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3rd 9-weeks Notes (The end is near…in more ways than one.)HW # Handouts

I. Spontaneous Processes and Entropy.A. A spontaneous process occurs without outside

intervention; it may either be fast or slow.1. Examples:

a) A ball rolls down a hill but never spontaneously rolls back

up the hill.

b) If exposed to air and moisture, steel rusts spontaneously but the reverse process does not

naturally occur.

c) A gas fills a container uniformly. It never spontaneously

collects at one end of the container.

d) Heat always flows from a hot object to a cold one. The

reverse process never occurs spontaneously.

e) Wood burns exothermically to form carbon dioxide and

water, but wood is not formed when the two compounds

are heated together.

f) At temperatures below 0 C, water freezes. At temperatures above 0 C, water melts.

2. Originally, chemists thought spontaneity was reflective of exothermic reactions only.

3. Scientists have since concluded that the common characteristic of all spontaneous reactions is an increase in entropy.

a) Entropy (S) is said to be a measure of randomness or

disorder of a system.* No accurate, simplistic definition exists.

b) Entropy describes the number of possible

arrangements that are available to a system existing in a given

state.

c) In nature, processes spontaneously proceed toward the

states that have the highest existing probabilities of existing.

d) Ssolid < Sliquid << Sgas

B. Positional Entropy.For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature.1. Solid CO2 and gaseous CO2

2. N2 gas at 1 atm and N2 at 1.0 x 10-2 atm

C. Predicting Entropy Changes.Predict the sign of the entropy change for each of the following processes.

1. Solid sugar is added to water to form a solution.

2. Iodine vapor condenses on a cold surface to form crystals.

II. Entropy and the Second Law of Thermodynamics.A. Reviewing the First Law of Thermodynamics.

1. The energy of the universe is constant; energy cannot be created or destroyed but changed from one form to another.

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2. The first law provides a means for accounting for energy, it gives no hint as to why a particular process occurs in a given direction.

B. The Second Law of Thermodynamics.1. This law states that in any spontaneous process there is always an increase in the entropy of the universe.

2. Energy in the universe is conserved, but entropy is not. The entropy of the universe is increasing.

Suniv = Ssys + Ssurr

3. To predict spontaneity, we must know the sign of Suniv

a. If Sunivthenthe entropy of the universe increases and the process is spontaneous in the

direction written.b. If Suniv < 1, then the entropy of the universe decreases and the process is spontaneous in the

opposite direction.

c. If Suniv = 0, then the process has no tendency to occur and the system is at equilibrium.

d. In a living cell, large molecules are assembled from smaller ones. Is this process consistent with the

second law?

III. The Effect of Temperature on Spontaneity.A. Consider the following:

H2O(l) ------> H2O(g)

Water is the system, everything else is the surroundings.

1. For liquid water to vaporize, heat must be added (endothermic). Explain in terms of entropy.

2. Explain condensation in terms of entropy.

B. Conclusions.1. Entropy changes in the surroundings are determined by heat flow and determines the sign.

a. Endothermic:

b. Exothermic:

2. The magnitude of Ssurr depends on the temperature.a. Endothermic:

b. Exothermic:

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3. The impact of the transfer of a given quantity of energy as heat to and from the surroundings will be greater at lower temperatures.

* Example:

4. Ssurr = _ H sys T

5. Determining Ssurr.* In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in the sulfide ores:

Sb2S3(s) + 3Fe(s) ----> 2Sb(s) + 3FeS(s) H = -125 kJ

Carbon is used as the reducing agent for oxide ores:

Sb4O6(s) + 6C(s) ----> 4Sb(s) + 6CO(g) H = 778 kJ

Calculate Ssurr for each of these at 25 C and 1 atm.

C. Summary.1. For an exothermic process, Ssurrpositive.

2. For an endothermic process, Ssurrnegative.

3.

IV. Free Energy.A. Free energy (G) is a function that combines the system’s

enthalpy and entropy.G = H - TS

B. The free energy change (G) is a measure of the spontaneity of a process and of the useful energy available from it.

G = H - TS

1. All of these quantities reflect the system (anytime in this chapter that a subscript is not noted, it is understood to be from the system’s perspective).

2. How does the equation relate to spontaneity?- G = - H + S and Ssurr = - H T T T

therefore

- G = - H + S = Ssurr + S = Suniv T T

3. Conclusion: A process (at constant T and P) is spontaneous in the direction in which the free energy decreases (-G means + Suniv).

4. Two functions predict spontaneity.a. The entropy of the universe (applies to all

processes).

b. Free energy (at constant T and P).* More useful to chemists.

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5.

6. Free Energy and Spontaneity.* At what temperature is the following process

spontaneous at 1 atm?

Br2(l) -----> Br2(g)

Ho = 31.0 kJ/mol and So = 93.0 J/K*mol

What is the normal boiling point of liquid Br2?

V. Entropy Changes in Chemical Reactions.A. Standard Molar Entropies, So

1. In thermodynamics it is the change in a certain function that is usually important.

* Absolute values for H and G cannot be determined.

2. The third law of thermodynamics states that at 0 K, the entropy of a pure crystal is equal to 0.

* Because this provides a starting point to compare all other entropies, an absolute entropy scale has

meaning.

3. Standard molar entropy, So, is the entropy of one mole of

a substance in its standard state, which occurs when a substance is heated from 0 K to 298 K at 1 atm of pressure.

B. Predicting Relative So Values of the System.1. Temperature changes.

* For a given substance, So increases as the temperature

rises.

2. Physical states and phase changes.* For a given substance, So increases as the substance

changes from a solid to a liquid to a gas (the change from

a liquid to a gas is greater than from a solid to a liquid).

3. Dissolution of a solid or a liquid.a. The entropy of a dissolved solid or liquid solute is greater than the entropy of the pure solute.

b. The type of solute and solvent and the nature of the solution process affects the overall entropy change.

4. Dissolution of a gas.* A gas always becomes more ordered when dissolved

in a liquid and gas.

5. Complexity of the element or compound.a. In general, difference in entropy values for

substances in the same phase are based on atomic size and

molecular complexity.

b. For elements within a periodic group, those with higher

molar masses have higher entropy.

c. For compounds, the chemical complexity increases as

the number of atoms (ions) in a compound increases,

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and so does the entropy.

d. For larger molecule, entropy increases with chain length.

6. Predicting the Sign of So.* Predict the sign of So for each of the following reactions. a) The thermal decomposition of solid calcium

carbonate:CaCO3(s) -----> CaO(s) + CO2(g)

b) The oxidation of SO2 in air:2SO2(g) + O2(g) -----> 2SO3(g)

7. Predicting Relative Entropy Values.* Choose the member with the higher entropy in each

of the following pairs, and justify your choice. a) 1 mol SO2(g) or 1 mol SO3(g)

b) 1 mol CO2(s) or 1 mol CO2(g)

c) 3 mol O2(g) or 2 mol O3(g)

d) 1 mol KBr(s) or 1 mol KBr(aq)

e) Sea water in midwinter at 2 C or in mid summer at 23 C

f) 1 mol HF(g) or 1 mol HI(g)

C. Calculating the Change in Entropy of a Reaction.1. Because entropy is a state function, the entropy change for a given reaction can be calculated by taking the difference between the standard entropy values of products and

those of the reactants.

Soreaction =npSoproductsnrSoreactants

2. Calculating So.a. Calculate So at 25 C for the reaction 2NiS(s) + 3O2(g) -----> 2SO2(g) + 2NiO(s)

Use the standard entropy values found in the appendix.

b. Calculate So for the reaction of aluminum oxide by hydrogen gas: Al2O3(s) + 3H2(g) -----> 2Al(s) + 3H2O(g)

Use the standard entropy values found in the appendix.

VI. Free Energy and Chemical Reactions.A. Standard Free Energy.

1. Standard free energy (Go) is the change in the free energy that will occur if the reactants in their standard states are converted to the products in their standard states.

2. The value of Go tells nothing about the rate of a reaction, only its eventual equilibrium position.

B. Calculating Go as a State Function.1. GoHoTSo

* Consider the reaction 2SO2(g) + O2(g) -----> 2SO3(g)

carried out at 25 C and 1 atm. Calculate Ho and So,

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then calculate Go. Use the appendix in this book for standard values.

