AP Chemistry Super Saturday

Review

I tried to include as much review materialas possible in this session. Work on practice

tests and review the Bozeman videos for other material.

AP Chemistry Super Saturday

Review5 Essentials

1. Know the basics – writing formulas,writing and balancing equations, dimensional analysis, atomic theory,acid-base theories (Arrhenius and BronstedLowry), VSEPR, kinetic molecular theory,and collision theory.

5 Essentials

2. Atomic and Molecular StructuresAtomic structures (like electron configurations)will help explain relationships on the periodictable which explains many physical and chemical properties. Molecular structures involves Lewisstructures and VSEPR to determine shapeswhich describes polarities thus describing intermolecular forces which describes manyphysical properties.

5 Essentials

3. Stoichiometric CalculationsBasic stoichiometry, limiting reactants,titration calculations, and empirical formulas.

4. Principles of chemical kinetics, equilibrium,And thermodynamicsKinetics- describes the speed in which substances reactEquilibrium – used to determine the extent ofa reaction or the composition at equilibriumThermodynamics – explains why chemical reactionshappen in terms of kinetic and potential energies

5 Essentials

5. Representation and InterpretationBe able to draw what is happening at the molecular level and read and interpret graphsand data tables

Net Ionic Equations

Graphic: Wikimedia Commons User Tubifex

Solubility Rules – AP Chemistry

All sodium, potassium, ammonium, and nitrate salts are soluble in water. Memorization of other “solubility rules” is beyond the scope of this course and the AP Exam.What dissociates (“breaks apart”) – Aqueous solutions of the following:

• All strong acids (HCl, HBr, HI, HNO3, H2SO4, and HClO4)

• Strong bases (group I and II hydroxides)

• Soluble salts

Write the net ionic for the following:

1. Solutions of lead nitrate and potassium chloride are mixed

2. Solutions of sulfuric acid and potassium hydroxide are mixed.

3. Solid sodium hydroxide is mixed with acetic acid

Big Idea #6:Chemical Equilibrium

2NO2(g) 2NO(g) + O2(g)

Sketch a graph of change in concentration vs. Time for the reaction above

2NO2(g) 2NO(g) + O2(g)

Be able to explain the variance in slope

Law of Mass Action

For the reaction:

Where K is the equilibrium constant, and is unitless

jA + kB lC + mD

kj

ml

BA

DCK

][][

][][

Product Favored EquilibriumLarge values for K signify the reaction is “product favored”

When equilibrium is achieved, most reactant has been converted to product

Reactant Favored EquilibriumSmall values for K signify the reaction is “reactant favored”

When equilibrium is achieved, very little reactant has been converted to product

Writing an Equilibrium Expression

2NO2(g) 2NO(g) + O2(g)

K = ???

Write the equilibrium expression for the reaction:

22

22

][

][][

NO

ONOK

Conclusions about Equilibrium Expressions

The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse

2NO2(g) 2NO(g) +

O2(g)

2NO(g) + O2(g) 2NO2(g)

22

22

][

][][

NO

ONOK

][][

][1

22

22'

ONO

NO

KK

Conclusions about Equilibrium Expressions

When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power.2NO2(g) 2NO(g) +

O2(g)

NO2(g) NO(g) + ½O2(g)

22

22

][

][][

NO

ONOK

][

]][[

2

2

1

22

11

NO

ONOKK

A) 0.584

B) 4.81

C) 0.416

D) 23.1

E) 0.208

If the equilibrium constant for A + B C is 0.208 then the equilibrium constant for 2C       2A + 2B is

Answer: D

Equilibrium Expressions Involving Pressure

For the gas phase reaction: 3H2(g) + N2(g) 2NH3(g)

))(( 3

2

22

3

HN

NHP

PP

PK

pressurespartialmequilibriuarePPP HNNH 223,,

np RTKK )(

Heterogeneous Equilibria

The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present

Write the equilibrium expression for the reaction: PCl5(s) PCl3(l) + Cl2(g)

Pure solid

Pure liquid

][ 2ClK

2Clp PK

The Reaction Quotient

For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action.

jA + kB lC + mD

kj

ml

BA

DCQ

][][

][][

Significance of the Reaction Quotient

If Q = K, the system is at equilibrium

If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved

If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

A) To the left.

B) To the right.

C) The above mixture is the equilibrium mixture.

D) Cannot tell from the information given.

If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a one-liter container, which direction would the reactioninitially proceed?

