Osmotic
Pressure:
a fascinating
behavior.
Yet it is the
result of a
very simple
tendency to
equalize the
concentrations
of solutions.
The critical part is the membrane!!!
A VERY practical application/consequence
of Osmotic Pressure.
Hypertonic soln
Hypotonic soln
Îcrenation
(shrivels)
Îhemolysis
(bursts)
– Cucumber placed in NaCl solution loses water to
shrivel up and become a pickle.
– Limp carrot placed in water becomes firm because
water enters via osmosis.
– Salty food causes retention of water and swelling of
tissues (edema).
– Water moves into plants through osmosis.
– Salt added to meat or sugar to fruit prevents bacterial
infection (a bacterium placed on the salt honey will
lose water through osmosis and die).
1
Osmotic Pressure
or π = (n/V) R T
πV=nRT
or π = M R T
π = ρ g h, where ρ = density of solution
g = 9.807 m s-2
h = height of column
π = ρ g h, where ρ = density of solution
g = 9.807 m s-2
h = height of column
If h = 0.17 m of a dilute aqueous soln with ρ = 1.00 g/cm3
π = [(1.00g/cm3)(10-3 kg/g)(106 cm3/m3)](9.807 m/s2)(0.17 m)
= 1.7 x 103 kg m-1 s-1 = 1.7 x 103 Pa
or = (1.7 x 103 Pa) / (1.013 x 105 Pa/atm) = 0.016 atm
A chemist dissolves 2.00 g of protein in 0.100 L of
water. The observed osmotic pressure is 0.021 atm at
25 oC. What is the MW of the protein?
2
Colloids
3
Hydrophilic and Hydrophobic Colloids
Focus on colloids in water.
“Water loving” colloids: hydrophilic.
“Water hating” colloids: hydrophobic.
Molecules arrange themselves so that hydrophobic
portions are oriented towards each other.
• If a large hydrophobic macromolecule (giant molecule)
needs to exist in water (e.g. in a biological cell),
hydrophobic molecules embed themselves into the
macromolecule leaving the hydrophilic ends to interact
with water.
•
•
•
•
Hydrophilic and Hydrophobic Colloids……..
• Typical hydrophilic groups are polar (containing C-O,
O-H, N-H bonds) or charged.
• Hydrophobic colloids need to be stabilized in water.
• Adsorption: when something sticks to a surface we say
that it is adsorbed.
• If ions are adsorbed onto the surface of a colloid, the
colloids appears hydrophilic and is stabilized in water.
• Consider a small drop of oil in water.
• Add to the water sodium stearate.
Hydrophilic and Hydrophobic Colloids
Hydrophilic and Hydrophobic Colloids
• Sodium stearate has a long hydrophobic tail
(CH3(CH2)16-) and a small hydrophobic head (-CO2-Na+).
• The hydrophobic tail can be absorbed into the oil drop,
leaving the hydrophilic head on the surface.
• The hydrophilic heads then interact with the water and
the oil drop is stabilized in water.
4
Removal of Colloidal Particles
• Colloid particles are too small to be separated by
physical means (e.g. filtration).
• Colloid particles may be coagulated (enlarged) until
they can be removed by filtration.
• Methods of coagulation:
– heating (colloid particles move and are attracted to each other
when they collide);
– adding an electrolyte (neutralize the surface charges on the
colloid particles).
– Dialysis: using a semipermeable membranes separate ions
from colloidal particles
Chapter 14
14.1
14.2
14.3
14.4
14.5
14.5
14.7
Chemical Kinetics
Factors that Affect Reaction Rates
Reaction Rates
Changes of Rate with Time
Reaction Rates and Stoichiometry
Concentration and Rate
Exponents in the Rate Law
Units of Rate Constants
Using Initial Rates to Determine Rate Laws
The Change of Concentration with Time
First-Order Reactions
Second-Order Reactions
Half-Life
Temperature and Rate
Reaction Mechanisms
Catalysis
C4H9Cl(aq) + H2O (l) Æ C4H9OH (aq) + HCl (aq)
Rate = −
∆[C4 H 9Cl ]
∆[C4 H 9OH ]
=+
∆t
∆t
Note the signs!
5
Consider the reaction
2 HI(g) Æ H2(g) + I2(g)
It’s convenient to define the rate as
rate = −
1 ∆[ HI ]
∆[ H 2 ]
∆[ I 2 ]
=+
=+
2 ∆t
∆t
∆t
And, in general for
aA
Rate = −
In fact, the instantaneous rate corresponds to d[A]/dt
+
bB
Æ
cC + dD
1 ∆[ A]
1 ∆[ B ] 1 ∆[C ] 1 ∆[ D]
=−
=
=
a ∆t
b ∆t
c ∆t
d ∆t
2 N2O5 = 4 NO2 + O2 (g)
Sample exercise 14.2
at T = 45 oC in carbon tetrachloride as a solvent
The decomposition of N2O5 proceeds according to the equation
2 N2O5 (g) Æ 4 NO2 (g) + O2 (g)
If the rate of decomposition of of N2O5 at a particular instant in a vessel
is 4.2 X 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 ?