2. Solving Go Using Hess’s Law.* Using the following data (at 25 C) Cdiamond(s) + O2(g) -----> CO2(g) Go = -397 kJ Cgraphite(s) + O2(g) -----> CO2(g) Go = -394 kJ

Calculate Go for the reactionCdiamond(s) -----> Cgraphite(s)

3. Standard Free Energy of Formation (Gfo).Go =npGfoproductsnrGforeactants

* Methanol is a high-octane fuel used in high-performance

racing engines. Calculate Go for the reaction 2CH3OH(g) + 3O2(g) -----> 2CO2(g) + 4H2O(g)

given the free energies of formation in this book.

4. Free Energy and Spontaneity.* A chemical engineer wants to determine the

feasibility of making ethanol (C2H5OH) by reacting water with

the ethylene (C2H4) according to the equation

C2H4(g) + H2O(l) -----> C2H5OH(l)

Is the reaction spontaneous under standard conditions?

VII. The Dependence of Free Energy on Pressure.A. The equilibrium position represents the lowest free energy

value available to a particular reaction.* Free energy changes throughout the course of a reaction

because it is pressure and concentration dependent.

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B. The Effects of Pressure.1. Consider 1 mole of a gas at a given T. Slarge volume > Ssmall volume or Slow pressure > Shigh pressure

2. A more detailed argument (beyond this course) states that

G = Go + RTln(P) where Gois the free energy of a gas at 1 atm, G is the free energy at pressure P, R is the universal gas constant, and T is the Kelvin temperature.3. This leads to the equation

G = Go + RTln(Q) where Q is the reaction quotient, T is the Kelvin temperature, R is the universal gas constant 8.314 J/mol K, Go is the free energy at 1 atm, and G is the free energy at a specified temperature.

4. Calculating Go.* One method for synthesizing methanol (CH3OH)

involves reacting carbon monoxide and hydrogen gases:

CO(g) + 2H2(g) -----> CH3OH(l) Calculate G at 25 C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0

atm are converted to liquid methanol.

C. The Meaning of G for a Chemical Reaction.1. A system achieves the lowest possible free energy by going to equilibrium, not by going to completion.

* Spontaneous reactions do not go to completion.

2. Illustration.

VIII. Free Energy and Equilibrium.* The Equilibrium Point.

1. The equilibrium point occurs at the lowest value of free energy available to the reaction.

2. Recall the following:a. If Q < K, then the reaction as written proceeds to the right (Q/K <1).

b. If Q > K, then the reaction as written proceeds to the left (Q/K > 1).

c. If Q = K, then the reaction as written is at equilibrium,

and there is no net reaction in either direction (Q/K = 1).

3. The sign of G and the magnitude of Q/K are related mathematically.

a. If Q/K < 1, then ln(Q/K) < 0; the reaction proceeds to the right (G < 0).

b. If Q/K > 1, then ln(Q/K) > 0; the reaction proceeds to the left (G > 0).

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c. If Q/K = 1, then ln(Q/K) = 0; the reaction is at equilibrium (G = 0).

4. Conclusions.a. For nonstandard conditions NOT at equilibrium,

G = Go + RTln(Q)

b. For standard conditions at equilibriumGo = -RTln(K)

5. Free Energy and Equilibrium I.* Consider the ammonia synthesis reaction

N2(g) + 3H2(g) <====> 2NH3(g)

where Go = -33 kJ per mole of N2 consumed at 25 C. For each of the following mixtures of reactants and products

at 25 C, predict the direction in which the system will shift to reach equilibrium.

a) PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00 x 10-2 atm

b) PNH3 = 1.00 atm, PN2 = 1.00 atm, PH2 = 1.00 atm

6. Free Energy and Equilibrium II.* The overall reaction for the corrosion (rusting) of iron

by oxygen is4Fe(s) + 3O2(g) <====> 2Fe2O3(s)

Using Hfo and So values in the appendix, calculate

the equilibrium constant for this reaction at 25 C.

IX. Free Energy and Work.A. For a spontaneous reaction, G is the maximum work

obtainable from the system.wmax = G

B. For a nonspontaneous process, G is the minimum work that must be done to the system to make a change happen.

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C. For Gsys = Hsys - TSsys, Gsys is the portion of the total energy change that does the work.* TSsys is given off as heat and is not usable.

D. In the real world, some free energy is always changed to heat and is thereby unharnessed. Therefore, no process is 100% efficient. 1. A spontaneous reaction will occur and can do work on the surroundings.

2. A nonspontaneous reaction will not occur unless the surroundings do work on it.

3. A reaction at equilibrium can no longer do work.

I. Galvanic Cells.A. The Galvanic Cell.

1. A galvanic cell is a device in which chemical energy is changed to electrical energy.

2. A galvanic cell uses a spontaneous redox reaction to produce a current that can be used to do work. The system does work on the surroundings.

3. A Review of Oxidation and Reduction and the Galvanic Cell.a. A redox reaction involves the transfer of electrons from the reducing agent to the oxidating agent.

b. Oxidation.* Involves a loss of electrons.

* Increase in oxidation number.

* “To get more positive.”

* Occurs at the anode of a galvanic cell.

c. Reduction.* Involves a gain of electrons.

* Decrease in oxidation number.

* “To get more negative.”

* Occurs at the cathode of a galvanic cell.

d. Identification of Redox Components.* Specify which of the following equations represents oxidation-reduction reactions, and indicate the oxidizing agent, the reducing agent, the species

being oxidized, and the species being reduced.

a) CH4(g) + H2O(g) ----> CO(g) + 3H2(g)

b) 2AgNO3(aq) + Cu(s) ----> Cu(NO3)2(aq) + 2Ag(s)

c) H+(aq) + 2CrO42-(aq) ----> Cr2O7

2-(aq) + H2O(l)

e. Half-Reaction Method for Balancing Redox Reactions.* Acidic Solution.

Cr(s) + NO3-(aq) ----> Cr3+(aq) + NO(g) [acidic]

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* Basic Solution.

MnO4-(aq) + I-(aq) ----> MnO4

2-(aq) + IO3-(aq) [basic]

f. Major Redox Points to Remember.* Any redox reaction can be treated as the sum of

the reduction and oxidation half-reactions.

* Mass (atoms) and charge are conserved in each half-

reaction.

* Electrons lost in one half-reaction are gained in the other.

* Even though the half-reactions are treated separately,

electron loss and electron gain occur simultaneously.

4. The Salt Bridge or the Porous Disk.* These devices allow ion flow to occur (circuit completion) without mixing the solutions. They are typically made of

sodium sulfate or potassium nitrate.

B. Cell Potential (Ecell)1. Cell potential (electromotive force, emf) is the driving force in a galvanic cell that pulls electrons from the reducing agent in one compartment to the oxidizing agent in the other.

2. The volt (V) is the unit of electrical potential.

3. Electrical charge is measured in coulombs (C).

4. Therefore, a volt is 1 joule of work per coulomb of charge transferred: 1 V = 1 J/C.

5. A voltmeter is a device which measures cell potential.

II. Standard Reduction Potentials.A. Standard Cell Potentials.

1. The measured potential of a voltaic cell is affected by changed in concentration of the reactants as the reaction proceeds and by energy losses due to heating of the cell and external circuit.

2. In order to compare the output of different cells, the standard cell potential (Eocell) is obtained at 298 K, 1 atm for gases, 1 M for solutions, and the pure solid for electrodes.

B. The Standard Hydrogen Electrode.1. This is considered the reference half-cell electrode, with a potential equal to 0.00 V.

2. It is obtained when platinum is immersed in 1 M H+(aq), through which H2(g) is bubbled.

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C. The Standard Electrode (Half-Cell) Potential (Ehalf-cell).1. A standard electrode potential always refers to the half-reaction written as a reduction.

Oxidized form + ne- -----> reduced form Eohalf-cell

2. Reversing a reaction changes the sign of the potential.Reduced form + ne- -----> oxidized form -Eohalf-cell

3. Eocell = Eoreduction + Eooxidation

4. As the potential increases in value (more positive), the tendency to occur increases (spontaneity occurs). Therefore, Eocell must be positive if the cell is galvanic (voltaic).

