, 2 N2O5(g) 4 NO2(g) + O2(g), the rate of formation of NO2(g) is

4.0 x 10-3

mol L-1

s-1

.

(a) Calculate the rate of disappearance of N2O5(g)

(b) Calculate the rate of appearance of O2(g).

Answers: (a) 2.0 x 10-3 mol L-1 s-1; (b) 1.0 x 10-3 mol L-1 s-1

Hint: Eq(3) gives

rate v 12 vN2O5

14 vNO2

vO2 so here

rate v 14 vNO2

= (1/4)( 4.0 x 10

-3 mol L

-1s

-1) = 1.0 x 10

-3 mol L

-1s

-1.

______________________________________________________________________________

2. The reaction 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) is found experimentally to be second

order in NO(g) and first-order in H2(g).

(a) Write the rate law for the reaction.

(b) What is the overall order of the reaction?

(c) What are the units for the rate constant k?

(d) If [NO] is doubled (while keeping [H2] constant), by what factor will the reaction rate

increase?

(e) If [H2] is doubled (while keeping [NO] constant), by what factor will the reaction rate

increase?

Answers: (a) rate = v = k [NO]2 [H2] ; (b) third order overall; (c) L

2 mol

-2 s

-1; (d) 4-fold; (e) rate

will double.

45

How to Determine the Rate Law from Experimental Data

There are two general methods to determine the rate law from experimental data. The first is the

method of initial rates which is appropriate when the reaction is relatively slow so that the initial

rate v0 may be determined as a function of initial concentrations. The second method, which

makes use of integrated rate equations, is applicable when the reaction is sufficiently fast so that

concentration versus time data may be collected over several half-lives. The half-life of a

substance is the time needed for its concentration to fall to one-half its initial concentration.

Method I: Method of Initial Rates

The reaction is assumed to have a rate law in the form nm BAk ][][v rate so that the initial rate

is given by

nm BAk 000 ][][v rate initial (1)

We want to find the order of the reaction m in

[A]0 and the order n in

[B]0 . Data is gathered for

experiments in which

[B]0 (the initial concentration of B) is kept constant and the initial rate v0 is

determined for different values of

[A]0 (the initial concentration of A). If

[B]0 is held constant

and

[A]0 is increased by the factor f, the rates for the two runs will be

nm BAk 001run ,0 ][][v (2)

nm BAfk 002run ,0 ][)][ (v (3)

Dividing Eq(3) by Eq(2) gives the ratio of the initial rates:

m

nm

nm

fBAk

BAfk

00

00

1run ,0

2run ,0

][][

][)][ (

v

v (4)

By comparing the factor f by which

[A]0 was increased with the ratio of the initial rates

fm

from

Eq(4), the reaction order in [A] can be determined as shown in the table below:

Table 1. Reaction Order Using the Method of Initial Rates

Reaction Order in [A] Eq(4) Ratio Initial Rate Result

zeroth order, m = 0

fm f

01 1run ,02run ,0 vv

first order, m = 1

fm f

1 f 1run ,02run ,0 vv f

second order, m = 2

fm f

2

1run ,0

2

2run ,0 vv f

Suppose that in going from run 1 to run 2,

[A]0 is doubled so f = 2. Then, for example, if the

initial rate is constant ( 1run ,02run ,0 vv ) the reaction is zeroth order in [A] (i.e., m = 0 in the rate

law). If 1run ,02run ,0 v 2v when

[A]0 is doubled, the reaction is first order in [A] (m = 1 in the

rate law). Finally, if 1run ,01run ,0

2

2run ,0 v 4 v2v when

[A]0 is doubled, the reaction is second

order in [A] (m = 2 in the rate law).

46

An analogous process may be used to find n, the order of the reaction in [B] using runs where

[A]0 is held constant but

[B]0 is varied.

In general, if the initial concentration of a substance is increased by a factor f (while all other

initial concentrations are held constant), the initial rate v0 will increase by f n if the reaction has

an order of n for the substance.

Experimentally, to find each initial rate v0 we need to make up a solution with the desired initial

concentrations,

[A]0 and [B]o. The concentration of a reactant (or product) is then measured as a

function of the time t. A graph of concentration versus time is made with the data, and the initial

rate v0 is found from the slope of the graph (which should be linear at this initial stage of the

reaction).

______________________________________________________________________________

Problem Using the Method of Initial Rates

1. The following data were obtained for the reaction A + B + C products:

Experiment

[A]0

[B]0

[C]0

Initial rate, v0

(mol L-1

s-1

)

1

1.25 x 10-3 M

1.25 x 10-3 M

1.25 x 10-3 M

0.0087

2 2.50 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 0.0174

3 1.25 x 10-3 M 3.02 x 10-3 M 1.25 x 10-3 M 0.0508

4 1.25 x 10-3 M 3.02 x 10-3 M 3.75 x 10-3 M 0.457

5 3.01 x 10-3 M 1.00 x 10-3 M 1.15 x 10-3 M ?

(a) Write the rate law for the reaction. Explain your reasoning in arriving at your rate law. [Hint:

Table 1 is useful here.]

(b) What is the overall order of the reaction?

(c) Determine the value of the rate constant. [Hint: Use your rate law from (a) and appropriate

data.]

(d) Use the data to predict the reaction rate for experiment 5.

Answers to Problem 1.

(a) 2

0

2

000 ][][][v CBAkrate .

In runs 1 and 2, [B]0 and [C]0 are constant; when [A]0 doubles (f = 2), r0 increases by

fn

= 2 = 21: thus 1

st order in [A]0.

In runs 1 and 3, [A]0 and [C]0 are constant; when [B]0 increases by the factor f = 2.416, v0

increases by a factor of

fn = 5.839 = (2.416)

2: thus 2

nd order in [B]0.

In runs 3 and 4, [A]0 and [B]0 are constant; when [C]0 increases by a factor of f = 3, v0

increases by a factor of

fn = 9 = 3

2: thus 2

nd order in [C]0.

(b) Fifth order overall.

(c) k = 2.85 x 1012 L4 mol-4 s-1. (Use the rate law from (a) above and the data from any run to

calculate k.

47

(d) rate = v0 = 0.0113 mol L-1 s-1.

______________________________________________________________________________

Method II: Use of the Integrated Rate Equations Rate equations are differential equations that in simple cases can be easily integrated using

standard methods of calculus. For example, if the rate law has the form

rate d[A]

dt k [A]

n (5)

the equation may be integrated (from t = 0 to t, and from [A] =

[A]0 to [A]t) for different reaction

orders in [A] (i.e., different values of n). The equations given in the table below are obtained for

n = 0, 1, and 2. The useful feature of these equations is that they may be written in straight-line

form, y = mx + b (where m = slope and b = y-intercept are constants).

Table 2. Reaction Order Using the Integrated Rate Equations

Order in

[A]

Rate

Law

Integrated Form,

y = mx + b

Straight

Line Plot

Half-Life

t1/2

zeroth

order

(n = 0)

rate = k [A]o=

k

[A]t = - k t +[A]o

[A]t vs. t

(slope = - k)

t1/ 2 [A]0

2k

first

order

(n = 1)

rate = k [A]1

ln[A]t = - k t +

ln[A]o

ln[A]t vs. t

(slope = - k)

t1/ 2 ln2

k

0.693

k

second

order

(n = 2)

rate = k [A]2

1

[ A]t

k t 1

[A]0

1

[A]t

vs. t

(slope = k)

t1/ 2 1

k[A]0

Experimental data is collected for a reaction over time to give the concentration of A ([A]t) at a

number of times t. These data points are then tested graphically against the various rate

equations. The reaction order is determined by deciding which of the plots (see Table 2) gives

the best straight line.

Note that this method requires data from a single experiment. However the data must be

collected over several half-lives. If too little data is collected, all the plots will give a straight

line, and it will not be possible to determine the reaction order. Recall that the half-life t1/2 is the

time it takes for the concentration to fall to one-half its initial value [A]0. If the reaction is slow,

several half-lives will be a great deal of time, and the Method of Initial Rates is more convenient.

Other Forms of the First-Order Integrated Rate Equation The first-order rate equation is frequently re-arranged to give

48

ln[A]t

[A]0

k t or ln

[A]0

[A]t

k t (6)

These forms are obtained easily from the integrated form of the first-order rate equation in the

table above by using the property of logarithms,

lna

b

ln(a) ln(b) (7)

Another property of logarithms is that if

ln(y) = x (8)

then

y = ex (9)

These relations lead to additional forms of the first order integrated rate equation:

[A]t

[A]0

e

k t or [A]0

[A]t

e

k t (10)

The forms given in Eqs (6) and (10) are useful in solving problems in which the fraction of

material left [A]t/[A]o is related to the decay time t of the material. Many chemical decay

processes and all nuclear decay reactions follow first-order kinetics.

Note that for first-order reactions, the half-life t1/2 is easily determined from the rate constant k,

and vice versa (see Table 2 above).

______________________________________________________________________________

Problems Using the Integrated Rate Equations

2. A pesticide decomposes following first-order kinetics.

(a) If the half-life of the pesticide is 12 years, what is the rate constant k for the decomposition

reaction?

(b) What fraction of the pesticide will be left after 36 years?

(c) What fraction of the pesticide will be left after 100 years?

(d) How many years will it take for 99.9% of the pesticide to decompose?


(a) k = 0.05775 yr-1 ≈ 0.058 yr-1; (b) 36 years = 3 half-lives. Thus 0.125 = 1/8 left; (c) 0.0031

(or 0.31%); (d) When 99.9% is gone, 0.1% is left, so the fraction left = 0.001. To reach this

point will take119.6 yr ≈ 120 yr.

______________________________________________________________________________

49

3. For a chemical reaction A+ 2 B C, a plot of 1/[A]t versus time t is found to give a straight

line with a positive slope.

(a) What is the order of the reaction? [Hint: See Table 2.]

(b) How could you determine the rate constant k for the reaction from your graph?

Answers to Problem 3:

(a) Second order; (b) Draw the best straight line through the points of the plot of 1/[A]t versus

time t . The slope of the line is equal to the rate constant k.

______________________________________________________________________________

4. The following data were obtained on the reaction 2 A B:

Time, s 0 5 10 15 20

[A], mol L-1 0.100 0.0141 0.0078 0.0053 0.004

(a) Plot the data and determine the order of the reaction.

