Chemistry 223 Equilibrium Lecture

8/20/2019 Chemistry 223 Equilibrium Lecture

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General Chemistry III

CH 223Prof. Thomas Greenbowe

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• CH 223 is the third course in the 3-term GeneralChemistry sequence.

• Math 111 is a pre-requisite for CH 222

• Math 112 is a pre-requisite for CH 223

• Concurrent registration in the General ChemistryLaboratory course, CH 229, is recommended butnot required.

• !"##$ &''() *+,-. eFCC 4- f ACFDC 4-6 a$$-A=> N@;


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All course information is posted on CanvasSyllabus **download the syllabus**

Announcements

Lecture scheduleLecture notes, exam scores, and more

http://canvas.uoregon.edu


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Required Course Materials

•

Textbook: Chemistry with ConnectPlus, Custom, byMartin Silberberg; McGraw-Hill. Same textbook used inCH221, CH222

•

Connect access code: comes with custom book, orpurchase on-line. ConnectPlus is not required.

• i>Clicker or i>Clicker 2

• Register i-clicker through Canvas by tomorrow

• approved non-graphing scientific calculator


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•

Hour exam 1, Wed. Aug. 26, 9:00 am 25%

• Hour exam 2, Wed. Sept. 2, 9:00 am 25%

•

Final exam, Thurs. Sept. 11, 9:00 am 30%•

Connect On-line homework 10%

• Clicker questions 5%

•

In class activities/paper HW 5%

Mark the dates for the exams on your calendar now!

AssessmentUsing a weighted average


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LearnSmart

•

LearnSmart is an adaptive learning system

designed to help you learn faster, study more

efficiently, and retain more knowledge for

greater success in this course.

•

LearnSmart focuses primarily on conceptual

understanding of the course material.

Homework assignments give you a chance to

apply your understanding to chemicalproblems.


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• LearnSmart focuses primarily on conceptual

understanding of the course material. (Homeworkassignments give you a chance to apply your

understanding to chemical problems.)

•

If it works . . .


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Access to Connect HW and LearnSmart

•

go directly to the Connect site for this course. The linkwill be posted on Blackboard later today.


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Greenbowe [email protected] 177 Onyx

Office Hours:• Tuesdays: 11:00 –11:50 p.m. 177 Onyx or 160E Klamath

• Thursdays: 11:00 –11:50 p.m. 177 Onyx or 160E Klamath

•

Or by appointment Or, drop in most afternoons


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Course GTF

Brandon Schabes

Office Hours:• TBA


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Any questions about the syllabus, or

what to expect this term?


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Gases in an Equilibrium State

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D namic Eq ilibri m


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Dynamic Equilibrium

•

Dynamic equilibrium is the condition wherein the rates ofthe forward and reverse reactions are equal.

• Once the reaction reaches equilibrium, the concentrations

of all the chemicals remain constant because thechemicals are being consumed and produced at thesame rate.

•

A system at equilibrium does NOT mean equal amounts ofreactants and products.


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The Concept of Dynamic Equilibrium

Chemical equilibrium occurs when a reaction

and its reverse reaction proceed at the same rate.


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Reaching equilibrium on the macroscopic and

molecular levels.


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The Equilibrium State

All reactions are reversible and under suitable conditionswill reach a state of equilibrium.

At equilibrium, the concentrations of products and reactantsno longer change because the rates of the forward and

reverse reactions are equal.At equilibrium: rateforward = ratereverse

Chemical equilibrium is a dynamic state because reactions

continue to occur, but because they occur at the same rate,no net change is observed on the macroscopic level.

Reaction equations are written with a double arrow to indicatethat a state of equilibrium exists.


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Table 16.7 Rate Laws for General Elementary Steps

Elementary Step Molecularity Rate Law

A product

2A product

A + B product

2A + B product

Unimolecular

Bimolecular

Bimolecular

Termolecular

Rate = k [A]

Rate = k [A]2

Rate = k [A][B]

Rate = k [A]2[B]

For elementary steps only , the reactioncoefficients become the exponents in the rate law.

This is because the elementary reactions tell us

about the actual collisions that take place.


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An Equilibrium System

At equilibrium, both the forward and reverse

reactions occur at the same rate, we write

the equation with a double arrow to indicate

an equilibrium system:

m=dP]1^ =md=]1^


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Given the reaction as an elementary step:

N2O

4(g ) ! 2NO

2(g )

Write the rate laws for the forward and reverse

reactions.

N2O

4! 2NO

2 : rate

f = k f N2O4!" #$

2NO2 % N

2O

4 : rate

r = k

r NO

2!" #$

2

Where k = rate constant and [ ] indicates themolar concentration.


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At equilibrium, rate forward = rate reverse.

k f N2O4!" #$ = kr NO2!" #$2

k f

kr=

NO2

!" #$2

eq

N2O

4!" #$eq

Under these circumstances, the equilibrium constant

for the reversible reaction may be written as follows:

K c =

NO2

!" #$2

eq

N2O

4!" #$

eq


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The Equilibrium Constant, 2 'X

The ratio of the rate constants is a constantat that temperature, and the expression

becomes

2 'X x# 3 # 4

pmd=q= pm=dPq

x

! 'Q R ST8#U

#

ST#8>U


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Initial and Equilibrium Concentration Ratios for

the N2O4-NO2 System at 200°C (473 K)

Initial Equilibrium

Expt [N2O4] [NO2] Q , [N2O4]eq [NO2]eq K =

1 0.1000 0.0000 0.0000 0.00357 0.193

2 0.0000 0.1000 ! 0.000924 0.0982

3 0.0500 0.0500 0.0500 0.00204 0.146

4 0.0750 0.0250 0.0833 0.00277 0.170

[NO2]2

[N2O4]