5. Spontaneous reactions occur between a substance and those below it.

6. Although some half-reactions must be manipulated with coefficients, NEVER MULTIPLY THE Eocell BY THE COEFFICIENT!!!

7. Refer to the Reduction Table in the book.

8. Problems.a. Galvanic Cells. * Consider a galvanic cell based on the reaction

Al3+(aq) + Mg(s) ----> Al(s) + Mg2+(aq)

Give the balanced cell reaction and calculate Eocell.

* Consider a galvanic cell based on the reaction

MnO4-(aq) + H+(aq) + ClO3

-(aq) ----> ClO4-(aq) + Mn2+(aq) +

H2O(l)

Give the balanced cell reaction and calculate Eocell.

b. Calculating an Unknown Eohalf-cell from Eocell. * A voltaic cell based on the reaction between aqueous Br2

and vanadium (III) ions has Eocell = 1.39 V:

Br2(aq) + 2V3+(aq) + 2H2O(l) ----> 2VO2+(aq) + 4H+(aq) + 2Br-(aq)

What is the standard electrode potential for the reduction of

VO2+ to V3+?

D. Line Notations.1. This is a short-hand abbreviation for a galvanic cell.

2. Key Parts:a. The components of the anode compartment are written to

the left of the cathode compartment.

b. Double vertical lines separate the half-cells and represents the

wire and salt bridge.

c. Within each half-cell, a single vertical line represents a phase

boundary. A comma separates half-cell components in the same

phase.

d. Half-cell components appear in the same order as in the half-

reaction, while electrodes appear at the extreme left and right

of the notation.

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3. Examples:

E. The Complete Description of a Galvanic Cell.1. A cell will always run spontaneously in the direction that produces a positive cell potential.

2. A complete description of a galvanic cell always includes four items:

a. The cell potential (always positive for a galvanic cell) and the

balanced cell reaction.

b. The direction of electron flow, obtained by inspecting the half- reactions and using the direction that gives a positive Eocell.

c. Designation of the anode and cathode.

d. The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if

none of the substances participating in the half-reaction is a conducting solid.

3. Diagramming Voltaic Cells.* In one compartment of a voltaic cell, a graphite rod dips into

an

acidic solution of K2Cr2O7 and Cr(NO3)3; in the other, a tin bar dips into a Sn(NO3)2 solution. A KNO3 salt bridge joins the

half- cells. The tin electrode is negative relative to the graphite. Diagram the cell, show balanced equations, and write the

cell notation.

4. Description of a Galvanic Cell.* Describe completely the galvanic cell based on the following

half- reactions under standard conditions:

Ag+ + e- -----> Ag Eocell = 0.80 V (1)Fe3+ + e- -----> Fe2+ Eocell = 0.77 V (2)

In addition, draw the cell and write the line notation.

III. Cell Potential, Electrical Work, and Free Energy.A. Cell Potential.

1. The work that can be accomplished when electrons are transferred (emf) is defined in terms of a potential difference (in volts) between two circuits.

emf = potential difference (V) = work (J) or 1 V = 1 J/C

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charge (C)

2. Work is viewed from the point of view of the system.

Therefore, E = -w/q or -w = qE3. The maximum work is defined as

wmax = -qEmax

* Maximum work is an impossibility. In any real, spontaneous process some energy is always wasted-the actual work

realized is always less than the calculated maximum.

4. The Faraday (F).a. The faraday is defined as the charge of 1 mole of electrons.

b. F = 96,485 C/mol e-

5. The purpose of a voltaic cell is to convert the free energy change of a spontaneous reaction into the KE of e- moving through an external circuit.

wmax = G

6. For a galvanic cell,

G = -nFEmax or Go = -nFEomax

a. If Ecell > 0, then G< 0 and the process is spontaneous.

b. If Ecell < 0, then G> 0 and the process is nonspontaneous.c. If Ecell = 0, then G= 0 and the process is at equilibrium.

7. Calculating Go for a Cell Reaction.* Using the Reduction Table in your book, calculate Go for the reaction

Cu2+(aq) + Fe(s) ----> Cu(s) + Fe2+(aq)

Is the reaction spontaneous?

8. Predicting Spontaneity.* Using the Reduction Table in your book, predict whether 1 M HNO3 will dissolve gold metal to form a 1 M Au3+ solution.

B. The Relationship to the Equilibrium Constant.1. Two relationships:

Go = -RTlnK Go = -nFEocell

2. Through substitution,

Eocell = RT/nF lnK

3. Calculating K and Go from Eocell.

* When cadmium metal reduces Cu2+ in solution, Cd2+ forms in addition to copper metal. If Go = -143 kJ, calculate K at 25

C. What would be Eo

cell in a voltaic cell that used this reaction?

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IV. Dependence of Cell Potential on Concentration.A. The Effects of Concentration on E.

1. Think Le Chatelier’s Principle!

2. For the cell reaction

2Al(s) + 3Mn2+(aq) ----> 2Al3+(aq) + 3Mn(s) Eocell = 0.48 V

predict whether Ecell is larger or smaller than Eocell for the

following cases.

a) [Al3+] = 2.0 M, [Mn2+] = 1.0 M

b) [Al3+] = 1.0 M, [Mn2+] = 3.0 M

B. The Nernst Equation.1. The Nernst equation gives the relationship between the cell potential and the concentrations of cell components.

E = Eo - RT/nF lnQat 25 C

E = Eo - 0.0591/n logQ

2. The Effects of Q.

a. When Q < 1 and [reactant] > [product], Ecell > Eocell.

b. When Q = 1 and [reactant] = [product], Ecell = Eocell.

c. When Q > 1 and [reactant] < [product], Ecell < Eocell.

3. A cell in which the concentrations are not in their standard states will continue to discharge until equilibrium is reached.

* At this point, Q = K and Ecell = 0.

4. At equilibrium, the components in the two cell compartments have the same free energy, and G = 0. The cell can no longer do work!

5. Problems.a. Consider a cell based on the reaction

Fe(s) + Cu2+(aq) ----> Fe2+(aq) + Cu (s)

If [Cu2+] = 0.30 M, what [Fe2+] is needed to increase Ecell by 0.25 V above Eo

cell at 25 C?

b. Describe the cell based on the following half-reactions:

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VO2+ + 2H+ + e- -----> VO2+ + H2O Eo = 1.00

V

Zn2+ + 2e- -----> Zn Eo = -0.76 V

where T = 25 C[VO2

+ ] = 2.0 M

[H+ ] = 0.50 M

[VO2+ ] = 0.01 M

[Zn2+ ] = 0.1 M

C. Concentration Cells.1. Concentration cells are constructed with the exact same half-reactions, with the exception of a difference in concentrations.

2. Voltages are typically small as electrons are transferred from the cell of higher concentration to the cell of lower concentration.

3. Eocell = 0.00 because conditions are not standard (not 1 M).

4. Calculating the Potential of a Concentration Cell.* A concentration cell is built using two Au/Au3+ half-cells. In half-cell A, [Au3+] = 7.0 x 10-4 M, and in half-cell B, [Au3+] = 2.5 x10-2 M. What is Ecell, and which electrode is negative?

V. Electrolysis.A. The Electrolytic Cell.

1. An electrolytic cell uses electrical energy to drive a nonspontaneous process.

2. The process is called electrolysis, which involves forcing a current through a cell to produce a chemical reaction for which the cell potential is negative.3. Everything is the same as a galvanic cell except the signs of the anode and cathode.

B. The Stoichiometry of Electrolysis.1. Faraday’s Law of Electrolysis: the amount of a substance produced at each electrode is directly proportional to the amount of electric charge flowing through the cell.

2. The SI Unit of current is the ampere (A).

1 ampere = 1 coulomb/second or 1 A = 1 C/s

3. Applying the Relationship Among Current, Time, and Amount of a

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Substance.* Using a current of 4.75 A, how many minutes does it take to

plate 1.50 g Cu onto a sculpture from a CuSO4 solution?

I. Pressure.A. The Barometer.

1. A barometer is a device to measure atmospheric pressure.

2. The barometer was invented in 1643 by a student of Galileo, Italian scientist Evangelista Torricelli.

3. This barometer is constructed by filling a glass tube with liquid mercury and inverting it in a dish of mercury.

* Diagram:

4. Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity.

a. The weather affects the pressure reading.

b. Altitude also affects the pressure. * At sea level, the height of the column of mercury is 760

mm.

