Chem 162- 2010 Final Exam Review

Chem 162-2012 Final exam review from 2010 final exam 1

Chem 162-2012 4/27 Review for Final Exam – Review of 2010 final exam Dr. Ed Tavss

CHAPTER TITLE REQUESTS High Priority 16 Acids and Bases 59 17 Acid-Base/Solubility Equilibria 71 19 Electrochemistry 49 14 Chemical Kinetics 29 20 Nuclear Chemistry 41 Low Priority 18 Thermochemistry 18 15 Chemical Equilibrium 25 22 Coordination Chemistry 22 13 Physical Properties of Solutions 13


CHAPTERS 16 & 17 pH [H+] 14 Basic 1x10-14 H2O + H2O ←→ H3O+ + OH- Kw = 1x10-14 7 Neutral (e.g. H2O) 1x10-7 Kw = [H+][OH-] = 1 x 10-14 pH + pOH = 14 Kw = Ka x Kcb = Kca x Kb = 1 x 10-14* *a & b = acid and base; ca & cb = conjugate acid and conjugate base

pKa + pKb = 14 0 Acidic 1x10o

pH units of 1 to 13 is what is commonly used, but pH’s as low as -1 or as high as 15 are possible. No acids in existence for <-1 or bases for >+15.

pH = -log[H+] pH [H+] [H+] = 10-pH pOH = -log[OH-] [OH-] = 10-pOH

pOH [OH-]

Henderson-Hasselbalch equation: pH = pKa + log([B]/[A]) or pOH = pKbase + log([Acid]/[Base]) where acid = acid or conjugate acid, and base = base or conjugate base. At equal conc. of conj. base and acid, pH = pKa of a buffer [H+] = Ka = 10-pH

pH + pO

H = 14

pH +

pO

H =

14

Kw =

[H+ ][

OH

- ] = 1

x 1

0-14 K

w = [H

+][OH

-] = 1 x 10-14

pOH = -log[OH-]

[OH-] = 10-pOH

pH = -log[H+]

[H+] = 10-pH


SEVEN STRONG ACIDS ET: Discuss strong acids and strong bases and the strengths of their conjugate bases and acids; practice both.

Very weak base (i.e., not a base; Ka Acid spectator ion)** ~106 HCl + H2O H3O+ + Cl-

~108 HBr + H2O H3O+ + Br-

~109 HI + H2O H3O+ + I-

H2SO4 + H2O H3O+ + HSO4-***

HNO3 + H2O H3O+ + NO3-

HClO4 + H2O H3O+ + ClO4-

HClO3 + H2O H3O+ + ClO3-)*

*Strong, but not common, acid. Generally not considered as a strong acid. **These conjugate bases are known as “nominal bases”, i.e., a base by definition, but effectively not a base.

***Although HSO4- is effectively not a base, it is effective as an acid. HSO4- + H2O ←→ H3O+ + SO4

2-

All other acids are weak acids, e.g., HA (acetic acid), H2SO3, RNH3+, BF3

Arrhenius acid: Anything that provides a proton, e.g., HA, H2SO3, RNH3

+ BL Acid: Anything tending to give up a proton, e.g., HA, H2SO3, RNH3

+ (Lewis Acid: Anything tending to react with an electron pair, e.g., BF3)

STRONG BASES (Most Group 1A and 2A hydroxides [not HOH])

e.g. Very weak acid (i.e., not an acid; Base spectator ion) LiOH OH- + Li+

NaOH OH- + Na+

Mg(OH)2 2OH- + Mg2+

Ba(OH)2 2OH- + Ba2+ All other bases are weak bases, e.g., RNH2, CO3

2-

Arrhenius Base: Anything that provides an OH- group, e.g., NaOH BL Base: Anything that tends to react with a proton, e.g., NaOH, RNH2, CO3

2- Lewis Base: Anything that has available non-bonding electrons, e.g., NaOH, RNH2, CO3

2-


RELATIONSHIP BETWEEN STRUCTURE AND STRENGTH OF ACID, H-Z

ET: Impossible to understand this topic based on H&P’s discussion

Let Z = atom which the H is attached to, which is O for oxyacids, or F, Cl, S, etc. for binary acids. Strength of acid is determined completely by [homolytic] strength of H-Z bond. (If the bond is weak, the acid is strong; if the bond is strong, the acid is weak.*)

*Bond strength correlates inversely with bond polarity. That is, the weaker the bond, the more polar the bond is. Some argue that it is the H-Z bond weakening, not the increased bond polarity, that results in increased acid strength; others argue that it is the increased bond polarity, not the bond weakening, that results in increased acid strength.

Two competing factors:

(1) For decreasing H-Z bond strength (i.e., increasing acidity). • Increasing strength of electron withdrawing effect of Z. • Increasing bond distance • Increasing anion size.

(All three factors are effective in dispersing the negative charge of the bonding electrons, and therefore stabilizing the incipient conjugate base.)

• Polarity of solvent (e.g. H2O) (H2O forms an ion-dipole bond with the partially positive H atom.

Therefore, increased polarity of the H helps to weaken the H-Z bond by pulling off the proton.

(2) For increasing H-Z bond strength (i.e., decreasing acidity). • Strength of electrostatic interaction of the Hδ+Zδ- bond. This is

governed by decreasing bond distance and small anion size. For these two reasons the H-F bond is stronger than the H-Cl bond.

For the comparison of strengths of most acids, the electron withdrawing effect of Z is the much more important factor; i.e., the greater the electronegativity (which directly corresponds to the electron withdrawing effect), the weaker the H-Z bond. The weaker the H-Z bond, the stronger the acid. (The electrostatic interaction of H-Z bonds for most acids [i.e., oxyacids] is too weak [because of the large distance between the charge centers] to be a significant factor in acid strength.) Hence, HOF is a stronger acid than HOCl. For the comparison of vertical group acids, both factors need to be taken into consideration. The more important factor of the two depends on the H-Z bond distance. The H-Z electrostatic (δ+ δ-) interaction of the two partial charges is the dominant factor when the atoms are close to each other (according to Coulomb’s law); the closer the two charges (e.g., H-F), the stronger the H-Z bond; the stronger the H-Z bond, the weaker the acid. Hence, although F has a significant electron-withdrawing effect, the electrostatic interaction is the dominant effect, resulting in H-F being a weak acid. In descending the vertical group acids in the periodic table (e.g., toward H-I), the electrostatic interaction becomes weaker due to the increasing distance between the atoms, and the size of the anions [increased size results in greater charge dispersal and therefore more stability] while the importance of the electrostatic interaction as a factor is diminished. Therefore, the strength of the electron withdrawing effect becomes more important while descending vertical group acids. Hence, HI is a strong acid due to a significant electron withdrawing effect of the iodine atom, and an insignificant electrostatic interaction between the distant H and I atoms.

** misleadingly referred to as bond strength in H&P


RELATIONSHIPS BETWEEN STRUCTURE AND STRENGTHS

OF ACIDS AND BASES • Vertical group binary acids: Acid strength increases as period

number (proton-halogen bond distance) increases. • All other acids: Acid strength increases as electronegativity

increases.


ET: First discuss equations, then reverse equations invert K, then Ka x Kb = 10-14 Write the equil. equations and identify K’s for the following reactions: Weak acid + water (e.g., acetic acid + H2O): HA + H2O ←→ H3O+ + A- 1 Assume Ka = ~1 x 10-5 Conjugate base + water (e.g., sodium acetate + H2O): A- + H2O ←→ HA + OH- 3 Kcb = Kw/Ka = (1 x 10-14)/(1 x 10-5) = 1 x 10-9 Conjugate base + H3O+ (e.g., sodium acetate + H3O+): A- + H3O+ → HA + H2O 2 K = 1/Ka = 1/(1 x 10-5) = 1 x 105 (or bring to completion; then solve either using HH or reverse + Ka) Weak acid + strong base (e.g., acetic acid + sodium hydroxide): HA + OH- → H2O + A- 4 K = 1/(Kw/Ka) = Ka/Kw = (1 x 10-5)/(1 x 10-14) = 1 x 109 (or bring to completion; then solve either using HH or reverse + Ka) Weak base + water (e.g., NH3 + H2O): RNH2 + H2O ←→ RNH3

+ + OH- Assume Kb = ~1 x 10-5 Conjugate acid + water (e.g., sodium acetate + H2O): RNH3

+ + H2O ←→ RNH2 + H3O+ Kca = Kw/Kb = (1 x 10-14)/(1 x 10-5) = (1 x 10-9) Conjugate acid + OH- (e.g., ammonium chloride + OH-): RNH3

+ + OH- → RNH2 + H2O K = 1/Kb = 1/(1 x 10-5) = 1 x 105 (or bring to completion; then solve either using HH or reverse + Kb) Weak base + strong acid (e.g., NH3 + H3O+): RNH2 + H3O+ → RNH3

+ + H2O K = 1/(Kw/Kb) = Kb/Kw = (1 x 10-5)/(1 x 10-14) = 1 x 109 (or bring to completion; then solve either using HH or reverse + Kb)


0

2

4

6

8

10

12

14

0 10 20 30 40

TITATION CURVE OF WEAK ACID WITH STRONG BASE

pH

Volume of Titrant (mL)

Completion or End Point or 9 for SB & WA Equivalence Point 7 for SB & SA or 5 for WB & SA Stoichiometric Point or Neutralization Point

Red HIn + In- Phenolphthalein Indicator Colorless

Half-way Point or Half-equivalency Point pH = pKa + log([A-]/[HA]) pH = pKa + log([X]/[X]) pH = pKa 5 = pKa Ka = 10-pH = 1 x 10-5

Weak acid

Strong base

Henderson-Hasselbalch equation pH = pKa + log([B]/[A]) At half-way point pH = pKa [H+] = Ka (= 10-pH)

HA + OH- ←→ H2O + A-

1:1 HA:A-

A-

HA

A titration is a procedure where a base is added to an acid, or an acid is added to a base until the stoichiometric amount of base or acid is added This is called the

Strong base

Weak base

Strong acid

Weak acid


SIX STRONG ACIDS ET: Discuss strong acids and strong bases and the strengths of their conjugate bases and acids; practice both.

Very weak base (i.e., not a base; Ka Acid spectator ion)** ~106 HCl + H2O H3O+ + Cl-

~108 HBr + H2O H3O+ + Br-

~109 HI + H2O H3O+ + I-

H2SO4 + H2O H3O+ + HSO4-***

HNO3 + H2O H3O+ + NO3-

HClO4 + H2O H3O+ + ClO4-

(HClO3 + H2O H3O+ + ClO3-)*

*Strong, but not common, acid. Generally not considered as a strong acid. **These conjugate bases are known as “nominal bases”, i.e., a base by definition, but effectively not a base.

***Although HSO4- is effectively not a base, it is effective as an acid. HSO4- + H2O ←→ H3O+ + SO4

2-

All other acids are weak acids, e.g., HA (acetic acid), H2SO3, RNH3+, BF3

Arrhenius acid: Anything that provides a proton, e.g., HA, H2SO3, RNH3

+ BL Acid: Anything tending to give up a proton, e.g., HA, H2SO3, RNH3

+ (Lewis Acid: Anything tending to react with an electron pair, e.g., BF3)

STRONG BASES (Most Group 1A and 2A hydroxides [not HOH])

e.g. Very weak acid (i.e., not an acid; Base spectator ion) LiOH OH- + Li+

NaOH OH- + Na+

Mg(OH)2 2OH- + Mg2+

Ba(OH)2 2OH- + Ba2+ All other bases are weak bases, e.g., RNH2, CO3

2-

Arrhenius Base: Anything that provides an OH- group, e.g., NaOH BL Base: Anything that tends to react with a proton, e.g., NaOH, RNH2, CO3

2- Lewis Base: Anything that has available non-bonding electrons, e.g., NaOH, RNH2, CO3

2-


Chem 162-2010 Final exam Acids and Bases - Chapters 16 & 17 Acid-base properties of salts concepts 3. Which of the following are BASIC salts? W. NH4I X. BaCl2 Y. KNO3 Z. NaF A. Z only B. Xand Y C. Y and Z D. W only E. W and X ET: Get rid of schmutz ions and determine pH of what’s remaining. NH4I → NH4

+ + I- I- is such a weak base that it doesn’t act basic. I- + H2O → NR But NH4

+ is a weak acid. NH4

+ + H2O ←→ NH3 + H3O+ Therefore, solution is acidic. BaCl2 → Ba2+ + 2Cl- Group I and II cations are not acidic. Ba2+ + H2O → NR The conjugate bases of strong acids are not basic. Cl- + H2O → NR Therefore, this salt is neutral in water. KNO3 → K+ + NO3

- Group I and II cations are not acidic. K+ + H2O → NR The conjugate bases of strong acids are not basic. NO3

- + H2O → NR Therefore, this salt is neutral in water. NaF → Na+ + F- Group I and II cations are not acidic. Na+ + H2O → NR F- is a weak conjugate base. F- + H2O ←→ HF + OH- Together, NaF provides a basic solution.