1 ∆[ N 2O5 ]
1 ∆[ NO2 ]
1 ∆[O2 ]
Rate = −
=+
=+
2
∆t
4 ∆t
1 ∆t
i.e. the rate of the reaction is 2.1 x 10-7 M/s
the rate of appearance of NO2 is 8.4 x 10-7 M/s
and the rate of appearance of O2 is 2.1 x 10-7 M/s
2 N2O5 = 4 NO2 + O2 (g)
∆t
min
0
{N2O5]
mol/L
184
{N2O5]
mol/L
0
2.33
184
2.08
319
1.91
526
1.67
867
1.35
1198
1.11
1877
0.72
∆[N2O5]
mol/L
- ∆[N2O5]/ ∆t
mol/L-min
- ∆[N2O5]/ ∆t
mol/L-min
2.33
184
0.25
1.36 x 10-3
So let’s try something rather arbitrary at this point.
0.17
1.26 x 10-3
Let’s divide the instantaneous, average rate by
0.24
1.16 x 10-3
0.32
0.94 x 10-3
0.24
0.72 x 10-3
0.39
0.57 x 10-3
2.08
135
319
1.91
207
526
1.67
341
867
[N2O5]
and/or
[N2O5]2
1.35
331
1198
1.11
679
1877
∆[N2O5]
mol/L
∆t
min
The information on the previous slide is a bit of a
nuisance, since the instantaneous rate keeps changing
—and you know how much we like constant values
or linear relationships!
at T = 45 oC in carbon tetrachloride as a solvent
Time
min
Time
min
0.72
6
2 N2O5 = 4 NO2 + O2 (g)
at T = 45
{N2O5]
mol/L
[N2O5]
avg
oC
It’s convenient to write this result in symbolic form:
in carbon tetrachloride as a solvent
∆[N2O5]
mol/L
- ∆[N2O5]/ ∆t
mol/L-min
Avg rate
/[N2O5]av
Avg rate
/[N2O5]av2
Rate = k [N2O5]
where the value of k is about 6.2 x 10-4
2.33
10-3
2.21
0.25
1.36 x
2.00
0.17
1.26 x 10-3
6.3 x 10-4
3.2 x 10-4
1.79
0.24
1.16 x 10-3
6.5 x 10-4
3.6 x 10-4
1.51
0.32
0.94 x 10-3
6.2 x 10-4
4.1 x 10-4
1.23
0.24
0.72 x 10-3
5.9 x 10-4
4.8 x 10-4
0.92
0.39
0.57 x 10-3
6.2 x 10-4
6.7 x 10-4
6.2 x
10-4
2.8 x
10-4
2.08
1.91
so that when [N2O5] = 0.221,
Rate = (6.2 x 10-4 )(0.221)
= 1.37 x 10-4 which is the ‘average rate’
we started with
1.67
1.35
1.11
0.72
In fact, we really should take into account the 2 in front of the
N2O5, in accordance with the rule we developed earlier.
Notice the nice
constant value!!!
This leads us to the general concept of Reaction Order
When Rate = k [reactant 1]m [reactant 2]n
we say the reaction is
m-th order in reactant 1
n-th order in reactant 2
Other reactions and their observed reaction orders
2 N2O5 = 4 NO2 + O2 (g)
Rate = k [N2O5]
!!!
CHCl3 (g) + Cl2 (g) Æ CCl4 (g) + HCl(g) Rate = k[CHCl3][Cl2]1/2
H2 (g) + I2 (g)
Æ 2 HI (g)
Rate = k[H2][I2]
and (m + n)-th order overall.
The order must be determined experimentally!!!
Be careful—because these orders are NOT related necessarily
We’ll see later that it depends on the Reaction Mechanism,
rather than the overall stoichiometry.
to the stoichiometry of the reaction!!!
Be careful: the measurement of the rate will always depend on
observations of the reactants or products and involves stoichiometry,
but the part on the right, the order, does not depend on the stoichiometry.