LeChatelier’s Principle

When a system atequilibrium is placed

understress, the system willundergo a change in

sucha way as to relieve

thatstress and restore a state of equilibrium.

Henry Le Chatelier

When you take something away from a system at equilibrium, the system shifts in such a way as to replace some of what you’ve taken away.

Le Chatelier Translated:

When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

•Do FRQ #1

Acid Equilibrium

and pHSøren Sørensen

Acid/Base Definitions Arrhenius Model

Acids produce hydrogen ions in aqueous solutions

Bases produce hydroxide ions in aqueous solutions

Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors

Acid Dissociation

HA H+ + A-

Acid Proton Conjugate

base

][

]][[

HA

AHKa

Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

Dissociation Constants: Strong Acids

AcidFormul

aConjugate

BaseKa

 Perchloric   HClO4   ClO4-   Very large 

 Hydriodic   HI   I-   Very large 

 Hydrobromic   HBr   Br-   Very large 

 Hydrochloric   HCl   Cl-   Very large 

 Nitric   HNO3   NO3-   Very large 

 Sulfuric   H2SO4   HSO4-   Very large 

 Hydronium ion   H3O+   H2O   1.0 

Dissociation Constants: Weak Acids

Acid FormulaConjugate Base

Ka

 Iodic   HIO3   IO3-   1.7 x 10-1 

 Oxalic   H2C2O4   HC2O4-   5.9 x 10-2 

 Sulfurous   H2SO3   HSO3-   1.5 x 10-2 

 Phosphoric   H3PO4   H2PO4-   7.5 x 10-3 

 Citric   H3C6H5O7   H2C6H5O7-   7.1 x 10-4 

 Nitrous   HNO2   NO2-   4.6 x 10-4 

 Hydrofluoric  HF   F-   3.5 x 10-4 

 Formic   HCOOH   HCOO-   1.8 x 10-4 

 Benzoic   C6H5COOH   C6H5COO-   6.5 x 10-5 

 Acetic   CH3COOH   CH3COO-   1.8 x 10-5 

 Carbonic   H2CO3   HCO3-   4.3 x 10-7 

 Hypochlorous   HClO   ClO-   3.0 x 10-8 

 Hydrocyanic   HCN   CN-   4.9 x 10-10 

Reaction of Weak Bases with Water

The base reacts with water, producing its conjugate acid and hydroxide ion:

CH3NH2 + H2O CH3NH3+ + OH- Kb = 4.38 x 10-4

4 3 3

3 2

[ ][ ]4.38 10

[ ]b

CH NH OHK x

CH NH

Kb for Some Common Weak Bases

Base FormulaConjugat

e AcidKb

Ammonia   NH3  NH4+  1.8 x 10-5 

 Methylamine  CH3NH2  CH3NH3+  4.38 x 10-4 

 Ethylamine  C2H5NH2  C2H5NH3+  5.6 x 10-4 

 Diethylamine  (C2H5)2NH  (C2H5)2NH2+  1.3 x 10-3 

 Triethylamine   (C2H5)3N   (C2H5)3NH+  4.0 x 10-4 

 Hydroxylamine  HONH2   HONH3+

   1.1 x 10-8 

 Hydrazine H2NNH2  H2NNH3+

   3.0 x 10-6 

 Aniline  C6H5NH2   C6H5NH3+

   3.8 x 10-10 

 Pyridine  C5H5N   C5H5NH+    1.7 x 10-9 

Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

Reaction of Weak Bases with Water

The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion:

B + H2O BH+ + OH-

[ ][ ]

[ ]b

BH OHK

B

(Yes, all weak bases do this – DO NOTtry to make this complicated!)Ex. Write the reaction of ammonia with water

Self-Ionization of Water

H2O + H2O H3O+ + OH-

At 25, [H3O+] = [OH-] = 1 x 10-7

Kw is a constant at 25 C:

Kw = [H3O+][OH-]

Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

Calculating pH, pOHpH = -log10(H3O+)

pOH = -log10(OH-)

Relationship between pH and pOH pH + pOH = 14

Finding [H3O+], [OH-] from pH, pOH

[H3O+] = 10-pH

[OH-] = 10-pOH

Calculate the pH of a 0.1M HCl solution?

