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Demo today ???. Colloids. 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.

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Demo today

Demo today ???


Demo today

Colloids


Demo today

  • 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.


Demo today

  • 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.


Demo today

Hydrophilic and Hydrophobic Colloids


Demo today

  • 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.


Demo today

  • 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


Demo today

Chapter 14 Chemical Kinetics

14.1Factors that Affect Reaction Rates

14.2Reaction Rates

Changes of Rate with Time

Reaction Rates and Stoichiometry

14.3Concentration and Rate

Exponents in the Rate Law

Units of Rate Constants

Using Initial Rates to Determine Rate Laws

14.4The Change of Concentration with Time

First-Order Reactions

Second-Order Reactions

Half-Life

14.5Temperature and Rate

14.5Reaction Mechanisms

14.7Catalysis


Demo today

Following the Progress of the Reaction A  B


Demo today

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

Note the signs!


Demo today

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


Demo today

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


Demo today

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 disappearance of N2O5 is 4.2 x 10-7 M/s

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


Demo today

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

Time∆t[N2O5] ∆[N2O5]- ∆[N2O5]/ ∆t

minmin mol/L mol/L mol/L-min

02.33

1842.08

3191.91

5261.67

8671.35

11981.11

18770.72


Demo today

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

Time∆t[N2O5] ∆[N2O5]- ∆[N2O5]/ ∆t

minmin mol/L mol/L mol/L-min

02.33

1840.251.36 x 10-3

1842.08

1350.171.26 x 10-3

3191.91

2070.241.16 x 10-3

5261.67

3410.320.94 x 10-3

8671.35

3310.240.72 x 10-3

11981.11

6790.390.57 x 10-3

18770.72


Demo today

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


Demo today

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

[N2O5][N2O5]av∆[N2O5]- ∆[N2O5]/ ∆tAvg rateAvg rate

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

2.33

2.210.251.36 x 10-36.2 x 10-42.8 x 10-4

2.08

2.000.171.26 x 10-36.3 x 10-43.2 x 10-4

1.91

1.790.241.16 x 10-36.5 x 10-4 3.6 x 10-4

1.67

1.510.320.94 x 10-36.2 x 10-4 4.1 x 10-4

1.35

1.230.240.72 x 10-35.9 x 10-4 4.8 x 10-4

1.11

0.920.390.57 x 10-36.2 x 10-4 6.7 x 10-4

0.72

Notice the nice constant value!!! which we’ll call ‘k’


Demo today

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.


Demo today

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!!!


Demo today

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.


Demo today

Let’s explore the results when 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 where [A]o = conc. at t = 0

orln [A]t = - kt + ln [A]o

orln ([A]t/[A]o = - kt

This is the expression of concentration vs time

for a First-Order Reaction


Demo today

Same information in a better format:

To give

these forms

of the

“integrated

rate law”:


Demo today

Consider First-Order Reactions

To give

these forms

of the

“integrated

rate law”:


Demo today

An example of the plots of concentration vs time

for a First-Order Reaction


Demo today

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,


Demo today

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


Demo today

Consider now Second-Order Reactions

Note

the

2 !


Demo today

Second-Order Reactions

  • We can show that the half life

  • A reaction also can have rate constant expression of the form

    rate = k[A][B],

    i.e., be second order overall, but be first order in A and in B.


Demo today

Recall for

1st order:

And for 2nd order:

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


Demo today

t1/2 = 1.73 sec

(‘2nd half-life’)

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

= 5.0 sec

(‘2nd half life’)

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

= 2.5 sec

(‘1st half-life’)

t1/2 = 0.693/0.4 = 1.73 sec


Demo today

MQ-1 122, WI 07:THE MEAN WAS 112.00 (64.00%)

THE MODE WAS 95.00 WITH 10 STUDENTS. High: 175

THE MEDIAN WAS 112.50 Low 38

THE STANDARD DEVIATION WAS 29.35

N = 472

THE POSSIBLE RANGE WAS FROM 0.00 TO 175.00

THE ACTUAL RANGE WAS FROM 38.00 TO 175.00 (21.7% TO 100.0%)

CUTS TAKEN AT 72.88 (APROX. 72.50) AND 26.86 (APROX. 27.50) PERCENTILES.

