Acids and Bases
Download
Report
Transcript Acids and Bases
Acids and Bases
1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon
dioxide gas
Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
2
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
base
base
acid
acid
acid
conjugate
acid
base
conjugate
base
3
Acid-Base Properties of Water
H+ (aq) + OH- (aq)
H2O (l)
autoionization of water
H
O
H
+ H
[H
O
H
]
H
+ H
O
-
H
base
H2O + H2O
acid
O
+
conjugate
acid
H3O+ + OHconjugate
base
4
The Ion Product of Water
H2O (l)
H+ (aq) + OH- (aq)
[H+][OH-]
Kc =
[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 250C
Kw = [H+][OH-] = 1.0 x 10-14
[H+] = [OH-]
Solution Is
neutral
[H+] > [OH-]
acidic
[H+] < [OH-]
basic
5
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
-14
K
1
x
10
w
-15 M
=
=
7.7
x
10
[OH-] =
[H+]
1.3
6
pH – A Measure of Acidity
pH = -log [H+]
[H+] = [OH-]
At 250C
[H+] = 1 x 10-7
pH = 7
acidic
[H+] > [OH-]
[H+] > 1 x 10-7
pH < 7
basic
[H+] < [OH-]
[H+] < 1 x 10-7
pH > 7
Solution Is
neutral
pH
[H+]
7
Other important relationships
pOH = -log [OH-]
[H+][OH-] = Kw = 1.0 x 10-14
-log [H+] – log [OH-] = 14.00
pH + pOH = 14.00
pH Meter
8
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
9
Strong Electrolyte – 100% dissociation
NaCl (s)
H 2O
Na+ (aq) + Cl- (aq)
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)
10
Weak Acids are weak electrolytes
H3O+ (aq) + F- (aq)
HF (aq) + H2O (l)
HNO2 (aq) + H2O (l)
H3O+ (aq) + NO2- (aq)
HSO4- (aq) + H2O (l)
H3O+ (aq) + SO42- (aq)
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)
Strong Bases are strong electrolytes
NaOH (s)
KOH (s)
H 2O
H 2O
Ba(OH)2 (s)
Na+ (aq) + OH- (aq)
K+ (aq) + OH- (aq)
H 2O
Ba2+ (aq) + 2OH- (aq)
11
Weak Bases are weak electrolytes
F- (aq) + H2O (l)
NO2- (aq) + H2O (l)
OH- (aq) + HF (aq)
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.
12
Strong Acid (HCl)
Weak Acid (HF)
13
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
Start 0.002 M
HNO3 (aq) + H2O (l)
End 0.0 M
0.0 M
0.0 M
H3O+ (aq) + NO3- (aq)
0.002 M 0.002 M
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
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.
Start 0.018 M
Ba(OH)2 (s)
End 0.0 M
0.0 M
0.0 M
Ba2+ (aq) + 2OH- (aq)
0.018 M 0.036 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6
14
Weak Acids (HA) and Acid Ionization Constants
HA (aq) + H2O (l)
HA (aq)
H3O+ (aq) + A- (aq)
H+ (aq) + A- (aq)
[H+][A-]
Ka =
[HA]
Ka is the acid ionization constant
Ka
weak acid
strength
15
16
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq)
H+ (aq) + F- (aq)
HF (aq)
Initial (M)
Change (M)
Equilibrium (M)
H+ (aq) + F- (aq)
0.50
0.00
0.00
-x
+x
+x
0.50 - x
x
x
x2
= 7.1 x 10-4
Ka =
0.50 - x
Ka
[H+][F-]
= 7.1 x 10-4
Ka =
[HF]
x2
= 7.1 x 10-4
0.50
[H+] = [F-] = 0.019 M
[HF] = 0.50 – x = 0.48 M
Ka << 1
0.50 – x 0.50
x2 = 3.55 x 10-4
x = 0.019 M
pH = -log [H+] = 1.72
17
Ionized acid concentration at equilibrium
percent ionization =
x 100%
Initial concentration of acid
For a monoprotic acid HA
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration
18
Molecular Structure and Acid Strength
H X
H+ + X-
The
stronger
the bond
The
weaker
the acid
HF << HCl < HBr < HI
acidity
increases
19
20
Definition of An Acid
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
••
••
H+ + OH••
acid base
H+ +
acid
••
H
N H
H
base
••
H O H
••
H
+
H N H
H
21
Lewis Acids and Bases
+
F B
••
H
F
N H
F
H
acid
base
F
F B
F
H
N H
H
No protons donated or accepted!
22
23
Chemical Kinetics
24
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
25
A
B
D[A]
rate = Dt
D[B]
rate =
Dt
26
red-brown
2Br- (aq) + 2H+ (aq) + CO2 (g)
Br2 (aq) + HCOOH (aq)
time
t 1< t 2 < t 3
393 nm
light
Detector
D[Br2] D Absorption
27
Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
D[Br2]
average rate = =Dt
tfinal - tinitial
instantaneous rate = rate for specific instance in time
28
2H2O2 (aq)
2H2O (l) + O2 (g)
PV = nRT
n
P=
RT = [O2]RT
V
1
[O2] =
P
RT
D[O2]
1 DP
rate =
=
RT Dt
Dt
measure DP over time
29
30
Reaction Rates and Stoichiometry
2A
B
Two moles of A disappear for each mole of B that is formed.