* At the peak of Mt. Everest, the column measures 270 mm Hg.

B. Units of Pressure.1. The open-ended manometer (barometer).

* Diagrams:

2. Mercury is used to measure pressure because of its high density. By way of comparison, the column of water required to measure a given pressure would be 13.5 times as high as a mercury column used for the same purpose.

3. Common Units of Pressure.* 1 atmosphere (atm) = 760 mm Hg

760 torr101,324 Pascals (Pa)101.325 kPa29.92 in Hg14.7 lb/in2

4. Pressure Conversions.* The pressure of a gas is measured as 49 torr. Represent

this pressure in both atmospheres and pascals.

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II. The Gas Laws of Boyle, Charles, and Avogadro.A. Boyle’s Law.

1. Named after Irish chemist Robert Boyle (1627-1691).

2. Using a closed-ended J-tube, Boyle found that gas volume is inversely proportional to its pressure.

3. With technology, Boyle’s law holds precisely only at very low pressures.

4. If a gas obey Boyle’s law exactly, it is said to be “ideal.”

5. Boyle’s Law: PV = k or P1V1 = P2V2.6. Boyle’s Law I.

* Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is found in the exhaust of

automobiles and power plants. Consider a 1.53-L sample of gaseous SO2 at a pressure of 5.6 x 103 Pa. If the pressure is changed to 1.5 x

104 Pa at a constant temperature, what will be the new volume

of the gas?

7. Boyle’s Law II.* In a study to see how closely gaseous ammonia obey Boyle’s

law, several volume measurements were made at various

pressures, using 1.0 mol NH3 gas at a temperature of 0 C. Using the

results listed below, calculate the Boyle’s law constant for NH3 at

the various pressures.

B. Charles’s Law.1. Named in honor of Jacques Charles (1746-1823).

* He was the first person to fill a hot air balloon with hydrogen gas

on the first solo balloon flight.

2. He found that the volume of a gas at constant pressure increases linearly with the temperature of the gas.

a. The temperature was originally plotted in degrees C.

b. A plot of the gases showed that all volumes extrapolated to the

same temperature, -273.2 C. * This temperature is also known as 0 Kelvin or absolute

zero.

* The Kelvin scale was devised by English physicist William Thomson, also known as Lord Kelvin, 50 years after

Charles’s conclusions.

3. Charles’s Law: V/T = k or V1/T1 = V2/T2* Temperature is in Kelvins.

4. Charles’s Law.* A sample of gas at 15 C and 1 atm has a volume of 2.58 L.

What volume will the gas occupy at 38 C and 1 atm?

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C. Avogadro’s Law.1. Named in honor of Italian chemist Amadeo Avogadro (1776-1856).

2. He stated that at the same temperature and pressure, equal volumes of different gases contain the same number of particles.

3. Avogadro’s Law: V/n = k or V1/n1 = V2/n2

4. At 1 atm of pressure and 273 K, 1 mol of any gas occupies 22.4 L.

5. Avogadro’s Law.* Suppose we have a 12.2-L sample containing 0.50 mol

oxygen gas (O2) at a pressure of 1 atm and a temperature of 25 C. If all

this O2 is converted to ozone (O3) at the same temperature and

pressure, what would be the volume of the ozone?

III. The Ideal Gas Law.A. Summary.

* Boyle’s Law: V = k/P (at constant T and n)

Charles’s Law: V = kT (at constant P and n)

Avogadro’s Law: V = kn (at constant T and P).

B. The Ideal Gas Law.1. Combining the above laws we obtain the ideal gas law:

2. R is the universal gas constant. * Values.

R = 0.08206 L atm/mol K = 8.314 L kPa/mol K = 62.4 L mm Hg/mol K3. Real gases behave as ideal gases as they approach low pressures (below 1 atm) and high temperatures.

4. Ideal Gas Law I.* A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0 C and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this sample of gas.

5. Ideal Gas Law II.* Suppose we have a sample of ammonia gas with a volume

of 3.5 L at a pressure of 1.68 atm. The gas is compressed to a

volume of 1.35 L at a constant temperature. Use the ideal gas law to calculate the final pressure.

18


6. Ideal Gas Law III. * A sample of methane gas that has a volume of 3.8 L at 5 C is

heated to 86 C at constant pressure. Calculate its new volume.

7. Ideal Gas Law IV.* A sample of diborane gas (B2H6), a substance that bursts

into flames when exposed to air, has a pressure of 345 torr at a temperature of -15 C and a volume of 3.48 L. If conditions

are changed so that the temperature is 36 C and the pressure is

468 torr, what will be the volume of the sample?

8. Ideal Gas Law V.* A sample containing 0.35 mol argon gas at a temperature of

13 C and a pressure of 568 torr is heated to 56 C and a pressure

of 897 torr. Calculate the change in volume that occurs.

IV. Gas Stoichiometry.A. STP.

1. STP stands for standard temperature and pressure.a. The temperature = 0 C = 273 K.

b. The pressure is 1 atm or an equivalent.

2. Molar volume under these condition for an ideal gas is 22.42 L.

3. Gas Stoichiometry I.* A sample of nitrogen gas has a volume of 1.75 L at STP.

How many moles of N2 are present?

4. Gas Stoichiometry II.* Quicklime (CaO) is produced by the thermal decomposition

of calcium carbonate (CaCO3). Calculate the volume of CO2 at

STP produced from the decomposition of 152 g CaCO3 by the

reaction

CaCO3(s) -----> CaO(s) + CO2(g)

19


5. Gas Stoichiometry III.* A sample of methane gas having a volume of 2.80 L at 25 C

and 1.65 atm was mixed with a sample of oxygen gas having a

volume of 35.0 L at 31 C and 1.25 atm. Them mixture was then

ignited to form carbon dioxide and water. Calculate the volume of

CO2 formed at a pressure of 2.50 atm and a temperature of 125

C.

B. Molar Mass of a Gas and Gas Density.1. Deriving the equations for molar mass of a gas and gas density:

2. Gas Density/Molar Mass.* The density of a gas was measured at 1.50 atm and 27 C

and found to be 1.95 g/L. Calculate the molar mass of the gas.

V. Dalton’s Law of Partial Pressure.A. Dalton’s Law.

1. Named in honor of English scientist John Dalton.a. He stated that gases mix homogeneously.

b. He also said that each gas in a mixture behaves as if it were the

only gas present.

c. Dalton’ Law: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone.

d. P total = P1 + P2 + P3 + ......

e. The fact that the pressure exerted by an ideal gas is not affected by the identity (composition) of the gas particle

reveal two things. * The volume of the individual gas particles must not be important.

* The forces among the particles must not be important.

2. Dalton’s Law I.* Mixtures of helium and oxygen are used in scuba diving

tanks to help prevent “the bends.” For a particular dive, 46 L He at

25 C and 1.0 atm and 12 L O2 at 25 C and 1.0 atm were pumped

into a tank with a volume of 5.0 L. Calculate the partial pressure

of each gas and the total pressure in the tank at 25 C.

B. Mole Fraction.1. A mole fraction () is the ratio of the number of moles of a given

20


component in the total number of moles in the mixture.

2. Arriving at the mole fraction equation:

3. Dalton’s Law II.* The partial pressure of oxygen was observed to be 1.56 torr

in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.

4. Dalton’s Law III.* The mole fraction of nitrogen in the air is 0.7808. Calculate

the partial pressure of N2 in air when the atmospheric pressure

is 760. torr.

5. Gas Collection over Water.* A sample of solid potassium chlorate (KClO3) was heated in a

test tube and decomposed by the following reaction:

2KClO3(s) -----> 2KCl(s) + 3O2(g)

The oxygen produced was collected by water displacement at 22

C at a total pressure of 754 torr. The volume of the gas collected was 0.650 L, and the vapor pressure of water at 22

C is 21 torr. Calculate the partial pressure of O2 in the gas

collected and the mass of KClO3 in the sample that was decomposed.