THE UPPER CUT CONTAINED 130 STUDENTS; THE LOWER, 130 STUDENTS.

TOTAL NUMBER OF DETECTED ERRORS: 0.

DISTRIBUTION OF SCORES

0.00 - 9.99 0

10.00 - 19.99 0

20.00 - 29.99 0

30.00 - 39.99 1

40.00 - 49.99 8

50.00 - 59.99 6

60.00 - 69.99 22

70.00 - 79.99 32

80.00 - 89.99 45

90.00 - 99.99 50

100.00 - 109.99 51

110.00 - 119.99 59

120.00 - 129.99 50

130.00 - 139.99 53

140.00 - 149.99 38

150.00 - 159.99 42

160.00 - 169.99 12

170.00 - 175.00 3

Correct answers:

BCDAE CCDDE CADBE BDABD AAADC ACDBA A

These students are in danger of failing!

Please See Me!!!


Demo today

Chapter 14 Chemical Kinetics

14.1Factors that Affect Reaction Rates

14.2Reaction Rates

Changes of Rate with Time

Reaction Rates and Stoichiometry

14.3Concentration and Rate

Exponents in the Rate Law

Units of Rate Constants

Using Initial Rates to Determine Rate Laws

14.4The Change of Concentration with Time

First-Order Reactions

Second-Order Reactions

Half-Life

14.5Temperature and Rate

14.5Reaction Mechanisms

14.7Catalysis


Demo today

First Order Reactions

Second Order Reactions


Demo today

Example Exercise 14.8

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

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

0.00.0100-4.610100

50.00.00787-4.845127

100.00.00649-5.038154

200.00.00481-5.337208

300.00.00380-5.573263

Is the reaction first or second order in NO2 ?

What is the rate constant for the reaction?


Second order processes

NO2(g)

NO (g) + 1/2 O2(g)

Second-Order Processes

The decomposition of NO2 at 300°C is described by the equation

and yields data comparable to this:


Second order processes1

Second-Order Processes

  • Graphing ln [NO2] vs.t yields:

  • The plot is not a straight line, so the process is not first-order in [A].


Second order processes2

Second-Order Processes

  • Graphing ln 1/[NO2] vs. t, however, gives this plot.

  • Because this is a straight line, the process is second-order in [A].


Demo today

Example of Second-Order Plots of conc vs time

NO2 (g)  NO (g) + ½ O2 (g) at 300 oC (See page 591)

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 .


Demo today

Similar problem based on reaction

2 C2F4 C4F8


Demo today

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


Demo today

General Order of reaction

First Order reactions

Second Order reactions

Integrated form of each

Half lives of each


Demo today

Take out a clean sheet of paper and print on it

your name and section number

and an indication it is “pop quiz # 2”.

Mon Rec, 2:30, 3:30

Josh Dettman115

120

Venuka Durani116

119

Xiangke Chen117

118

  • Wed Rec, 2:30, 3:30

  • Brian Culp109

  • 112

  • Jason Hoy110

  • 113

  • Luyuan Zhang111

  • 113


Demo today

Take out a clean sheet of paper and print on it

your name and section number

and an indication it is “pop quiz # 2”.

Tue Rec

Dan Poole122

Nick Selner123

Jim Bowsher124

Yu-Kay Law 129

Nadia Casillas 128

Thur Rec

Dan Poole126

Nick Selner127

Jim Bowsher125

Yu-Kay Law 121


Demo today

  • A certain reaction has the form A  B. At 400 K and

  • [A]0 = 2.8 x 10 -3 M, data for a plot of [A] vs t were collected.

  • It was then found that a plot of 1/[A] vs t yielded a

  • straight line with a slope of 3.60 x 10-2 L∙mol -1 ∙s -1 .