CHEM 162-2000 FINAL EXAM CHAPTERS 16 & 17 - ACID AND BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS

44. Methylamine, CH3NH2 is a weak base with Kb = 3.6x10-4 at 25oC. What is the equilibrium constant for the following reaction at 25oC? CH3NH2(aq) + H+(aq) ←→ CH3NH3

+(aq) A. 2.8x1011 B. 3.6x1010 C. 1.7x108 D. 3.6x103 E. 2.8x103 ET note: This is the same reaction as in the table: CH3NH2(aq) + H3O+(aq) ←→ CH3NH3

+(aq) + H2O. The right side is Kca.

CH3NH2(aq) + H+(aq) ←→ CH3NH3+(aq)

CH3NH2(aq) + H3O+(aq) ←→ CH3NH3+(aq) + H2O

The right side of the equation is Kca. K = 1/Kca Kw = Kb x Kca Kca = Kw/Kb K = 1/(Kw/Kb) = Kb/Kw = (3.6 x 10-4)/(1 x 10-14) = 3.6 x 1010


CHEM162-2010 HOURLY EXAM II + ANSWERS CHAPTERS 16 & 17 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CALCULATIONS

45. How many moles of solid NH4Cl must be dissolved in 1.00L of a 0.950M NH3(aq) solution in order to prepare a buffer with a pH of 9.60. Kb(NH3) = 1.8x10-5 A. 2.3 mol B. 1.6 mol C. 0.75 mol D. 0.43 mol E. 0.34 mol ET Note: The solution consists of NH3, NH4Cl and H2O. Think HH equation. pH 9.60 pOH = 4.40 -log[OH-] = 4.40 [OH-] = 4.0 x 10-5 NH3(aq) + H2O ←→ NH4

+(aq) + OH-(aq)

Initial 0.950 0 0 Change Equilibrium 4.0 x 10-5 NH3(aq) + H2O ←→ NH4

+(aq) + OH-(aq)

Initial 0.950 Y 0 Change -X +X +X Equilibrium 0.950-X Y+X 4.0 x 10-5 NH3(aq) + H2O ←→ NH4

+(aq) + OH-(aq)

Initial 0.950 Y 0 Change -4.0 x 10-5 +4.0 x 10-5 +4.0 x 10-5 Equilibrium 0.950-(4.0 x 10-5) Y + (4.0 x 10-5) 4.0 x 10-5 Henderson-Hasselbach: pH = -log Kca + log([A-]/[HA]) pH = -log (Kw/Kb) + log([A-]/[HA]) 9.60 = -log((1x10-14)/(1.8 x 10-5)) + log(0.950/Y) X = 0.4295


CHEM 162-2010 FINAL EXAM Chapters 16 & 17a - Acid & Base Equilibria Acid and base equilibria calculations

37. Phosphoric acid, H3PO4, is a polyprotic acid. If 0.30 mol H3PO4 and 0.60 mol KOH are reacted in aqueous solution, which one of the following ions will have the highest concentration at equilibrium? A. H3O+ B. OH- C. H2PO4

- D. PO4

3- E. HPO4

2- ET note: (1) Strong bases bring the reaction to completion. (2)The Ka’s become much, much smaller in going from K1 to K2 to K3. K3 is essentially zero. 1st reaction: Bring to completion; H3PO4 is the limiting reactant. H3PO4 + OH- → H2PO4

- + H2O Initial 0.30 0.60 0 0 Change -0.30 -0.30 +0.30 Equilibrium 0 0.30 0.30 2nd reaction: H2PO4

- + OH- → HPO42- + H2O

Initial 0.30 0.30 0 0 Change -0.30 -0.30 +0.30 Equilibrium 0 0 0.30 3rd reaction: HPO4

2- + H2O → H3O+ + PO43-

Initial 0.30 0 0 Change -X +X +X Equilibrium 0.30-X +X +X ([H3O+][PO4

3-])/[HPO42-] = K3

([X][X])/[0.30-X] = K3 In the third reaction, K3 is so small due to the formation of a triply charged substance (i.e., unstable), that essentially nothing happens. Hence, X is very, very small A. No. H3O+ only forms in the third step, but since K3 is so small virtually no H3O+ will form. B. No. The OH- gets completely used up in the first and second steps. C. No. The H2PO4

- that is formed in the first step is completely used up in the second step. D. No. The PO4

3- forms only in the third step, but since third step Ka’s of polyprotic acids are soooo small, there is virtually no PO4

3- formed. E. Yes. The HPO4

2- that is formed in the second step is virtually untouched in the third step because 3rd Ka’s are very, very small.


RELATIONSHIPS BETWEEN STRUCTURE AND STRENGTHS OF ACIDS AND BASES

• Vertical group binary acids: Acid strength increases as period

number (proton-halogen bond distance) increases. • All other acids: Acid strength increases as electronegativity

increases.


CHEM 162-2010 FINAL EXAM CHAPTERS 16 & 17 - ACID AND BASE EQUILIBRIA RELATIONSHIP BETWEEN ACID AND BASE STRUCTURE AND STRENGTH

26. Arrange the following acids in order of increasing acid strength. HBrO HClO3 HClO2 HClO HIO A. HClO < HBrO < HIO < HClO2 < HClO3 B. HClO3 < HClO2 < HClO < HIO < HBrO C. HIO < HBrO < HClO < HClO2 < HClO3 D. HClO < HClO2 < HClO3 < HBrO < HIO E. HClO3 < HClO2 < HClO < HBrO < HIO ET note: All oxyacids All five are oxyacids, so the strength depends on the relative weakness of the bonds, which is caused by differences in electronegativity. Electronegativity ranking: F > O > Cl > Br > I HIO is the weakest acid. HBrO is a little stronger than HIO because Br is stronger in electronegativity than I. HClO is a little stronger than HBrO because Cl is stronger in electronegativity than Br. HClO2 is a little stronger than HClO because it contains two electronegative oxygens. HClO3 is a little stronger than HClO2 because it contains three electronegative oxygens.


Chem 162-2010 Final Exam + Answers Chapters 16 & 17 - Applications of Aqueous Equilibria (of Acids and Bases) Buffers (including Henderson-Hasselbalch equation) Concepts

18. A buffer is formed by adding 1.0 mol of CH3COOH and 1.0 mol CH3COO- in a 1.0 L container. The pH of the buffer is 4.74. Which of the following would make the smallest change in the pH? A. Add 0.50 mol of NaOH to 1.0L of the buffer B. Add 0.50 mol of HCl to 1.0L of the buffer C. Add 49L of water to 1.0L of the buffer D. Add 0.50 mol of CH3COOH to 1.0 L of the buffer. E. Add 0.50 mol of CH3COO- to 1.0L of the buffer. ET: Addition of any acid or base will change the pH, some more than others. But, based on the HH equation, addition of H2O dilutes the [A-] and [HA] equally, so the log term will still drop out.

Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) A. False. Addition of a strong base will react with the HA, resulting in a small increase in pH. B. False. Addition of a strong acid will react with the A-, resulting in a small decrease in pH. C. True. In the Henderson-Hasselbalch equation, 1.0 mol of CH3COOH and 1.0 mol CH3COO- in 1.0L results in the following: pH = pKa + log([1]/[1]). In this case, since the log of 1/1 drops out, the pH = the pKa of the acetic acid. If the buffer were diluted 50 fold, then the equation would be pH = pKa + log([0.02]/[0.02]). The log of 0.02/0.02 drops out again, and the pH still equals the pKa, i.e., the pH doesn’t change at all upon dilution of the buffer. (Only the buffer capacity changes.) D. False. Addition of a weak acid will react with the A-, resulting in a very small decrease in pH. E. False. Addition of a weak base will react with the HA, resulting in a very small increase in pH.


0

2

4

6

8

10

12

14

0 10 20 30 40

TITATION CURVE OF WEAK ACID WITH STRONG BASE

pH

Volume of Titrant (mL)

Completion or End Point or 9 for SB & WA Equivalence Point 7 for SB & SA or 5 for WB & SA Stoichiometric Point or Neutralization Point

Red HIn + In- Phenolphthalein Indicator Colorless

Half-way Point or Half-equivalency Point pH = pKa + log([A-]/[HA]) pH = pKa + log([X]/[X]) pH = pKa 5 = pKa Ka = 10-pH = 1 x 10-5

Strong base

Henderson-Hasselbalch equation pH = pKa + log([B]/[A]) At half-way point pH = pKa [H+] = Ka (= 10-pH)

HA + OH- ←→ H2O + A-

1:1 HA:A-

A-

HA

A titration is a procedure where a base is added to an acid, or an acid is added to a base until the stoichiometric amount of base or acid is added This is called the

Strong base

Weak base

Strong acid

Weak acid


CHAPTERS 16 & 17 CHEM 162-2010 Final exam Chapters 16 & 17B - Applic. of Acid & Base Equilibria (Buffers & Titrations) Titrations and Indicators

1. Aniline, one of the first synthetic dyes, has a Kb of 7.4 x 10-10. A student is planning to titrate a solution containing aniline to determine its concentration. In picking an indicator, what should the student consider (aside from the color of the dye)? A. The choice of indicator will depend on the initial concentration. B. The indicator should change color at or near pH 9. C. The indicator should change color at or near pH 3. D. The indicator should change color at or near pH 7. E. The indicator should change color at or near pH 6. ET note: Problem can be done quantitatively or qualitatively. Quantitatively involves bringing the reaction to completion and then back again. Qualitatively involves realizing that 10-10 is a very,very weak base, and estimating that the equivalence point would be at ~ pH 3 on the titration diagram.

Weak bases are generally titrated with strong acids. When a strong acid titrates a strong base the stoichiometric point is approximately 7. When a strong acid titrates a weak base, e.g., NH3 (Kb = 1.8 x 10-5), the stoichiometric point is approximately 5. Aniline is a very weak base (Kb = 7.4 x 10-10). Hence the stoichiometric point should be well below 5. The only option is 3.


CHAPTER 16

PRECIPITATION REACTIONS (HEAVY METAL IONS FORMING

PRECIPITATES)

1A 2A 3A 4A 5A 6A 7A 8A

He

Li+ Be B C N O F Ne

Na+ Mg 3B 4B 5B 6B 7B 8B 8B 8B 1B 2B Al Si P S2- Cl- Ar

K+ Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br- Kr

Rb+ Sr Y Zr Nb Mo Te Ru Rh Pd Ag+

? Cd In Sn Sb Te I- Xe

Cs+ Ba2+ La Hf Ta W Re Os Ir Pt Au Hg22+ Tl Pb2+ Bi Po At- Rn

Chem 162-2012 Final exam review from 2010 final exam

20

PRECIPITATION REACTIONS Concept: Soluble salts: All group IA metal ions and NH4

+ All nitrates (NO3

-), acetates (CH3CO2-),

perchlorates (ClO4-)

All Cl-, Br-, I-* (except Ag+, Pb2+, Hg22+)

All IO3- and IO4

- (except Pb2+, Hg22+, Ba2+)

All SO42- (except Pb2+, Hg2

2+, Ba2+, Ca2+) *Note absence of F-

Insoluble salts: Ag+, Pb2+, Hg2

2+, Ba2+ Hydroxides (OH-) Sulfides (S2-) Carbonates (CO3

2-) Chromates (CrO4

2-) Phosphates (PO4

3-)

HIG

H P

RIO

RIT

Y

LOW

PR

IOR

ITY


21

CHEM 162-2001 HOURLY EXAM III Chapter 16 SOLUBILITY PRODUCT

5. What is the pH of an aqueous solution that is saturated with Mn(OH)2 (Ksp = 1.9 x 10-13)? A. 9.86 B. 8.53 C. 10.23 D. 11.20 E. 7.56 ET: Begin with Ksp equilibrium equation. Solve as usual.

Mn(OH)2(s) ←→ Mn2+(aq) + 2 OH-(aq) Initial Y 0 0 Change -X +X +2X Equilibrium Y-X +X +2X [Mn2+][OH-]2 = Ksp X x (2X)2 = 1.9 x 10-13 4X3 = 1.9 x 10-13 X = 3.625 x 10-5 2X = 7.251 x 10-5 pOH = -log[OH-] pOH = -log[7.251 x 10-5] pOH = 4.14 pH = 14 - 4.14 = 9.86


22

CHEM-2010 FINAL EXAM + ANSWERS CHAPTER 16 - EQUILIBRIA SOLUBILITY PRODUCT 29. A solution is initially 0.10M in Mg2+(aq) and 0.10M in Fe2+(aq). Solid NaOH is slowly added. What is the concentration of Fe2+ when Mg(OH)2 first precipitates? Ksp(Mg(OH)2 = 6.0 x 10-10; Ksp(Fe(OH)2) = 7.9 x 10-16 A. 4.1 x 10-6 B. 2.6 x 10-9 C. 3.8 x 10-11 D. 1.3 x 10-7 E. 6.7 x 10-5 ET: Selective precipitation problem. Note that selective precipitation is the main factor in the Qualitative Inorganic Analysis Scheme at the end of chapter 16.

