Upload
maggie-fulk
View
219
Download
1
Tags:
Embed Size (px)
Citation preview
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