reaction rates ppt

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Unit 3

Chemical Kinetics andChemical Equilibrium

Reaction Rates

Rate Laws

First and Second Order Reactions

Chemical Equilibrium

Equilibrium Constants

LeChatelier¶s Principle



Reaction Rates

Questions to consider:

What makes ³superglue´ bond instantly

while Elmer¶s glue does not?

What factors determine how quickly foodspoils?

Why do ³glow sticks´ last longer when

stored in the freezer?

How do catalytic converters removevarious pollutants from car exhaust?



Reaction Rates

These types of questions can be answered

using chemical kinetics.

The study of the speed or rate at which

chemical reactions occur



Reaction Rates

The rate of a chemical reaction is affected

by many factors, including:

concentration of reactantsas concentration of reactants

increases the rate of reaction generally

increases



Reaction Rates


by many factors (cont):

reaction temperature food spoils more quickly at room

temperature than in a refrigerator

bacteria grow faster at RT than at

lower temperatures



Reaction Rates



presence of a catalyst

a substance that increases the rate of a reaction without being consumed in

the reaction

Enzymes

biological catalysts proteins that increase the rate of

biochemical reactions



Reaction Rates



surface area of solid or liquid reactantsor catalysts

as surface area increases the rate of

reaction generally increases



Reaction Rates

The speed of an object or event is the

change that occurs in a given time interval.

Speed of a car = change in distancetime interval

= (d

(t

Remember, the term change always refers to

final value minus initial value.



Reaction Rates

Similarly, the rate (or speed) of a reaction can

be determined:

Rate = change in concentration (or moles) of producttime interval

Rate = ( (conc. or moles)

(t



Reaction Rates

Consider the chemical reaction:

A B

Time = 0.

10. mol A

t = 20. min

5.0 mol A

5.0 mol B

t = 40. min

2.0 mol A

8.0 mol B



Reaction Rates

The reaction rate for a chemical reaction

can be expressed as either:

the increase in concentration (or number of moles) of a product as a function of

time.

the decrease in concentration (or number of moles) of a reactant as a function of

time

OR



Reaction Rates

In this reaction:

Average rate of

appearance of B = change in # of moles of Bchange in time

= ( (mol B)

( t

We can calculate the average rate for any

time interval involved in the reaction.



Reaction Rates

If we consider the rate of appearance of B

over the first 20 minutes of reaction:

Average rate of appearance of B = ( (mol B)

( t

= 5.0 mol B ± 0.0 mol B20. min ± 0. min

= 0.25 mol/min



Reaction Rates

The average rate of appearance of B during

the second 20 minutes of the reaction:

Avg. rate = 8.0 mol B ± 5.0 mol B

40. min ± 20. min

= 0.15 mol/min

Notice that the average rate of reaction

decreases over the course of the reaction.



Reaction Rates

The rate of a reaction can also be

expressed as the disappearance of A as a

function of time.

For this particular reaction, when 1 mole of

B is formed, 1 mole of A must disappear.

A B



Reaction Rates

Time (B (A

Interval (t (t

0 ± 20.0 min 0.25 mol

min

20.0 ± 40.0 min 0.15 mol

min

Notice: (B/(t = - (A/(t



Reaction Rates

We don¶t want to report two different values

for the rate of the reaction.

For reactions with 1:1 stoichiometry:

Avg. rate = ( (moles product)

( t

= - ( (moles reactant)

( t



Reaction Rates

For most reactions, the reaction rate is

expressed as a change in concentration of

a particular reactant or product

Average Rate = ( [Product] = - ( [Reactant]

( t ( t

where [Product] = concentration of product[Reactant] = concentration of reactant

units: M / sec or M / min

M = molarity = moles/liter



Reaction Rates

On the exam, you will be expected to find

the average rate of reaction for a specific

time interval when given the concentration

or number of moles of either reactants or

products as a function of time.



Reaction Rates

Example: Given the following data, what is

the average rate of the following reaction over

the time interval from 54.0 min to 215.0 min?

