OsmoticPressure:a fascinatingbehavior.

Yet it is theresult of avery simpletendency toequalize theconcentrationsof solutions.

The critical part is the membrane!!!

A VERY practical application/consequenceof Osmotic Pressure.

Hypertonic soln

crenation (shrivels)

Hypotonic soln

hemolysis (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).

Osmotic Pressure π V = n R T

or π = (n/V) R T 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?

ColloidsColloids

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.

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 Chemical Kinetics

14.1 Factors that Affect Reaction Rates14.2 Reaction Rates

Changes of Rate with TimeReaction Rates and Stoichiometry

14.3 Concentration and RateExponents in the Rate LawUnits of Rate ConstantsUsing Initial Rates to Determine Rate Laws

14.4 The Change of Concentration with TimeFirst-Order ReactionsSecond-Order ReactionsHalf-Life

14.5 Temperature and Rate14.5 Reaction Mechanisms14.7 Catalysis

C4H9Cl(aq) + H2O (l) C4H9OH (aq) + HCl (aq)

t

OHHC

t

ClHCRate

][][ 9494

Note the signs!

In fact, the instantaneous rate corresponds to d[A]/dt

Consider the reaction 2 HI(g) H2(g) + I2(g)

It’s convenient to define the rate as

t

I

t

H

t

HIrate

][][][

2

1 22

And, in general for

aA + bB cC + dD

t

D

dt

C

ct

B

bt

A

aRate

][1][1][1][1

Sample exercise 14.2

The decomposition of N2O5 proceeds according to the equation

2 N2O5 (g) 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 ?

t

O

t

NO

t

ONRate

][

1

1][

4

1][

2

1 2252

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)

at T = 45 oC in carbon tetrachloride as a solvent

Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t min min mol/L mol/L mol/L-min

0 2.33

184 2.08

319 1.91

526 1.67

867 1.35

1198 1.11

1877 0.72

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t min min mol/L mol/L mol/L-min

0 2.33184 0.25 1.36 x 10-3

184 2.08135 0.17 1.26 x 10-3

319 1.91207 0.24 1.16 x 10-3

526 1.67341 0.32 0.94 x 10-3

867 1.35331 0.24 0.72 x 10-3

1198 1.11679 0.39 0.57 x 10-3

1877 0.72

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!

So let’s try something rather arbitrary at this point.

Let’s divide the instantaneous, average rate by

[N2O5] and/or [N2O5]2

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

{N2O5] [N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t Avg rate Avg ratemol/L avg mol/L mol/L-min /[N2O5]av /[N2O5]av

2

2.332.21 0.25 1.36 x 10-3 6.2 x 10-4 2.8 x 10-4

2.082.00 0.17 1.26 x 10-3 6.3 x 10-4 3.2 x 10-4

1.911.79 0.24 1.16 x 10-3 6.5 x 10-4 3.6 x 10-4

1.671.51 0.32 0.94 x 10-3 6.2 x 10-4 4.1 x 10-4

1.351.23 0.24 0.72 x 10-3 5.9 x 10-4 4.8 x 10-4

1.110.92 0.39 0.57 x 10-3 6.2 x 10-4 6.7 x 10-4

0.72Notice the nice

constant value!!!

It’s convenient to write this result in symbolic form:

Rate = k [N2O5]

where the value of k is about 6.2 x 10-4

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

In fact, we really should take into account the 2 in front of theN2O5, in accordance with the rule we developed earlier.

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

and (m + n)-th order overall.

Be careful—because these orders are NOT related necessarily

to the stoichiometry of the reaction!!!

2 N2O5 = 4 NO2 + O2 (g) Rate = k [N2O5] !!!

CHCl3 (g) + Cl2 (g) CCl4 (g) + HCl(g) Rate = k[CHCl3][Cl2]1/2

H2 (g) + I2 (g) 2 HI (g) Rate = k[H2][I2]

Other reactions and their observed reaction orders

The order must be determined experimentally!!!

We’ll see later that it depends on the Reaction Mechanism,rather than the overall stoichiometry.

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 = - (Δ[N2O5] / Δ t = - d[N2O5] / dt = k [N2O5]

or, in general for A products

Rate = - Δ[A] / Δt = 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

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

BCDCC DBCEB CDBBE CAACA AEEDD ECBBC EBC

These people needto talk with

Dr. Mathews

Chapter 14 Chemical Kinetics

14.1 Factors that Affect Reaction Rates14.2 Reaction Rates

Changes of Rate with TimeReaction Rates and Stoichiometry

14.3 Concentration and RateExponents in the Rate LawUnits of Rate ConstantsUsing Initial Rates to Determine Rate Laws

14.4 The Change of Concentration with TimeFirst-Order ReactionsSecond-Order ReactionsHalf-Life

14.5 Temperature and Rate14.5 Reaction Mechanisms14.7 Catalysis

Consider First-Order Reactions

][][][

Akdt

Ad

t

ARate

tA

Adtk

A

Adt0

][

][ 0 ][

][

ktA

Aor

ktAAor

AktA

t

t

t

0

0

0

][

][ln

]ln[]ln[

]ln[]ln[To givethese formsof the“integrated rate law”:

An example of the plots of concentration vs timefor a First-Order Reaction

The Change of Concentration with TimeThe Change of Concentration with TimeHalf-LifeHalf-Life• 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,

kkt

693.0ln21

21

The Change of Concentration with TimeThe Change of Concentration with TimeFor a First-Order ReactionFor a First-Order Reaction

The identical length of thefirst and second half-life

is a SPECIFIC characteristicof First-Order reactions

Consider now Second-Order Reactions

2][][][

Akdt

Ad

t

ARate

tA

Adtk

A

Adt0

][

][ 20 ][

][

0][

1

][

1

Akt

A t

Second-Order ReactionsSecond-Order Reactions• We can show that the half life

• 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.

0A1

21

kt

t1/2 = 0.693/0.4 =

1.73 sect1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1

= 2.5 sec

t1/2 = [(0.4)(0.5)] -1

= 5.0 sec

Is this first or second order in the reactant? What is k?

Recall for 1st order:

kkt

693.0ln21

21

And for 2nd order:

0A1

21

kt

Example Exercise 14.8

NO2 (g) NO (g) + ½ O2 (g) at 300 oC Page 540

Time/s [NO2] ln[NO2] 1/[NO2]

0.0 0.0100 -4.610 10050.0 0.00787 -4.845 127100.0 0.00649 -5.038 154200.0 0.00481 -5.337 208300.0 0.00380 -5.573 263

Is the reaction first or second order in NO2 ?What is the rate constant for the reaction?

Example of Second-Order Plots of conc vs time

NO2 (g) NO (g) + ½ O2 (g) at 300 oC Page 540

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 .

Similar problem based on reaction2 C2F4 C4F8

Similar problem based on reaction C2F4 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 ?

1/Ct = 1/Co + kt

= 1/(0.100 M) + (0.0448 M -1 s-1 )(205 s)

= 19.2 M -1 or Ct = 5.21 x 10 -2 M

General Order of reaction

First Order reactions

Second Order reactions

Integrated form of each

Half lives of each

14.3 The Change of Concentration with TimeFirst-Order ReactionsHalf-LifeSecond-Order Reactions

14.4 Temperature and RateThe Collision ModelActivation EnergyThe Orientation FactorThe Arrhenius Equation

14.5 Reaction MechanismsElementary StepsMultistep MechanismsRate Laws of Elementary StepsRate Laws of Multistep MechanismsMechanisms with and Initial Fast Step

Week Five Chemical Kinetics (cont)