  • Write an expression for the rate law.

  • rate = k[A]2

  • b) Write an expression for the integrated rate law.

  • 1/[A] = 1/[A]0 + kt

  • c) What is the rate constant for the reaction?

  • 3.60 x 10-2 L∙mol -1 ∙s -1

  • d) What is the half-life for the reaction, as given?


Demo today

  • A certain reaction has the form A  B. At 400 K and

  • [A]0 = 5.6 x 10 -3 M, data for a plot of [A] vs t were collected.

  • It was then found that a plot of 1/[A] vs t yielded a

  • straight line with a slope of 1.80 x 10-2 L∙mol -1 ∙s -1 .

  • Write an expression for the rate law.

  • rate = k[A]2

  • b) Write an expression for the integrated rate law.

  • 1/[A] = 1/[A]0 + kt

  • c) What is the rate constant for the reaction?

  • 1.80 x 10-2 L∙mol -1 ∙s -1

  • d) What is the half-life for the reaction, as given?


Demo today

Remember this figure? Let’s use it for another point.

The ‘initial rate’

is very useful in

determining the

order of a reaction.

It is the only time

when there are no

products present!


Demo today

The initial rate of a reaction was measured for several different starting concentrations of A and B, and the results are as follows:

SAMPLE EXERCISE 14.6 Determining a Rate Law from Initial Rate Data

Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

Solution

Analyze: We are given a table of data that relates concentrations of reactants with initial rates of reaction and asked to determine

(a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table.

Plan: (a) We assume that the rate law has the following form: Rate = k[A]m[B]n, so we must use the given data to deduce the reaction orders m and n. We do so by determining how changes in the concentration change the rate. (b) Once we know m and n, we can use the rate law and one of the sets of data to determine the rate constant k. (c) Now that we know both the rate constant and the reaction orders, we can use the rate law with the given concentrations to calculate rate.

Solve: (a) As we move from experiment 1 to experiment 2, [A] is held constant and [B] is doubled. Thus, this pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0).


Demo today

SAMPLE EXERCISE 14.6 continued

In experiments 1 and 3, [B] is held constant so these data show how [A] affects rate. Holding [B] constant while doubling [A] increases the rate fourfold. This result indicates that rate is proportional to [A]2 (that is, the reaction is second order in A). Hence, the rate law is

This rate law could be reached in a more formal way by taking the ratio of the rates from two experiments:

Using the rate law, we have

2n equals 1 under only one condition:

We can deduce the value of m in a similar fashion

Using the rate law gives


Demo today

SAMPLE EXERCISE 14.6continued

Because 2m = 4, we conclude that

(b) Using the rate law and the data from experiment 1, we have

(c) Using the rate law from part (a) and the rate constant from part (b), we have

Because [B] is not part of the rate law, it is irrelevant to the rate, provided that there is at least some B present to react with A.

Check: A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see

if we can correctly calculate the rate. Using data from experiment 3, we have

Thus, the rate law correctly reproduces the data, giving both the correct number and the correct

units for the rate.


Demo today

SAMPLE EXERCISE 14.7 Using the Integrated First-Order Rate Law

The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1

at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0  10–7 g/cm3. Assume that the average temperature of the lake is 12°C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to drop to

3.0  10–7 g/cm3?

Solve: (a) Substituting the known quantities into Equation 14.13, we have

We use the ln function on a calculator to evaluate the second term on the right, giving

To obtain [insecticide]t = 1 yr, we use the inverse natural logarithm, or ex, function on the calculator:

Note that the concentration units for [A]t and [A]0 must be the same.

Solution

Analyze: We are given the rate constant for a reaction that obeys first-order kinetics, as well as information

about concentrations and times, and asked to calculate how much reactant (insecticide) remains after one year. We must also determine the time interval needed to reach a particular insecticide concentration. Because the exercise gives time in (a) and asks for time in (b), we know that the integrated rate law, Equation 14.13, is required.