Mg(OH)2(s) ←→ Mg2+ + 2OH-

Fe(OH)2(s) ←→ Fe2+ + 2OH- Mg(OH)2(s) ←→ Mg2+ + 2OH-

Initial Y 0 0 Change -X +X +2X Equilibrium 0.10 2X [Mg2+][OH-]2 = Ksp 0.10 x (2X)2 = 6.0 x 10-10 X = 3.872 x 10-5 2X = 7.744 x 10-5 = [OH-] to just begin precipitation of Mg(OH)2 Fe(OH)2(s) ←→ Fe2+ + 2OH-

Initial Y 0 0 Change Equilibrium 0.10 +2X [Fe2+][OH-]2 = Ksp 0.10 x (2X)2 = 7.9 x 10-16 X = 4.44 x 10-8 2X = 8.88 x 10-8 = [OH-] to just begin precipitation of Fe(OH)2 So Fe(OH)2 begins to precipitate first. Stop the titration when [OH-] just reaches 7.744 x 10-5. Determine the concentration of Fe2+ in solution when [OH-] just reaches 7.744 x 10-5M [Fe2+][OH-]2 = Ksp [Y] x ((7.744 x 10-5)2) = (7.9 x 10-16) Y = [Fe2+] = 1.317 x 10-7


23

CHEM 162-2010 Final exam Chapter 16 - Transition Metals and Coordination Chemistry Complex Ion Equilibria

28. 0.40 moles of AgNO3 and 2.5 moles of S2O32- are dissolved in 1.00L of aqueous solution. A

coordination complex forms. Ag+(aq) + 2S2O3

2- ←→ [Ag(S2O3)2]3-(aq) Kf = 1.7 x 1013 Calculate [Ag+] in this solution. A. 8.1 x 10-15M B. 5.3 x 10-15M C. 1.7 x 10-13M D. 2.0 x 10-12M E. 1.7 x 10-13M ET: Typical complex ion problem using small K rule (because of large K) and then right-to-left rule.

Ag+(aq) + 2S2O32-(aq) ←→ [Ag(S2O3)2]3-(aq)

Initial 0.40 2.5 0 Change -X -2X +X Equilibrium 0.40-X 2.5-2X +X This is going to give a quadratic equation. Use the large K rule. Bring the reaction to completion, and then back to equilibrium. Ag+ is the limiting reactant. Ag+(aq) + 2S2O3

2-(aq) ←→ [Ag(S2O3)2]3-(aq)

Initial 0.40 2.5 0 Change -0.40 -0.80 +0.40 Equilibrium 0 1.70 +0.40 Now go back to equilibrium. Since it’s more convenient for the student to go from left to right, reverse the equation, and invert K. [Ag(S2O3)2]3- ←→ Ag+ (aq) + 2S2O3

2-(aq)

New initial 0.40 0 1.70 Change -X +X +2X Equilibrium 0.40-X +X 1.70 + 2X ([Ag+][S2O3

2-]2)/[Ag(S2O3)2]3- = 1/Kf ([X][1.70+2X]2)/[0.40-X] = 1/(1.7x1013) = 5.88 x 10-14 To avoid a quadratic equation, use the small K rule. ([X][1.70]2)/[0.40] = 5.88 x 10-14 X = 8.14 x 10-15 = [Ag+]


24

CHAPTER 19 BALANCING REDOX EQUATIONS

(Follow rules in descending order of priority)

(1) Write net reaction. (2) Identify pairs of substances involved in oxidation (OXLEA) and pairs of substances involved in reduction half-reactions. (3) Balance (in the following order) -- the atoms (not O or H); -- O, using H2O; -- H, using H+; -- charge, using electrons; -- electrons of each half-cell reaction, using half-equation coefficients. (4) Add the two half-cell equations. (5) Get rid of spectator ions/spectator molecules/spectator electrons. (6) Adjust H+ with OH- if basic solution is requested. (7) Verify that all atoms and charges are balanced.


25

for lecture 20CHAPTER 19 - ELECTROCHEMISTRY

ELECTRODE REDUCTION POTENTIALS Some Selected Standard Electrode Reduction Potentials ET: In this table all reactions are written as reductions. Eo, volt (red’n) F2 + 2 e- → 2 F- +2.87 H2O2 + 2 H+ + 2 e- → 2 H2O +1.78* MnO4

- + 8H+ + 5e- → Mn2+ + 4H2O +1.51 Au3+ + 3e- → Au +1.50 PbO2 + 4 H+ + 2 e- → Pb2+ + 2 H2O +1.46 Cl2 + 2 e- → 2 Cl- +1.36 Cr2O7

2- + 14 H+ + 6 e- → 2 Cr3+ + 7 H2O +1.33 O2 + 4 H+ + 4 e- → 2 H2O +1.23* Br2 + 2e- → 2Br- +1.09 Ag+ + e- → Ag +0.80 Fe3+ + e- → Fe2+ +0.77 MnO4

- + 2 H2O + 3 e- → MnO2 + 4 OH- +0.60 I2 + 2 e- → 2 I- +0.54 Cu2+ + 2 e- → Cu +0.34 Cu2+ + e- → Cu+ Ref. electrode: +0.16 2H+ + 2 e- → H2 1M H+ & 1atm H2 0.00 Fe3+ + 3e- → Fe -0.036 Pb2+ + 2 e- → Pb -0.13 Ni2+ + 2 e- → Ni -0.23 Cd2+ + 2e- → Cd -0.40 Cr3+ + e- → Cr2+ -0.50 Zn2+ + 2 e- ----> Zn -0.76 2 H2O + 2 e- → H2 + 2 OH- -0.83* Al3+ + 3e- → Al -1.66 Mg2+ + 2 e- → Mg -2.37 Na+ + e- → Na -2.71 Ba2+ + 2e- → Ba -2.91 Cs+ + e- → Cs -3.03 * H2O half-cell reactions

RED

UC

TIO

N

OX

IDA

TIO

N


26

ELECTROCHEMISTRY FORMULAS

Eo

cell = Eoredn + Eo

oxidn Ecell = Eo

cell -(RT/nF)ln(Q) = Nernst equation (Ecell = Eo

cell – ((8.314 x 298.15)/(n x 96485)) x (2.303 x log(Q))

Ecell = Eocell -(0.0592/n)log(Q) = Nernst equation

At equilibrium: 0 = Eocell -(0.0592/n)log(K)

Eocell = (0.0592/n)log(K)

∆G = ∆Go + RT ln(Q) At equilibrium: 0 = ∆Go + RTln(K) ∆Go = -RTln(K) ∆Go = -nFEo

cell ∆G = -nFEcell wmax = ΔG Definitions: 1 ampere = 1 coulomb of charge/second 1 mole of electrons caries a charge of 1 Faraday = 96485 coulombs Calculations : (1) Amperes x ((Coulombs/sec)/Ampere) x sec x (1 mol e-/96485 coulomb) x (mol subst./mol e-) = mol substance or (2) (amperes x seconds)/(96485 x electrons) = mol substance


27

LINE NOTATION

• Salt bridge, represented by a double vertical line, in middle. • Anode components on left; cathode components on right o Anode and cathode electrode substance on far left and far right, respectively. o All species of each half-cell reaction listed between its electrode and the salt bridge. - If species are in same phase, separate by commas. - If species are in different phases, separate by single vertical lines.


28

LINES OF DEFENSE AGAINST RUST - Paint the iron to exclude air. - Coat with a less active material, e.g., Sn

- Galvanizing: Coat with a more active material, e.g., Zn (galvanized iron) Forms a thin, hard ZnO coating, impervious to air. - Stainless steel: Stainless steel is an alloy of iron with small

amounts of chromium and nickel, which form oxide coatings which protect the iron.

- Sacrificial anode: Fe connected to a chunk of more active material, e.g., Zn, Mg, Al, directly or with a wire.


29

LINE NOTATION

• Salt bridge, represented by a double vertical line, in middle. • Anode components on left; cathode components on right o Anode and cathode electrode substance on far left and far right, respectively. o All species of each half-cell reaction listed between its electrode and the salt bridge. - If species are in same phase, separate by commas. - If species are in different phases, separate by single vertical lines.


30

CHEM 162-2010 FINAL EXAM CHAPTER 19 - ELECTROCHEMISTRY GALVANIC CELLS & REDUCTION-POTENTIAL CONCEPTS

4. In the cell diagram Pt(s)│Br2(l)│Br-(aq)║Cl-(aq)│Cl2(g)│Pt(s) (1) The cathode reaction is Br2(l) + 2e- → 2Br-(aq). (2) Platinum is part of the cell as an inert conductor. (3) The cell potential will change as the concentration of Cl-(aq) changes. A. Only (1) is true. B. Only (1) and (3) are true. C. Only (2) is true. D. All three are true. E. Only (2) and (3) are true. ET: Explain line diagram as anode on the left and cathode on the right.

(1) False. In a line diagram, the cathode is on the right side. It is the reaction of Cl2(l) + 2e- → 2Cl-(aq). (2) True. Platinum is part of the cell as an inert conductor. (3) True. In all voltaic cells, the cell potential decreases as the reaction goes forward. The reaction going forward means that the Br- concentration decreases, while the Cl- concentration increases. Eventually the cell potential becomes zero, and the reaction stops. This can be demonstrated mathematically by plugging numbers into the Nernst equation for this reaction. At first, the value of E is the difference between the voltages of the two half-cells. But as the concentration of Cl- increases (and correspondingly the concentration of Br- decreases), the value of E becomes smaller. E = 0.29 – ((0.0592/1) x log ([Cl-]/[Br-]))


31

CHEM 162-2010 FINAL EXAM CHAPTER 19 - ELECTROCHEMISTRY REDUCTION POTENTIAL CALCULATIONS 15. Given the following data: AgCl(s) + e- → Ag(s) + Cl- Eo = 0.22 volts Zn2+(aq) + 2e- → Zn(s) Eo = -0.76 volts The voltaic cell represented by: Zn(s)│Zn2+(0.0010M)║Cl-(?M)│AgCl(s)│Ag(s) has a measured cell potential of 1.17 volts. What is the molarity of Cl- in this cell? A. 0.0010M B. 0.0050M C. 0.0190M D. 0.0500M E. 1.00M ET: Need to understand line diagrams. Straightforward filling in Nernst equation with one unknown, the [Cl-].

The left side of line diagrams is the anode, and the right side is the cathode. Eo Zn → Zn2+ + 2e- +0.76 2(AgCl + 1e- → Ag + Cl-) +0.22 Zn + 2AgCl → Zn2+ + 2Ag + 2Cl- +0.98V Zn(s) + 2AgCl(s) → Zn2+ + 2Ag(s) + 2Cl- 0.0010M XM Ecell = Eo

cell – (0.0592/n)logQ Ecell = Eo

cell – (0.0592/n)log([Zn2+][Cl-]2) 1.17 = 0.98 – (0.0592/2) x log([0.0010][X2]) X = 0.0195 = [Cl-]


32

BALANCING REDOX EQUATIONS

(Follow rules in descending order of priority)

(1) Write net reaction. (2) Identify pairs of substances involved in oxidation (OXLEA) and pairs of substances involved in reduction half-reactions. (3) Balance (in the following order) -- the atoms (not O or H); -- O, using H2O; -- H, using H+; -- charge, using electrons; -- electrons of each half-cell reaction, using half-equation coefficients. (4) Add the two half-cell equations. (5) Get rid of spectator ions/spectator molecules/spectator electrons. (6) Adjust H+ with OH- if basic solution is requested. (7) Verify that all atoms and charges are balanced.


33

CHEM 162-2010 FINAL EXAM CHAPTER 19 - ELECTROCHEMISTRY OXIDATION-REDUCTION REACTION CONCEPTS

7. Consider the unbalanced half reaction: S2O3

2- +OH- → SO42- + H2O + e-

When the above half reaction is balanced with whole number coefficients, which of the following are the correct coefficients for OH- and SO4

2-? OH- SO4

2- A. 8 3 B. 5 5 C. 4 6 D. 10 2 E. 3 7 ET: Balancing a half-cell equation in basic solution

S2O32- → SO4

2- Balance the S’s. S2O3

2- → 2(SO42-)

Balance the oxygens with H2O. S2O3

2- + 5H2O → 2(SO42-)

Balance the hydrogens with H+. S2O3

2- + 5H2O → 2(SO42-) + 10H+

Balance the charge with electrons. S2O3

2- + 5H2O → 2(SO42-) + 10H+ + 8e-

Make it basic by adding OH- to both sides. S2O3

2- + 5H2O + 10OH-→ 2(SO42-) + 10H+ + 8e- + 10OH-

Simplify the equation. S2O3

2- + 5H2O + 10OH-→ 2(SO42-) + 10H2O + 8e-

Get rid of spectators. S2O3

2- + 10OH-→ 2(SO42-) + 5H2O + 8e-


34

Chem 162-2010 Final Exam + Answers Chapter 19 - Electrochemistry Free energy and cell potential calculations/Nernst equation, concentration cell

9. The Eo for the reaction Cr2O7

2-(aq) + 6Fe2+(aq) + 14H+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) is +0.56V at 25oC. Calculate the equilibrium constant for the reaction. A. 7.1x1056 B. 1.4x103 C. 2.9 x 109 D. 37.8 E. 2.5x1028 Eo

cell = +(0.0592/n)log(K) ET note: We know Eo

cell. We just need to calculate “n” to find “K”. We need to find the value of n, the number of electrons transferred. Cr2O7

2- → Cr3+ Cr2O7

2- → 2Cr3+ Cr2O7

2- → 2Cr3+ + 7H2O Cr2O7

2- + 14H+ → 2Cr3+ + 7H2O Cr2O7

2- + 14H+ + 6e- → 2Cr3+ + 7H2O Fe2+ → Fe3+ Fe2+ → Fe3+ + 1e- 6(Fe2+ → Fe3+ + 1e-) Cr2O7

2- + 14H+ + 6e- → 2Cr3+ + 7H2O 6(Fe2+ → Fe3+ + 1e-) Cr2O7

2- + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O n = 6 0.56 = +(0.0592/6) x log(K) K = 7.125 x 1056


35

CHEM 162-2010 EXAM III + ANSWERS CHAPTER 19 ELECTROCHEMISTRY FREE ENERGY AND CELL POTENTIAL CALCULATIONS/NERNST EQUATION ET: Very similar to previous problem except that here we have to calculate Eo.