(answer =1)

A Weak Acid Equilibrium Problem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?(answer=4.52)

A Weak Base Equilibrium Problem

What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? (answer=9.48)

Acid-Base Properties of Salts

To determine if a salt is acidic or basic, determine the stronger parent.Examples:KClNH4ClNaC2H3O2

NaClKNO3

Acid-Base Properties of SaltsIf both parents are weak:

Type of Salt Examples

Comment pH of solution

Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid

NH4C2H3O2

NH4CN

Cation is acidic,Anion is basic

See below

IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic

IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic

IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

Buffered Solutions

A solution that resists a change in pH when either hydroxide ions or protons are added.

Buffered solutions contain either:A weak acid and its saltA weak base and its salt

Acid/Salt Buffering Pairs

Weak AcidFormula

of the acidExample of a salt of the

weak acid Hydrofluoric  HF   KF – Potassium fluoride 

 Formic   HCOOH   KHCOO – Potassium formate 

 Benzoic   C6H5COOH   NaC6H5COO – Sodium benzoate

 Acetic   CH3COOH   NaH3COO – Sodium acetate 

 Carbonic   H2CO3   NaHCO3 - Sodium bicarbonate

 Propanoic   HC3H5O2    NaC3H5O2  - Sodium propanoate

 Hydrocyanic   HCN   KCN - potassium cyanide 

The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

Base/Salt Buffering PairsThe salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

BaseFormula of the base

Example of a salt of the weak acid

Ammonia   NH3  NH4Cl - ammonium chloride

 Methylamine

 CH3NH2 CH3NH2Cl – methylammonium

chloride

 Ethylamine  C2H5NH2 C2H5NH3NO3 -  ethylammonium

nitrate

 Aniline  C6H5NH2  C6H5NH3Cl – aniline hydrochloride

 Pyridine  C5H5N    C5H5NHCl – pyridine hydrochloride

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)

pH

Titration of an Unbuffered Solution

A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NaOH (0.10 M)

pH

Titration of a Buffered Solution

A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH

Comparing Results

Graph

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45

mL 0.10 M NaOH

pH

Buffered

Unbuffered

Henderson-Hasselbalch Equation

][

][log

][

][log

acid

basepK

HA

ApKpH aa

][

][log

][

][log

base

acidpK

B

BHpKpOH bb

This is an exceptionally powerful tool that can be used in your problem solving.

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)

pH

Title:

A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

Endpoint is above pH 7

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NaOH (0.10 M)

pH

Title:

A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

Endpoint is at pH 7

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters HCl (0.10 M)

pH

Title:

A solution that is 0.10 M NaOH is titrated with 0.10 M HCl

Endpoint is at pH 7 It is important to

recognize that titration curves are not always increasing from left to right.

1

2

3

4

5

6

7

8

9

10

11

12

13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NH3 (0.10 M)

pH

Title:

A solution that is 0.10 M HCl is titrated with 0.10 M NH3

Endpoint is below pH 7

Solubility

Equilibria

Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate.

Graphic: Wikimedia Commons user PRHaney

Solving Solubility Problems

For the salt AgI at 25C, Ksp = 1.5 x 10-16

Answer = solubility of AgI in mol/L = 1.2 x 10-8 M

Solving Solubility Problems

For the salt PbCl2 at 25C, Ksp = 1.6 x 10-

5

Answer = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

Solving Solubility with a Common Ion

For the salt AgI at 25C, Ksp = 1.5 x 10-16

What is its solubility in 0.05 M NaI?

Answer = solubility of AgI in mol/L = 3.0 x 10-15 M

Big Idea #5: Spontaneity,

Entropyand Free Energy

Spontaneity, Entropy

and Free Energy

G = H - TS

Spontaneous Processes and Entropy

First Law• “Energy can neither be created nor destroyed"

• The energy of the universe is constant

Spontaneous Processes• Processes that occur without outside intervention

• Spontaneous processes may be fast or slow–Many forms of combustion are fast

–Conversion of diamond to graphite is slow

Which of the following reactions is spontaneous?