VI. The Kinetic Molecular Theory of Gases.A. KMT.

1. The kinetic molecular theory (KMT) is a simple model that attempts to explain the properties of an ideal gas.

2. From HONORS CHEM: Kinetic Theory Revisited.* Recall that gases consist of hard, spherical particles, usually atoms or molecules.

a. Because the particles are so small and the distances between

them are so great, their individual volumes are insignificant.

21


* The volume of a gas is mostly empty space.

b. Property: Gases are highly compressible.

* Also, recall that there are no attractive or repulsive forces between gas particles.

a. Gases are free to move inside their containers.

b. Property: A gas expands until it takes the volume and shape of

its container.

* Next, recall that gas particles move rapidly in constant random motion.

a. The particles travel in straight paths and move independently of

each other.

b. Only when a particle collides with a container wall or another gas

particle does it deviate from its straight-line path.

* In addition, recall that collisions between gas particle are elastic, meaning that the total kinetic energy remains constant and that the kinetic energy is transferred without loss from one particle to another.

* Finally, recall that the average kinetic energy of a collection of gas particles is directly proportional to the Kelvin temperature of the gas.

3. Assumptions.a. The particles are so small compared with the distance

between them that the volume of the individual particles can be

assumed to be negligible (zero).

b. The particles are in constant motion. The collisions of the particles with the wall of the container are the cause of the pressure exerted by the gas.

c. The particles are assumed to exert no forces on each other; the

are assumed to neither attract nor to repel each other.

d. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin

temperature of the gas.

4. Pressure is defined as the force exerted by particles colliding with a a container wall. This is not the same as atmospheric pressure.

5. Real gases do not conform to these assumptions because they have finite volumes and do exert forced on each other.

B. Justification of the gas laws using KMT.1. Pressure and Volume (Boyle’s Law).

a. Law: For a given sample of gas at constant T and n, if the volume

of a gas is decreased, the pressure increase.

b. KMT: Since a decrease in volume means that the gas particles

will hit the container wall more often, the pressure should increase.

2. Volume and Temperature (Charles’s Law).a. Law: At constant P and n, the volume of a gas is directly proportional to the Kelvin temperature.

b. KMT: When a gas is heated to a higher temperature, the speeds

of its molecules increase and thus hit the walls more often with

more force. The only way to keep the pressure constant in this

situation is to increase the volume of the container.

3. Volume and Number of Moles (Avogadro’s Law).a. Law: The volume of a gas at constant P and T depends

directly on the number of gas particles present.

b. KMT: An increase in the number of gas particles at the same

temperature would cause the pressure to increase if the volume

were held constant. The only way to return the pressure to its

22


original value is to increase the volume.

c. It is important to recognize that the volume of a gas depends

only on the number of gas particle present.

d. The individual volumes of the particles are not a factor because

the particle volumes are so small compared to the distances

between the particles.

4. Mixture of Gases (Dalton’s Law).a. Law: The total pressure exerted by a mixture of gases is the sum of the pressures of the individual gases.

b. KMT: All gases are independent of one another and the volumes

of the individual particles are unimportant. Thus the identities

of the gas particles do not matter.

5. Deriving the Ideal Gas Law.a. Derivation:

b. The experimental and theoretical equations for the ideal gas law

are virtually identical, giving confidence in the KMT.

6. The Meaning of Temperature.a. Temperature is a measure of the energy of molecular

motion; a transfer of KE from collisions of higher energy objects to collections of lower energy particles.

b. Equation:

7. Root Mean Square Velocity.

a. The root mean square velocity (urms) is the square root of the

average of the squares of the individual velocities of gas particles.

b. Derivation:

c. Calculate the root mean square velocity for the atoms in a sample of helium gas at 25 C.

VII. Effusion and Diffusion.A. Effusion.

1. Effusion is the process by which a gas escapes from its container through a tiny hole into an evacuated tube.

2. Thomas Graham (1805-1869), a Scottish chemist, found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

3. Graham’s Law of Effusion:

4. Effusion Rates.

23


* Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the

enrichment process to produce fuel for nuclear reactors.

5. KMT: The effusion rate for a gas depends directly on the average

velocity of its particles.a. The faster the gas particles are moving, the more likely

they are to pass through the effusion orifice.

b. Prediction:

B. Diffusion.* Diffusion is the movement of one gas through another and uses the same equation as effusion.

VIII. Real Gases.A. Deviations.

1. Molecules have finite molecular volumes.

2. Interactions exist between molecules.

3. When volume decreases, condensation often occurs.

4. Gases behave nearly ideally under ordinary conditions of low pressure and high temperature.

B. Reasons for Deviations.1. Graph:

2. When Pext > 10 atm, deviations occur.

3. Intermolecular attractions lower molecular speeds and lessen the force of collision with the container wall.

a. Increasing the external pressure causes this to occur.

b. Lowering temperature yields the same effect.

c. Both cases may cause condensation.

d. Because the volume decreases in both cases, the PV/RT is

lowered.

4. Molecular Volume.a. As external pressure increases, free volume decreases and molecular volume becomes a significant factor,

outweighing the importance of the effects of intermolecular attractions.

b. The “V” used in the ratio PV/RT still refers to the volume of the

container, not the volumes of the gas molecules.

c. Because “V” is artificially too large, the ratio PV/RT increases.

5. The van der Waals Equation.a. This equation adjusts pressure “up” and volume “down.”

b. The equation:

c. The a and b variables are the van der Waals constants. * They are positive values specific to each gas.

* They are obtained from a table.

* a is related to molar mass and relate with the strength of the

intermolecular attractions.

* b is a measure of actual molecular volume.

24


I. Types of Attractions.A. Intramolecular Attractions.

1. These are forces within a molecule or polyatomic ion.

2. They are also referred to as bonding forces.

3. These influence chemical properties.

B. Intermolecular Attractions.1. These are forces between molecules, ions, or atoms.

2. They influence physical properties.

II. A Comparison of States of Matter.A. KMT.

1. For a gas, the energy of attraction is less than the energy of motion. The particles, therefore, are far apart.

2. For a liquid, the energy of attraction is stronger, but KE allows for movement. The particles, therefore, are closer, but with motion.

3. For a solid, the energy of attraction is greater than the energy

25


of motion. The particles, therefore, as arranged in a fixed, organized pattern.

B. Shape/Volume.1. For a gas, the molecules conform to the container in both shape and volume.

2. For a liquid, the molecules conform to the container shape; the volume is limited by surface.

3. For a solid, the compound has a defined shape and volume.

C. Compressibility.1. For a gas, the compressibility is high because of the great distances between particles.

2. For a liquid, the compressibility is very low.

3. For a solid, the compressibility is virtually nonexistent.D. Ability of Flow.

1. For a gas, this ability is very high.

2. For a liquid, this ability is moderate.

3. For a solid, this ability is almost none.

III. A Look at Phase Changes for Water.* Diagram:

IV. Types of Intermolecular Attractions.A. Introduction.

1. Bonding forces are relatively strong because they involve higher charges that are closer together (covalent radius).

2. Intermolecular forces are relatively weak because they typically involve charges that are far apart (van der Waals radius).

B. Ion-Dipole Forces.1. This occurs when an ion is attracted to a polar molecule.

2. Example:

C. Dipole-Dipole Forces.1. This occurs between two polar molecules.

2. Example:

D. Hydrogen Bonding.1. This is a special, very strong type of dipole-dipole force.

2. This occurs only when a hydrogen is bonded to a nitrogen, oxygen, or fluorine atom.

3. Example:

26


4. Drawing Hydrogen Bonds between Molecules.

E. Ion-Induced Dipole Forces.1. This occurs when an ion is attracted to a distorted nonpolar molecule.

2. The electron density of the nonpolar molecule is shifted as a result of a charged particle.

3. Example:

F. Dipole-Induced Dipole.1. This occurs between a polar molecule and a distorted nonpolar molecule.

2. Example:

G. Dispersion(London) Forces.

1. These forces often occur when all particles collide.

2. This force is also related to molar mass because heavier molecules tend to have more electrons capable of being polarized and distorted.