11. Given Eo(V) Co2+ + 2e- → Co(s) -0.28 Cd2+ + 2e- → Cd(s) -0.40 What is the equilibrium constant for Cd(s) + Co2+(aq) ←→ Cd2+(aq) + Co(s) A. 2.1 x 105 B. 1.1 x 104 C. 3.3 x 10-4 D. 4.5 x 10-6 E. 3.1 x 104 ET: Straightforward; calculate Eo

cell, n, and then find K

Cd(s) → Cd2+(aq) + 2e- ℰ° = +0.40V Co2+(aq) + 2e- → Co(s) ℰ° = -0.28V Cd(s) + Co2+(aq) ←→ Cd2+(aq) + Co(s) ℰ°cell = 0.12V Eo

cell = (0.0592/n)log(K) 0.12 = (0.0592/2) x log(K) K = 1.13x104


36

Chem 162-2010 Final Exam + Answers Chapter 19 - Electrochemistry Galvanic cells, line notation & reduction potential

16. If the standard reduction potentials for the couples Cu2+/Cu, Ag+/Ag and Fe2+/Fe are +0.34V, +0.80V and -0.44V respectively, which is the strongest reducing agent? A. Cu B. Ag C. Ag+ D. Fe2+ E. Fe ET: Discuss “Standard Reduction Potential” table as being for reductions, which means “oxidizing agents”, but “reducing agents” is for oxidations. Standard reduction potential table: Ag+ + e- → Ag Eo = +0.80V Cu2+ + 2e- → Cu Eo = +0.34V Fe2+ + 2e- → Fe Eo = -0.44V A strong reducing agent means that the substance is easily oxidized. So change the table to a standard oxidation potential table. Fe → Fe2+ + 2e-. Eo = +0.44V Cu → Cu2+ + 2e- Eo = -0.34V Ag → Ag+ + e- Eo = -0.80V Fe is the most easily oxidized and therefore the strongest reducing agent.


37

anio

ns

catio

ns

anio

ns

catio

ns


38

Chem 162-2010 Final exam + answers Chapter 19- Electrochemistry Galvanic cells, line notation & reduction potential concepts

17. Which one of the following statements is false in a galvanic cell? A. Sustained electron flow cannot occur without the salt bridge or a porous disk. B. Electrons flow from the anode to the cathode through an external wire. C. The electrode compartment in which oxidation occurs is called the anode. D. Cations flow from the salt bridge to the anode compartment. E. It is a device for converting chemical energy to electrical energy. ET: Students need to be familiar with the galvanic and electrolytic cells and their differences. A. True. The salt bridge contains ions to neutralize the buildup of charges at

the electrodes. If there is no salt bridge, then the charges at the electrodes will build up, making the system thermodynamically unstable, resulting in the reaction stopping.

B. True. Oxidation is loss of electrons at the anode. The electrons flow from the anode to the cathode through an external wire.

C. True. Oxidation is loss of electrons at the anode. D. False. Negative charge is built up at he cathode. Therefore, cations flow

from the salt bridge to the cathode compartment in order to neutralize the build up of this negative charge

E. True. The chemicals change spontaneously, by way of the electrons going through an external wire. On the other hand, an electrochemical cell is one which converts electrical energy (from a battery) into chemical energy (increasing the Gibbs free energy of the products).


39

Chem 162-2010 Final exam + answers Chapter 19- Electrochemistry Free energy and cell potential calculations/Nernst equation, concentration cell calculations

19. Given Br2(l) + 2e- → 2Br-(aq) Eo = 1.065V O2(g) + 4H+(aq) + 4e- → 2H2O(l) Eo = 1.229 V Calculate ∆Go for the reaction below: 2Br2(l) + 2H2O(l) → 4Br-(aq) + O2(g) + 4H+(aq) A. +31.7 kJ B. -31.7 kJ C. -15.8 kJ D. -42.2 kJ E. +63.3 kJ ET note: Straightforward. Use equation “ΔGo = -nFEo

cell”; calculate n and Eocell, then find ΔGo. ΔGo = -nFEo

cell Eo(V)

2(Br2(l) + 2e- -> 2Br-(aq)) 1.065 2H2O(l)→ O2(g) + 4H+(aq) + 4e- -1.229 2Br2(l) + 2H2O(l) ←→ 4Br-(aq) + O2(g) + 4H+(aq) -0.164 ΔGo = -4 x 96485 x (-0.164) = 6.33x104 J = 63.3 kJ


40

Chem 162-2010 Final Exam + Answers Chapter 19 – Electrochemistry Electric work/amperes

31. What mass of copper will be deposited at the cathode if a current of 0.15A is passed through a solution of CuSO4 for 45 seconds? A. 2.22 x 10-3g B. 1.03 x 107g C. 2.07 x 107g D. 1.11 x 10-3g E. 3.50 x 10-5g

Cu2+ + 2e- → Cu

ET note: Straightforward; calculate mol and then convert to mass.

(amperes x seconds)/(96485 x electrons) = mol substance (0.15 x 45)/(96485 x 2) = mol Cu = 3.498 x 10-5 mol Cu 3.498 x 10-5 x 63.546g/mol = 2.22 x 10-3 g Cu


41

CHAPTER 14 FORMULAS ET: Rate = speed. Using a diagram with Y being 100 lbs and 110 lbs, and days 0 to 5, show that the increase in weight = slope = (Y2 – Y1)/(X2 – X1) = +2lb/day. Decrease in weight = -increase in weight = -slope = -(Y2 – Y1)/(X2 – X1) = -2lb/day. ET: Point out that table in middle of page contains the key formulas in kinetics. Discuss these formulas at beginning of recitation, focusing on 1o reaction.

C12H22O11(sucrose) + H2O → 2 C6H12O6(glucose)

Rate of sucrose disappearance = -(∆ [sucrose])/(∆ time) = -([sucrosef] - [sucrosei])/(tf - ti) Rate of glucose appearance = 2 x Rate of sucrose disappearance

aA + bB → cC + dD General rate of reaction = -(1/a)(∆[A]/∆t) = -(1/b)(∆[B]/∆t) = (1/c)(∆[C]/∆t) = (1/d)(∆[D]/∆t)

Determination of order of reaction through method of initial rates: For a single reactant: Rate 1 = k[A1]m Rate 2 = k[A2]m (Rate 1/Rate 2) = (k[A1]m)/(k[A2]m) = ([A1]/[A2])m

For two reactants: Rate1 = k[A1]m[B1]n Rate2 = k[A2]m[B1]n (Rate1/Rate2) = (k[A1]m[B1]n)/(k[A2]m[B1]n) = (A1/A2)m

Integrated Reaction Differentiated rate law k Order Reaction Rate** RATE LAW*** (y = mx + b) Half-life* units 0 AvgRate = -(C2-C1)/(t2-t1) InstRate = k[C]o =k [C]t = -kt + [C]o t1/2 = [C]o/2k M1s-1

1 AvgRate = -(C2-C1)/(t2-t1) InstRate = k[C]1 ln[C]t = -kt + ln[C]o t1/2 = 0.693/k Mos-1

2 AvgRate = -(C2-C1)/(t2-t1) InstRate = k[C]2 1/[C]t = kt + 1/[C]o t1/2 = 1/(k[C]o) M-1s-1 ** Rate of appearance = +slope; rate of disappearance = -rate of appearance = -slope. *** Differentiated Rate Law may be for more than one component, e.g., Rate = k[C]1[D]2

*Half-lives: For a zero order reaction, each successive half-life is ½ the time of the preceding one. For a first order reaction, each successive half-life is equal in time to the preceding one. For a second order reaction, each successive half-life is double time of the preceding one. Arrhenius equation: k = Ae(-Ea/RT) ln k = -Ea/RT + ln A; ln k = ((-Ea/R) x (1/T)) + ln A

A = frequency factor = combination of steric factor and collisional frequency Ea = energy of activation ln k2 - ln k1 = (-Ea/RT2) - (-Ea/RT1) ln (k2/k1) = -(Ea/R)[(1/T2) - (1/T1)]


42

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5

SUC

RO

SE C

ON

CEN

TRA

TIO

N (M

)

TIME (s)

X1,Y1

X2,Y2

Rate Plot ET: Discuss formation of plot from data; then discuss definitions

C (M) Time (s) 1.0 0 0.9 0.1 0.8 0.3 0.7 0.4 0.6 0.7 0.5 1.0 0.4 1.5 0.3 2.3 0.2 4.0 Rate = k[C]2 k = 1

Rate = -(Y2 - Y1)/(X2 - X1) Note 1,2,3,4 Rate = -(0.8-1.0)/(0.25-0) = +0.80M/s Note2

Rate = -(Y2 - Y1)/(X2 - X1) Rate = -(0.3- 0.5)/(2.33-1.0) = 0.15M/s

Note1 Rate: Change in concentration of a reactant or product per change in time. 2 Average rate: Rate calculated between two points. Note rate faster at beginning than end due to the concentration of C12H22O11 being greater at the beginning. 3 Instantaneous rate: Rate calculated at one point. 4 Initial rate: Instantaneous rate at time close to zero, to avoid the complication of reverse reactions. 5 Rate law: Relationship between rate of reaction and concentration of reactants

C12H22O11 + H2O → 2C6H12O6 (hypothetical example)

X1,Y1

X2,Y2

Hypothetically: Reaction is 2nd order with respect to C12H22O11 and zero order with


43

PRIORITIES IN SOLVING KINETICS PROBLEMS

FIND REACTION ORDER • Use method of initial rates - If initial rates are available* *If initial rates are not available they can be calculated, but that is probably not worth the time on an exam.

• Evaluation of half-life data • Identifying integrated rate law straight line graph - From actual graph - From description of graph • Recognizing k units • Plugging numbers into equations to get a constant “k”.

(the “constant-constant” method) - Integrated rate law equation - Differentiated rate law equation - Half-life equation

FIND THE RATE CONSTANT

• After identifying reaction order, plug numbers into - Differentiated rate law equation or - Integrated rate law equation or - Half-life equation • Slope or negative slope of integrated rate law graph

Note: Diff. rate law equation contains conc. and rate terms.

Integ. rate law and t1/2 equations contain conc. and time terms. EVERYTHING ELSE

• Use table of formulas for everything else.


44

COLLISION THEORY, INCLUDING ARRHENIUS EQUATION

Collision theory is a model that explains reaction rates. o Molecules must collide to react. o Frequency of molecular collisions is important o Molecules must have the correct steric orientation.

o Molecules must have a minimum energy (≥activation energy) o Rate of reactions increase with temperature

ET: Possible way to discuss Activation Energy is with molecular models, e.g., I- + CH3Br → CH3I + Br-

Arrhenius equation: k = Ae(-Ea/RT)

Rate = k[A]m[B]n The “k” in the Arrhenius equation is the same as the “k” in the rate equ.

ln k = -Ea/RT + ln A


45

REQUIREMENTS FOR HYPOTHETICAL MECHANISMS

1. The sum of the calculated elementary stoichiometry steps must be equal to the experimentally determined stoichiometric equation.

2. The calculated rate law must be consistent with the experimentally determined rate law.

3. No elementary step can have a molecularity greater than three (even three is rare).


46

CHEM 162-2010 FINAL EXAM CHAPTER 14 - KINETICS RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIVES CALCULATIONS REACTION RATE

41. The instantaneous rate of the second-order decomposition of A is 2.70x10-3M/s when [A] = 0.60M. What is the instantaneous rate of this reaction when [A] = 0.20M? A. 6.0x10-4M/s B. 4.5x10-4M/s C. 3.0x10-4M/s D. 8.0x10-4M/s E. 1.2x10-3M/s ET: First thing to do is to find k. Go from differentiated rate law to k to differentiated rate law.