CH3OH (aq) + HCl (aq) CH3Cl (aq) + H2O (l)

Time (min) [HCl] (M)

0.0 1.85

54.0 1.58

107.0 1.36

215.0 1.02



Reaction Rates

Given: [HCl]54 min = 1.58 M

[HCl]215 min = 1.02 M

Find: avg. rate of disappearance of HCl

Avg. rate = - ( [HCl]

( t

= - (1.02 M - 1.58 M)215 min - 54 min

= 0.0035 M / min



Reaction Rates

Example: Calculate the average reaction ratefor the reaction A B during the first 60.0minutes using the following data:

Time [A]

0.0 min 1.50 M20.0 min 1.00 M

40.0 min 0.80 M60.0 min 0.75 M80.0 min 0.70 M



Reaction Rates

So far, all reactions have had a one-to-one

stoichiometry.

What happens when the coefficients arenot all 1?

2 A 3B



Reaction Rates

Consider the following reaction:

2 HI (g) H2 (g) + I2 (g)

Time mol mol mol

(min) HI H2 I2

0.0 2.00 0.0 0.0

10.0 1.50

20.0 1.00

30.0 0.75



Reaction Rates

Calculate the change in HI and H2 as a functionof time for the first 20.0 minutes of reaction:

2 HI (g) H2 (g) + I2 (g)

(HI (H2

Time mol mol mol (t (t

(min) HI H2 I2 (mol/min)

0.0 2.00 0.0 0.0

10.0 1.50 0.25 0.25

20.0 1.00 0.50 0.50

30.0 0.75 0.75 0.75



Reaction Rates

The average reaction rate must be

numerically the same, regardless of whether

you express it as the rate of appearance of

product or the rate of disappearance of

reactant.

HI disappears twice as fast as H2 appears. To

make the rates equal:

Rate = - 1 ( [HI] = ( [H2]

2 (t ( t



Reaction Rates

In general, for a reaction:

a A + b B c C + d D

the rate of the reaction can be found by:

Rate = - 1 ([A] = - 1 ([B] = 1 ([C] = 1 ([D]

a (t b (t c (t d (t



Reaction Rates

Rate = - 1 ([A] = - 1 ([B] = 1 ([C] = 1 ([D]

a (t b (t c (t d (t

This equation can be used to establish therelationship between rate of change of one

reactant or product to another reactant or

product.

You have to be able to do this on the

test, too!



Reaction Rates

Example: How is the rate of disappearance of

N2O5 related to the rate of appearance of NO2

in the following reaction?

2 N2O5 (g) 4 NO2 (g) + O2 (g)



Reaction Rates

Example: If the rate of decomposition of N2O5

in the previous example at a particular instant

is 4.2 x 10-7M /s, what is the rate of

appearance of NO2

?

2 N2O5 (g) 4 NO2 (g) + O2 (g)

Given: - ([N2O5] = 4.2 x 10-7 M /s

( t



Reaction Rates

2 N2O5 (g) 4 NO2 (g) + O2 (g)

Rate = - 1 ([N2O5] = 1 ([NO2]

2(

t 4(

t

So:

([NO2] = - 4

([N2O5]

( t 2 ( t

= 2 x 4.2 x 10-7 M /s = 8.4 x 10-7 M/s



Reaction Rates

Recall that the average reaction rate

changes during the course of the reaction.

Until now, we have calculated average

reaction rates.

The reaction rate at a particular time (not

time interval) is called the instantaneous

reaction rate.



Reaction Rate

The instantaneous reaction rate is found by

determining the slope of a line tangent to

the curve at the particular time of interest.

Fortunately (for you), you won¶t have to do

this on the exam or HW!



Rate Laws

Consider the data presented earlier for the

disappearance of HCl as a function of time

for the following reaction.


Time (min) [HCl] (M)

0.0 1.8554.0 1.58

107.0 1.36

215.0 1.02



Rate Laws

The average reaction rate decreases with

time.

The reaction slows down as the

concentration of reactants decreases.


Time (min) [HCl] (M) Avg. Rate (M /min)

0.0 1.8554.0 1.58 0.0050

107.0 1.36 0.0042

215.0 1.02 0.0031



Rate Laws

In general, the rate of any reaction depends

on the concentration of reactants.

The way in which the reaction rate varieswith the concentration of the reactants can

be expressed mathematically using a rate

law.

