1 acids and bases copyright © the mcgraw-hill companies, inc. permission required for reproduction...
Post on 19-Dec-2015
215 Views
Preview:
TRANSCRIPT
1
Acids and Bases
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon dioxide gas
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Bases
3
A Brønsted acid is a proton donorA Brønsted base is a proton acceptor
acidbase acid base
acidconjugate
basebase conjugate
acid
4
O
H
H + O
H
H O
H
H H OH-+[ ] +
Acid-Base Properties of Water
H2O (l) H+ (aq) + OH- (aq)
H2O + H2O H3O+ + OH-
acid conjugate
base
base conjugate
acid
autoionization of water
5
H2O (l) H+ (aq) + OH- (aq)
The Ion Product of Water
Kc =[H+][OH-]
[H2O][H2O] = constant
Kc[H2O] = Kw = [H+][OH-]
The ion-product constant (Kw) is the product of the molar concentrations of H+ and OH- ions at a particular temperature.
At 250CKw = [H+][OH-] = 1.0 x 10-14
[H+] = [OH-]
[H+] > [OH-]
[H+] < [OH-]
Solution Is
neutral
acidic
basic
6
What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M?
Kw = [H+][OH-] = 1.0 x 10-14
[H+] = 1.3 M
[OH-] =Kw
[H+]
1 x 10-14
1.3= = 7.7 x 10-15 M
7
pH – A Measure of Acidity
pH = -log [H+]
[H+] = [OH-]
[H+] > [OH-]
[H+] < [OH-]
Solution Is
neutral
acidic
basic
[H+] = 1 x 10-7
[H+] > 1 x 10-7
[H+] < 1 x 10-7
pH = 7
pH < 7
pH > 7
At 250C
pH [H+]
8
pOH = -log [OH-]
[H+][OH-] = Kw = 1.0 x 10-14
-log [H+] – log [OH-] = 14.00
pH + pOH = 14.00
Other important relationships
pH Meter
9
The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?
pH = -log [H+]
[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M
The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?
pH + pOH = 14.00
pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60
pH = 14.00 – pOH = 14.00 – 6.60 = 7.40
10
Strong Electrolyte – 100% dissociation
NaCl (s) Na+ (aq) + Cl- (aq)H2O
Weak Electrolyte – not completely dissociated
CH3COOH CH3COO- (aq) + H+ (aq)
Strong Acids are strong electrolytes
HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)
H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)
11
HF (aq) + H2O (l) H3O+ (aq) + F- (aq)
Weak Acids are weak electrolytes
HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)
HSO4- (aq) + H2O (l) H3O+ (aq) + SO4
2- (aq)
H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)
Strong Bases are strong electrolytes
NaOH (s) Na+ (aq) + OH- (aq)H2O
KOH (s) K+ (aq) + OH- (aq)H2O
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O
12
F- (aq) + H2O (l) OH- (aq) + HF (aq)
Weak Bases are weak electrolytes
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Conjugate acid-base pairs:
• H3O+ is the strongest acid that can exist in aqueous solution.
• The OH- ion is the strongest base that can exist in aqeous solution.
13
Strong Acid (HCl) Weak Acid (HF)
14
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Start
End
0.002 M
0.002 M 0.002 M0.0 M
0.0 M 0.0 M
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6
15
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Weak Acids (HA) and Acid Ionization Constants
HA (aq) H+ (aq) + A- (aq)
Ka =[H+][A-][HA]
Ka is the acid ionization constant
Ka
weak acidstrength
16
17
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]
= 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50 0.00
-x +x
0.50 - x
0.00
+x
x x
Ka =x2
0.50 - x= 7.1 x 10-4
Ka x2
0.50= 7.1 x 10-4
0.50 – x 0.50Ka << 1
x2 = 3.55 x 10-4 x = 0.019 M
[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72
[HF] = 0.50 – x = 0.48 M
18
percent ionization = Ionized acid concentration at equilibrium
Initial concentration of acidx 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0
x 100% [HA]0 = initial concentration
19
Molecular Structure and Acid Strength
H X H+ + X-
The stronger the bond
The weaker the acid
HF << HCl < HBr < HI
acidityincreases
20
21
Arrhenius acid is a substance that produces H+ (H3O+) in water
A Brønsted acid is a proton donor
A Lewis acid is a substance that can accept a pair of electrons
A Lewis base is a substance that can donate a pair of electrons
Definition of An Acid
H+ H O H••••
+ OH-••••••
acid base
N H••
H
H
H+ +
acid base
N H
H
H
H+
22
Lewis Acids and Bases
N H••
H
H
acid base
F B
F
F
+ F B
F
F
N H
H
H
No protons donated or accepted!