If you are provided with the rate and concentration then use the “Differentiated Rate Law”. The Differentiated Rate Law for a second order reaction is: Rate = k[C]2. Rate = k[C]2

2.7 x 10-3 = k[0.60]2 k = 0.0075 The rate constant doesn’t change no matter what the rate is or no matter what the concentration is. Hence, use the same equation to determine the rate: Rate = k[C]2 Rate = 0.0075 x [0.20]2

Rate = 3.0 x 10-4


47

CHEM-2010 FINAL EXAM + ANSWERS CHAPTER 14 - KINETICS RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIVES CONCEPTS INTEGRATED RATE LAW

ET: First order plot (ln[A]t on y-axis, vs. t on x-axis) can be used to find the rate constant. If a plot of ln[A] vs time gives a straight line, then the reaction is first order. ln[A]t = -kt + ln[A]o Y = mx + b k = -the slope X1 = 0 Y1 = -1.47 X2 = 1195 Y2 = -3.60 slope = (Y2-Y1)/(X2-X1) = (-3.60 –(-1.47))/(1195-0) = -1.78 x 10-3s-1 -slope = k = 1.78 x 10-3s-1 B = Answer


48

Very complex problem Chem 162-2010 final exam Chapter 14A - Kinetics Rates,rate constants, reaction orders, half-lives calculations 10. Consider the reaction N2(g) + 3H2(g) → 2NH3(g). This reaction proceeds by the rate law -(∆P(N2))/∆t = kP(N2), where k = 1.25x10-3s-1. Starting with 0.45 atm of N2 and 1.55 atm of H2 and no NH3, what will the total pressure be after 145s have elapsed? [Hint: total pressure is the sum of all the partial pressures at that time.] A. 0.375 atm B. 1.85 atm C. 1.33 atm D. 0.075 atm E. 0.650 atm ET: The reaction is first order. We have the value of k. We have the initial pressure and the time, so use the integrated rate law to find the final pressure. The difficult part of this problem is realizing that the integrated rate law should be dealing with the limiting reactant, which is N2 in this case. N2 + 3H2 → 2NH3 Initial 0.45 atm 1.55 atm 0 atm Change -X -3X +2X 145 second pressures 0.45-X 1.55-3X 2X ln[C]t = -kt + ln[C]o ln[C]t = (-1.25x10-3 x 145s) + ln[0.45atm] [C]t = 0.375M 0.45 – X = 0.375 X = 0.075 atm N2 + 3H2 → 2NH3 Initial 0.45 atm 1.55 atm 0 atm Change 0.075 atm 3 x 0.075 = 0.225 atm 2 x 0.075 = 0.150 atm 145 second pressures 0.375 atm N2 + 3H2 → 2NH3 Initial 0.45 atm 1.55 atm 0 atm Change 0.075 atm 0.225 atm 0.150 atm 145 second pressures 0.375 atm 1.325 atm 0.150 atm Total pressure after 145 seconds = 0.375 + 1.325 + 0.150 = 1.85 atm


49

CHEM 162-2010 FINAL EXAM Chapter 14A - Kinetics Rates,rate constants, reaction orders, half-lives calculations

43. Consider the reaction 2A + B → 3C + D. The rate of reaction was measured as a function of concentration. From the data below, calculate the rate law. Exp. [A] M [B] M Initial rate M/s 1 0.300 0.100 1.23 x 10-4 2 0.400 0.100 2.91 x 10-4 3 0.300 0.300 1.11 x 10-3 A. 4.09x10-3[A][B] B. 1.37x10-2[A]2[B]

C. 0.455[A]3[B]2 D. 1.37[A]2[B]2 E. 0.0409[A][B]2 Use initial rate method to find rate law. Use simultaneous equations to cancel unknowns. Rewrite table to line up initial rates. Experiment [A] (M) [B] (M) Initial rate (M/s)

1 0.300 0.100 1.23 x 10-4 2 0.400 0.100 2.91 x 10-4 3 0.300 0.300 11.1 x 10-4

Rate = k[A]m[B]n Use the initial rate method. We can proceed by getting rid of “A” in order to calculate “n”. Divide experiment 3 by experiment 1: 11.1 x 10-4 = k[0.300]m[0.300]n

1.23 x 10-4 = k[0.300]m[0.100]n 9.02 = 3n n = 2.0 We can proceed by getting rid of “B” in order to calculate “m”. Divide experiment 2 by experiment 1: 2.91 x 10-4 = k[0.400]m[0.100]n

1.23 x 10-4 = k[0.300]m[0.100]n 2.37 = 1.33m m = 3.0 Rate = k[A]3[B]2 This must be “C”, but we can go further to calculate “k” Plug any set of numbers into the rate equation to find k. Exp. 1: 1.23x10-4 = k[0.300]3[0.100]2 k = 0.456 Rate = 0.456[A]3[B]2


50

Chem 162-2010 Final Exam + Answers Chapter 14 – Kinetics Reaction mechanisms, activation energy & catalysts calculations Mechanisms 30. Given the following mechanism, what is the expected rate law? 2A ←→ B fast equilibrium A + B → C slow C → D fast A. rate = k[A][B] B. rate = k[A]3 C. rate = k[B]1/2 D. rate = k[A]3[C] E. rate = k[A]2[B]1 Rate always dependent only on the slow step; this is a typical rate law for a component in the slow step being a reactive intermediate.

Rate = k[A][B] But B is an unstable intermediate. Therefore, replace it using the fast equilibrium equation. Rate of forward reaction = kf[A]2 Rate of reverse reaction = kr[B] Rate of forward reaction = Rate of reverse reaction kf[A]2 = kr[B] [B] = ((kf/kr)[A]2) Rate = k[A][B] Rate = k[A]((kf/kr)[A]2) Rate = k[((kf/kr)[A])][A]2 Rate = k’[A]3


51

CHEM 162-2010 EXAM I CHAPTER 14 - KINETICS REACTION MECH., ACTIVATION ENERGY & CATALYSTS CALCULATIONS

13. The rate constant for the reaction CO(CH2COOH)2(aq) → CO(CH3)2(aq) + 2CO2(g) is 4.75 x 10-4s-1 at 20.0oC and 1.63 x 10-3s-1 at 30.0oC. Find Ea for this reaction. A. 1.32 kJ B. 615 kJ C. 91.0 kJ D. 898 kJ E. 10.9 kJ Using the Arrhenius equation: k1 = 4.75 x 10-4s-1 T1 = 20oC = 293K k2 = 1.63 x 10-3s-1 T2 = 30oC = 303K ln (k2/k1) = -(Ea/R)[(1/T2) - (1/T1)] ln ((1.63x10-3)/(4.75 x 10-4)) = -(Ea/8.314)[(1/303) - (1/293)] Ea = 9.10x104 J = 91.0 kJ/mol


52

CHEM 162-2010 FINAL EXAM


53

CHEM 162-2010 FINAL EXAM CHAPTER 14 - KINETICS REACTION MECH., ACTIVATION ENERGY & CATALYSTS CONCEPTS REACTION MECHANISMS

The addition of a catalyst to a chemical reaction 34. The addition of a catalyst to a chemical reaction. A. Increases the concentration of products at equilibrium. B. Increases the fraction of reactant molecules with a given kinetic energy C. Provides an alternate mechanism with a lower activation energy. D. Lowers the enthalpy change of the overall reaction. E. Lowers the activation energy of the forward reaction without affecting the

activation energy of the reverse reaction. ET: Use catalysis diagram to answer most of these questions. Point out that the catalyst also changes the frequency factpr in the Arrhenius equation. The frequency factor, “A” should really be called the frequecy-steric factor. How the molecules interact changes when using a catalyst. The rate of the reaction can speed up due to a change in the molecules orientations. A. False. A catalyst has no effect on the concentration of reactants or

products, i.e., it doesn’t change the equilibrium constant. It just serves to speed up the reaction.

B. False. Only an increase in temperature increases the fraction of reactant molecules with a given kinetic energy.

C. True. A catalyst provides an alternate mechanism with a lower activation energy. It is as if a person walked through a tunnel in the mountain (low activation energy) as opposed to climbing over the mountain (high activation energy) to get to the other side.

D. False. A catalyst does not affect the enthalpy of the reactants or the products. If it did, then the equilibrium constant would change, but the equilibrium constant doesn’t change due to using a catalyst.

E. False. It lowers the activation energy of the forward and reverse reactions.


54

CHAPTER 20 - SELECTED EQUATIONS FOR NUCLEAR CHEMISTRY Radioactivity, or radioactive decay, is the spontaneous change of the nuclei of certain atoms, accompanied by the emission of subatomic particles and/or high-frequency electromeagnetic radiation.

CAPTURED OR EMITTED PARTICLES/RAYS Alpha particles 42α = 42He Sum of protons and neutrons = mass number (neutrons and protons are nucleons) Beta particles*** β-= 0-1β = 0

-1e Number of protons (or charge) = atomic number

Gamma ray** 00γ

Positron particle β+= 01e Neutron 1

0n Proton 1

1H **Nuclei have energy levels just as electrons do. When a nucleus in an excited state drops to a lower energy level, the energy is released in the form of a gamma ray. Atoms that emit gamma rays are high energy and are identified with an asterisk. ***An electron can be emitted or captured. Capture: An inner core electron can “fall” into the nucleus to combine with a proton and form a neutron. 0-1e + 11H → 10n. An outer shell electron then drops to fill the vacancy, releasing an X-ray. 125

53I + 0-1e → 12552Te + 00γ

RADIOACTIVE DECAY RATE Reaction Differential Integrated Order Rate Law rate law Half-life* 1 Rate = λN ln(Nt/No) = -λt t1/2 = 0.693/λ** ln(Nt) = -λt + ln(No)

Rate = rate of radioactive decay = activity (“A”) = disappearance of atoms/sample per second = disintegrations of atoms/sample per second = disintegrations per second = decays per second λ = the decay constant N = concentration in molarity, number of atoms/sample, molecules/sample, moles/sample, g/sample, %, % of a sample, emission counts/sample, disintegrations/sample, rate of decay/sample

Note similarity to equations from KINETICS chapter: KINETICS

Reaction Differential Integrated Order Rate Law rate law Half-life* 1 Rate = k[A] ln(At) = -kt + ln(Ao) t1/2 = 0.693/k * For a first order reaction, each successive half-life is equal to the preceding one. **Unlike chemical reactions, k is not dependent on T, or anything else. MASS DEFECT Calc. nuclear mass (= nucleon mass) = Actual nuclear mass + mass defect (Calc’d mass of neutrons + protons) (actual mass of nucleus) (binding energy) E = mc2 J = kg x (3.00 x 108 ms-1)2

or 931.5 MeV/u

Units in 1o reactions cancel, so are irrelevant.


55

Effects of Radiation Somatic Damage: damage to organism itself: sickness or death Genetic Damage: damage to genetic machinery: affects offspring Biological effects depend on: (1) Energy of Radiation Measured in rad = 10-2J/kg tissue (2) Penetrating effect of Radiation Gamma: Highly penetrating Beta: penetrates about 1 cm Alpha: stopped by skin (3) Ionizing ability of radiation. This describes the ability of the radiation to remove electrons from biological molecules, forming ions. Generally detrimental to health. Gamma: penetrates well, but cause only occasional ionization Alpha particles: not very penetrating, but very effective at causing ionization if inside the body. Beta particles: penetrating; not as likely to ionize Gamma rays: Very penetrating; not very ionizing


56

CHEM 162-2010 FINAL EXAM CHAPTER 20 - THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS

20. What is the product of the following nuclear reaction? 236

92U → 4 10n + 13653I + ?

A. 9638Sr

B. 9039Y

C. 9640Zr

D. 9841Nb

E. 9639Y

ET: Simple math

236

92U → 4 10n + 13653I + 96

39? 236

92U → 4 10n + 13653I + 96

39Y


57

BELT OF STABILITY - The narrow band of stable isotopes (dark blue and light blue circles) is sometimes called the peninsula of stability in a sea of instability. Any nuclide in the sea of instability (pink circles) will decay in such a way that the nucleus can come ashore onto the peninsula. Only 279 of the ~2000 known nuclides are stable. - In three “Belt of Stability” slides is a discussion of the decay of a uranium nuclide and two cerium nuclides. Relative to a stable reference atom, the uranium nuclide has too many protons, which makes it unstable. Relative to a stable cerium nuclide, the cerium nuclides either have too high or too low a ratio of neutrons to protons, which make them unstable. Too high a ratio is unstable because neutrons, by themselves, are unstable, having only a 12.6 minute half-life. Too low a ratio is unstable because of the proton-proton interaction. The perfect ratio is when the neutrons form a stable complex with the protons, without any extra protons or any extra neutrons. The modes of decay are presented. These modes are typical of corresponding nuclides that contain either too many protons, or a non-optimum ratio of neutrons to protons.

Too many protons and neutrons in this region; Protons > 83 therefore, decay by alpha emission Lose protons and neutrons concurrently

Neutron to proton ratio too high in this region; therefore, becomes stable through decay by beta emission

Neutron to proton ratio too low in this region; therefore, decay by positron emission and/or electron capture

Instability due to too many protons. Mode of decay to decrease number of protons: alpha decay Reference point: 209 (126 neutrons)

83 protonsBi. e.g., 238

92U → 42α + 23490Th

Result: Going in correct direction (90 is approaching 83 protons).

