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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 crenation (shrivels).

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slide1

Osmotic

Pressure:

a fascinating

behavior.

Yet it is the

result of a

very simple

tendency to

equalize the

concentrations

of solutions.

slide5

Hypertonic soln

  • crenation

(shrivels)

  • Hypotonic soln
  • hemolysis

(bursts)

slide6

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).
slide7

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

slide8

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

slide9

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?

slide19

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.
slide21

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.
slide23

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.
slide25

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
slide27

Chapter 14 Chemical Kinetics

14.1 Factors that Affect Reaction Rates

14.2 Reaction Rates

Changes of Rate with Time

Reaction Rates and Stoichiometry

14.3 Concentration and Rate

Exponents in the Rate Law

Units of Rate Constants

Using Initial Rates to Determine Rate Laws

14.4 The Change of Concentration with Time

First-Order Reactions

Second-Order Reactions

Half-Life

14.5 Temperature and Rate

14.5 Reaction Mechanisms

14.7 Catalysis

slide32

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

It’s convenient to define the rate as

And, in general for

aA + bB  cC + dD

slide33

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 ?

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

slide34

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

slide35

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 0.25 1.36 x 10-3

184 2.08

135 0.17 1.26 x 10-3

319 1.91

207 0.24 1.16 x 10-3

526 1.67

341 0.32 0.94 x 10-3

867 1.35

331 0.24 0.72 x 10-3

1198 1.11

679 0.39 0.57 x 10-3

1877 0.72

slide36

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

slide37

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

{N2O5] [N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t Avg rate Avg rate

mol/L avg mol/L mol/L-min /[N2O5]av /[N2O5]av2

2.33

2.21 0.25 1.36 x 10-36.2 x 10-4 2.8 x 10-4

2.08

2.00 0.17 1.26 x 10-36.3 x 10-4 3.2 x 10-4

1.91

1.79 0.24 1.16 x 10-36.5 x 10-4 3.6 x 10-4

1.67

1.51 0.32 0.94 x 10-36.2 x 10-4 4.1 x 10-4

1.35

1.23 0.24 0.72 x 10-35.9 x 10-4 4.8 x 10-4

1.11

0.92 0.39 0.57 x 10-36.2 x 10-4 6.7 x 10-4

0.72

Notice the nice constant value!!!

slide38

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 the

N2O5, in accordance with the rule we developed earlier.

slide39

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!!!

slide40

Other reactions and their observed reaction orders

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]

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.

slide41

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

slide43

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 need

to talk with

Dr. Mathews

slide44

Chapter 14 Chemical Kinetics

14.1 Factors that Affect Reaction Rates

14.2 Reaction Rates

Changes of Rate with Time

Reaction Rates and Stoichiometry

14.3 Concentration and Rate

Exponents in the Rate Law

Units of Rate Constants

Using Initial Rates to Determine Rate Laws

14.4 The Change of Concentration with Time

First-Order Reactions

Second-Order Reactions

Half-Life

14.5 Temperature and Rate

14.5 Reaction Mechanisms

14.7 Catalysis

slide45

Consider First-Order Reactions

To give

these forms

of the

“integrated

rate law”:

slide47

The Change of Concentration with Time

Half-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,
slide48

The Change of Concentration with Time

For a First-Order Reaction

The identical length of the

first and second half-life

is a SPECIFIC characteristic

of First-Order reactions

slide50

Second-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.

slide51

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

= 5.0 sec

t1/2 = 0.693/0.4 = 1.73 sec

t1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1

= 2.5 sec

slide52

Recall for

1st order:

And for 2nd order:

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

slide53

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 -4.610 100

50.0 0.00787 -4.845 127

100.0 0.00649 -5.038 154

200.0 0.00481 -5.337 208

300.0 0.00380 -5.573 263

Is the reaction first or second order in NO2 ?

What is the rate constant for the reaction?

slide54

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 .

slide57

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

slide58

General Order of reaction

First Order reactions

Second Order reactions

Integrated form of each

Half lives of each

slide62

Week Five Chemical Kinetics (cont)

14.3 The Change of Concentration with Time

First-Order Reactions

Half-Life

Second-Order Reactions

14.4 Temperature and Rate

The Collision Model

Activation Energy

The Orientation Factor

The Arrhenius Equation

14.5 Reaction Mechanisms

Elementary Steps

Multistep Mechanisms

Rate Laws of Elementary Steps

Rate Laws of Multistep Mechanisms

Mechanisms with and Initial Fast Step