23
24
Chemical Kinetics
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
25
Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -[A]t
rate = [B]t
[A] = change in concentration of A over time period t
[B] = change in concentration of B over time period t
Because [A] decreases with time, [A] is negative.
26
A B
rate = -[A]t
rate = [B]t
27
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nmlight
Detector
[Br2] Absorption
red-brown
t1< t2 < t3
28
Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
average rate = -[Br2]t
= -[Br2]final – [Br2]initial
tfinal - tinitial
slope oftangent
slope oftangent slope of
tangent
instantaneous rate = rate for specific instance in time
29
2H2O2 (aq) 2H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RTnV
[O2] = PRT1
rate = [O2]t RT
1 Pt=
measure P over time
30
31
Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = [B]t
rate = -[A]t
12
aA + bB cC + dD
rate = -[A]t
1a
= -[B]t
1b
=[C]t
1c
=[D]t
1d
32
Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = -[CH4]
t= -
[O2]t
12
=[H2O]
t12
=[CO2]
t
33
The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x + y)th order overall
34
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
rate = k [F2][ClO2]
35
F2 (g) + 2ClO2 (g) 2FClO2 (g)
rate = k [F2][ClO2]
Rate Laws
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant (not product) concentrations.
• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
1
36
Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8
2- (aq) + 3I- (aq) 2SO42- (aq) + I3
- (aq)
Experiment [S2O82-] [I-]
Initial Rate (M/s)
1 0.08 0.034 2.2 x 10-4
2 0.08 0.017 1.1 x 10-4
3 0.16 0.017 2.2 x 10-4
rate = k [S2O82-]x[I-]y
Double [I-], rate doubles (experiment 1 & 2)
y = 1
Double [S2O82-], rate doubles (experiment 2 & 3)
x = 1
k = rate
[S2O82-][I-]
=2.2 x 10-4 M/s
(0.08 M)(0.034 M)= 0.08/M•s
rate = k [S2O82-][I-]
37
Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions
Order Rate LawConcentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1[A]
=1
[A]0
+ kt
[A] = [A]0 - kt
t½ln 2k
=
t½ =[A]0
2k
t½ =1
k[A]0
38
Exothermic Reaction Endothermic Reaction
The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.
A + B AB C + D++
39
Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g) 2NO2 (g)
N2O2 is detected during the reaction!
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
40
2NO (g) + O2 (g) 2NO2 (g)
Mechanism:
41
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
The molecularity of a reaction is the number of molecules reacting in an elementary step.
• Unimolecular reaction – elementary step with 1 molecule
• Bimolecular reaction – elementary step with 2 molecules
• Termolecular reaction – elementary step with 3 molecules
42
Unimolecular reaction A products rate = k [A]
Bimolecular reaction A + B products rate = k [A][B]
Bimolecular reaction A + A products rate = k [A]2
Rate Laws and Elementary Steps
Writing plausible reaction mechanisms:
• The sum of the elementary steps must give the overall balanced equation for the reaction.
• The rate-determining step should predict the same rate law that is determined experimentally.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.
43
Sequence of Steps in Studying a Reaction Mechanism
44
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1: NO2 + NO2 NO + NO3
Step 2: NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2
45
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
Ea k
ratecatalyzed > rateuncatalyzed
Ea < Ea′
Uncatalyzed Catalyzed
)/( RTEaeAk
46
In heterogeneous catalysis, the reactants and the catalysts are in different phases.
In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.