238

92U

Unstable nuclide due to high neutron to proton ratio relative to stable nuclide. Reference 140 (82 neutrons)

58 protonsCe is approx. the average of the 4 stable cerium nuclides. Rewrite: 82n

58pCe; neutron:proton ratio = 1.41 Example of unstable isotope: 148 (90 neutrons

58 protonsCe Rewrite: 90n

58pCe; neutron:proton ratio = 1.55 Mode of decay to approach reference (i.e., to decrease neutron to proton ratio) is electron emission. 148

58Ce → 0-1e + 14859La

Rewrite: 90n58pCe → 0-1e + 89

59Pr; neutron:proton ratio = 1.51

14858Ce (unstable high n:p ratio)

14058Ce

(stable)Unstable nuclide due to low neutron to proton ratio relative to stable nuclide. Reference 140 (82 neutrons)

58 protonsCe is approx. the average of the 4 stable cerium nuclides. Rewrite: 82n

58pCe; neutron:proton ratio = 1.41 Example of unstable isotope: 130 (72 neutrons)

58 protonsCe Rewrite: 72n

58pCe; neutron:proton ratio = 1.24 Mode of decay to approach reference (i.e., to increase neutron to proton ratio) is beta emission or electron capture. Beta emission: 130

58Ce → 0+1e + 13057La

Rewrite: 72n58pCe → 0+1e + 73

57La; neutron:proton ratio = 1.28 Result: Going in correct direction Electron capture: 130

58Ce + 0-1e → 13057La

Rewrite: 72n58pCe + 0-1e → 73

57La; neutron:proton ratio = 1.28 Result: Going in correct direction

13058Ce (unstable low

n:p ratio)

20983Bi


58

CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 20 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS

24. Which of the following statements regarding nuclear reactions is false. A. Nuclear reactions involve substantially greater energy changes than

ordinary chemical reactions. B. Nuclear decay follows first order kinetics. C. Nuclear reactions do not require conservation of mass. D. Nuclear reactions are typically the same for different isotopes of the same

element. E. The rate of nuclear decay is not temperature dependent. ET: Show “Belt of stability” diagram to discuss “D”. Also, mention that nothing affects the rate of nuclear decay. A. True. I don’t know if all nuclear reactions involve substantially greater

energy changes than ordinary chemical reactions, but the nuclear reactions involving splitting or fusion of nuclei provide much greater energy than ordinary chemical reactions.

B. True. Nuclear decay always follows first order kinetics. C. True. Virtually all nuclear reactions involve going from a higher energy

state into a lower energy state, losing mass and releasing energy. D. False. Typically one isotope is stable while another one is not. The

unstable ones radiate particles or energy, while changing into another nuclide.

E. True. I don’t know why, but temperature cannot change the rate of nuclear decay. One would think that, based on collision theory, the number of atoms that can reach the energy of activation would increase with increasing temperature, but apparently this isn’t so.


59

0

1

2

3

4

5

6

7

8

9

10

0 50 100 150 200 250 300

Bin

ding

ene

rgy

per n

ucle

on (M

eV)

Mass Number

FISSION & FUSION BINDING ENERGY 56

26Fe (atom most stable per nucleon)

Fission e.g., 235

92U + 10n → 14156Ba + 92

36Kr + 3 10n (capable of chain reaction due to extra neutrons formed) Critical mass: Mass of reactant needed to achieve the critical state to have a chain reaction. In an atomic bomb resulting from the explosive fission of U-235, a supercritical mass is achieved suddenly when two sub-critical pieces are brought together.

Nuclear reactor: Binding energy creates electricity Control rods: Absorbs excess neutrons

Fusion e.g., 21H + 31H → 42He + 10n (requires a temperature > 40,000,000K for the positively charged nuclei to move fast enough to overcome the plus plus repulsion. A fusion bomb (“hydrogen bomb”) uses a fission explosion (uranium or plutonium) to provide the high temperature needed to initiate fusion.) H-H fusion occurs in the sun.

ET: Show stability of 56-Fe vs reactants with typical simple elementary step curve .


60

Chem 162-2010 Final Exam + Answers Chapter 20 - The Nucleus Radioactive decay particles and rays

6. The stable isotopes of lead are Pb-204, Pb-206, and Pb-208. The unstable isotope Pb-214 is most likely to undergo A. electron capture B. neutron emission C. α particle emission D. β particle emission E. positron emission ET: Unstable isotopes decay in a way that will make them more stable. Pb-214 tries to become more like Pb-204, -206 and -208, with regard to the neutron to proton ratio. ET: Discuss fission, fusion and critical mass in above Fe-56 slide. Except for hydrogen (1

1H), which has 0 neutrons and 1 proton, stable isotopes have a ratio of approximately 1 neutron:1proton, to 1.5 neutrons:1 proton. Pb-204 has a ratio of (122/82=) 1.49 neutrons to 1 proton. Pb-206 has a ratio of (124/82=) 1.51 neutrons to 1 proton. Pb-208 has a ratio of (126/82=) 1.54 neutrons to 1 proton Pb-214 has a ratio of (132/82=) 1.61 neutrons to 1 proton Clearly, Pb-214 is the least stable, with a high neutron:proton ratio. A. 214

82 + 0-1e → 21481Tl neutron:proton = 133/81 = 1.64. This ratio is going

in the wrong direction. B. 214

82Pb → 10n + 21382Pb neutron:proton = 131/82 = 1.60. This ratio is going

in the correct direction. C. 214

82Pb → 42He + 21080Tl neutron:proton = 130/80 = 1.63. This ratio is

going in the wrong direction. D. 214

82Pb → 0-1e + 21483Hg neutron:proton = 131/83 = 1.58. This ratio is going

in the correct direction. E. 214

82Pb → 0+1e + 21481Tl neutron:proton = 133/81 = 1.64. This ratio is going

in the wrong direction. Neutron emission and beta particle emission are both going in the correct direction, but beta particle emission, not neutron emission, is what actually happens.


61

CHEM-2010 FINAL EXAM + ANSWERS CHAPTER 20 - NUCLEAR CHEMISTRY NUCLEAR BINDING ENERGY 2. What is the binding energy for 237

93Np? 1p = 1.0073u, 1n = 1.0087u, 1e- = 5.49x10-4u, the atomic mass of 23793Np =

237.0481u; 1.66 x 10-24g = 1u (The problem should have said binding enery of 237

93Np per atom.) A. 2.89 x 10-10J B. C. D. E. ET: Calculated nuclear mass = Actual nuclear mass + Mass defect. Mass defect can be converted into binding energy (E=mc2). ET: “u” is atomic mass units, amu’s, or “Daltons” or grams/mol. Note that sometimes mass defect or binding energy is asked per mole, per atom, or per nucleon. Calculated Np nuclear mass = Actual nuclear mass + Mass defect 93 11H + 144 10n = 237

93Np + mass defect Calculated nuclear mass = (93 x 1.0073) + (144 x 1.0087) = 238.932u Actual nuclear** mass = actual atomic mass – mass of electrons = 237.0481u – (93 x (5.49x10-4)) = 236.997u **Nuclear mass is the mass of the atom without its electrons.

Calculated nuclear mass = Actual nuclear mass + Mass defect 238.932u = 236.997u + mass defect mass defect = 1.935u 1.66 x 10-24g = 1u 1.935u x (1.66 x 10-24g/u) x (1 kg/1000g) = 3.212 x 10-27 kg E = mc2 E = 3.212 x 10-27 kg x (3 x 108m/s)2 = 2.891 x 10-10 kgm2/s2 x (1J/((kgm2/s2)) = 2.89 x 10-10J


62

CHEM 162-2010 FINAL EXAM CHAPTER 20 - THE NUCLEUS RADIOACTIVE DECAY KINETICS

40. The specific rate constant, λ, for the radioactive decay of 11B is 0.0490s-1. What mass of a 0.500g sample of 11B will remain after 28.0s? A. 0.250g B. 0.127g C. 0.305g D. 0.404g E. 0.0625g ET: The order and rate constant are already determined. Given time, k and initial concentration, find final concentration with the integrated rate law.

[No] = 0.500g [Nt] = ? t = 28s k = 0.0490s-1 ln(Nt) = -kt + ln(No) ln(Xg/sample) = (-0.0490 x 28) + ln(0.500g/sample) X = 0.1268g


63

CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 20 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS

47. A sample of wood from an Egyptian tomb has a 14C activity of 7.07 dpm (disintegrations per minute) per gram of C, while a sample of freshly cut wood has an activity of 15.3 dpm per gram of C. Calculate the age of the wood. The half life of 14C is 5730 years. A. 5100 years B. 3600 years C. 6400 years D. 7200 years E. 9200 years ET: Carbon-14 dating to determine length of time the wood is dead. Solve problem qualitatively, then quantitatively first finding k, then using the integrated rate law with initial and final concentrations and k to find t.

Qualitative approach: The half-life of the 14C is 5730 years. Since going from a concentration of 15.3 dpm to 7.7 dpm is one half-life, and the concentration is 7.07 dpm, then we have gone to slightly more than one half-life. Hence, the age of the wood should be slightly greater than 5730 years. 6400 years is a reasonable guesstimate. Quantitative approach: t1/2 = 0.693/λ 5730 = 0.693/λ λ = 0.00012094 Use the Integrated Rate Law, ln(At/Ao) = -λt ln(Nt) = -λt + ln(No) ln(7.07) = -0.00012094 x t + ln(15.3) t = 6383 years


64

Effects of Radiation Somatic Damage: damage to organism itself: sickness or death Genetic Damage: damage to genetic machinery: affects offspring Biological effects depend on: (1) Energy of Radiation Measured in rad = 10-2J/kg tissue (2) Penetrating effect of Radiation Gamma: Highly penetrating Beta: penetrates about 1 cm Alpha: stopped by skin (3) Ionizing ability of radiation. This describes the ability of the radiation to remove electrons from biological molecules, forming ions. Generally detrimental to health. Gamma: penetrates well, but cause only occasional ionization Alpha particles: not very penetrating, but very effective at causing ionization if inside the body. Beta particles: penetrating; not as likely to ionize Gamma rays: Very penetrating; not very ionizing


65

CHEM 162-2010 FINAL EXAM CHAPTER 20 - THE NUCLEUS MISCELLANEOUS

14. The most dangerous feature of radioactivity is A. the amount of gamma radiation emitted. B. the solubility of most isotopes. C. the long half-lives of radioisotopes. D. the ionizing of important biological compounds. E. the penetrating power making it hard to contain. ET: Discuss reference table. A. Rarely, if ever, does a radioactive nuclei emit gamma radiation. B. Although the isotope might be soluble, it is not necessarily dangerous. C. Long half-lives contribute to danger, but if the emitted particle or ray is not

dangerous, then the radioactivity is not dangerous. D. Radiation can ionize molecules in living matter. Even slight exposure can

damage DNA and cause birth defects, leukemia, bone cancer, and other forms of cancer.

E. Highly penetrating radioactive isotopes can be contained in containers such as Pb.


66

CHAPTER 18 - SPONTANEITY, ENTROPY AND FREE ENERGY First law of thermodynamics: Energy can neither be created nor destroyed. Second law of thermodynamics: The entropy of the universe always increases in a spontaneous process. Third law of thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. Entropy describes the number of arrangements (positions and/or energy levels) available to a system. Entropy is (officially called chaos, disorder, randomness) positional freedom, or positional availability, or disorder, the number of microstates or number of energy levels or being unconfined positionally or escape from positional confinement or freedom from positional confinement ΔGo = ΣnpΔGo

f(products) - ΣnrΔGof(reactants)

ΔHo = ΣnpΔHof(products) - ΣnrΔHo

f(reactants) ΔSo

system = ΣnpSoproducts - ΣnrSo

reactants

ΔSsystem = ΣnpSproducts - ΣnrSreactants (used for ΔSsystem anytime) ΔSsystem = ΔHsystem/T (used for ΔSsystem only when at equilibrium)

Trouton’s rule: At normal B.P., ΔSvaprization ≈ 87J mol-1K-1 ΔSsurr = ΔHsurr/T = -ΔHsystem/T (used for ΔSsurr anytime) ΔSuniv = ΔSsys + ΔSsurr = ΔSsys – (ΔHsys/T) Four ways to find ΔGo: (1) ΔGo = ΔHo - TΔSo (or ΔG = ΔH - TΔS) (2) ΔGo = ΣnpΔGo

f(products) - ΣnrΔGof(reactants)

(3) ΔGTo = ΔG1

o + ΔG2o (i.e., adding equations [Hess’s law])

(4) ΔGo = -RT ln(K) ΔG = ΔGo + RT ln(Q) (Q can be any pressure or concentration)

wmax = ΔG lnKeq = ((-∆Ho/R) x 1/T) + ∆So/R Plot straight line: Y = mX + b ln(K2/K1) = -(∆Ho/R)((1/T2) – (1/T1)): van’t Hoff equation ln(P2/P1) = -(∆Hvapn

o/R)((1/T2) – (1/T1)): Clausius-Clapeyron equation


67

Chem 162-2010 Final Exam + Answers Chapter 18 – Thermodynamics Free energy concepts

8. What is the physical significance of ∆G? A. The amount of heat transferred at constant temperature. B. The maximum work that a system can do. C. The temperature dependence of ∆H. D. The degree of disorder of the system. E. The temperature dependence of ∆S. ET: ∆G and work are both energy. When ∆G does work, the energy isn’t all converted into work. This is a conceptually difficult question. A. False. Heat transferred is ΔH, not ΔG. B. True. ΔG = workmax C. False. Not related. ∆G is highly temperature dependent; ∆H is only

slightly temperature dependent. D. False. The degree of disorder is ΔS, not ΔG. E. False. Not related. ∆G is highly temperature dependent; ∆S is only

slightly temperature dependent.