Let’s explore the results for the result
Rate = k [N2O5]
This can be expressed as
Rate = - (∆[N2O5] / ∆ t = - d[N2O5] / dt = k [N2O5]
or, in general for
A Æ products
Rate = - ∆[A] / ∆t = d[A] / dt = k [A]
rearrangement and integration from time = 0 to t = t gives the result
ln[A]t - ln[A]o = -kt
or
ln [A]t = -kt + ln [A]o
or
ln ([A]t/[A]o = - kt
This is the expression of concentration vs time
for a First-Order Reaction
7
Chapter 14
High: 175/175
Lo: 10/175
Mean 106 (60.6%)
DISTRIBUTION OF SCORES
0.00 - 9.99
0
10.00 - 19.99
1
20.00 - 29.99
0
30.00 - 39.99
4
40.00 - 49.99
4
50.00 - 59.99 22
60.00 - 69.99 32
70.00 - 79.99 37
80.00 - 89.99 48
90.00 - 99.99 61
100.00 - 109.99 61
110.00 - 119.99 51
120.00 - 129.99 52
130.00 - 139.99 31
140.00 - 149.99 37
150.00 - 159.99 25
160.00 - 169.99 12
170.00 - 175.00 2
These people need
to talk with
Dr. Mathews
14.1
14.2
14.3
14.4
14.5
14.5
14.7
Chemical Kinetics
Factors that Affect Reaction Rates
Reaction Rates
Changes of Rate with Time
Reaction Rates and Stoichiometry
Concentration and Rate
Exponents in the Rate Law
Units of Rate Constants
Using Initial Rates to Determine Rate Laws
The Change of Concentration with Time
First-Order Reactions
Second-Order Reactions
Half-Life
Temperature and Rate
Reaction Mechanisms
Catalysis
BCDCC DBCEB CDBBE CAACA AEEDD ECBBC EBC
Consider
First-Order Reactions
Rate = −
∫
[ A ]t
[ A ]0
∆[ A]
d [ A]
=>
= −k[ A]
∆t
dt
An example of the plots of concentration vs time
for a First-Order Reaction
t
d [ A]
= − k ∫ dt
0
[ A]
To give
these forms
of the
“integrated
rate law”:
ln[ A]t = − kt + ln[ A]0
or ln[ A]t − ln[ A]0 = − kt
⎡ [ A] ⎤
or ln ⎢ t ⎥ = − kt
⎣[ A]0 ⎦
The Change of Concentration with Time
The Change of Concentration with Time
HalfHalf-Life
For a FirstFirst-Order Reaction
• Half-life is the time taken for the
concentration of a reactant to drop to half its
original value.
• That is, half life, t1/2 is the time taken for [A]0
to reach ½[A]0.
• Mathematically,
t1 = −
2
The identical length of the
first and second half-life
is a SPECIFIC characteristic
of First-Order reactions
ln 1
k
2 = 0.693
k
8
Second-Order Reactions
Consider now
Rate = −
∫
SecondSecond-Order Reactions
∆[ A]
d [ A]
=>
= −k[ A]2
∆t
dt
[ A ]t
[ A ]0
• We can show that the half life
t1 =
2
t
d [ A]
= −k ∫ dt
2
0
[ A]
1
k [A ]0
• A reaction can have rate constant expression
of the form
rate = k[A][B],
i.e., is second order overall, but has first order
dependence on A and B.
1
1
= kt +
[ A]t
[ A]0
Recall for
1st order:
t1 = −
2
t1/2 = [(0.4)(0.5)] -1
= 5.0 sec
ln 1
k
2 = 0.693
k
And for 2nd order:
t1 =
2
1
k [A ]0
Is this first or second order in the reactant? What is k?
t1/2 = 0.693/0.4 =
1.73 sec
t1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1
= 2.5 sec
Example Exercise 14.8
NO2 (g) Æ NO (g) + ½ O2 (g) at 300
oC
Example of Second-Order Plots of conc vs time
Page 540
Time/s
[NO2]
ln[NO2]
1/[NO2]
0.0
50.0
100.0
200.0
300.0
0.0100
0.00787
0.00649
0.00481
0.00380
-4.610
-4.845
-5.038
-5.337
-5.573
100
127
154
208
263
NO2 (g) Æ NO (g) + ½ O2 (g) at 300 oC Page 540
Is the reaction first or second order in NO2 ?
What is the rate constant for the reaction?
Since graph (b) gives a straight line, it is a second-order reaction in NO2 .
i.e. rate = k[NO2]2 .
And the slope is k, where k = 0.534 M -1 s -1 .
9
Similar problem based on reaction
2 C2F4 Æ C4F8
Similar problem based on reaction
C2F4 Æ 1/2 C4F8
which we are told is 2nd order in the reactant,
with a rate constant of 0.0448 M -1 s-1
or 0.0448 L mol -1 s -1 at 450 K.
If the initial concentration is 0.100 M, what
will be the concentration after 205 s ?
General Order of reaction
First Order reactions
Second Order reactions
Integrated form of each
1/Ct = 1/Co + kt
= 1/(0.100 M) + (0.0448 M -1 s-1 )(205 s)
Half lives of each
= 19.2 M -1 or Ct = 5.21 x 10 -2 M
10
Week Five
14.3
14.4
14.5
Chemical Kinetics (cont)
The Change of Concentration with Time
First-Order Reactions
Half-Life
Second-Order Reactions
Temperature and Rate
The Collision Model
Activation Energy
The Orientation Factor
The Arrhenius Equation
Reaction Mechanisms
Elementary Steps
Multistep Mechanisms
Rate Laws of Elementary Steps
Rate Laws of Multistep Mechanisms
Mechanisms with and Initial Fast Step
11