3. Shape is also a factor because more points of attraction means more polarizability.

4. Example:

F. Predicting the Type and Relative Strength of Intermolecular Forces.* In each of the following pairs, identify all the intermolecular forces present and select the substance with the higher boiling point:

a. MgCl2 or PCl3

b. CH3NH2 or CH3F

c. CH3OH or CH3CH2OH

d. Hexane or cyclohexane

e. CH3Br or CH3Cl

f. CH3CH2OH or CH3OCH3

g. C2H6 or C3H8

V. Properties of a Liquid.A. Surface Tension.

1. Diagrams:

27


2. Surface tension is the energy required to increase the surface area by a given amount ( in J/m2).

3. To reduce the instability of the entire substance at the surface, a liquid will reduce the number of particles exposed to the surface.

* The particles form a tight “skin” at the surface.

4. The stronger the forces between the particles in a liquid, the greater the surface tension.

B. Capillarity.1. Capillarity is the ability of a substance to naturally “rise up” a tube against gravity.

2. Diagrams:

C. Viscosity.1. Viscosity is the resistance to flow.

2. The viscosity of a liquid decreases with increasing temperature.

3. Molecular shape also plays a role in determining a liquid’s viscosity.

* Longer molecules have more contact points for attractive forces

and have higher viscosities.

VI. Properties of Solids.A. Crystalline Solids.

1. Crystals have lattice structures made of unit cells.a. A lattice is an array of points that forms a regular pattern.

b. A unit cell is the simplest arrangement of points that, when

repeated in all directions, gives the lattice.

c. Diagram:

d. The coordination number of a particle in a crystal is the number of nearest neighbors surrounding it.

2. Types of cubic unit cells.a. Simple Cubic. * Diagram:

b. Body-Centered Cubic. * Diagram:

c. Face-Centered Cubic.

28


* Diagram:

3. Packing Efficiency.a. Packing efficiency is the percentage of the available

volume of a crystal is occupied by “spheres.”

b. Simple cubic packing efficiency is 52% (48% air) and is very rare.

c. Body-centered cubic packing efficiency is 68% and is more common. * Examples:

d. Hexagonal closest packing efficiency is 74% (the most). * Examples:

e. Cubic closest packing efficiency is 74% (the most) and is based

on the face-centered cubic unit cell. * Examples:

4. Types and Properties of Solids.a. Atomic Network. * Structural Unit: Atom.

* Type of Bonding: Directional covalent bonds.

* Diagram:

* Typical Properties: Hard, high melting point, insulator.

* Examples:

b. Atomic Metallic. * Structural Unit: Atom.

* Type of Bonding: Nondirectional covalent bonds involving electrons that are delocalized

throughout the crystal.

* Diagram:

* Typical Properties: Wide range of hardness, wide range of melting points, conductor.

* Examples:

c. Atomic Group 8A. * Structural Unit: Atom.

* Type of Bonding: London dispersion forces.

* Diagram:

* Typical Properties: Very low melting points.

* Examples:

29


d. Molecular. * Structural Unit: Molecule.

* Type of Bonding: Polar molecules- dipole/dipole interactions;

Nonpolar molecules- London dispersion forces.

* Diagram:

* Typical Properties: Soft, low melting, insulator.

* Examples:

e. Ionic. * Structural Unit: Ion.

* Type of Bonding: Ionic.

* Diagram:

* Typical Properties: High melting point, insulator.

* Examples:

B. Amorphous Solids.* They have an irregular lattice structure with many defects.

VII. Quantitative Aspects of Changes of State.A. The Heating-Cooling Curve.

1. Diagram:

2. (*) indicates a change in heat accompanied by a change in temperature that equals the change in KE as the most probable

speed changes.

3. (#) indicates a change in heat accompanied by a constant temperature (constant KE), which is associated with a change

in PE, as the average distance between molecules changes.

B. The Equilibrium Nature of Phase Changes.1. Liquid-gas equilibria.

a. This is viewed for a CLOSED CONTAINER in a VACUUM.

b. The rate of condensation = rate of vaporization.

30


c. Vapor pressure is the pressure exerted by a vapor in a closed

system. * Vapor pressure increases with temperature as the KE increases.

* Vapor pressure increases with weaker intermolecular attractions.

d. The Clausius-Clapeyron Equation relates vapor pressure and

temperature. * Equation:

* The vapor pressure of ethanol is 115 torr at 34.9 C. If Hvap

of C2H5OH is 40.5 kJ/mol, calculate the temperature (in C) when the vapor pressure is 760 torr.

* At 34.1 C, the vapor pressure of water is 40.1 torr. What is

the vapor pressure at 85.5 C? The Hvap of water is 40.7 kJ/mol.

e. Boiling can only occur in an open container and occurs when the

vapor pressure equals the external pressure. * The normal boiling point of a substance is boiling at

standard pressure (1 atm).

2. Solid-liquid equilibria.a. The rate of melting = rate of freezing.

b. The normal boiling point of a substance is melting at standard

pressure (1 atm).

3. Solid-gas equilibria.* The rate of sublimation = rate of deposition.

VIII. Phase Diagrams.A. Diagram 1.

31


B. Diagram 2.

IX. The Properties of Mixtures.A. Introduction.

1. A mixture is defined as two or more substances physically mixed together but not chemically combined.

2. A solution is a homogeneous mixture with one phase.

B. Types of Solutions: Intermolecular Forces and the Prediction ofSolubility.1. Definitions.

a. A solute is

b. A solvent is

c. Miscibility is

d. Solubility is

e. To be dilute means

f. To be concentrated means

2. The Solvation Process.a. Solute particles must separate.

b. Some solvent particles must separate.

c. Solute and solvent particles must mix together.

d. Whether or not a solution occurs depends on the relative strengths of the solute-solute, solvent-solvent, and solute- solvent intermolecular forces.

3. Intermolecular attractions present in solutions.*

32


4. Rule: “ Like dissolve like.”* Molecular size may also be a factor.

5. Gas-liquid solutions.a. Solute-solute attractions are very weak.

b. Solute-solvent forces are weak, but are the major factor in gas

solubility in a liquid.

6. Gas solutions and solid solutions.a. Gas-gas: All gases are infinitely soluble in one another. * Example:

b. Gas-solid: The gas occupies the spaces between the closely packed particles. * Example:

c. Solid-solid: These are called alloys. * Substitutional: structural substitution. ** Examples:

* Interstitial: fill in the spaces within the structure. ** Example:

C. Energy Changes in the Solution Process.1. Heats of Solution.

a. Explanation.

b. Diagrams:

2. Heats of Hydration: Ionic Solids in Water.a. Hsoln = Hsolute Hhydr

b. Hhydr is always exothermic (negative) because the ion-dipole

force is greater than the hydrogen bonding in water.

c. The charge density, the ratio of charge to volume, affects Hhydr.

33


* The trend down a group is to decrease Hhydr because a larger ion means a larger radius between particles,

therefore less energy is required to break the attraction.

* The trend across a period is to increase Hhydr as charge increases.

3. The Solution Process and the Tendency Toward Disorder.a. Entropy is a measure of a system’s disorder.

b. Because the solution process occurs naturally, which creates

more disorder, entropy is increased.

D. Solubility as an Equilibrium Process.1. Definitions.

a. An unsaturated solution means

b. A saturated solution means

c. A supersaturated solution means

2. Factors Affecting Solubility.a. Temperature. * Solubility increases with temperature if the solution

process is endothermic (Hsoln > 0).

* Solubility decreases with temperature if the solution process

is exothermic (Hsoln < 0).

* Generally, ionic solids increase their solubility in water as temperature increases.

* Generally, gases decrease their solubility in water as temperature increases.

b. Pressure.

* Pressure changes have little effect on solid and liquid solubility.

* Henry’s Law states that the solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas

(Pgas) above the solution.

* In a soft-drink plant, the partial pressure of CO2 gas inside a

bottle of cola is adjusted to 4 atm at 25 C. What is the solubility of CO2 under these conditions? Henry’s law

constant for CO2 in water = 3 x 10-2 mol/L atm at 25 C.

E. Quantitative Ways of Expressing Concentration.1. Molarity (M)

2. Molality (m)

3. Parts by mass

4. Parts by volume

5. Mole fraction (X)

6. What is the molality of a solution prepared by mixing 32.0 g

34


CaCl2 with 271 g water?

7. Calculate the ppm by mass of calcium in a 3.50 g pill that contains 40.5 mg Ca.

8. The label on a 0.750-L bottle Italian Chianti indicates “11.5% alcohol by volume.” What volume is alcohol (in L) does it contain?