An equation that shows how the reactionrate depends on the concentration of the

reactants



Rate Laws

For a generalized chemical reaction:

w A + x B y C + z D

the general form of the rate law is:

Rate = k[A]m [B]n

where k = rate constant

m, n = reaction order



Rate Laws

Rate Constant (k) a proportionality constant that relates the

concentration of reactants to the reaction

rate

Reaction Order the power to which the concentration of

a reactant is raised in a rate law

Overall reaction order The sum of all individual reaction orders



Rate Laws

Rate laws must be determined

experimentally.

Measure the instantaneous reaction rate

at the start of the reaction (i.e. at t = 0) for various concentrations of reactants.

You CANNOT determine the rate law by

looking at the coefficients in the balancedchemical equation!



Rate Laws

First Order Reaction

Overall reaction order = 1

Rate = k[A]

Expt [A] (M) Rate (M/s)

1 0.50 1.00

2 1.00 2.00

3 2.00 4.00



Rate Laws

Second Order Reaction


Rate = k[A]2


1 0.50 0.50

2 1.00 2.00

3 1.50 4.50



Rate Laws

Third Order Reaction


Rate = k[A]3


1 0.50 0.25

2 1.00 2.00

3 1.50 6.75



Rate Laws

Zero Order Reaction


Rate = k[A]0 = k


1 0.50 2.00

2 1.00 2.00

3 1.50 2.00



Rate Laws

REMEMBER

Rate laws must be determined

experimentally.

Determine the instantaneous reaction

rate at the start of the reaction (i.e. at t =

0) for various concentrations of

reactants.

You CANNOT determine the rate law by

looking at the coefficients in the balanced

chemical equation!



Rate Laws

To determine the rate law from

experimental data,

identify two experiments in which the

concentration of one reactant has beenchanged while the concentration of the

other reactant(s) has been held constant

determine how the reaction rate changedin response to the change in the

concentration of that reactant.



Rate Laws

To determine the rate law from

experimental data (cont)

Repeat this process using another set of

data in which the concentration of thefirst reactant is held constant while the

concentration of the other one is

changed.



Rate Laws

Example: The initial reaction rate of the

reaction A + B C was measured for

several different starting concentration of A

and B. The following results were obtained.Determine the rate law for the reaction.

Expt # [A] (M) [B] (M) Initial rate (M /s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 8.0 x 10-5

3 0.200 0.100 16.0 x 10-5



Rate Laws

Rate = k [A]m [B]n

Compare experiments 1 and 2 to find n:

[A] = constant[B] = doubles

Expt # [A] (M) [B] (M) Initial rate (M /s)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 8.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Rate doubles:8.0 x 10-5 = 2.0

4.0 x 10-5

x 2x 2 x 2x 2

[2]n = 2.0n = 1

Rate = k[A]m[B]1



Rate Laws

Rate = k [A]m [B]n

Compare experiments 1 and 3 to find m:

[A] = doubles[B] = constant

Expt # [A] (M) [B] (M) Initial rate (M /s)1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 8.0 x 10-5

3 0.200 0.100 16.0 x 10-5

Rate quadruples:16.0 x 10-5 = 4.0

4.0 x 10-5

x 2x 2 x 4x 4

[2]m = 4.0n = 2

Rate = k[A]2[B]



Rate Laws

You can also solve this using algebra:

Rate = k [A]m [B]n

Compare experiments 1 and 3 to find m:

Rate 2 = k [0.200 M]m [0.100 M]n = 16.0 x 10-5 =4.0

Rate 1 k [0.100 M]m [0.100 M]n 4.0 x 10-5

[0.200 M]m = 4.0

[0.100 M]m

2m = 4.0 only if m = 2

[2.00]m = 2.0

Rate = k[A]2[B]n



Rate Laws

You can also solve this using algebra:

Rate = k [A]m [B]n

Compare experiments 1 and 2 to find n:

Rate 2 = k [0.100 M]m [0.200 M]n = 8.0 x 10-5 = 2.0

Rate 1 k [0.100 M]m [0.100 M]n 4.0 x 10-5

[0.200 M]n = 2.0

[0.100 M]n

2n = 2.0 only if n = 1

[2.00]n = 2.0

Rate = k[A]2[B]

reaction rates ppt

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