Plan: (a) We are given k = 1.45 yr–1, t = 1.00 yr, and [insecticide]0 = 5.0  10–7 g/cm3, and so Equation 14.13

can be solved for 1n[insecticide]t. (b) We have k = 1.45yr–1, [insecticide]0 = 5.0  10–7 g/cm3, and [insecticide]t

= 3.0  10–7 g/cm3, and so we can solve Equation 14.13 for t.


Demo today

SAMPLE EXERCISE 14.7 continued

(b) Again substituting into Equation 14.13, with [insecticide]t = 3.0  10–7 g/cm3, gives

Solving for t gives

PRACTICE EXERCISE

The decomposition of dimethyl ether, (CH3)2O, at 510°C is a first-order process with a rate constant of

6.8  10–4s–1:

If the initial pressure of (CH3)2O is 135 torr, what is its partial pressure after 1420 s?

Check: In part (a) the concentration remaining after 1.00 yr (that is,1.2  10–7 g/cm3) is less than the original

concentration (5.0  10–7 g/cm3), as it should be. In (b) the given concentration (3.0  10–7 g/cm3) is greater

than that remaining after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is a

reasonable answer.

Answer: 51 torr


Demo today

SAMPLE EXERCISE 14.8 Determining Reaction Order from the Integrated Rate Law

The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300°C,

Is the reaction first or second order in NO2?

Solution

Analyze: We are given the concentrations of a reactant at various times during a reaction and asked to

determine whether the reaction is first or second order.

Plan: We can plot ln[NO2] and 1/[NO2] against time. One or the other will be linear, indicating whether the

reaction is first or second order.


Demo today

SAMPLE EXERCISE 14.8continued

Solve: In order to graph ln[NO2] and 1/[NO2] against time, we will first prepare the following table from

the data given:


Demo today

Figure 14.8 Kinetic data for decomposition of NO2. The reaction is NO2(g) NO(g)

+ 1/2O2(g), and the data were collected at 300°C. (a) A plot of [NO2] versus time is not linear, indicating that the reaction is not first order in NO2. (b) A plot of 1/[NO2] versus time is linear, indicating that the reaction is second order in NO2.

SAMPLE EXERCISE 14.8continued

As Figure 14.8 shows, only the plot of 1/[NO2] versus time is linear. Thus, the reaction obeys a second-order rate law: Rate = k[NO2]2. From the slope of this straight-line graph, we determine that k = 0.543 M–1 s–1 for the disappearance of NO2.

PRACTICE EXERCISE

Consider again the decomposition of NO2 discussed in the Sample Exercise. The reaction is second order in NO2 with k = 0.543 M–1s–1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining concentration after 0.500 h?

Answer: Using Equation 14.14, we find[NO2] = 1.00  10–3M


Demo today

Using ‘Initial Rates’ to determine order of reaction.


Demo today

Chemical Kinetics (cont)

14.3The Change of Concentration with Time

First-Order Reactions

Half-Life

Second-Order Reactions

14.4Temperature and Rate

The Collision Model

Activation Energy

The Orientation Factor

The Arrhenius Equation

14.5Reaction Mechanisms

Elementary Steps

Multistep Mechanisms

Rate Laws of Elementary Steps

Rate Laws of Multistep Mechanisms

Mechanisms with and Initial Fast Step


Demo today

Note the DRAMATIC effect of temperature on k


Demo today

Temperature and Rate

The Collision Model eg H2 + I2


Demo today

The Collision Model

  • The more molecules present, the greater the probability of collision and the faster the rate.

  • Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

  • The higher the temperature, the more energy available to the molecules and the faster the rate.

  • In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.


Demo today

Activation Energy

  • Arrhenius: molecules must posses a minimum amount of energy to react. Why?

    • In order to form products, bonds must be broken in the reactants.

    • Bond breakage requires energy.

  • Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.


Demo today

Activation Energy


Demo today

Activation Energy

  • Consider the rearrangement of acetonitrile:

    • In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.

    • The energy required for the above twist and break is the activation energy, Ea.

    • Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.


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