[Hint: See Table 2. Fill in the table below and make three graphs on graph paper (or using a

computer graphing program): (1) Zeroth-order graph: plot [A]t versus t; (2) First-order graph:

plot ln[A]t versus t; Second-order graph: plot 1/[A]t versus t .]

Time, s [A], mol L-1 ln[A] 1/[A], L mol-1

0 0.1000

5 0.0141

10 0.0078

15 0.0053

20 0.0040

(b) Determine the rate constant.

[Hint: See Table 2. How does the slope m relate to the rate constant k for the reaction order you

determined in part (a)?]


(a) You should find that the plot of 1/[A]t versus t gives the best straight line, so the reaction is

second order;

(b) k = 12 L mol-1 s-1.

50

Data Table and Plots for Problem 4

Time, s [A], mol L-1 ln[A] 1/[A], L mol-1

0 0.1000 -2.303 10

5 0.0141 -4.262 71

10 0.0078 -4.854 128

15 0.0053 -5.240 189

20 0.0040 -5.521 250

51

How to Work from a Mechanism to a Rate Law

Once you have experimentally determined the rate law, the next task to to find a mechanism

consistent with the rate law. Few reactions occur in a single step. Usually the mechanism is a

series of unimolecular and bimolecular steps that add to give the net equation for the reaction.

Mechanism 1 is a possible mechanism for the reaction

2 A + B C :

52

A + B

k1 k1

AB (fast to pre-equilibrium)

A + AB

k2 C (slow)

_______________________

Net equation:

2 A + B C

Here AB is an intermediate in the reaction, since it forms in an early step but is consumed in a

later step.

The rate of any step in the mechanism is found using the equation

(rate of step) vstep kstep (product of conc. of reactants in the step) (1)

Eq(1) gives the rates for all the steps of the mechanism above:

Rate of first step forward (bimolecular step)

v1 k1 [A][B]

Rate of first step backward (unimolecular step)

v-1 k1 [AB]

Rate of second step forward (bimolecular step)

v2 k2 [A][AB]

The rate law consistent with a the mechanism is found by assuming that the reaction rate equals

the rate of the slow step:

rate = v = vslow step (2)

Since in this example, the second step is slow

rate = v = vslow step = v2 k2 [A][AB] (3)

AB is an intermediate, so we are not finished yet. We need to re-express [AB] in terms of

concentrations of species that appear in the net equation. This is easy to do since the first step is

fast to pre-equilibrium. This tells us that v1 = v-1, so

k1 [A][B] k1 [AB]

Solving this equation for [AB] gives

[AB]k1 [A][B]

k1 (4)

Substituting Eq(4) into Eq(3) gives the rate law consistent with Mechanism 1:

rate = v = v2 k2 [A] k1 [A][B]

k1

or, equivalently,

53

rate = v =k2 k1 [A]2[B]

k1 k [A]

2[B] (5)

where k = k2k1/k-1. Thus Mechanism 1 predicts that the reaction is second order in A, first order

in B, and third order overall.

Mechanism 2 is another possible mechanism for the reaction:

A + B

k1 AB (slow)

A + AB

k2 C (fast)

_______________________

Net equation:

2 A + B C

In this case the first step is slow, and Eq(2) gives

rate = v = vslow step = v1 k1 [A][B] (6)

There are no intermediates in Eq(6), so we are finished. Thus Mechanism 2 predicts that the

reaction is first order in A, first order in B, and second order overall.

If either Eq (5) or Eq(6) is consistent with the experimental rate law, it is a possible mechanism

for the reaction. However, there may be other mechanisms that are also consistent with the

experimental rate law.

Enzyme Catalysis Catalysts are substances that react in an early step in a reaction mechanism, and are regenerated

in a later step (how do they differ from intermediates?). Enzymes are biological catalysts. Many

enzyme-catalyzed reactions have been found to follow the Michaelis-Menten Mechanism:

E + S

k1 k1

ES (fast to pre-equilibrium)

ES

k2 E + P (slow)

_______________________

Net equation: S

P

Here S is the substrate, E the enzyme, ES the enzyme-substrate complex, and P the product. The

problem below refers to this mechanism.

1. (a) What is the catalyst in the Michaelis-Menten Mechanism? ____________

(b) What intermediate (if any) is in the Michaelis-Menten Mechanism? _____________

(c) Derive the rate law for an enzyme reaction that follows the Michaelis-Menten

Mechanism. Express the rate law in a form involving no concentrations of intermediates.

(Hint: The method is analogous to that used for Mechanism 1 above.)

Answers: (a) E ; (b) ES ; (c) rate = v = k [E][S] where k = k2k1/k-1.

54

How to Solve Equilibrium Problems

Equilibrium problems are a common type of problem in Chemistry which involve the

equilibrium constant K. More specifically, Kc is used when the equilibrium constant is written in

terms of concentrations, and Kp is appropriate for a gas reaction if partial pressures of gases are

given.

For a reaction

a A + b B c C + d D

Kc is defined to be

Kc [C]

c[D]

d

[A]a[B]b

(at equilibrium)

where all concentrations are equilibrium concentrations. The double arrow in the equation

indicates the reaction goes in both directions (and never reaches completion). At equilibrium, the

rates of the forward and reverse reactions are equal, and there is no further change in

concentration. Kc is a constant for a given reaction at a given temperature.

Methods for solving equilibrium problems are best shown with examples.

______________________________________________________________________________

Example 1

Calculate the equilibrium constant Kc at 25 oC for the reaction

2 NOCl(g) 2 NO(g) + Cl2(g)

using the following information. In one experiment 2.00 mol of NOCl is placed in a 1.00 -L

flask, and the concentration of NO after equilibrium is achieved is 0.66 mol/L.

Method:

(a) Calculate the initial concentration of NOCl from the information given.

(b) Form an equilibrium table and fill in all known quantities. Represent unknown quantities

with variables. Use the information in the table to calculate the concentration of NOCl and Cl2 at

equilibrium.

(c) Form the Kc equation and fill in all known information. Calculate Kc.

Solution:

Step(a): The initial concentration of NOCl is

[NOCl]0 = 2.0 mol

1.00 L 2.00 M

Step(b): The equilibrium table is:

55

Balanced Equation 2 NO(g) + Cl2(g)

Initial Concentrations (M) 2.00 0 0

Change (M) - 2x 2x x

Equilibrium Concentrations (M) (2.00 - 2x) 2x = 0.66 x

From the table we have

[NO] = 2x = 0.66 M

x = 0.33 M

Thus

[NOCl] = (2.00 - 2x) M = (2.00 - 2(0.33)) M = 1.34 M

[Cl2] = x = 0.33 M

Step(c): We now have the information we need to calculate Kc:

Kc [NO]

2[Cl2 ]

[NOCl]2

at equilibrium

(0.66)

2(0.33)

(1.34)2 0.080

______________________________________________________________________________

Example 2

When solid ammonium carbamate sublimes, it dissociates completely into ammonia and carbon

dioxide according to the equation

N2H6CO2(s) 2 NH3(g) + CO2(g)

At 25 oC, experiment shows that the total pressure of the gases in equilibrium with the solid is

0.116 atm. What is the equilibrium constant Kp?

Method:

The steps are:

(a) Form an equilibrium table and fill in all known quantities. Represent unknown quantities

with variables.

(b) Use the table together with the rule that the sum of the partial pressures equals the total

pressure to find the partial pressures of NH3 and CO2 at equilibrium.

(c) Form the Kp equation and fill in all known information. Calculate Kp.

Solution:

Step (a): The equilibrium table is:

Balanced Equation N2H6CO2 2 NH3(g) + CO2(g)

Initial Partial Pressures (atm) (solid!) 0 0

Change (atm) not applicable 2x x

Equilibrium Partial Pressures (atm) not applicable 2x = p(NH3) x = p(CO2)

56

Step (b): The total pressure at equilibrium (given as 0.116 atm) equals the sum of the partial

pressures. The last line of the equilibrium table allows us to express p(NH3) and p(CO2) in terms

of x. Thus we have

pTotal = p(NH3) + p(CO2)

0.116 atm = 2x + x = 3x

x = 0.03867 atm

It follows that

p(NH3) = 2 x = 2 (0.0387 atm) = .07734 atm

p(CO2) = x = 0.03867 atm

Step (c): Now we have the information needed to calculate Kp:

Kp pNH3

2 pCO2

at equilibrium

(0.07734)2(0.03867) 2.3110

4

______________________________________________________________________________

Example 3

A sample of 0.0020 moles of F2 was sealed into a 2.0 L reaction vessel and heated to 1000 K to

study the dissociation into F atoms:

F2 2 F

At this temperature, Kc = 1.210-4

. What are [F2] and [F] at equilibrium? What is the percent

dissociation of F2?

Method:

The steps are:

(a) Calculate the initial concentration of F2 from mole and volume information given.

(b) Form an equilibrium table and fill in all known quantities.

(c) Form the Kc equation and fill in all known information. Solve for the unknown.

(d) Calculate results asked for in the problem.

Solution:

Step (a): The initial concentration of F2 is

[F2]0 = 0.0020 mol

2.0 L

0.0010 M

Step (b): The equilibrium table is given below. Note that as the reaction proceeds, reactants are

consumed and products are formed in proportion to their stoichiometric coefficients.

57

Balanced Equation F2 2 F

Initial Concentration (M) 0.0010 0

Change (M) - x 2x

Equilibrium Concentration (M) 0.0010 - x 2x

Step (c): Next we form the Kc equation and fill in equilibrium values. Note that [F] = (2x) and

that the entire quantity is squared:

Kc 1.2 104

[F]2

[F2 ]

(at eq)

2x 2

0.0010 x

4x2

0.0010 x

(0.0010 - x)(1.210-4

) = 4 x2

Next we rearrange the equation above to quadratic form, a x2 + b x + c = 0:

4 x2 + 1.210

-4 x - 1.210

-7 = 0

In our case,

a = 4, b = 1.210-4

, and c = - 1.210-7

and, using the quadratic formula, we have:

x b b

2 4ac

2a(1.210

4) (1.2 10

4)

2 4(4)(1.2 10

7)

2(4)

x 1.2 10

4 1.93410

6

8

This gives x+ = 1.5910-4

and x- = - 1.8910-4

. Clearly, only the positive root makes sense

physically; there is no such thing as a (-) concentration! Thus x = x+ = 1.5910-4

.