1 D[A]
rate = 2 Dt
aA + bB
D[B]
rate =
Dt
cC + dD
1 D[A]
1 D[B]
1 D[C]
1 D[D]
rate = ==
=
a Dt
b Dt
c Dt
d Dt
31
Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)
D[CH4]
D[CO2]
1 D[O2]
1 D[H2O]
rate = =
==
Dt
Dt
Dt
2 Dt
2
32
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
33
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 = k [F2][ClO2]
Rate quadruples
y=1
34
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.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
35
Determine the rate law and calculate the rate constant for the
following reaction from the following data:
S2O82- (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
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
36
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life
t½ =
[A]0
2k
t½ = ln 2
k
1
t½ =
k[A]0
37
A+B
Exothermic Reaction
+
AB+
C+D
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of
energy required to initiate a chemical reaction.
38
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
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
39
2NO (g) + O2 (g)
2NO2 (g)
Mechanism:
40
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.
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
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
41
Rate Laws and Elementary Steps
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
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.
42
Sequence of Steps in Studying a Reaction Mechanism
43
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
44
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
k A e( Ea / RT )
Ea
Uncatalyzed
k
Catalyzed
ratecatalyzed > rateuncatalyzed
Ea′ < Ea
45
In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
•
Haber synthesis of ammonia
•
Ostwald process for the production of nitric acid
•
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
•
Acid catalysis
•
Base catalysis
46
Catalytic Converters
catalytic
CO + Unburned Hydrocarbons + O2 converter
CO2 + H2O
catalytic
2NO + 2NO2 converter
2N2 + 3O2
47
Enzyme Catalysis
48
Binding of Glucose to Hexokinase
49
Enzyme Kinetics
D[P]
rate =
Dt
rate = k [ES]
50
All of the following may be true concerning catalysts
and the reaction which they catalyze EXCEPT
a. catalysts are not used up by the reaction
b. catalysts lower the activation energy
c. catalysts increase the rate of the reverse
reaction
d. catalysts shift the reaction equlibrium to the
right
51
As the temperature is increased in an exothermic
gaseous reaction, all of the following increase
EXCEPT
a. reaction rate
b. rate constant
c. activation energy
d. all of the above
52
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
53
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 reaction
b. first order reaction
c. second order reaction
d. third order reactin
54
55
Chemical Equilibrium
56
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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)
NO2
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)
57
N2O4 (g)
2NO2 (g)
equilibrium
equilibrium
Start with N2O4
Start with NO2 & N2O4
equilibrium
Start with NO2
58
constant
59
N2O4 (g)
K=
[NO2]2
[N2O4]
aA + bB
K=
[C]c[D]d
2NO2 (g)
= 4.63 x 10-3
cC + dD
Law of Mass Action
[A]a[B]b
60
K=
[C]c[D]d
aA + bB
cC + dD
[A]a[B]b
Equilibrium Will
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
61
Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase.
N2O4 (g)
Kc =
2NO2 (g)
[NO2]2
Kp =
[N2O4]
2
PNO
2
PN2O4
In most cases
Kc Kp
aA (g) + bB (g)
cC (g) + dD (g)
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)
62
Homogeneous Equilibrium
CH3COOH (aq) + H2O (l)
[CH3COO-][H3O+]
Kc′ =
[CH3COOH][H2O]
CH3COO- (aq) + H3O+ (aq)
[H2O] = constant
[CH3COO-][H3O+]
= Kc′ [H2O]
Kc =
[CH3COOH]
General practice not to include units for the equilibrium
constant.
63
The equilibrium constant Kp for the reaction
2NO2 (g)
2NO (g) + O2 (g)
is 158 at 1000K. What is the equilibrium pressure of O2 if the
PNO = 0.400 atm and PNO = 0.270 atm?
2
Kp =
2
PNO
PO2
2
PNO
2
PO2 = Kp
2
PNO
2
2
PNO
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
64
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases.
CaCO3 (s)
[CaO][CO2]
Kc′ =
[CaCO3]
Kc = [CO2] = Kc′ x
[CaCO3]
[CaO]
CaO (s) + CO2 (g)
[CaCO3] = constant
[CaO] = constant
Kp = PCO2
The concentration of solids and pure liquids are not
included in the expression for the equilibrium constant.
65
Le Châtelier’s Principle
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.
• Changes in Concentration
N2 (g) + 3H2 (g)
2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
66
Le Châtelier’s Principle
• Changes in Concentration continued
Remove Add
aA + bB
Add Remove
cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
left
right
right
left
67
Le Châtelier’s Principle
• Changes in Temperature
Change
Increase temperature
Decrease temperature
N2O4 (g)
Exothermic Rx
Endothermic Rx
K decreases
K increases
K increases
K decreases
2NO2 (g)
colder
hotter
68
Le Châtelier’s Principle
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift
equilibrium.
69
Le Châtelier’s Principle - Summary
Change Equilibrium
Constant
no
Change
Shift Equilibrium
Concentration
yes
Pressure
yes*
no
Volume
yes*
no
Temperature
yes
yes
Catalyst
no
no
*Dependent on relative moles of gaseous reactants and products
70
As the temperature is increased, the equilibrium of
gaseous reaction will always:
a. shift to the right
b. shift to the left
c. remain constant
d. the answer cannot be determined from
the information given
71
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 zero
d. All reactants will start to move forward at
equilibrium
72
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]
73