• H2(g) + I2(g) ↔ 2HI Kc=49• Br2 + Cl2 ↔ 2BrCl Kc=6.9

• HF + H2O ↔ F- + H30+ Kc=6.8x10-4

Second Law of Thermodynamics

"In any spontaneous process there is always an increase in the entropy of the universe"

Ssolid < Sliquid << Sgas

For reactions at constant temperature:G0 = H0 - TS0

Calculating Free Energy Method #1

Calculating Free Energy: Method #2

An adaptation of Hess's Law:Cdiamond(s) + O2(g) CO2(g) G0 = -397 kJCgraphite(s) + O2(g) CO2(g) G0 = -394 kJ

CO2(g) Cgraphite(s) + O2(g) G0 = +394 kJ

Cdiamond(s) Cgraphite(s) G0 =

Cdiamond(s) + O2(g) CO2(g) G0 = -397 kJ

-3 kJ

Calculating Free Energy Method #3

Using standard free energy of formation (Gf0):

0 0 0(products) (reactants)p f r fG n G n G

Gf0 of an element in its standard state is zero

Free Energy and Equilibrium

Equilibrium point occurs at the lowest value of free energy available to the reaction system

At equilibrium, G = 0 and Q = K

G0 KG0 = 0 K = 1G0 < 0 K > 1G0 > 0 K < 1

Bond Energy

• Breaking bonds require energy (+)• Forming bonds releases energy (-)

• If the reaction A + B → C is exothermic, which is larger – the energy needed to break the bonds or the energy released when forming the bonds?

•TRY FRQ #2

Big Idea #4: Kinetics, Rates,

andRate Laws

Reaction RateThe change in concentration of a reactant or product per unit of time

12

12 ][][

tt

ttimeatAttimeatARate

t

ARate

][

2NO2(g) 2NO(g) + O2(g)Reaction Rates:

2. Can measure appearance of products

1. Can measure disappearance of reactants

3. Are proportional stoichiometrically

2NO2(g) 2NO(g) + O2(g)Reaction Rates:4. Are equal to

the slope tangent to that point

[NO2]

t

5. Change as the reaction proceeds, if the rate is dependent upon concentration2[ ]

constantNO

t

Rate Laws

Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.

Integrated rate laws express (reveal) the relationship between concentration of reactants and time

The differential rate law is usually just called “the rate law.”

Writing a (differential) Rate Law

2 NO(g) + Cl2(g) 2 NOCl(g)

Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:

Experiment

[NO](mol/L)

[Cl2]

(mol/L)

RateMol/L·s

1 0.250 0.250 1.43 x 10-6

2 0.500 0.250 5.72 x 10-6

3 0.250 0.500 2.86 x 10-6

4 0.500 0.500 11.4 x 10-6

Writing a Rate LawPart 1 – Determine the values for the exponents in the rate law:

Experiment

[NO](mol/L)

[Cl2]

(mol/L)

RateMol/L·s

1 0.250 0.250 1.43 x 10-6

2 0.500 0.250 5.72 x 10-6

3 0.250 0.500 2.86 x 10-6

4 0.500 0.500 1.14 x 10-5

In experiment 1 and 2, [Cl2] is constant while [NO] doubles.

R = k[NO]x[Cl2]y

The rate quadruples, so the reaction is second order with respect to [NO] R = k[NO]2[Cl2]y

Writing a Rate LawPart 2 – Determine the value for k, the rate constant, by using any set of experimental data:

Experiment

[NO](mol/L)

[Cl2]

(mol/L)

RateMol/L·s

1 0.250 0.250 1.43 x 10-6

R = k[NO]2[Cl2]

261.43 10 0.250 0.250mol mol mol

x kL s L L

6 3 25

3 3 2

1.43 109.15 10

0.250

x mol L Lk x

L s mol mol s

Writing a Rate LawPart 3 – Determine the overall order for the reaction.

R = k[NO]2[Cl2]

Overall order is the sum of the exponents, or orders, of the reactants

2 + 1 = 3

The reaction is 3rd order

Determining Order withConcentration vs. Time data

(the Integrated Rate Law)

Zero Order:

First Order:

Second Order:

linearisionconcentratvstime .

linearisionconcentratvstime )ln(.

linearisionconcentrat

vstime1

.

Solving an Integrated Rate Law

Time (s) [H2O2] (mol/L)

0 1.00

120 0.91

300 0.78

600 0.59

1200 0.37

1800 0.22

2400 0.13

3000 0.082

3600 0.050

Problem: Find the integrated rate law and the value for the rate constant, kA graphing calculator with linear regression analysis greatly simplifies this process!!