68

Chem 162-2010 Final Exam Chapter 18-Thermodynamics Free energy and equilibria concepts

12. Under standard conditions, calcium reacts readily with chlorine gas. What conclusions may be drawn from this fact? A. Keq > 1 and ∆Go < 0 B. Keq < 1 and ∆Go > 0 C. Keq > 1 and ∆Go = 0 D. Keq < 1 and ∆Go < 0 E. Keq > 1 and ∆Go > 0 ET: If the reaction goes to the right from standard conditions, then ∆Go < 0, and the [Ca2+] and [Cl-] concentration increases. Ca + Cl2 ←→ Ca2+ + 2Cl- 1M 1atm 1M 1M If the reaction goes to the right spontaneously under standard state conditions, then the concentration of products will increase from the initial 1M, and the reactants will decrease, and therefore, Keq >1. If the reaction goes to the right spontaneously under standard state conditions then ∆Go < 0.


69

CHEM 162-2010 HOURLY EXAM III + ANSWERS CHAPTER 18: THERMODYNAMICS FREE ENERGY AND EQUILIBRIA CALCULATIONS

22. Calculate ∆G for the following reaction at 400K when the pressures of NH3(g) and CO2(g) are 1.00 atm and 2.00 atm, respectively

NH2CONH2(s) + H2O(l) ←→ 2NH3(g) + CO2(g) ∆Go = 27.6 kJ A. 31.3 kJ B. 27.6 kJ C. 34.9 kJ D. 25.3 kJ E. 29.9 kJ ET: Increasing concentration on the right makes the reaction less spontaneous.

NH2CONH2(s) + H2O(l) ←→ 2NH3(g) + CO2(g) ∆G = ? 1.00atm 2.00atm ΔG = ΔGo + RTln(Q) ΔG = 27600 + (8.314 x 400 x (ln(2/(1)2))) ΔG = 2.99 x 104 J = 29.9 kJ


70

CHEM 162-2010 HOURLY EXAM III + ANSWERS CHAPTER 18: THERMODYNAMICS FREE ENERGY AND EQUILIBRIA CALCULATIONS

27. Given the data below, what is the equilibrium constant at 500K for the following reaction? PCl3(g) + Cl2(g) ←→ PCl5(g) Compound ∆Hf

o(kJ/mol) So(Jmol-1K-1) PCl3 -287.0 311.7 PCl5 -374.9 364.5 Cl2 0 223.0 A. 2.8 B. 2800 C. 1.96 D. 0.51 E. 1.00 ET: Use ΔHo - TΔSo to find ΔGo; then use ΔGo = -RT ln(K) to find K.

PCl3(g) + Cl2(g) ←→ PCl5(g) ∆Ho -287000 0 -374900 So 311.7 223.0 364.5 ∆Go = ∆Ho - T∆So ∆Ho = -374900 – (-287000 + 0) = -8.79x104

∆So = 364.5 – (311.7 + 223.0) = -170.2 ∆Go = (-8.79x104) – (500 x -170.2) = -2.8 x 103

∆Go = -RTln(K) (-2.8 x 103) = ((-8.314) x 500 x (ln(K))) K = 1.96


71

CHEM 162-2010 FINAL EXAM + ANSWERS CHAPTER 18 THERMODYNAMICS FREE ENERGY AND EQUILIBRIA CALCULATIONS 33.

33. The Keq for the reaction 2SO2(g) + O2(g) ←→ 2SO3(g) is 910 at 900K and 0.38 at 1170K. What is ∆Ho for this reaction? A. 252 kJ B. -252 kJ C. -0.252 kJ D. -7.57x105 kJ E. 75.7 kJ ET: The van’t Hoff equation is the quantitative counterpart to LeChatelier’s principle for the effect on K after raising the temperatue of an endothermic/exothermic reaction. K1 = 910 T1 = 900K K2 = 0.38 T2 = 1170K van’t Hoff equation: ln(K2/K1) = (∆Ho/R)(1/T1 – 1/T2) ln(0.38/910) = (∆Ho/8.314)(1/900 – 1/1170) ∆Ho = -2.52 x 105 J/mol = -2.52 x 102 kJ/mol


72

CHEM 162-2010 FINAL EXAM Chapter 18 - Thermodynamics Entropy concepts

39. All of the following processes would be expected to have positive ∆So, except A. C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) B. N2H4(l) → N2(g) + 2H2(g) C. CaCO3(s) → CaO(s) + CO2(g) D. (NH4)2SO4(s) → 2NH4

+(aq) + SO42-(aq)

E. 2NH3(g) + CO2(g) → NH2CONH2(aq) + H2O(l) ET: Entropy is positional freedom. “E” isn’t positional freedom; it is positional confinement.

A. Going from a solid to a liquid and gas is positive ∆So. B. Going from a liquid to gases is positive ∆So. C. Going from a solid to a solid and gas is positive ∆So. D. Going from a solid to a solution is positive ∆So. E. Going from a gas to a solution and liquid is negative ∆So.


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CHAPTER 15 - CHEMICAL EQUILIBRIUM

A + B ←→ C + D (gases or solutes, only) Law of mass action (or molar interaction): Kc = ([C][D])/([A][B]) 2A + 3B ←→ 4C + 1D A + A + B + B + B ←→ C + C + C + C + D Kc = ([C] • [C] • [C] • [C] • [D])/( [A] • [A] • [B] • [B] • [B]) Kc = ([C]4[D])/([A]2[B]3) eA + fB ←→ gC + hD Kc = ([C]g[D]h)/([A]e[B]f) e.g., 2NO + O2 2NO2 Kc = ([NO2]2)/([NO]2[O2]) K A B [B]/[A] = K1 C D [D]/[C] = K2 A + C B + D ([B][D])/([A][C]) = K Addition of ([B]/[A]) + ([D]/[C]) = (([B][C] + [A][D]))/([A][C]) Equations K(s) Add Multiply Reverse Invert Double Square Halve Square root Kpinatm = KcinM(RT)∆ngas


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UNDERSTANDING EQUILIBRIA DYNAMIC* EQUILIBRIUM

(*not static equilibrium) ET: At equilibrium, the forward and reverse reactions proceed at equal rates and the concentrations of reactants and products remain constant.

Mountain INITIALLY Slum sad k1 = 0.01/s k-1 = 0.10/s town with Palacial happy 0 sad town with people/mi2 1100 happy people/mi2 Forward rate pre-equilibrium = k1 x reactant conc. = 0.01s-1 x 1100 = 11 people/mi2s-1 Reverse rate pre-equilibrium = k-1 x reactant conc. = 0.10s-1 x 0 = 0 people/mi2s-1 Mountain AT EQUILIBRIUM Slum sad k1 = 0.01/s k-1 = 0.10/s town with Palacial happy 100 sad town with people/mi2 1000 happy people/mi2 Equilibrium reached

Macroscopically: You continuously see 1000 happy people on the left and 100 sad people on the right. Microscopically: You also see a constant equal exchange of sad people and happy people going across the bridge, a “dynamic equilibrium”. That is, the rate forward equals the rate reverse.

Rate forward = k1 x [reactant] = 0.01s-1 x 1000 = 10 people/mi2s-1 Rate reverse = k-1 x [product] = 0.10s-1 x 100 = 10 people/mi2s-1

Also note:

K = [Products]/[Reactants] = 100/1000 = 0.1 K = k1/k-1 = 0.01/0.10 = 0.1


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Le Chatelier’s Principle: H&P: “When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change.” Le Chatelier’s Concept - Relieve stress ET: If a stress (change) is placed on a system at equilibrium, the position of equilibrium will shift in a direction to counteract reduce the applied stress.


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CHEM162-2010 HOURLY EXAM II + ANSWERS CHAPTER 15 - NON ACID-BASE CHEMICAL EQUILIBRIUM NON-ACID-BASE EQUILIBRIUM CALCULATIONS 35. Solid ammonium chloride decomposes when heated according to the following reaction: NH4Cl(s) ←→ NH3(g) + HCl(g) A sample of ammonium chloride is heated in an evacuated container. At equilibrium, the total pressure in the container is 1.5x 10-4 atm. Calculate Kp for this reaction. A. 7.8x10-5 B. 2.3x10-8 C. 3.5x10-6 D. 1.5x10-4 E. 5.6x10-9 NH4Cl(s) ←→ NH3(g) + HCl(g) NH4Cl(s) ←→ NH3(g) HCl(g)

Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [NH3][HCl] = Kp [X][X] = Kp But what is the value of X? X + X = 0.00015 2X = 0.00015 X = 0.000075 Kp = X2 = [0.000075]2 = 5.6 x 10-9


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CHEM-2010 FINAL EXAM + ANSWERS CHAPTER 15 - NON ACID AND BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS 21. For the reaction

21. 2BrCl(g) ←→ Br2(g) + Cl2(g) Kc = 32 at 500K What would happen if a 1.0L bulb were filled with 0.10mol BrCl(g), 0.50mol Br2(g), and 0.50mol Cl2(g)? A. Some BrCl will react spontaneously to form additional Br2 and Cl2. B. Some of the Br2 and Cl2 will react spontaneously to form additional BrCl. C. Nothing will happen because the initial concentrations correspond to equilibrium concentrations. D. All of the BrCl will be consumed. E. All of the Br2 will be consumed. 2 BrCl(g) ←→ Br2(g) + Cl2(g) 0.10 0.50 0.50 Kc = ([Br2][Cl2])/([BrCl]2) ([0.50][0.50])/([0.10]2) = 25 = Q When Q is less than Kc, then the numerator is too small, so the reaction will go to the right to make the numerator larger and to reach equilibrium. A. True B. False. The reaction will shift to the right, not to the left. C. False. The initial concentrations do not correspond to equilibrium concentrations. D. False. The reaction will shift to the right, but no component is ever totally consumed. E. False. The concentration of Br2 will increase, not decrease.


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CHEM 162-2010 FINAL EXAM CHAPTER 15 - NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS

46. Given: C(s) + 2H2(g) ←→ CH4(g) Kp = 0.262 at 1270K What is Kc for the above reaction? A. 20.5 B. 19.6 C. 12.9

D. 27.3 E. 16.3 Kp = Kc(RT)∆ngas 0.262 = Kc(0.08214 x 1270)-1 Kc = 27.3


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CHAPTER 22CCoooorrddiinnaattiioonn CCoommppoouunndd FFuunnddaammeennttaallss

Ligand (L: ligo, to bind): A substance (neutral or anionic) having a lone electron pair that bonds to a transition metal cation (i.e., Lewis base-Lewis acid interaction). Usually a transition metal

Complex ion: Transition metal cation + ligands Complex ion is sometimes called a complex. Complex ion is usually a cation, but could be neutral or anionic. The metal is usually, but not always, a transition metal;.

Coordination compound (Transition) metal cation + ligands + counterions Complex ion (in brackets) The complex ion may be positively or negatively charged or uncharged; the counterions have the opposite charge; the purpose of the counterions is to make a zero charge overall; if the complex ion isn’t charged then there are no counterions.

Coordination compound dissociation in aqueous solution: [Co(NH3)5Cl]Cl2 → [Co(NH3)5Cl]2+ + 2Cl- Alfred Werner (1893) theorized two types of Cl bonding in a single molecule to explain why AgNO3 formed a precipitate with only two of the three chlorine atoms. This was the beginning of Coordination Chemistry. Ligands are much more tightly attached to the metal cation than are the counterions. The ligands are tightly attached to the metal cation through covalent bonds (actually coordinate covalent bonds [both electrons of the bond coming from one atom]), whereas the counterions move freely around the solution (weaker ion-dipole bonds with the solvent).

- AgNO3 will not react with the covalently bonded Cl; it will only react with the above two chloride counterions.

Coordination number: The number of bonds between the metal ion and the ligands (6 in the above case; therefore, octahedral structure)


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LLiiggaanndd TTyyppeess Unidentate – Bonds to metal ion using only “one tooth” (even if it has

several teeth) – Include: NH3, Cl-, NO2-, CN-, OH-, H2O

: C ≡ N: cyano-N or cyano-O, depending on which atom is bonded to the metal ion . . . . [ : O ― N = O:]- . . . .

nitrito-N or nitrito-O, depending on which atom is bonded to the metal ion ET: Note that NO2

- and CN- have a choice of two binding sites, but they are still both monodentate because they can only bond with one site at a time. – Bidentate - e.g., ethylenediamine (“en”), H2NCH2CH2NH2

oxalato, -O2CCO2-

binds with two teeth onto a metal cation to form a chelate The ligand must be large enough (4 or 5 atoms long) to form a 5 or 6 atom ring with the metal cation. – Polydentate – binds with more than one tooth onto a metal

cation to form a chelate The ligand must be large enough (4 or 5 atoms long) to form a 5 or 6 atom ring with the metal cation.

e.g., 2-(O2C)2NCH2CH2N(CO2)22- EDTA

(ethylenediaminetetraacetato)

Chelating (<Gk. “claw”), forming a ring


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SOME COMMON LIGANDS

Formula Name as Ligand Abbreviation Neutral Molecules NH3 Ammine CH3NH2 Methylamine NO Nitrosyl CO Carbonyl Anions F- Fluoro Cl- Chloro Br- Bromo I- Iodo OH- Hydroxo NO2

- Nitrito-N ONO- Nitrito-O CN- Cyano SCN- Thiocyanato-S NCS- Thiocyanato-N Polydentate H2NCH2CH2NH2 Ethylenediamine en [OOCCOO]2- Oxalato ox [(OOCCH2)2NCH2CH2N(CH2COO)2]4- Ethylenediaminetetraacetato EDTA


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RULES FOR NAMING COORDINATION COMPOUNDS (1) Name cation before anion. (2) Name ligands before metal ion. (3) Name ligands alphabetically. Ignore prefixes. (Note: When writing a formula from a name, place anionic ligands before neutral ones.)