Step (d): Now we are ready to calculate the results asked for in the problem. For this we go

back to the last line of our equilibrium table, and use the value of x calculated above to obtain

[F2] = (0.0010 - x) = (0.0010 - 1.5910-4

) = 8.410-4

[F] = 2 x = 2 (1.5910-4

) = 3.210-4

The equation for percent dissociation is:

% dissociation = quantity dissociated

initial amount

100%

In this case,

58

% dissociation of F2 = x

[F2 ]0

100%=

1.5910-4

0.0010

100%

= 15.9 % = 16 %

______________________________________________________________________________

Approximate Methods

In many cases calculations may be simplified by making approximations. For example, if Kc

is very small (i.e., < 5.010-5

) and the initial reactant concentration c0 is fairly large (i.e., c0 >

1000Kc ), to a good approximation

(c0 x) c0

Use of this type of approximation eliminates the need to solve quadratic equations in many cases.

Making Use of Le Chatelier’s Principle

Le Chatelier’s Principle is very useful in predicting how a system at equilibrium will respond

to a change. It states that when a system at equilibrium is disturbed, the equilibrium shifts so as

to undo, in part, the effect of the disturbance.

There are three common ways an equilibrium may be disturbed:

Change in the concentration (or partial pressure) of one of the reactants or products.

Change in the temperature.

Change in the volume of the container.

Effect of Changes in Concentration (or Partial Pressure)

If a system at equilibrium is disturbed by the addition of a reactant (or the removal of a

product), Le Chatelier‟s principle predicts that the equilibrium will shift right. Shifting right will

use up some of the added reactant (or replace some of the removed product), and therefore “undo,

in part” the disturbance.

Similarly, if the disturbance is the removal of a reactant (or the addition of a product), Le

Chatelier‟s principle predicts that the equilibrium will shift left. Shifting left will replace some

of the removed reactant (or use up some of the added product), and therefore “undo, in part” the

disturbance.

The same conclusions may be reached by considering the values of Q and K. When the

equilibrium is disturbed, the reaction quotient Q changes so that it is no longer equal to K. If the

resulting Q is greater than K, the reaction will proceed backwards until once again Q equals K.

Conversely, if after the disturbance Q is less than K, the reaction will shift forwards until Q

equals K. In all cases, the shifts stop when Q = K.

Since concentrations of solids are constants and do not appear in expressions for Q or K,

removing or adding some solid does not cause shifts. However, shifts in the equilibrium do

change the amount of solid present!

59

Effect of Changes in Temperature

If a reaction is endothermic (∆H > 0), heat is absorbed in the forward reaction and released in

the backward reaction; thus in endothermic reactions, heat behaves like a reactant. Increasing

the temperature (adding heat) shifts the reaction right, since that is the direction which absorbs

heat and “undoes, in part” the disturbance. Similarly, decreasing the temperature of an

endothermic reaction shifts the reaction left.

Exothermic reactions (∆H < 0) release heat in the forward direction, and absorb heat in the

reverse direction; thus heat acts like a product in exothermic reactions. Increasing the

temperature (adding heat) shifts the reaction left since that is the direction that absorbs heat and

“undoes, in part” the disturbance. Conversely, lowering the temperature shifts it right.

The equilibrium constant K is temperature dependent and the shifts above change its value.

Reaction Type Role of heat

Endothermic (∆H > 0) reactants + heat products K K

Exothermic (∆H < 0) reactants products + heat K K

Effect of Changes in the Volume of the Container

When the volume of a reaction vessel is decreased, the partial pressures of all gases in the

container increase so the total pressure increases. Following Le Chatelier‟s principle, the

reaction shifts to reduce the total pressure since that “undoes, in part” the disturbance. This

means that the shift is in the direction which contains the fewest moles gas.

Similarly, if the volume of the reaction vessel is increased, the total pressure decreases, and

the shift is in the direction which contains the most moles gas.

In either case, if both sides of the equation have the same number of moles gas, the change in

the volume of the container has no effect on the equilibrium.

Effect of the Addition of a Catalyst

Catalysts speed up the rate at which equilibrium is obtained, but have no effect on the

magnitude of K or Q. They increase both the forward and backward rate of reaction. Since

catalysts do not appear in the net equation for a reaction, they are not involved in the expressions

for K or Q.

______________________________________________________________________________

Exercises

In the problems below, for each change given in the first column of the table, use Le Chatelier's

principle to predict

• the direction of shift of the equilibrium.

• the effect on the quantity in the third column.

______________________________________________________________________________

1. For the following reaction

5 CO(g) + I2O5(s) I2(g) + 5 CO2(g) ∆Ho = -1175 kJ

for each change listed, predict the equilibrium shift and the effect on the indicated quantity.

60

Change

Direction

of Shift

(; ; or no change)

Effect on

Quantity

Effect

(increase,

decrease,

or no change)

(a) decrease in volume Kc

(b) raise temperature amount of CO(g)

(c) addition of I2O5(s) amount of CO(g)

(d) addition of CO2(g) amount of I2O5(s)

(e) removal of I2(g) amount of CO2(g)

Answers: # 1(a) , no change; (b) , increase; (c) no change, no change; (d) , increase; (e) ,

increase.

______________________________________________________________________________

2. Consider the following equilibrium system in a closed container:

Ni(s) + 4 CO(g) Ni(CO)4(g) ∆Ho = - 161 kJ

In which direction will the equilibrium shift in response to each change, and what will be the

effect on the indicated quantity?

Change

Direction

of Shift

(; ; or no change)

Effect on

Quantity

Effect

(increase,

decrease,

or no change)

(a) add Ni(s) Ni(CO)4(g)

(b) raise temperature Kc

(c) add CO(g) amount of Ni(s)

(d) remove Ni(CO)4(g) CO(g)

(e) decrease in volume Ni(CO)4(g)

(f) lower temperature CO(g)

(g) remove CO(g) Kc

Answers: # 2(a) no change, no change; (b) , decrease; (c) , decrease; (d) , decrease; (e) ,

increase; (f) , decrease; (g) , no change.

61

Conjugate Acid-Base Pairs Arranged by Strength

The stronger the acid, the weaker the conjugate base. The stronger the base, the weaker the

conjugate acid.

Conjugate bases of diprotic acids are often atypical (see entries in italics for examples).

ACID BASE

Strength of Acid Name Formula Formula Name Strength of Base

STRONG

ACIDS

perchloric acid HClO4 ClO4– perchlorate ion Neutral Anion

sulfuric acid H2SO4 HSO4– hydrogensulfate ion

Moderately Strong

Acid!

hydroiodic acid HI I– iodide ion

Neutral

Anions

hydrobromic acid HBr Br– bromide ion

hydrochloric acid HCl Cl– chloride ion

nitric acid HNO3 NO3– nitrate ion

Strong Acid hydronium ion H3O+ H2O water Neutral

Moderately

Strong

Acid!

hydrogensulfate ion HSO4– SO4

2– sulfate ion Neutral Anion!

WEAK ACIDS

Acid strength

INCREASES as you go

UP the column.

hydrofluoric acid HF F– fluoride ion

WEAK BASES

Base strength

INCREASES as

you go DOWN the

column.

nitrous acid HNO2 NO2– nitrite ion

acetic acid HC2H3O2 C2H3O2– acetate ion

carbonic acid H2CO3 HCO3– hydrogencarbonate

ion

hydrosulfuric acid H2S HS– hydrogensulfide ion

ammonium ion NH4+ NH3 ammonia

hydrocyanic acid HCN CN– cyanide ion

Basic Anion! hydrogencarbonate

ion

HCO3– CO3

2– carbonate ion

Weak Acid methylammonium

ion

CH3NH3+ CH3NH2 methylamine

Neutral water H2O OH–

hydroxide ion Strong Base

Basic Molecule ammonia NH3 NH2

– amide ion

STRONG

BASES Neutral

Molecules

hydrogen H2 H– hydride ion

methane CH4 CH3– methide ion

Strong Base!

hydroxide ion OH– O2– oxide ion

62

Acid-Base Calculations

The Ion-Product Constant for Water, Kw

Water undergoes ionization to a small extent:

H20(l) H+(aq) + OH

–(aq)

The equilibrium constant for the reaction is the ion-product constant for water Kw:

Kw [H][OH

]1.010

14 (1)

This is a key equation in acid-base chemistry. Note that the product of [H+] and [OH

–] is a

constant at a given temperature (Eq(1) value is for 25oC). Thus as the hydrogen ion concentration

of a solution increases, the hydroxide ion concentration decreases (and vice versa).

The pH scale is widely used to report the molar concentration of hydrogen ion H+(aq) in aqueous

solution. The pH of a solution is defined as

pH log10[H] (2)

Similarly, pOH and pKw are defined as

pOH log10[OH] (3)

pKw log10(Kw)14.00 (4)

If you take the log10 of both sides of Eq(1), multiply the resulting equation by (-1), and use the

definitions of pH, pOH and pKw above, the result is the very useful equation

pH + pOH = pKw = 14.00 (5)

Equations (2) and (3) above may be solved for [H+] and [OH

–] respectively to give

[H]10

pH (6)

[OH]10

pOH (7)

(Here we use the well known rule that if

log10 y x , then

y 10x

.) In practice, the pH scale is

only used when [H+(aq)] is less than 1.0 M.

Acidic, basic, and neutral solutions can be distinguished as shown below:

63

Type of Solution pH [H+] Color of litmus

Acidic < 7.00 > 1.0 107 pink

Neutral = 7.00 = 1.0 107 in between

Basic > 7.00 < 1.0 107 blue

pH and [H+] Calculations for Strong Acids and Bases

By definition, strong acids and bases are 100% ionized in water solution. Ionization of a strong

acid gives rise to H+ ions, and ionization of a strong base produces OH

– ions. The equilibrium

constant for a strong acid or strong base is undefined, since the reaction the ionization is

complete. There is no equilibrium!

In nearly all cases of practical interest the [H+] for a strong acid (or the [OH

–] for a strong base)

is determined completely by the stoichiometry of the reaction. Once the [OH–] or pOH is known

for a base, the [H+] or the pH of the base may be calculated using Eq(1) and/or Eq(5).

Exercises

1. Complete the following table:

pH [H+] pOH [OH

–]

Acidic,

basic, or

neutral?