(Click here to download my Rate Laws program for theTi-83 and Ti-84)

Time vs. [H2O2]Time (s)

[H2O2]

0 1.00

120 0.91

300 0.78

600 0.59

1200 0.37

1800 0.22

2400 0.13

3000 0.082

3600 0.050

y = ax + b a = -2.64 x 10-4

b = 0.841r2 = 0.8891r = -0.9429

Regression results:

Time vs. ln[H2O2]

Time (s) ln[H2O2]

0 0

120 -0.0943

300 -0.2485

600 -0.5276

1200 -0.9943

1800 -1.514

2400 -2.04

3000 -2.501

3600 -2.996

Regression results:

y = ax + b a = -8.35 x 10-4

b = -.005r2 = 0.99978r = -0.9999

Time vs. 1/[H2O2]

Time (s)

1/[H2O2]

0 1.00

120 1.0989

300 1.2821

600 1.6949

1200 2.7027

1800 4.5455

2400 7.6923

3000 12.195

3600 20.000

y = ax + b a = 0.00460b = -0.847r2 = 0.8723r = 0.9340

Regression results:

And the winner is… Time vs. ln[H2O2]

1. As a result, the reaction is 1st order

2. The (differential) rate law is:

3. The integrated rate law is:

4. But…what is the rate constant, k ?

][ 22OHkR

02222 ]0ln[]ln[ HktOH

Finding the Rate Constant, k

Method #2: Obtain k from the linear regresssion analysis.

Now remember:

k = -slope

k = 8.35 x 10-4s-1

Regression results:

y = ax + b a = -8.35 x 10-4

b = -.005r2 = 0.99978r = -0.999902222 ]0ln[]ln[ HktOH

141032.8 sxslope

Rate Laws SummaryZero Order First Order Second Order

Rate Law Rate = k Rate = k[A] Rate = k[A]2

Integrated Rate Law

[A] = -kt + [A]0 ln[A] = -kt + ln[A]0

Plot that produces a straight line

[A] versus t ln[A] versus t

Relationship of rate constant to slope of straight line

Slope = -k Slope = -k Slope = k

Half-Life

1

[ ]versus t

A

0

1 1

[ ] [ ]kt

A A

01/ 2

[ ]

2

At

k 1/ 2

0.693t

k 1/ 2

0

1

[ ]t

k A

Reaction Mechanism

The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.

The sum of the elementary steps must give the overall balanced equation for the reaction

The mechanism must agree with the experimentally determined rate law

Rate-Determining Step

In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction.The experimental rate law must agree with the rate-determining step

Identifying the Rate-Determining Step

For the reaction:2H2(g) + 2NO(g) N2(g) +

2H2O(g)The experimental rate law is:

R = k[NO]2[H2]Which step in the reaction mechanism is the rate-determining (slowest) step?

Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g)Step #2 N2O(g) + H2(g) N2(g) + H2O(g)

Step #1 agrees with the experimental rate law

Identifying Intermediates

For the reaction:2H2(g) + 2NO(g) N2(g) + 2H2O(g)

Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)

Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g)Step #2 N2O(g) + H2(g) N2(g) + H2O(g)

2H2(g) + 2NO(g) N2(g) + 2H2O(g)

N2O(g) is an intermediate

Collision ModelKey Idea: Molecules must collide to react.However, only a small fraction of collisions produces a reaction. Why?

Collision ModelCollisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy).Colliding particles must be correctly oriented to one another in order to produce a reaction.

Factors Affecting RateIncreasing temperature always increases the rate of a reaction.

Particles collide more frequently Particles collide more energeticallyIncreasing surface area increases

the rate of a reaction

Increasing Concentration USUALLY increases the rate of a reaction

Presence of Catalysts, which lower the activation energy by providing alternate pathways

•TRY FRQ #3

Big Idea #2 Intermolecular Forces

Relative Magnitudes of Forces

The types of bonding forces vary in their strength as measured by average bond energy.

Covalent bonds (400 kcal/mol)

Hydrogen bonding (12-16 kcal/mol )

Dipole-dipole interactions (2-0.5 kcal/mol)

London forces (less than 1 kcal/mol)

Strongest

Weakest

London Dispersion Forces

The temporary separations of charge that lead to the London force attractions are what attract one nonpolar molecule to its neighbors.

Fritz London1900-1954

London forces increase with the size of the molecules.

Synonyms: “Induced dipoles”, “dispersion forces”, and “dispersion-interaction forces”

London Dispersion Forces

Dipole-Dipole

• Forces of attraction between two polar molecules

• Permanent dipoles

Hydrogen Bonding

Special type of dipole-dipole in which hydrogen bonds with N, O, F.