- For neutral ligands use name of molecule Exceptions: H2O, NH3, CO, and NO (see table I below).

(4) Anionic ligands (AL) - Add “o” to root name of anionic ligands (see table I below)

(5) Anionic complex ions (ACI) - Use Latin name of metal if possible, with “ate” (see Table II).

(6) Prefixes to ligands - Use di-, tri-, tetra-, etc. for the number of monodentate ligands, not mono. - Use bis-, tris-, tetrakis-, etc. for the number of polydentate ligands.

(7) Roman numeral of metal - Indicate oxidation state of metal

Table I - Names of Some Common Unidentate Ligands Neutral Molecules H2O Aqua NH3 Ammine CH3NH2 Methylamine CO Carbonyl NO Nitrosyl Anions F- Fluoro Cl- Chloro Br- Bromo I- Iodo OH- Hydroxo CN- Cyano NO2

- Nitrito

Table II - Latin Names Used for Some Metal Ions in Anionic Complex ions Fe Ferrate Cu Cuprate Pb Plumbate Ag Argentate Au Aurate Sn Stannate


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FORMAT OF ISOMER SLIDES • Structural isomers Coordination isomers Linkage isomers • Stereoisomers Geometrical isomerism Double-bonded structures (for Orgo) Square planar structures Tetrahedral structures (not possible) Octahedral structures Optical isomerism Square planar structures (not possible) Tetrahedral structures Octahedral structures - Monodentate ligands


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Chapter 22 25. In the reaction CaO(s) + CO2(g) ←→ CaCO3(s), A. CaO is acting as a Lewis acid because the resulting salt is basic. B. CaO is acting as a Lewis base because it donates an electron pair. C. CO2 is acting as a Lewis acid because it donates an electron pair. D. CO2 is acting as a Lewis base because it accepts an electron pair. E. CO2 is acting as a Lewis acid because CaO is a Bronsted-Lowry base. O O

║ │ Ca2+ O2- + C → Ca2+ O—C ║ ║ O O CO2 is acting as a Lewis acid. CaO is acting as a Lewis base. A. False, because CaO is acting as a Lewis base. B. True. CaO is acting as a Lewis base because the oxide donates an electron pair. C. False. Although CO2 is acting as a Lewis acid, the suggested mechanism is incorrect. Lewis acids don’t donate electron pairs. They receive electron pairs. D. False. CO2 is acting as a Lewis acid. E. False. Although CO2 is acting as a Lewis acid, CaO is not acting as a Bronsted-Lowry base. A Bronsed-Lowry base is a proton acceptor. The CaO is not accepting protons in this reaction.

CHAPTER 22 38. Which of the following pairs are not structural isomers? A. [Co(NO2)(NH3)5]2+ [Co(ONO)(NH3)5]]2+ B. [Cr(SO4)(NH3)5]Cl [CrCl(NH3)5]SO4 C. [Co(NH3)6][CrCl6] [Cr(NH3)6[CoCl6] D. [CoCl2(NH3)4]Br [CoBrCl(NH3)4]Cl

E. [Co(en)(H2O)4]3+ [Co(en)2(H2O)2]3+ A. are structural isomers. They are linkage isomers. B. are structural isomers. They are coordination isomers. C. are structural isomers. They are coordination isomers. D. are structural isomers. They are coordination isomers. E. are not structural isomers. They are not isomers.


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CHAPTER 22 48. What is the correct name for the compound [Co(NO2)2(en)2]NO3? [en = ethylenediamine] A. Cobaltodinitro-N-diethylenediamine nitrate B. diethylenediaminedinitrocobalt(III)nitrate C. bi(ethylenediamine)bisnitrocobaltate(III)nitrate D. diethylenediaminedinitrito-N-cobalt(I) nitrate

E. bis(ethylenediamine)dinitrito-N-cobalt(III)nitrate A. False. The metal is always listed at the end of the cation; also, more than one ethylenediamine is always preceded by bis, tris, tetrakis, etc, not di, tri, tetra, etc. Also, the term for NO2 in complex ions is nitrito, not nitro. B. False. More than one ethylenediamine is always preceded by bis, tris, tetrakis, etc, not di, tri, tetra, etc. Also, the term for NO2 in complex ions is nitrito, not nitro. Also, the nitrito group needs a description of whether it is connected at the N or O. C. False. More than one ethylenediamine is always preceded by bis, tris, tetrakis, etc, not “bi”. Nitro is preceded by “di”, not “bis”. Also, the term for NO2 in complex ions is nitrito, not nitro. Also, the Co group in a cationic complex ion is called “cobalt”, not “cobaltate”; “Cobaltate” is used in an anionic complex ion. D. False. More than one ethylenediamine is always preceded by bis, tris, tetrakis, etc, not di, tri, tetra, etc. E. True CHAPTER 22 49. The correct formula of triamminediaquachlorocobalt(III) iodide is A. [CoCl(NH3)2(H2O)3]I B. [CoCl(NH3)3(H2O)3]I C. [CoCl(NH3)3(H2O)2]I3

D. [CoCl(NH3)3(H2O)2]I2 E. [Co3Cl(NH3)3(H2O)2]I (+3) + (-1) = +2 The molecule needs two iodides to balance the +2 charge in the complex ion.


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CHAPTER 12 FORMULAS

ET note: Not everything goes through moles, e.g., volume to grams, grams to volume, atoms to molecules, molecules to atoms. CHAPTER 5 - STOICHIOMETRY

moles A

↕ moles B

atoms molecules (or atoms if particle is an atom)

MB or VB gB or MWB

VB

MA or VA

VA gA or MWA

MOLES ARE AT THE CENTER (FOR CONVERSIONS, WORK THROUGH MOLES)

MW

(D = g/V)

molA = MALA moles = PAVA/RTA for gases

(molecules x atoms/molecule = atoms)

(molesA = gA/MWA)

(molesB = gB/MWB)

(D = g/V)

(moles x Avog. No. = molecules (or atoms))

%A

%B

Empirical formula

Molecular f l

(MW/EW) x Emp form =

moles = g/MW PV = nRT moles = M x V P1V1/n1T1=P2V2/n2T2 g/MW = M x V 6.022x1023molecule/mole D = g/L

MWA or AWA = gA/moleA molesA = gA/MWA

molA = MALA moles = PAVA/RTA for gases


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FORMULAS gsolute + gsolvent = gsolution mLsolute + mLsolvent ≠ mLsolution Mass percent = grams of solute/100 g solution Mole percent = moles of solute/100 moles solution Molarity = moles of solute/L of solution ET: Molarity is temperature dependent, but molality is not. Difficult but important interconversion Molality = moles of solute/kilogram of solvent PPM = grams of solute/1,000,000 grams solution Volume percent = volume of solute/100 mL of solution Proof = 2 x Vol. %; e.g., 2 x 40 mL/100 mL solution = 80 proof Mole fraction: XA = nA/(nA + nB) XA + XB = 1 Particle fraction: PXA = inA/(inA + inB) PXA + PXB = 1

Raoult’s law: Don’t use: Psoln = XsolventPosolvent

Use: Psoln = PXsolventPosolvent

For two volatile components: Psoln = PXsolventAPosolventA + PXsolventBPo

solventB PXsolvent = insolvent/(insolvent + insolute) PXA + PXB = 1 van’t Hoff factor, “i”; i = moles of particles in solution/moles of solute dissolved Boiling-point elevation: ΔT = Tf - Ti = Kbimsolute Note that H&P uses: ΔT = Tf(solution) – Tf(solvent) = Kbimsolute; my formula and H&P’s formula are identical.

Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Note that H&P uses: ΔT = Tf(solution) – Tf(solvent) = -Kfimsolute; my formula and H&P’s formula are identical.

Osmotic pressure: πV = inRT* * Begin problem with this formula, not the equivalent π = iMsoluteRT

ΔHsoln = ΔHsolute-solutebondbreaking + ΔHsolvent-solventbondbreaking + ΔHsolute-solventbondforming

Henry’s Law: SA=kPA; MA = Solubility of dissolved gas in solution “S” may be any unit of concentration (depending on the units of “k”), e.g., X, M, %.

ET note: Not everything goes through moles, e.g., volume to grams, grams to volume, atoms to molecules, molecules to atoms.


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ENERGETICS OF SOLUTIONS AND SOLUBILITIES

Relative Strength

• Electrostatic interaction (ions) 1000 (= charge-charge interaction (“strong”)

= ionic interaction) Coulomb’s Law: F = k(Q1 x Q2)/r2

• Ion-dipole interaction 100-500(?)

• Hydrogen bonding 30 Y = N, O or F (“moderate”) Y ― H Y

• Dipole-dipole interaction 10

(polar molec. & polar molec.) (“moderate”)

• London forces (with all molecules) 1* (non-polar molecules (“weak”)

& non-polar molec.)

Relatively strong bonds

Relatively weak bonds


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COLLOIDS A colloid is a dispersion in which the dispersed matter has one or more dimensions (length, width, or thickness) in the range from about 1nm to 1000 nm. The particles don’t settle out (suspension) but they aren’t dissolved into individual molecules either (solution).

Particle Fluid Tyndall Example Phase Precipitation Size appearance Effect** Solution NaCl in H2O Homogen No <1nm Clear No Colloid* Starch in H2O, milk, Varies No 1nm to 1000 nm Usually cloudy Yes gelatin, fog, steam, hair spray Suspension Sand in H2O Heterogen Yes >1000 nm Cloudy Yes *Also called “colloidal suspension” or “colloidal dispersion”. **The Tyndall effect is the scattering of light by colloidal particles, which makes a colloidal dispersion distinguishable from a true solution. Colloids occur in any phase. Gas phase: aerosol Liquid phase: emulsion Solid phase: sol


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CHEM 162-2010 FINAL EXAM Chapter 12 - Properties of Solutions Energetics of solutions and solubility concepts 23. Which compound below would be expected to be the least soluble in water? A. CH3CH2CH2I B. CH3CH2CH2NH2 C. CH3CH2-O-CH3 D. CH3CH2COOH E. CH3CH2COOCH3 The molecule that would be the most soluble would be the one that forms the strongest bonds with water. B, C, D and E can all form hydrogen bonds with water. However, A is not capable of forming hydrogen bonds with water. It forms a slightly weaker dipole-dipole bond with water, and is therefore the least soluble. Chem 162-2010 Final Exam + Answers Chapter 12 – Physical Properties of Solutions Solution concentrations calculations

42. What is the molality of 2.158M C6H12O6(aq) whose density is 1.1434g/cm3? A. 1.777m

B. 2.860m C. 2.158m D. 1.887m E. 2.467m 2.158 mol C6H12O6/Lsoln → ?mol C6H12O6/kgsolv First do numerator, then denominator. Numerator is already done: 2.158 mol → 2.158 mol Denominator: Plan: Lsoln → gsoln → gsolv → kgsolv 1L solution x 1.1434g/0.001L = 1143.4g solution gsolute + gsolvent = gsolution gsolute = 2.158mol x 180.18g/mol = 388.83 g 388.83gsolute + gsolvent = 1143.4gsolution gsolvent = 754.57g = 0.75457kg Combine numerator with denonimator: 2.158mol/0.75457kg = 2.86 molsolute/kgsolvent = 2.86molal


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CHEM 162-2010 FINAL EXAM CHAPTER 14 - PROPERTIES OF SOLUTIONS COLLIGATIVE PROPERTIES (BOILING POINT, FREEZING POINT, OSMOTIC PRESSURE CHANGES) CALCULATIONS 32. When 0.64g of the heart stimulant epinephrine was dissolved in 36.00g of benzene, the normal freezing point of the resulting solution was found to be 5.03oC. What is the approximate molar mass of epinephrine? Tf of benzene = 5.53oC; Kf for benzene = 5.12oC-kg/mol A. 180 g/mol B. 970 g/mol C. 510 g/mol D. 390 g/mol E. 96 g/mol Freezing-point depression: ΔT = Tf - Ti = -Kfimsolute Tf - Ti = -Kfimolsolute/kgsolvent Tf - Ti = -Kfi((masssolute)/(MWsolute))/kgsolvent 5.03 – 5.53 = -5.12 x 1 x ((0.64/MW)/0.036) MW = 182g/mol 51. What color is your exam booklet?

A. white B. yellow C. blue D. pink

Chem 162- 2010 Final Exam Review

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