(a) 5.4 x 10–4

(b) 7.8 x 10-10

(c) 10.75

(d) 5.00

Answers:

(a) pH = 3.27; pOH = 10.73; [OH–] = 1.85 x 10

–11 = 1.9 x 10

–11, acidic (since pH < 7).

(b) pH = 4.89, [H+] = 1.3 x 10

–5, pOH = 9.11, acidic (since pH < 7).

(c) [H+] = 1.8 x 10

-11, pOH = 3.25, [OH

–] = 5.6 x 10

–4, basic (since pH > 7).

(d) pH = 9.00, [H+] = 1.0 x 10

–9, [OH

–] = 1.0 x 10

–5, basic (since pH > 7).

2. Calculate the pH of a 0.0430 M HNO3 solution.

64

Answer:

Since HNO3 is a strong acid, the nitric acid solution will be 100% ionized. Thus [H+] = [NO3

–] =

0.0430 M. The pH = 1.37 (use Eq(2)).

3. Calculate the pH of a 0.020 M Ba(OH)2(aq) solution.

Answer:

Since Ba(OH)2 is a strong base it is 100% ionized. Note that ionization gives 2 OH– ions for each

mole of Ba(OH)2. Thus [OH–] = 2 x 0.020 M = 0.040 M. Eq(3) gives pOH = 1.40. Then using

Eq(5), pH = 12.60.

pH and [H+] Calculations for Weak Acids and Bases

Weak acids and bases are usually less than 5% ionized. The equilibrium constant for a weak acid

equilibrium is the acid ionization constant Ka, and for a weak base equilibrium is the base

ionization constant Kb.

A typical monoprotic weak acid equilibrium can be written in two forms, the second of which

emphasizes the Brønsted acid-base nature of the reaction:

HA H+(aq) + A

–(aq)

HA + H2O H3O+(aq) + A

–(aq) (8)

In Eq(9) the Brønsted acid HA donates a proton H+ to the Brønsted base H2O to form H3O

+ and

the conjugate base A–. The acid ionization constant (using the second form) is

Ka [H3O

][A]

[HA] (9)

A typical weak base equilibrium is

B + H2O BH+(aq) + OH

–(aq) (10)

In Eq(10) the Brønsted base B accepts a proton H+ from the Brønsted base H2O to form the

conjugate acid BH+ and OH

–. The base ionization constant is

65

Kb [BH][OH]

[B] (11)

Exercises

4. Calculate (a) the pH and (b) the percent ionization of a 0.250 M HC2H3O2 solution.

Ka(HC2H3O2) = 1.8 x 10-5

. (The formula for acetic acid may also be written as CH3COOH.)

HINT: Begin by filling out the equilibrium table below.

Balanced Equation HC2H3O2 H+

+ C2H3O2–

Initial Concentration (M)

Change (M)

Equilibrium Concentration (M)

Answer:


+ C2H3O2–

Initial Concentration (M) 0.250 0 0

Change (M) - x x x

Equilibrium Concentration (M) 0.250 - x x x

(a)

Ka [H][C2H3O2

]

[HC2H3O2 ]1.810

5

x 2

0.250 - x

x2

0.250 . This approximation is OK if the %

ionization is < 5%; it is in this case -- see answer to (b) below. Thus x2 = 4.5 x 10

-6; x = 2.12 x 10

-

3 = [H

+]. pH = 2.67.

(b)

% ionization = x

0.250

100%

2.12103

0.250

100% 0.85%.

5. Calculate the pH of a 0.600 M solution of methylamine CH3NH2. Kb = 4.4 x 10–4

.

HINT: Methylamine is a weak base. First write the equation for the reaction following the pattern

of Eq(10). Then fill out the equilibrium table below.

Balanced Equation CH3NH2 CH3NH3+ + OH

–


Change (M)


Answer:

Since CH3NH2 is a weak base, the balanced equation for the reaction is CH3NH2 + H2

CH3NH3+ + OH

–.

66

Balanced Equation CH3NH2

CH3NH3

+ + OH

–


Change (M) - x x x


Kb [BH][OH]

[B]

[CH3NH3

][OH]

[CH 3NH 2]

x2

0.600 x

x2

0.600 4.4 10

4. Thus x = 1.62 x 10

-2 =

[OH–], and pOH = 1.79. It follows from Eq(5) that pH = 12.21. NOTE: The approximation used

is OK since the % ionization is 2.7% (i.e., less than 5 %).

6. The pH of a 0.10 M solution of a weak base is 9.67. What is the Kb of the base?

Answer:

The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is

given below.

Balanced Equation B BH+ + OH

–


Change (M) - x x x


At equilibrium, [OH–] = [BH

+] = x. Use the pH to calculate the [OH

–] at equilibrium (which is

the value of x). Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33. Thus

[OH]10

pOH10

4.33 4.6810

5 x .

82522

102.210.0

)1068.4(

10.010.0[B]

]][OH[BH

x

x

xKb . The approximation is OK

since the % ionization is well under 5%.

Relationship between Ka for a Weak Acid and Kb for its Conjugate Base

The relationship between Ka for a weak acid HA and Kb for its conjugate base A– is

Ka(HA)

Kb(A–) = Kw = 1.0 x 10

-14 (12)

If we define pKa = - log10(Ka) and pKb = - log10(Kb), the logarithmic form of Eq(12) is

pKa(HA) + pKb(A–) = pKw = 14.00 (13)

67

The stronger the acid, the larger the Ka and the smaller the pKa. Likewise the stronger the base,

the larger the Kb and the smaller the pKb. Eqs(12) and (3) show that as the Ka increases (and the

pKa decreases), the Kb decreases (and the pKb increases). These equations give quantitative

support to the statement “the stronger the acid, the weaker the conjugate base.”

The justification for Eq(12) follows from the equations below. Recall that if Eq(1) + Eq(2) =

Eq(3), then K1

K2 = K3.

Eq(1), Weak Acid: HA + H2O H3O+(aq) + A

–(aq)

Ka(HA) [H 3O

][A]

[HA]

Eq(2), Conjugate Base: A–(aq) + H2O HA(aq) + OH

–(aq)

Kb(A)

[HA][OH]

[A]

___________________________

Eq(3) = Eq(1) + Eq(2) 2 H20(l) H3O+(aq) + OH

–(aq)

Kw [H3O][OH

]

Relationship between Kb for a Weak Base and Ka for its Conjugate Acid

Analogous equations to Eqs(12) and (13) above can be written the relationship between Kb for a

weak base B and Ka for its conjugate acid HB+:

Kb(B)

Ka(BH+) = Kw = 1.0 x 10

-14 (14)

pKb(B) + pKa(BH+) = pKw = 14.00 (15)

The equations below provide justification for these results:

Eq(1), Weak Base: B + H2O BH+(aq) + OH

–(aq)

Kb(B)[BH][OH]

[B]

Eq(2), Conjugate Acid: BH+(aq) + H2O H3O

+(aq) + B(aq)

Ka(BH+)

[H3O][B]

[BH+]

___________________________

Eq(3) = Eq(1) + Eq(2) 2 H20(l) H3O+(aq) + OH

–(aq)

Kw [H3O][OH

]

___________________________________________________________________________

Exercise

7. Use the following acidity constants to help answer the questions below:

Ka(HC2H3O2) = 1.8 x 10 – 5

; Ka(HCN) = 4.9 x 10 – 10

; Ka(HCOOH) = 1.7 x 10 - 4

(a) Which of the three acids is the weakest? ________________

(b) Which of the following bases is the strongest: C2H3O2-, CN

- , or HCOO

- ? __________

68

(c) What is the pKa of HCN? _____________

(d) What is the Kb for CN- ?____________

Answers:

(a) HCN; (b) CN - ; (c) 9.31; (d) 2.04 x 10

-5

2.0 x 10-5

.

_____________________________________________________________________

Strong and Weak Acids and Bases

Definitions

An Arrhenius acid is a compound that contains hydrogen and releases hydrogen ions (H+)

in water. A Brønsted acid is a proton donor (where a proton is a hydrogen ion, H+). A

Lewis acid is an electron pair acceptor.

An Arrhenius base is a compound that produces hydroxide ions (OH–) in water. A

Brønsted base is a proton acceptor. A Lewis base is an electron pair donor.

Strong Acids

Strong acids are 100% ionized in aqueous solution to form the hydronium ion, H3O+ (also

written as H+(aq)) and an anion. For example, HCl in water ionizes completely:

HCl + H2O H3O+(aq) + Cl–(aq) [goes to completion]

(or, equivalently, HCl + water H+(aq) + Cl–(aq) [goes to completion])

There are very few strong acids, but they are extremely important in chemistry since they are

excellent sources of H+(aq), a highly reactive ion!

Strong Acid Examples: HCl (hydrochloric acid), HBr (hydrobromic acid), HI (hydroiodic acid),

HNO3 (nitric acid), H2SO4 (sulfuric acid), HClO4 (perchloric acid), and a small number of non-

metallic oxides which react with water to give a strong acid (eg., SO3(g) + H2O H2SO4).

Weak Acids

Most acids are weak. Weak acids are typically less than 5% ionized in water; thus the

predominant species is the un-ionized form. Since relatively small amounts of H+(aq) are

formed, weak acids are not very reactive. Typical weak acid ionizations in water are

HC2H3O2 + H2O H3O+(aq) + C2H3O2–(aq)

(or, equivalently, HC2H3O2 + water H+(aq) + C2H3O2–(aq))

SO2(g) + H2O H2SO3 H+(aq) + HSO3–(aq)

(or, equivalently, SO2(g) + 2 H2O H3O+(aq) + HSO3–(aq))

69

In each case above, reaction proceeds only to a very limited extent; typically over 95% of the

weak acid remains un-ionized! Since the predominant form is un-ionized, chemists do not split

up weak acids into ions when writing an ionic equation.

Weak Acid Examples:

Molecular compounds with an acidic hydrogen: HC2H3O2 = CH3COOH (acetic acid), HF

(hydrofluoric acid), HNO2 (nitrous acid), HCN (hydrocyanic acid), C6H5COOH (benzoic

acid).

Non-metallic oxides: SO2(g) (sulfur dioxide), CO2(g) (carbon dioxide), and NO(g) (nitrogen

oxide). A few non-metallic oxides such as SO3(g) (sulfur trioxide) and N2O5(g) (dinitrogen

pentoxide) give strong acids when dissolved in water. Acidic oxides of S and N are major

contributors to acid rain.