Hydrogen bonding between ammonia and water

Hydrogen Bonding in DNA

N O

OH

OP

O

OH

OHN

N

NNH2

NO

OH

OP

O

OH

OH

NH

O

O

CH3

T A

Thymine hydrogen bonds to Adenine

Boiling point as a measure of intermolecular attractive forces

TRY FRQ #4

Definition: Half of the distance between nuclei in covalently bonded diatomic molecule Radius decreases across a period

Increased effective nuclear charge due to decreased shielding

Radius increases down a group Each row on the periodic table adds a “shell” or energy level to the atom

Atomic Radius

Big Idea #1: Periodic Trends

Table of Atomic

Radii

Period Trend:Atomic Radius

Increases for successive electrons taken from the same atom

Tends to increase across a period

Electrons in the same quantum level do not shield as effectively as electrons in inner levels

Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove

Tends to decrease down a group Outer electrons are farther from the

nucleus and easier to remove

Ionization Energy

Definition: the energy required to remove an electron from an atom

Ionization Energy: the energy required to remove an electron from an atom

Increases for successive electrons taken from the same atom

Tends to increase across a periodElectrons in the same quantum level do not shield as effectively as electrons in inner levels Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove

Tends to decrease down a group

Outer electrons are farther from the nucleus

Table of 1st Ionization Energies

Periodic Trend:Ionization

Energy

Electronegativity

Definition: A measure of the ability of an atom in a chemical compound to attract electrons

o Electronegativity tends to increase across a periodo As radius decreases, electrons get closer to the bonding atom’s nucleus

o Electronegativity tends to decrease down a group or remain the sameo As radius increases, electrons are farther from the bonding atom’s nucleus

Periodic Table of Electronegativities

Periodic Trend:Electronegativi

ty

Summary of Periodic Trends

Ionic Radii

Cations

Positively charged ions formed when

an atom of a metal loses one or more electrons Smaller than the corresponding

atom

Anions

Negatively charged ions formed

when nonmetallic atoms gain one or more electrons Larger than the corresponding

atom

Table of Ion Sizes

Determine the element

Answer: Hg

Determine the element

Answer: Na and Mg

•TRY FRQ #5

Electrochemistry

Electrochemistry Terminology #1

Oxidation – A process in which an element attains a more positive oxidation state

Na(s) Na+ + e-

Reduction – A process in which an element attains a more negative oxidation state

Cl2 + 2e- 2Cl-

Electrochemistry Terminology #2

Gain Electrons = Reduction

An old memory device for oxidation and reduction goes like this… LEO says GER

Lose Electrons = Oxidation

Electrochemistry Terminology #4

Anode The electrode

where oxidation occurs

CathodeThe electrode

where reduction occurs

Memory device:

Reductionat the

Cathode

Galvanic (Electrochemical) Cells

Spontaneous redox processes

have:A positive cell potential, E0

A negative free energy change, (-G)

G=-nFE

Zn - Cu Galvanic

Cell

Zn2+ + 2e- Zn E = -0.76V

Cu2+ + 2e- Cu E = +0.34V

From a table of reduction potentials:

Zn - Cu Galvanic

Cell

Cu2+ + 2e- Cu E = +0.34V

The less positive, or more negative reduction potential becomes the oxidation…

Zn Zn2+ + 2e- E = +0.76V

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Line Notation

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

An abbreviated representation of an electrochemical cell

Anodesolution

Anodematerial

Cathodesolution

Cathodematerial

| |||

Calculating G0 for a Cell

0 (2 )(96485 )(1.10 )coulombs Joules

G mol emol e Coulomb

G0 = -nFE0

n = moles of electrons in balanced redox equation

F = Faraday constant = 96,485 coulombs/mol e-

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

0 212267 212G Joules kJ

Concentration Cell

Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

Both sides have the same

components but at different

concentrations.

???

Concentration Cell

Both sides have the same

components but at different

concentrations.

The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction)

Zn Zn2+ (0.10M) + 2e- (oxidation)

???

CathodeAnode

Zn2+ (1.0M) Zn2+

(0.10M)

Electrolytic

Processes

A negative cell potential, (-E0)

A positive free energy change, (+G)

Electrolytic processes are NOT spontaneous. They have:

Solving an Electroplating Problem

Q: What mass of copper is plated out when a current of 10 amps is passed for 30 minutes through a solution of Cu2+?

(Amp=C/sec)

10 C

Cu2+ + 2e- Cu

1800sec 1 mol e-

96 485 C

1 mole Cu

2 mol e-

63.5 g Cu

1 mole Cu

= 5.94g Cu

sec

•Good Luck on the Exam!! Try your best. You have worked hard and will do great! I am very proud of each one of you.

• Mrs. L