9. A sample of rubbing alcohol contains 142 g isopropanol (C3H7OH) and 58.0 g water. What are the mole fractions of alcohol and water?

10. Converting Units of Concentration. * Hydrogen peroxide is a powerful oxidizing agent that is used in concentrated solution in rocket fuel systems and in dilute solution as a hair bleach. An aqueous solution H2O2 is 30% by mass and has a density of 1.11 g/mL. Calculate its

a. Molality

b. Mole fraction of H2O2

c. Molarity

F. Colligative (Collective) Properties of Solutions.1. These properties of a solution are always compared against the pure solvent.

2. Nonvolatile nonelectrolytic (ideal) solutions.a. Vapor Pressure Lowering (P). * This solution’s vapor pressure is always lower than the

pure solvent.

* Surface particles consist of some nonvolatile solute particles

that have replaced some solvent particles. Therefore, not as

many solvent particles are permitted to become a vapor.

* Raoult’s Law for ideal solutions:

* Calculate the expected vapor pressure at 25 C for a solution

prepared by dissolving 158.0 g of common table sugar (sucrose,

molar mass = 342.3 g/mol) in 643.5 cm3 of water. At 25 C, the

density of water is 0.9971 g/cm3 and the vapor pressure is

23.76 torr.

35


* Predict the vapor pressure of a solution prepared by mixing

35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g water

at 25 C. The vapor pressure of pure water at 25 C is 23.76 torr.

b. Boiling Point Elevation (Tb). * This results from the increases pressure (P).

* The boiling point of the solution is always higher than the pure

solvent under these conditions.

* Equations:

* The boiling point of ethanol is 78.5 C. What is the boiling point of a solution of 3.4 g vanillin (M = 152.14 g/mol) in

50.0 g ethanol? (Kb of ethanol = 1.22 C/m).

* A solution was prepared by dissolving 18.00 g glucose in 150.0 g

water. The resulting solution was found to have a boiling point

of 100.34 C. Calculate the molar mass of glucose. Glucose is a

molecular solid that is present as individual molecules in solution.

c. Freezing Point Depression (Tf). * The freezing point of the solution is always lower than the

pure solvent under these conditions.

* Solute particles hinder the rate at which the solvent particles

freeze.

* Equations:

* What mass of ethylene glycol (M = 62.1 g/mol), the main component in antifreeze, must be added to 10.0 L water

to produce a solution for use in a car’s radiator that freezes

36


at -10.0 F (-23.3 C)? Assume the density of water is exactly

1 g/mL.

* A chemist is trying to identify a human hormone, which controls metabolism, by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing-point depression was determined to be 0.240 C. Calculate the molar mass of the hormone.

d. Osmotic Pressure. * Osmotic pressure () is the pressure that results from the inability of solute particles to cross a semipermiable membrane; the pressure required to prevent the net

movement of solvent across the membrane.

* Diagrams:

* Equations:

* To determine the molar mass of a certain protein, 1.00 x 10-3 g

of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found

to be 1.12 torr at 25 C. Calculate the molar mass of the protein.

* What concentration of sodium chloride in water is needed to

produce an aqueous solution isotonic with blood ( = 7.70 atm

at 25 C)?

37


3. Volatile nonelectrolytic solutions.a. When a solution contains two volatile components, both contribute to the total vapor pressure, but not necessarily

in equal amounts.

b. Equal amounts of liquid do not produce equal amounts of vapor.

c. A solution is prepared by mixing 5.81 g acetone (C3H6O, molar

mass = 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass =

119.4 g/mol). At 35 C, this solution has a total vapor pressure of

260 torr. Is this an ideal solution? The vapor pressures of pure

acetone and pure chloroform at 35 C are 345 and 293 torr, respectively.

4. Colligative Properties of Electrolytic Solutions.a. Most electrolytic solutions are not ideal.

b. A vant Hoff factor (i) attempts to adjust the equations.

c. i =

d. Equations:

G. Colloids.1. A suspension contains very large particles that settle out of solution.

2. A solution contains very small solute particles that do not separate from the solvent.

3. A colloid contains intermediate particles (solutelike) distributed throughout a dispersing (solventlike) substance.

4. Evaluating colloids.a. The Tyndall effect measures the scattering of light by a colloidal suspension.

38


b. Brownian motion measures the change in speed and direction of

colloidal particles because of the action of the dispersing medium.

5. Types of Colloids.

I. Electromagnetic Radiation.A. The Nature of Light.

1. The primary characteristics of a wave.a. Wavelength () is the distance between two consecutive peaks

or troughs in a wave.

b. Frequency () is the number of waves (cycles) per second that

pass a given point in space.

c. Diagram:

39


d. All types of electromagnetic radiation (waves), a breakdown of

light based on wavelength and frequency, travel at the speed of

light (c). * The speed of light (c) = 2.9979 x 108 m/s.

* The electromagnetic spectrum:

e. Wavelength and frequency are inversely related. * = c where is the wavelength in meters, is the frequency in cycles per second, and c is the speed of light. * In the SI system, cycles is understood, and the unit per second becomes 1/s, or s-1, which is called hertz (Hz).

2. Frequency of Electromagnetic Radiation.* The brilliant red colors seen in fireworks are because of the emission of light with wavelengths around 650 nm when

strontium salts such as Sr(NO3)2 and SrCO3 are heated. Calculate the frequency of red light of wavelength 6.50 x 102 nm.

B. Some Distinguishing Properties of Wave.1. Refraction is the bending of a light wave (a change in angle) when it strikes a boundary.

2. Diffraction is the bending of a light wave around a boundary.

3. Light must be a wave because it possesses these properties.

II. The Nature of Matter.A. Light vs. Matter.

1. Blackbody Radiation.a. When a solid (specifically a particle) is heated, it emits light waves. * Examples:

b. German Max Planck (1858-1947) said that a hot, glowing object

emits or absorbs a specific amount of energy to produce this

light.

c. This specific energy is said to be quantized, or specific, lost or

gained only in integer multiples of h.

d. E = nh where h is Planck’s constant (6.626 x 10-34 J s), is the frequency of the electromagnetic spectrum absorbed or

emitted, E is the change in energy for a system, and n is a whole-

number integer (1, 2, 3, ...).

e. This implies that light energy of matter is not continuous (like a

rainbow), but that it is absorbed or emitted only at specific quantized energy states.

f. Is matter separate form light (energy)?

2. The Energy of a Photon (a packet of energy). * The blue color in fireworks is often achieved by heating copper(I)

chloride (CuCl) to about 1200 C. Then the compound emits blue

light having a wavelength of 450 nm. What is the increment of

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energy (the quantum) that is emitted at 4.50 x 102 nm by CuCl?

3. The Photoelectric Effect.a. When light strikes a monochromatic plate, an electrical

current flows (like a solar calculator).

b. Light must transfer momentum to matter, like particles do; light

is behaving like a particle!

c. Confusing feature of the Photoelectric Effect. * A threshold (minimum) frequency is required to knock an electron free from a metal. Wave theory associates the

light’s energy with the wave amplitude (intensity), not its

frequency (color), so it predicts that an electron should be knocked

free when the metal absorbs enough energy from any color of

light.

* Current flows the moment that light of high enough frequency

shines on the metal, regardless of its intensity. The wave theory predicted that in dim light there should be a time

lag before current flowed, while the electrons absorbed

enough energy to break free.

4. Albert Einstein’s Photon Theory.a. Einstein proposed that radiation is particulate, occurring as quanta of electromagnetic energy (packets), later called

photons.

b. Einstein solved the mysteries of the Photoelectric Effect. * A beam of light is composed of enormous numbers of

photons. Light intensity (brightness) is related to the number of photons emitted per unit of time, not their energy. One electron is freed from the metal when one photon of a

certain minimum energy is absorbed.

* An electron is freed the moment it absorbs a photon of enough

energy, not when it gradually accumulates energy from many

photons of lower energy.