Most cations are acidic. Examples include H+, ammonium ion (NH4

+), and amine-type

cations such as CH3NH3+. Numerous metallic cations are acidic; the more acidic metal

cations are those such as Be+2

, Al+3

, and Fe3+

which have a high charge to radius ratio.

Anions of type HX- which are conjugate to strong or moderately strong acids H2X. An

example is HSO4- (conjugate base of H2SO4) which is a moderately strong acid. This is a

tiny class!!

Strong Bases

Strong bases are 100% ionized in aqueous solution to form the hydroxide ion, OH–, and a

cation. There are very few strong bases, but they are extremely important in chemistry since they

are excellent sources of OH–(aq), a highly reactive ion! Typical ionization reactions are

NaOH(s) + water Na+(aq) + OH–(aq) [goes to completion]

Na2O(s) + H2O 2 Na+(aq) + 2 OH–(aq) [goes to completion]

Strong Base Examples:

Alkali metal hydroxides and the more soluble alkaline earth hydroxides: NaOH(s) (sodium

hydroxide), KOH(s) (potassium hydroxide), Ba(OH)2(s) (barium hydroxide).

Alkali metal oxides and the more soluble alkaline earth oxides: Na2O(s) (sodium oxide),

K2O(s) (potassium oxide), BaO(s) (barium oxide).

The less soluble hydroxides and oxides of the alkaline earth cations are weak bases. Since

solubility increases for these compounds as you go down Column II, the hydroxides and

oxides of Ba2+ and Sr2+ are generally considered strong bases, while those for Ca2+ are on

the borderline between strong and weak due to their limited solubility in water.

Weak Bases

The vast majority of bases are weak. Much like weak acids, weak bases are typically less

than 5% ionized. Since their water solutions contain low concentrations of OH–(aq), they are not

very reactive. Examples of weak base ionization reactions include

NH3 + H2O NH4+(aq) + OH–(aq)

CH3NH2 + H2O CH3NH3+(aq) + OH–(aq)

70

Cu(OH)2(s) + water Cu2+(aq) + 2 OH–(aq)

CuO(s) + H2O Cu2+(aq) + 2 OH–(aq)

In each case above, reaction proceeds only to a very limited extent; typically over 95% of the

weak base remains un-ionized! Weak bases are therefore not split up into ions when writing

ionic equations.

Weak Base Examples

Metal hydroxides and metal oxides other than those listed as strong bases above. Examples

include Mg(OH)2(s) (magnesium hydroxide), MgO(s) (magnesium oxide), and Cu(OH)2(s)

(copper(II) hydroxide).

Ammonia (NH3) and amine-type bases such as CH3NH2. Typical amine-type bases have an

ammonia-type structure, but with one or more of the H atoms replaced by a hydrocarbon

group.

Most anions are basic. Common examples include HCO3–(aq) (hydrogen carbonate or

bicarbonate ion) which is present in sodium bicarbonate (NaHCO3(s)) and CO32–(aq)

(carbonate ion), a constituent of calcium carbonate (CaCO3(s)).

Neutral Compounds and Ions

So many compounds and ions are acidic or basic, you may wonder whether anything is

neutral! Examples of neutral substances include:

Water, H2O.

Hydrocarbons, alcohols, sugars, starch, and many other organic molecules.

The cations present in the strong hydroxide and oxide bases: Na+, K+, Ba2+, etc.

Most anions produced upon ionization of the strong acids: Cl–, Br–, I–, NO3–, SO42–.

Predicting the pH of Salt Solutions To determine the pH of a salt solution, you need to consider whether the salt cations or anions (or both) hydrolyze. Hydrolysis is a Brønsted acid-base reaction of an ion with water to give excess H

+ or OH

–. All acidic or basic ions hydrolyze.

CATIONS Most cations are acidic!! Hydrolysis of acidic cations gives rise to H

+ (also written as H3O

+). Examples include:

(a) Amine-type cations such as NH4

+, CH3NH3

+, etc.

NH4

+ + H2O NH3 + H3O

+

CH3NH3

+ + H2O CH3NH2 + H3O

+

(b) Metal ions with a high charge to radius ratio such as H

+, Be

2+, Al

+3, Cr

+3, and Fe

+3.

71

Here it is useful to think of a 2-step process: (1) hydration (universal for ions)

Al+3

+ 6 H2O Al(H2O)63+

(aq)

(2) hydrolysis Al(H2O)6

3+ + H2O Al(OH)(H2O)5

2+ + H3O

+

The exceptions are neutral (no cations are basic!!) Neutral ions do not hydrolyze. Examples of neutral cations are metal ions which are relatively large for their charge:

Li+, Na

+, K

+, Rb

+, Cs

+, Ba

+2, Sr

+2, Ca

+2 (cations associated with the most basic

hydroxides)

ANIONS Most anions are basic!! Hydrolysis of a basic anions gives rise to OH

–:

acetate: C2H3O2

– + H2O HC2H3O2 + OH

–

cyanide: CN

– + H2O HCN + OH

–

bicarbonate: HCO3

– + H2O H2CO3 + OH

– 2O + CO2(g) + OH

–

Most of the exceptions are neutral. Neutral ions do not hydrolyze. They are all conjugate bases of strong, or moderately strong, acids. The most common neutral anions are

Cl–, Br

–, I

–, NO3

–, ClO4

–, and SO4

2–.

A few anions are acidic. They typically contain hydrogen and are conjugate bases of strong, or moderately strong, acids. Examples of acidic anions are HSO4

– (conjugate to H2SO4), HC2O4

–

(conjugate to H2C2O4 = oxalic acid), and HC4H4O6– (conjugate to H2C4H4O6 = tartaric acid).

They hydrolyze in water:

HSO4– + H2O SO4

2– + H3O

+

Buffers

A buffer is a solution that resists changes in pH. The pH of a buffer changes very little when

small amounts of a strong acid or strong base are added to the buffer.

A buffer consists of approximately equal amounts of a conjugate weak acid/weak base pair in

equilibrium with each other. Strong acids and their conjugate bases don‟t produce a buffer since

strong acid ionization is complete: there is no equilibrium!

72

Acidic Buffers

In an acidic buffer both the weak acid and conjugate base are present initially in roughly equal

concentrations. The equilibrium is

HA H+ + A

–

weak acid conjugate base

For a weak acid,

Ka [H ][A– ]

[HA] so

pKa pH log[A]

[HA]

. Thus

pH pKa log[A]

[HA]

: this is

the Henderson-Hasselbalch equation.

These equations are useful for estimating the pH of a buffer. When [A-] = [HA], it follows that

Ka = [H+] and pH = pKa.

Acidic Buffer Example: Acetic acid (HC2H3O2), sodium acetate (NaC2H3O2)

NaC2H3O2 is a soluble salt that dissolves in water to give Na+ and C2H3O2

- ions. Na

+ ions are

neutral spectator ions, so can be ignored. The C2H3O2- ions provide the conjugate base for the

buffer. The buffer equilibrium is

HC2H3O2 H+ + C2H3O2

–

acetic acid acetate ion

For acetic acid, Ka(HC2H3O2) = 1.8 x 10-5

so pKa = -log10Ka = 4.74. Thus when [HC2H3O2] =

[C2H3O2–], pH = pKa = 4.74.

Basic Buffers

In a basic buffer both the weak base and conjugate acid are present initially in roughly equal

concentrations. The equilibrium is

B + H2O BH+ + OH

–

weak base conjugate acid

For a weak base,

Kb [BH][OH– ]

[B] so

pKb pOH log[BH]

[B]

=

pKb (14.00pH) log[BH]

[B]

. Thus

pH 14.00pKb log[BH]

[B]

. Since pKa = 14.00 – pKb,

pH pKa (BH) log

[B]

[BH]

. This is once again the Henderson-Hasselbalch equation.

73

These equations are useful for estimating the pH of a buffer. When [BH+] = [B], Kb = [OH

–].

Consequently, pOH = pKb for the buffer. It follows that pH = 14.00 – pKb = pKa(BH+) for the

buffer.

Basic Buffer Example: Ammonia (NH3), ammonium chloride (NH4Cl)

NH4Cl is a soluble salt that dissolves in water to give NH4+ and Cl

– ions. Cl

– ions are neutral

spectator ions, so can be ignored. The NH4+ ions provide the conjugate acid for the buffer. The

buffer equilibrium is

NH3 + H2O OH– + NH4

+

ammonia ammonium ion

For ammonia, Kb(NH3) = 1.8 x 10-5

so pKb = -log10Kb = 4.74. Thus when [NH3] = [NH4+], pOH

= 4.74 and pH = pKa(NH4+) = 14.00 – 4.74 = 9.26.

Neutral Buffers

Neutral buffers have a pH close to 7.00. A good example is a NaH2PO4/Na2HPO4 buffer. Since

Na+ ions are neutral spectator ions, this is a dihydrogen phosphate/hydrogen phosphate (H2PO4

–

/HPO42–

) buffer. The buffer equilibrium is

H2PO4– H

+ + HPO4

2–

dihydrogen phosphate ionhydrogen phosphate ion

Here

Ka [H ][HPO4

2–]

[H2PO4

]

= 6.2 x 10–8

. If [H2PO4-] = [HPO4

2-], Ka = [H

+] and pH = pKa = 7.21.

Response of a Buffer to the Addition of a Strong Acid or a Strong Base

Added Acid

When a small amount of a strong acid such as HCl is added to a buffer, the H+ from the acid

reacts with the basic part of the buffer to give more of the acidic part of the buffer. The reaction

is assumed to go 100%. The new concentration of the acidic part of the buffer (increased from

the initial value) and the new concentration of the basic part of the buffer (decreased from the

initial value) are then used to calculate the pH of the buffer.

Added Base

When a small amount of a strong base is added to a buffer, the OH– reacts with the acidic part of

the buffer to give more of the basic part of the buffer. The reaction is assumed to go 100%. The

new concentration of the acidic part of the buffer (decreased from the initial value) and the new

74

concentration of the basic part of the buffer (increased from the initial value) are then used to

calculate the pH of the buffer.

Exercises

1. Identify which of the following mixed systems could function as a buffer solution. For each

system that can function as a buffer, write the equilibrium equation for the conjugate acid/base

pair in the buffer system

(a) KF/HF (b) NH3/NH4Br (c)KNO3/HNO3 (d) Na2CO3/NaHCO3

Answer:

(a) Buffer: HF H+ + F

–

(b) Buffer: NH3 + H2O NH4+ + OH

– (or NH4

+ H

+ + NH3)

(c) Not a buffer since HNO3 is a strong acid. It is 100% ionized.