• Haber synthesis of ammonia
• Ostwald process for the production of nitric acid
• Catalytic converters
• Acid catalysis
• Base catalysis
47
Catalytic Converters
CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic
converter
2NO + 2NO2 2N2 + 3O2
catalyticconverter
48
Enzyme Catalysis
49
Binding of Glucose to Hexokinase
50
rate = [P]t
rate = k [ES]
Enzyme Kinetics
51
All of the following may be true concerning catalysts and the reaction which they catalyze EXCEPT
a. catalysts are not used up by the reactionb. catalysts lower the activation energyc. catalysts increase the rate of the reverse reactiond. catalysts shift the reaction equlibrium to the right
52
As the temperature is increased in an exothermic gaseous reaction, all of the following increase EXCEPT
a. reaction rateb. rate constantc. activation energyd. all of the above
53
Which of the following changes to a reaction will always increase rate constant for that reaction?
a. decreaseing the temperature b. increasing the temperature
c. increasing the concentration of the reactants
d. increasing the concentration of the catalysts
54
When a radioactive isotope undergoes nuclear decay, the concentration of the isotope decreases exponentially with constant half live. It can be determined from this that radioactive decay is a
a. zeroth order reactionb. first order reactionc. second order reactiond. third order reactin
55
56
Chemical Equilibrium
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
57
Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
• the rates of the forward and reverse reactions are equal and
• the concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
Chemical equilibrium
N2O4 (g)
H2O (g)
2NO2 (g)
NO2
58
N2O4 (g) 2NO2 (g)
Start with NO2 Start with N2O4 Start with NO2 & N2O4
equilibrium
equilibrium equilibrium
59
constant
60
N2O4 (g) 2NO2 (g)
= 4.63 x 10-3K = [NO2]2
[N2O4]
aA + bB cC + dD
K = [C]c[D]d
[A]a[B]bLaw of Mass Action
61
K >> 1
K << 1
Lie to the right Favor products
Lie to the left Favor reactants
Equilibrium Will
K = [C]c[D]d
[A]a[B]baA + bB cC + dD
62
Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) 2NO2 (g)
Kc = [NO2]2
[N2O4]Kp =
NO2P 2
N2O4P
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT)n
n = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
In most cases
Kc Kp
63
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
Kc =′[CH3COO-][H3O+][CH3COOH][H2O]
[H2O] = constant
Kc = [CH3COO-][H3O+]
[CH3COOH]= Kc [H2O]′
General practice not to include units for the equilibrium constant.
64
The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2
2NO2 (g) 2NO (g) + O2 (g)
Kp = 2PNO PO
2
PNO2
2
PO2 = Kp
PNO2
2
PNO2
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
65
Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
[CaCO3] = constant[CaO] = constant
Kc = [CO2] = Kp = PCO2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
[CaO][CO2][CaCO3]
Kc =′
[CaCO3][CaO]
Kc x′
66
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Le Châtelier’s Principle
• Changes in Concentration
N2 (g) + 3H2 (g) 2NH3 (g)
AddNH3
Equilibrium shifts left to offset stress
67
Le Châtelier’s Principle
• Changes in Concentration continued
Change Shifts the Equilibrium
Increase concentration of product(s) left
Decrease concentration of product(s) right
Decrease concentration of reactant(s)
Increase concentration of reactant(s) right
left
aA + bB cC + dD
AddAddRemove Remove
68
Le Châtelier’s Principle
• Changes in Temperature
Change Exothermic Rx
Increase temperature K decreases
Decrease temperature K increases
Endothermic Rx
K increases
K decreases
colder hotter
N2O4 (g) 2NO2 (g)
69
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner
Le Châtelier’s Principle
70
Le Châtelier’s Principle - Summary
Change Shift EquilibriumChange Equilibrium
Constant
Concentration yes no
Pressure yes* no
Volume yes* no
Temperature yes yes
Catalyst no no
*Dependent on relative moles of gaseous reactants and products
71
As the temperature is increased, the equilibrium of gaseous reaction will always:
a. shift to the rightb. shift to the leftc. remain constantd. the answer cannot be determined from the
information given
72
All of the following are true concerning a reaction at equilibrium EXCEPT:
a. the rate of the forward reaction equals the rate of the reverse reaction
b. There is no change in the concentrations of both the products and the reactants
c. The activation energy has reached zerod. All reactants will start to move forward at
equilibrium
73
What is the equilibrium expression for the following reaction
CaCO3(s)CaO(s)+ CO2(g)
a. K=[CO2]b. K=[CaO] [CO2]c. K=[CaO] [CO2]/ [CaO]d. K= [CO2]/ [CaO]
top related