5. In 1922, American Arthur Compton (1892-1962) performed experiments involving collisions of X rays and electrons that showed that photons do actually have a resting mass, a property of matter!

a. This is the Compton Effect.

b. Equations:

6. Summary.a. Energy is quantized. It can occur only in discrete units

called quanta.

b. Electromagnetic radiation, which was previously thought to exhibit only wave properties, seems to show certain characteristics of particulate matter as well. * This is wave-particle duality or the dual nature of

light.

B. Wave-Particle Duality of Matter and Energy.1. From Einstein, we have the following:

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2. French physicist Louis de Broglie (1892-1987) proposed in 1923 that if waves of energy have some properties of particles, perhaps particles of matter have some properties of waves.

a. This is similar to a guitar string (a particle) producing sound (a

wave.)

b. De Broglie wave equation:

c. Calculations of Wavelengths. * Compare the wavelength for an electron (mass = 9.11 x

10-31 kg) traveling a speed of 1.0 x 107 m/s with the for a ball

(mass = 0.10 kg) traveling 35 m/s.

3. In 1927, Davisson and Germer verified De Broglie’s concept when

an electron was seen to exhibit wave properties of interference and diffraction.

III. The Atomic Spectrum of Hydrogen.* The Atomic Spectrum.

1. The atomic spectrum (light emitted) was thought to be continuous.

a. Again, this is similar to a rainbow.

b. All of the wavelengths of visible light are present.

c. Diagram:

2. When the emission spectrum of hydrogen in the visible region is passed through a prism, only a few lines are seen.

a. Diagram:

b. These lines correspond to discrete wavelengths of specific (quantized) energy. * Only certain energies are allowed for the electron in the hydrogen atom.

c. The hydrogen emission spectrum is called a line spectrum.

IV. The Bohr Model.A. The Quantum Model.

1. Niels Bohr (1885-1962) proposed in 1913 that the electron in a hydrogen atom moves around the nucleus only in certain allowed circular orbits.

a. Diagram:

b. The atoms has stationary states, called energy levels, of specific

energy around the nucleus.

c. Electrons can move to other energy levels by absorbing (jumping

to higher energy levels) or emitting (jumping to lower energy

levels) photons of specific (quantized) energy. * Energy levels farther from the nucleus more “unstable”

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and therefore more “energetic.”

* As we will see, the closer to the nucleus an energy level is, the

more stable it is and the less energetic it is.

d. The lowest (first) energy level of an atom is called ground state.

e. This model only works for one-electron atoms!

2. The most important equation to come from Bohr’s model is the expression for the energy levels available to the electron in the hydrogen atom:

E = -2.178 x 10-18 J (Z2/n2)

in which n is an integer (the larger the value of n, the larger is the orbit radius) and Z is the nuclear charge.

a. Bohr was able to calculate hydrogen atom energy levels that

exactly matched the values obtained by experiment.

b. The negative sign means that the energy of the electron bound

to the nucleus is lower than it would be if the electron were at

an infinite distance from the nucleus (n = infinity), where there

is no interaction and the energy is zero:

* Again, the closer to the nucleus, the more stable, and the

less energetic (less meaning a negative energy value).

c. Electrons moving from one energy level to another. * Equation:

3. Energy Quantized in Hydrogen.

* Calculate the energy required to excite the hydrogen electron

from level n = 1 to level n = 2. Also calculate the wavelength of

light that must be absorbed by a hydrogen atom in its ground

state to reach this excited state.

4. Electron Energies.* Calculate the energy required to remove the electron from a hydrogen atom in its ground state.

5. Although Bohr’s model fits the energy levels for hydrogen, it is a fundamentally incorrect model for the hydrogen atom, mainly because electrons do not travel in circular orbits.

V. The Quantum-Mechanical Model of the Atom.A. The Schrodinger Equation.

1. The Schrodinger equation is an extremely complex equation used to describe the 3-D quantum mechanical model of the hydrogen atom.

2. The hydrogen electron is visualized as a standing wave (a stationary wave as one found on a musical instrument) around the nucleus.

a. The circumference of a particular orbit would have to correspond to a whole number of wavelengths.

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b. Diagrams:

c. This is consistent with the fact that only certain electron energies are allowed.

B. The Heisenberg Uncertainty Principle.1. The Heisenberg uncertainty principle states that it is impossible to know simultaneously the exact position and velocity of a particle.

2. Stated mathematically:

C. The Physical Meaning of a Wave Function.1. The square of the function indicates the probability of finding an electron near a particular point in space.

2. The probability distribution, or electron density diagram, can be plotted to give the most probable region where an electron may be located.

* Examples:

3. The definition most often used by chemists to describe the size of the hydrogen 1s orbital is the radius of the sphere that encloses 90% of the total electron distribution.

VI. Quantum Numbers.A. Quantum Numbers.

1. A quantum number describes various properties of an atomic orbital or probable electron location.

2. Types.a. The principle quantum number (n) refers to the energy

level or integral values: 1, 2, 3, etc.

b. The angular momentum (azimuthal) quantum number (l) is

related to the shape of atomic orbitals. * The integral values are from 0 to n - 1.

* l = 0 is called s; l = 1 is called p; l = 2 is called d; l = 2 is called f.

c. The magnetic quantum number (ml) is related to the orientation

of the orbital in space. * The integral values are between l and -l, including zero.

3. Quantum Numbers for the First Four Levels of Orbitals in the Hydrogen Atom.

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VII. Electron Spin and the Pauli Principle.A. Electron Spin

1. Although the first three quantum numbers describe atomic orbitals, one more is needed to describe a specific electron.

* This is called spin, and is not a property of the orbital.

2. The spin quantum number (ms) indicates the “direction” an electron spins and can have one of two possible values, +1/2 or -1/2.

* Notation:

B. Pauli Exclusion Principle.1. American Wolfgang Pauli stated that no two electrons in the same atom can have the same set of four quantum numbers.

a. This is the Pauli exclusion principle.

b. Example:

2. An atomic orbital can hold a maximum of two electrons and they must have opposing spins.

3. The electrons of an atom in its ground state occupy the orbitals of lowest energy.

VIII. Polyelectronic Atoms.A. Electrostatic Effects.

1. In one electron systems, the only electrostatic force is the nucleus- electron attractive force.

2. Many-electron systems (polyelectronic atoms).a. Nucleus-electron attractions.

b. Electron-electron repulsions.

3. For polyelectronic atoms, one major consequence of having attractions and repulsions is the splitting of energy levels into sublevels of differing energies.

a. The energy of an orbital in a many-electron atom depends on its

n value (size) and secondly on its l value (shape).

b. This means that energy levels may split into s, p, d, and f sublevels, depending on the n value.

B. The Effects of Electrostatic Interactions on Orbital Energies.1. Recall that, by definition, an atom’s energy has a negative value.

a. A more stable orbital has a larger negative energy than a less

stable one.

b. Stronger attractions make the orbital lower in energy (larger

negative number).

c. Repulsions make the system higher in energy (smaller negative

number).

2. Major Conclusions.a. A greater nuclear charge (Z) lowers orbital energies. * Why?

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b. Electron-electron repulsions raise orbital energy. * Why?

c. Electrons in outer orbitals (higher n) are higher in energy. * Why?

d. Electrons that have a finite probability distribution near the nucleus (penetration) have lower energy (more negative). * Es < Ep < Ed < Ef.

* Why?

XI. Periodic Trends in Atomic Properties.A. Effective Nuclear Charge (Zeff)

1.What is Zeff?

2. What is the group trend, with explanation?

3. What is the period trend, with explanation?

B. Atomic Radius1. What is an atomic radius?



4. Where are the exceptions and why do they occur?

C. Ionization Energy1. What is ionization energy?

2.What type of ion is produced?

3. Is the process endothermic or exothermic?

4. Electrons are easily removed until what occurs?



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7. Why is each successive IE greater than the previous IE?


D. Electron Affinity1. What is electron affinity?



4. How does EA1 compare to EA2?


E. Metallic Behavior1. What defines metallic behavior?



F. Problems.1. Trends in Ionization Energy.

* The first ionization energy for phosphorus is 1060 kJ/mol, and

that for sulfur is 1005 kJ/mol. Why?

2. Ionization Energies.* Consider atoms with the following electron configurations:

1s22s22p6

1s22s22p63s1

1s22s22p63s2

Which atom has the largest first ionization energy, and which one has the smallest second ionization energy? Explain your choices.

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