(d) Buffer: HCO3– H

+ + CO3

2–

2. What is the pH of a 1 L solution containing 0.240 mol HC2H3O2 and 0.180 mol NaC2H3O2?

Ka(HC2H3O2) = 1.8 x 10-5

. HINT: Begin by filling out the equilibrium table below.


+ C2H3O2–


Change (M)


Answer:


+ C2H3O2–

Initial Concentration (M) 0.240 0 0.180

Change (M) - x x x

Equilibrium Concentration (M) 0.240 - x x 0.180 + x

Ka [H][C2H3O2

]

[HC2H3O2 ]1.810

5

x (0.180 x)

(0.240 - x)

x (0.180)

0.240. This approximation is OK if the %

ionization is < 5%; it is in this case. Thus x = 2.4 x 10-5

= [H+]. pH = 4.62.

3. What would be the pH of the buffer from #2 above if 0.010 mol of HCl were added per liter of

the buffer?

75

Answer:

The new initial concentrations are:

[C2H3O2-] = (0.180 - 0.010) M = 0.170 M (basic part decreases)

[HC2H3O2] = (0.240 + 0.010) M = 0.250 M (acidic part increases)

Thus

Ka [H][C2H3O2

]

[HC2H3O2 ]1.810

5

x (0.170 x)

(0.250 - x)

x (0.170)

0.250. Thus x = [H

+] = 2.647 x 10

-5;

pH = 4.58. Note that the pH has decreased very little!!

4. What would be the pH of the buffer from #2 above if 0.010 mol NaOH were added per liter of

the buffer?

Answer:

The new initial concentrations are:

[C2H3O2-] = (0.180 + 0.010) M = 0.190 M (basic part increases)

[HC2H3O2] = (0.240 - .010) M = 0.230 M (acidic part decreases)

Thus

Ka [H][C2H3O2

]

[HC2H3O2 ]1.810

5

x (0.190 x)

0.230 - x

x (0.190)

0.230. Thus x = [H

+] = 2.179 x 10

-5;

pH = 4.66. Note that the pH has increased very little!!

5. What would be the pH of 0.010 M HCl without the buffer?

Answer:

HCl is a strong acid and is therefore 100% ionized. Thus [H+] = 0.010M, so the pH = 2.00.

Note how much lower the pH is than it was with the buffer in #3 above (pH = 2.00 here versus

4.58 in #3 above)!

6. What would be the pH of 0.010 M NaOH without the buffer?

Answer:

NaOH is a strong base and is therefore 100% ionized. Thus [OH–] = 0.010M, so the pOH = 2.00.

Thus pH = 14.00 – pOH = 12.00.

Note how much higher the pH is than it was with the buffer in #4 above (pH = 12.00 here versus

4.66 in #4 above)!

76

Predicting Solubility

Solubility problems are equilibrium problems. The reactant in a solubility equilibrium is a

slightly soluble salt and the equilibrium constant for the reaction is the solubility product

constant, Ksp. Note that since the reactant is a solid, its concentration does not appear in the Ksp

expression. For example, the solubility equilibrium and Ksp for the salt SrF2(s) are

SrF2(s) Sr2+

(aq) + 2 F –

(aq) Ksp = [Sr2+

][F –]

2 = 2.0 x 10

-10

The molar solubility of a salt in water can be formed by setting up an equilibrium table and

solving for x. The solubility is the same quantity expressed in g/L (rather than M = mol/L).

______________________________________________________________________________

Example 1: Calculate (a) the molar solubility and (b) the solubility of SrF2(s) in water.

Solution: We first set up an equilibrium table:

Balanced Equation SrF2(s) Sr2+

(aq) + 2 F –

(aq)

Initial Concentrations (M) co 0 0

Change (M) - x x 2x

Equilibrium Concentrations (M) (co - x) x 2x

To determine the solubility we use the equilibrium concentrations from the table and the Ksp

value given above, and solve for x:

Ksp = [Sr2+

][F –]

2 = (x)( 2x)

2 = 4x

3 = 2.0 x 10

-10

x3 = 5.0 x 10

-11

x = 3.7 x 10-4

= molar solubility

To find the solubility we use the molar mass to convert the molar solubility to g/L:

? g/ L 3.7104 mol SrF2

1 L

108.62 g SrF 2

1 mol SrF2

4.010

2 SrF2 /L

______________________________________________________________________________

Predicting Precipitation

Away from equilibrium, the ion product Qsp may be defined. As always, Q has the same form as

K, but typically involves non-equilibrium concentrations. Precipitation can be predicted by

comparing Qsp with Ksp:

Relationship Solution Type Result

Qsp < Ksp Unsaturated More salt can dissolve without ppt forming

Qsp = Ksp Saturated No more salt can dissolve

Qsp > Ksp Supersaturated Salt will precipitate until Qsp = Ksp

77

Chemical Separations

If two precipitates are possible, the least soluble salt will precipitate first (the one with the

smaller Ksp). For example, if you add NaOH(aq) to a solution containing equal amounts of Ca2+

and Mg2+

, Mg(OH)2(s) will precipitate before Ca(OH)2(s) since Ksp(Mg(OH)2) = 1.2 x 10-11

is

less than Ksp(Ca(OH)2) = 8.0 x 10-6

. The reason is that for Mg(OH)2, Qsp > Ksp will occur at lower

[OH-] since the Ksp is smaller. If the Ksp values for two salts differ widely, the salts may be

separated by fractional precipitation.

Reduction in Solubility due to the Common Ion Effect

Addition of an ion “common” to a solubility equilibrium will reduce solubility. This can be

predicted in a qualitative way using Le Chatelier‟s Principle. For example, adding fluoride ion,

F–(aq), to the SrF2(s) equilibrium above will shift it left. The shift will increase the amount of

SrF2(s) in solid form, and thus decrease solubility.

The new solubility can be calculated as illustrated in the following example.

______________________________________________________________________________

Example 2. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?

Again we set up an equilibrium table, but now we have an initial concentration of fluoride ion:

Balanced Equation SrF2(s) Sr2+

(aq) + 2 F –

(aq)

Initial Concentrations (M) co 0 0.10

Change (M) - x x 2x

Equilibrium Concentrations (M) (co - x) x (0.10 + 2x)

We next use the equilibrium concentrations in the table and the Ksp value given above, and solve

for x. Note that since x is small, (0.10 + 2x) 0.10.

Ksp = [Sr2+

][F –]

2 = (x)( 0.10 + 2x)

2 (x)( 0.10)

2 = 0.010 x = 2.0 x 10

-10

x = 2.0 x 10-8

= molar solubility

______________________________________________________________________________

Note that as predicted by the common ion effect, this solubility is much lower than what we

calculated in Example 1 for pure water!

Increase in Solubility due to addition of a Species that Reacts with an Ion

The solubility of a salt will increase if a species is added which reacts with one of its ions. Once

again this is an example of shifts predicted by Le Chatelier‟s Principle. For example if the anion

reacts with the added substance, the concentration of the anion will be reduced. Thus the

equilibrium will shift right to “undo, in part, the disturbance”. The shift right will reduce the

amount of salt in solid form, and thereby increase its solubility. There are two common

examples of this phenomena:

(1) Reaction of the basic anion of a salt with a strong acid. Salts such as Mg(OH)2(s), BaCO3(s),

and NiS(s) are much more soluble in a strong acid solution than in water. On the other hand,

the solubility of AgCl(s) does not increase when 6 M HNO3 is added.

78

(2) Reaction of the acidic cation (Lewis acid) of a salt with a Lewis base to form a soluble

complex ion. Salts such as AgCl(s) and Cu(OH)2(s) are much more soluble in 6 M NH3(aq)

solution than in pure water due to the formation of Ag(NH3)2+(aq) and Cu(NH3)4

2+(aq)

complex ions respectively. Likewise, amphoteric hydroxides such as Al(OH)3(s) and

Zn(OH)2(s) are soluble in 6 M NaOH(aq) due to the formation of Al(OH)4-(aq) and

Zn(OH)42-

(aq) ions.

Thermodynamics and Equilibrium: Important Equations

Second Law of Thermodynamics

Entropy S is a measure of the disorder of a system: the greater the disorder the greater the

entropy. The second law of thermodynamics tells us the direction of spontaneous (natural)

change in the universe. The law states that in any spontaneous change the entropy of the

universe increases, and in an equilibrium process it remains unchanged: Suniverse 0 (where the

equality applies at equilibrium). A process for which Suniverse 0 is impossible (much like

going backwards in time)!

Gibbs Free Energy

A much more convenient criteria for spontaneity which can be derived from the second law

involves changes in Gibbs free energy G, where GH – TS. A change is spontaneous (product-

favored) in the direction written at a constant temperature T and constant pressure p if ∆G for the

change is negative. Mathematically, for a spontaneous change we must have

G p,T 0 (1)

At equilibrium, G p,T 0 . If G p,T 0 , the change cannot occur in the direction written

(but will be spontaneous in the opposite direction).

There are a number of ways to calculate ∆G . Since GH – TS, when the temperature T is

constant

G H TS (2)

Go H

o TS

o (3)

In these equations (and in all which follow), the temperature is in Kelvin. At standard conditions

(superscript “o” means standard conditions):

Go nprod

products

G f

oprod nreact

reactants

G f

oreact (4)

Ho nprod

products

H f

oprod nreact

reactants

H f

oreact (5)

So nprod

products

So

prod nreact

reactants

So

react (6)

79

Another method of calculating ∆G follows from the connection between the criteria for

spontaneous change, G p,T 0 , and our earlier Q < K criteria. Recall that when Q < K,

reaction is predicted to proceed forward until Q = K. It can be shown that

G RT ln K RT ln Q (7)

Note that this equation predicts (as required) that when Q < K, ∆G < 0. Also, when Q = K, ∆G =

0.

By definition at standard conditions all reactants and products are in their standard states:

they are pure if solid or liquid, and at 1 M concentration if dissolved. It follows that at standard

conditions, Q = 1 and ln Q = 0. Thus, Eq(7) reduces to

Go RT lnK (8)

where, as always, the “o” superscript means standard conditions. Rearranging Eq(8) gives us a

convenient equation for calculating K:

ln K G

o

RT or K e

Go / RT (9)

Finally, Eq(7) and Eq(8) may be combined to give

G Go RT lnQ (10)

The table below summarizes our equations for G under various conditions. Note how the

more specialized equations are easily derived from those for the general case.

General Case

G RT ln K RT ln Q

G Go RT ln Q

G H TS

Equilibrium Conditions

Q K

G 0

H TS

Standard Conditions

Q 1

G Go; H H

o; S S

o

Go RT ln K

Go H

o TS

o

Third Law of Thermodynamics

You may have noticed the difference in notation in the So equation, Eq(6), as compared to

that in Eq(4) and Eq(5). The reason for the difference is that entropy values are absolute. They

are based on the third law of thermodynamics which says that the entropy of a perfect crystalline

substance is zero at the absolute zero of temperature (0 K). We therefore assign the value of zero

80

entropy to perfect crystals at absolute zero. All non-perfect crystals and all substances (including

pure elements) at temperatures above 0 K have positive entropy. Entropy increases with

temperature, since disorder increases when the temperature is raised.

In contrast the zero of enthalpy and zero of free energy are undefined. We follow the

convention of setting the heat of formation H f

o and free energy of formation Gf

o of a pure

element in its standard state equal to zero.

How to Balance Equations for Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change.

Oxidation numbers are either real charges or formal charges which help chemists keep track of

electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device.

Oxidation cannot occur without reduction. In a redox reaction the substance which is

oxidized contains atoms which increase in oxidation number. Oxidation is associated with

electron loss (helpful mnemonic: LEO = Loss of Electrons, Oxidation). Conversely, the

substance which is reduced contains atoms which decrease in oxidation number during the

reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of

Electrons, Reduction).

Chemists often talk about oxidizing and reducing agents. Be careful with these terms!

An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a

reducing agent is a substance that reduces another reactant: it itself is oxidized. A

disproportionation reaction is a reaction in which the same element is both oxidized and reduced.

How to Assign Oxidation Numbers: The Fundamental Rules

Rules for assigning oxidation numbers are as follows:

• The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is zero.

• The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of

Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is -2.

• The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if an

ion.

• The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in

compounds is +2. The oxidation number of F is -1 in all its compounds.

• The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and the

ionic hydrides, such as NaH (where H = -1).

• The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O = 0)

and peroxides, such as H2O2 or Na2O2, where O = -1.

• For other elements, you can usually use rule (3) to solve for the unknown oxidation number.

Examples:

NO(g) has O = -2, so N = +2.

NO2(g) has O = -2, so N = +4.

SO42-

has O = -2. Thus S + 4(-2) = -2. Solving the equation gives S = -2 + 8 = +6.

K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(-2) = 0; 2 Cr = 12; Cr = +6.

81

How to Balance Redox Reactions Using the Method of Half-Reactions

Oxidation-reduction reactions are often tricky to balance without using a systematic

method. We shall use the method of half-reactions which is outlined in detail below.

Method in Acidic (or Neutral) Solution

Suppose you are asked to balance the equation below:

NO2– + MnO4

– NO3– + Mn+2 (in acid solution)

Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to

know which is which):

NO2– 3

–

MnO4– Mn+2

Next, balance for atoms. First do this for atoms other than O and H. (Both equations above are

already balanced for N and Mn, so no change is needed in this example.) Then balance for O

atoms by adding H2O to the reaction side deficient in O:

H2O + NO2– 3

–

MnO4– Mn+2 + 4 H2O

This leaves H atoms unbalanced. In acidic (or neutral) solution, balance for H atoms by adding

H+ to the side deficient in H:

H2O + NO2– 3

– + 2 H+

8 H+ + MnO4– Mn+2 + 4 H2O

The next step is to balance for charge. To do this, add electrons (e-) to the more positive side:

H2O + NO2- 3

- + 2 H+ + 2 e-

5 e- + 8 H+ + MnO4- Mn+2 + 4 H2O

Now you need to multiply the equations by appropriate factors so that the number of electrons

lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the

reduction half-reaction (GER):

5 x [ H2O + NO2- 3

- + 2 H+ + 2 e- ]

2 x [ 5 e- + 8 H+ + MnO4- Mn+2 + 4 H2O ]

Then, sum the above equations to obtain

5H2O + 5NO2- + 10 e- + 16H+ + 2MnO4

- 3- + 10H+ + 10 e- + 2Mn+2 + 8H2O

82

Finally, simplify by subtracting out species that are identical on both sides. Our final balanced

redox equation is

5 NO2- + 6 H+ + 2 MnO4

- 3- + 2 Mn+2 + 3 H2O

Check this equation to confirm that it is balanced for atoms and balanced for charge.

Method in Basic Solution

Suppose you are asked to balance the equation below:

I- + MnO4- 2

+ MnO2 (in basic solution)

Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to

know which is which):

I- 2

MnO4- MnO2

Next, balance for atoms. First do this for atoms other than O and H:

2 I- 2

MnO4- MnO2

Then balance for O atoms by adding H2O to the reaction side deficient in O:

2 I- 2

MnO4- MnO2 + 2 H2O

This leaves H atoms unbalanced. In basic solution (just as in acidic or neutral solution) first

balance for H atoms by adding H+ to the side deficient in H:

4 H+ + MnO4

- MnO2 + 2 H2O

In basic solution, follow this step by neutralizing the H+; do this by adding an equivalent amount

of OH- to both sides of the equation.

4 OH- + 4 H

+ + MnO4

- MnO2 + 2 H2O + 4 OH-

Then form water on the side which has both H+ and OH

- (recall that H

+ + OH

- H2O): in this

case we form 4 H2O on the left:

4 H2O + MnO4- MnO2 + 2 H2O + 4 OH-

Next simplify the water by subtracting 2 H20 from both sides. The half-reactions are now:

2 I– 2

83

2 H2O + MnO4– MnO2 + 4 OH

–

At this point the equations should be balanced for atoms. The next step is to balance for

charge. To do this, add electrons (e–) to the more positive side:

2 I– 2

+ 2 e–

3 e– + 2 H2O + MnO4

– MnO2 + 4 OH–

Now you need to multiply the equations by appropriate factors so that the number of electrons

lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the

reduction half-reaction (GER):

3 x [ 2 I- I2 + 2 e- ]

2 x [ 3 e- + 2 H2O + MnO4- MnO2 + 4 OH- ]

Sum the equations to obtain

6 I- + 6 e- + 4 H2O + 2 MnO4- I2

+ 6 e- MnO2 + 8 OH-

Finally, simplify by subtracting out species that are identical on both sides:

6 I- + 4 H2O + 2 MnO4- I2

+ 2 MnO2 + 8 OH-

Check our final equation above to confirm that it is balanced for atoms and balanced for charge.

Exercises:

Balance the following redox reactions. In each case

• (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction

half-reaction.

• (b) give the balanced net reaction.

• (c) identify the oxidizing agent and the reducing agent.

______________________________________________________________________________

1. Cl2(g) + S2O32-(aq) Cl-(aq) + SO42-(aq) in acid solution.

Answers:

(a) S2O32-(aq) + 5 H20 2 SO42-(aq) + 10 H+(aq) + 8 e- (oxidation half-reaction – LEO);

Cl2(g) + 2 e- 2 Cl-(aq) (reduction half-reaction – GER).

(b) S2O32-(aq) + 5 H20 + 4 Cl2(g) 2 SO42-(aq) + 10 H+(aq) + 8 Cl-(aq)

(c) S2O32-(aq) is the reducing agent; Cl2(g) is the oxidizing agent.

______________________________________________________________________________

84

2. O3(g) + Br-(aq) O2(g) + BrO-(aq) in basic solution.

Answers:

(a) Br - (aq) + H20 + 2 OH-(aq) BrO-(aq) + 2H2O + 2 e- or, after simplifying,

Br - (aq) + 2 OH-(aq) BrO-(aq) + H2O + 2 e- (oxidation half-reaction – LEO);

O3(g) + 2 H2O + 2 e- O2(g) + H2O + 2 OH-(aq) or, after simplifying,

O3(g) + H2O + 2 e- O2(g) + 2 OH-(aq) (reduction half-reaction – GER).

(b) Br - (aq) + O3(g) BrO-(aq) + O2(g)

(c) Br - (aq) is the reducing agent; O3(g) is the oxidizing agent.

______________________________________________________________________________

3. Balance the reaction, Br2(l) Br-(aq) + BrO3-(aq) in basic solution. Hint: this is a

disproportionation reaction!

Answer: 6 Br2(l) + 12 OH-(aq) 10 Br-(aq) + 2 BrO3-(aq) + 6 H2O

Factors Affecting Corrosion

Much of the current concern in this country and throughout the world about the need to

replace the "infrastructure" is related to the rusting of iron. It has been estimated that 25% of the

annual steel production in the United States goes towards the repair and replacement of metals

lost by corrosion. Bridges are in need of being replaced, steel rods in the cement of older

buildings and parking ramps are deteriorating, and the rusting of many machine parts causes

concern for equipment reliability.

What is Corrosion?

Corrosion is a general term which refers to the undesired oxidation of metals to oxides

and other compounds. When the metal is iron, the corrosion process is called rusting. While

iron is structurally strong, the iron oxides formed as the products of rusting are not. Other metals

such as copper and silver become tarnished from surface oxidation of the metal. Billions of

dollars must be spent annually to replace materials that have corroded. The greatest damage is

from the rusting of iron and steel. Thus prevention of the corrosion of iron is a topic of immense

practical importance.

The surface of metals can provide a voltaic (galvanic) cell if there is a difference in

potential between two or more points on the metal. A galvanic cell can be established in the

presence of water vapor and impurities in the metal or stressed areas created in the process of

85

machining something as simple as a nail.

The metal which is oxidized (corroded) may go into solution as ions. For example, in the

case of iron, Fe2+

ions can result: Fe(s) Fe2+

(aq) + 2e -. Air can then further oxidize the Fe

2+

ions to rust or Fe2O3.n H2O. Corrosion is often localized and limited to small areas on the

surface of the metal.

Half-Reactions in the Corrosion of Iron

Since corrosion reactions are oxidation-reductions reactions, they may be resolved into two

half-reactions. The oxidation half-reaction represents the oxidation of the metal. In the case of

iron it is

Fe(s) Fe2+

(aq) + 2e - Eoox = 0.44 v (1)

The reduction half-reaction involved in corrosion in an aqueous environment depends on the pH

and on the availability of oxygen. Possibilities, with their standard half-cell reduction potentials,

are

2 H2O + 2 e- H2(g) + 2 OH-(aq) Eored = - 0.83 v (- 0.42 v at pH=7) (2)

2 H+ + 2 e- H2(g) Eored = 0.00 v (- 0.42 v at pH=7) (3)

O2(g) + 2 H2O + 4 e- 4 OH-(aq) Eored = 0.40 v (0.81 v at pH=7) (4)

O2(g) + 4 H+ + 4 e- 2 H2O Eored = 1.23 v (0.81 v at pH=7) (5)

Note that the reduction potential of each of these half-reactions gets more positive as the [H+]

increases and more negative as it decreases. (Why? Use Le Chatelier's Principle!)

Which one of these reduction half-reactions actually occurs may be deduced experimentally.

If Eq(2) or Eq(3) is the reduction half-reaction, bubbles of H2 gas will be visible in the reaction

mixture. Also note that Eq(2) and Eq(4) both have OH- as a product which will turn

phenolphthalein indicator pink.

Water in contact with air contains sufficient dissolved oxygen to provide the reactants for

Eq(4). Thus the net reaction for rusting under normal conditions is obtained by combining Eqs(1)

and (4) to give

2 Fe(s) + O2(g) + 2 H2O 2 Fe(OH)2(s) (6)

The iron(II) hydroxide formed is then further oxidized to iron(III) hydroxide:

4 Fe(OH)2(s) + O2(g) + 2 H2O 4 Fe(OH)3(s) (7)

Iron(III)hydroxide is an idealized formula, often written as Fe2O3.3H2O. Variable amounts of

water may be removed from it to form the material commonly called rust, Fe2O3(H2O)x.

Methods of Preventing Corrosion

86

So what can we do to prevent corrosion? Surely we have found ways to prevent it, or at least to

slow it down.

(3) We can plate a metal with an inert material, like chromium, provided there are no cracks in

the plating to allow corrosion to begin. The tin coating of iron on “tin” cans acts in a similar

way. However, once the surface of a tin can is scratched, rusting occurs rapidly with iron

acting as the anode and the tin as the cathode.

(4) We can coat with another metal, like zinc, which oxidizes more easily than iron-based metals

and so is "sacrificed" to prevent corrosion of the metal coated. Zn(s) acts as a “sacrificial

anode” and is oxidized preferentially to iron; thus no rusting occurs even if the Zn(s) coating

is scratched. “Galvanized” iron is iron coated with Zn(s).

(5) Some metals form their own protective coating which prevents or slows down corrosion.

Aluminum is very easily oxidized and when exposed to air forms a very stable coating of

Al2O3(s) which protects the Al(s) underneath from corrosion. Copper becomes coated with

CuCO3(s) which gives it a green “patina” and protects it somewhat from further corrosion.

(6) There are also additional protective coatings, like the paint used on titanium alloys in the

racing bicycles, which prevent rapid corrosion and slow ultimate deterioration.

(7) Annealing (heating of material followed by slow cooling) also slows down the rate of

corrosion.

What Factors Increase the Rate of Corrosion?

On the other hand, salt increases the rate of corrosion. Think of the bridges and cars

attacked by ocean spray, or by putting salt on highways in the areas of the country where snow

and ice need to be melted in the winter. As is well known, acid rain also greatly increases the

rate of corrosion. In addition, places where a metal has been worked or stressed corrode most

rapidly; examples include the head or tip of a nail, or where a nail is bent.

Use of Indicators to Determine Products of Corrosion

Two indicators may be used to give evidence of corrosion, or protection from corrosion.

The first is phenolphthalein, an acid/base indicator which is colorless in pHs below 8.8 and pink

in pHs above that reading. The second is hexacyanoferrate(III) ion in the form of

K3Fe(CN)6(aq), an indicator which turns a deep blue in the presence of Fe2+

due to the reaction

Fe2+(aq) + K+(aq) + Fe(CN)63- KFe[Fe(CN)6](s) (8)

(from iron oxidation) (yellow solution) (deep blue precipitate)

These indicators enable us to detect products of Eqs(1)-(5) above. You will find them useful in lab!

Electrolysis

Many nonspontaneous reactions may be forced to occur electrolytically. All that is

required are two electrodes, a battery, and a conducting solution. Graphite pencil leads work

well as electrodes since they are excellent conductors and are non-reactive. To produce an

87

electrolytic cell, graphite electrodes are hooked up to the terminals of a nine-volt battery and then

dipped in an electrolyte solution in a Petri dish.

A battery is a voltaic cell and thus has a positive cell potential (Ecell). The spontaneous

reaction occurring in the battery pumps electrons out of the negative battery terminal and pulls

electrons into the positive battery terminal. Thus reduction in the solution in the Petri dish

occurs at the electrode attached to the (-) terminal, and oxidation occurs at the electrode attached

to the (+) terminal.

The cell potential of the solution being electrolyzed is always negative since the reaction

is nonspontaneous. If the battery is to be successful in forcing this nonspontaneous reaction to

occur, the battery voltage must be large enough to overcome the negative electrolytic cell

potential so that the sum of the battery potential and electrolytic cell potential is greater than zero.

In practice, some "overvoltage" is also required. A nine volt battery gives sufficient voltage to

electrolyze most solutions of interest.

Predicting Electrolysis Products

To predict the products of electrolysis, begin by writing down all the possible reactants

that are present in significant amounts in the solution being electrolyzed (we assume here that the

electrodes themselves are inert). A possible reactant that it is easy to forget about is water!

Water can be either oxidized or reduced (or both!):

Reduction: 2 H2O + 2 e- H2(g) + 2 OH-(aq) Eored = - 0.83 v (1)

( - 0.42 v at pH = 7)

Oxidation: 2 H2O O2(g) + 4 H+(aq) + 4 e- Eoox = - 1.23 v (2)

( - 0.81 v at pH = 7)

Note that both of these reactions have a gaseous product, but the reduction of water produces

OH -

(aq) while the oxidation of water produces H+(aq).

Other possible reactants are the dissolved anions and cations in the solution. Metal ions

are good candidates for reduction while halide ions are good candidates for oxidation:

2 Cl-(aq) Cl2(g) + 2 e- Eoox = - 1.36 v (3)

2 Br-(aq) Br2(l) + 2 e- Eoox = - 1.07 v (4)

2 I-(aq) I2(s) + 2 e- Eoox = - 0.53 v (5)

All three halogens are easily detected. Chlorine, Cl2(g), is recognized by the observation of gas

bubbles and its distinctive odor. Bromine dissolved in water gives a yellowish color, while in

cyclohexane it turns orange. Iodine dissolved in water gives a reddish-brown color, while it turns

lavender in cyclohexane. If sufficient iodine is produced, crystals of I2(s) will form on the anode:

these look metallic.

If more than one reactant is available for reduction, the one with the highest Eored potential

88

is most likely to be reduced. Similarly, if more than one reactant is available for oxidation, the

one with the most positive (least negative) Eoox potential is most likely to be oxidized. However,

if two possible reduction reactions have similar Eored values, kinetic effects may become the

deciding factor; either reduction reaction, or both, may occur. The same applies if two possible

oxidation reactions have Eoox values which are close. Careful observation and testing of products

is essential to determine which oxidation and reduction reactions have occurred.

______________________________________________________________________________

Exercises

1. A student has run an electrolysis reaction of an aqueous solution of MgBr2(aq) in a petri dish

using inert electrodes and a 9 volt battery. Observations are made and the following data is

recorded:

At the electrode attached to the (-) terminal of the battery, bubbles are observed. A white

precipitate forms around the electrode.

At the electrode attached to the (+) terminal of the battery, a small amount of bubbles are also

observed. The solution near the electrode has turned yellowish. When a few drops of the

yellowish liquid are mixed with cyclohexane, an orange layer forms at the top of the mixture.

When a few drops of phenolphthalein are added to the petri dish, a magenta pink color is

observed near the electrode attached to the (-) terminal of the battery.

Account for all the observations above. Write equations for the redox reactions which occurred

and account for the colors, precipitate, and bubbles.

Answer:

At the cathode (the (-) terminal) the possible reductions are:

2 H2O + 2 e- H2(g) + 2 OH-(aq) Eored = - 0.83 v ( - 0.42 v at pH = 7)

Mg+2

+ 2 e - Mg(s) Eored = - 2.37 v

At the anode (the (+) terminal) the possible oxidations are:

2 Br – (aq) Br2(l) + 2 e- E

oox = - 1.07 v

2 H2O O2(g) + 4 H+(aq) + 4 e- Eoox = - 1.23 v ( - 0.81 v at pH = 7)

The reduction reaction which occurs is usually the one with the more (+) Eored, and the oxidation

reaction is usually the one with the more (+)Eoox. Thus here we expect that H2O is reduced at the

cathode and Br -

(aq) is oxidized at the anode. The data indicates, however, that some H2O is

also oxidized at the anode. This is not surprising since the Eoox values are close.

The bubbles at the (-) terminal are due to H2(g) and the white precipitate is Mg(OH)2(s). The

89

magenta pink color at the (-) electrode is due to phenolphthalein which turns that color in the

presence of OH –

(aq), one of the products of the reduction of H2O.

The small amount of bubbles at the (+) electrode are O2(g). The other product at the (+)

electrode is Br2 which is yellowish in water but turns orange when dissolved in an organic

solvent such as cyclohexane. The cyclohexane layer is on top since cyclohexane is less dense

than water.

______________________________________________________________________________

2. Suppose that molten MgBr2(l) is electrolyzed. What reactions would take place, and what is

the minimum emf required for reaction?

Answer: No water is present in a molten salt, so only the salt ions are available to react. Thus,

Mg+2

would be reduced: Mg+2

+ 2 e - Mg(s) Eored = - 2.37 v

(e) Br – would be oxidized: 2 Br

– (aq) Br2(l) + 2 e

_ E

oox = - 1.07 v

The minimum emf required is equal to - Eocell = - (E

ored + E

oox) = - (- 2.37 v + - 1.07 v) = + 3.44

v

Sources: http://spiepho.sbc.edu

GENERAL CHEMISTRY - · PDF file1 GENERAL CHEMISTRY 1. Scientific Notation, Significant...

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