;,$$/' 4# 'S9'1.-'#0;4&+3&40' "'X/

2

eq[NO2]

[N2O4]eq


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Table 17.1 Initial and Equilibrium Concentration Ratios for

the N2O4-NO2 System at 200°C (473 K)

Initial Equilibrium

Expt [N2O4] [NO2] Q , [N2O4]eq [NO2]eq K=

1 0.1000 0.0000 0.0000 0.00357 0.193 10.4

2 0.0000 0.1000 ! 0.000924 0.0982 10.4

3 0.0500 0.0500 0.0500 0.00204 0.146 10.4

4 0.0750 0.0250 0.0833 0.00277 0.170 10.4

[NO2]2

[N2O4]

2

eq[NO2]

[N2O4]eq


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The Equilibrium Constant

• Consider the generalized reaction

k

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If we start with N2O4(g) in a sealed bottle and raise the temperature:

MX3.&.71.3-


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K eq and the extent of reaction

K reflects a particular ratio of product concentrations toreactant concentrations for a reaction.

A small value for K indicates that the reaction yields littleproduct before reaching equilibrium. The reaction favors

the reactants.

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A large value for K indicates that the reaction reachesequilibrium with very little reactant remaining. The

reaction favors the products.


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,V90V 31 (V' 13223497M D')( 6')0E9D') @ E'@0W37 @(

'Q5929DE95&X

a) Reactants are being consumed and productsare being formed.

b)

The rate of the reaction is zero.

c) The rate of the forward and reverse reactionsare equal.

d) Only products are present.

e) The rate of the forward reaction is greater thanthe rate of the reverse reaction.


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,V90V 31 (V' 13223497M D')( 6')0E9D') @ E'@0W37 @(

'Q5929DE95&X

a) Reactants are being consumed and productsare being formed.

b)

The rate of the reaction is zero.

c) The rate of the forward and reverse reactionsare equal.

d) Only products are present.

e) The rate of the forward reaction is greater thanthe rate of the reverse reaction.


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The Equilibrium Constant Expression

•

Consider the generalized reaction

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Write the equilibrium constant expression for the followingreaction:

CH 4( g )+2 H 2S ( g )!CS 2( g )+4 H 2( g )


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A. K c =

[CH 4][ H

2S ]

2

[CS 2][ H

2]4

C. K c =

[CS 2] [ H

2S ]

4

[CH 4] [ H

2]2

D. K c =

[CS 24] [ H

2]4

[CH 4] [ H

2S ]

2B. K

c =

[CH 4] [CS

2]

[ H 2S ]

2[ H

2]4

CH 4( g )+2 H 2S ( g )!CS 2( g )+4 H 2( g )


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A. K c =

[CH 4][ H

2S ]

2

[CS 2][ H

2]4

C. K c =

[CS 2] [ H

2S ]

4

[CH 4] [ H

2]2

D. K c =

[CS 24] [ H

2]4

[CH 4] [ H

2S ]

2B. K

c =

[CH 4] [CS

2]

[ H 2S ]

2[ H

2]4

CH 4( g )+2 H 2S ( g )!CS 2( g )+4 H 2( g )


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K eq and the extent of reaction

K reflects a particular ratio of product concentrations toreactant concentrations for a reaction.

A small value for K indicates that the reaction yields littleproduct before reaching equilibrium. The reaction favors

the reactants.

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0'-9'14031'L

A large value for K indicates that the reaction reachesequilibrium with very little reactant remaining. The

reaction favors the products.


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CH3COOH(aq) + H2O(l) ! H3O+

(aq) + CH3COO –

(aq) Keq = 1.8 x 10

–5 at 25ºC

How far does the reaction proceed?Is K eq large or small?

Given a solution of acetic acid, at equilibrium is there

more CH3COOH or more CH3COO – ?


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CH3COOH(aq) + H2O(l) H3O+

(aq) + CH3COO –

(aq) Keq = 1.8 x 10

–5 at 25ºC

How far does the reaction proceed?

• Small value implies that equilibrium favors the

reactants. In 0.10 M CH3COOH(aq), 99% of acetic

acid is present as CH3COOH molecules.

•

Acetic acid is a weak acid because it is only

partially dissociated.

Th E ilib i C t t t ll th


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The Equilibrium Constant tells us the

extent of the reaction

]40 DCC "^ 2 + x PL= S ACWPK

=<=

]1^ i d=

]1^=<=

d]1^

]40 DCC "^ 2 + x =LP S ACPB

=


58/75

Which of the following best describes a system at

equilibrium with a Kc of 5 x 104

? A + B!

C + D

A. The system is nearing completion and will end soon

B. The reaction proceeds hardly at all, mostly reactants

presentC. Appreciable concentrations of both reactants and

products are presentD.

There are mostly products present with some

reactants also presentE.

The is no way to know from the information given


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Which of the following best describes a system at

equilibrium with a Kc of 5 x 104

? A + B!

C + D

A. The system is nearing completion and will end soon

B. The reaction proceeds hardly at all, mostly reactants

presentC. Appreciable concentrations of both reactants and

products are presentD. There are mostly products present with some

reactants also presentE.

The is no way to know from the information given

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


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2(g) 2(g) (g)

At time 0 s, there are only reactants in the mixture.Is this system at equilibrium?

[H2] = 8, [I2] = 8, [HI] = 0

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


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2(g) 2(g) (g)

[H2] = 6, [I2] = 6, [HI] = 4

At time 16 s, there are both reactants and products in themixture. Are both the forward reaction and reverse reaction

taking place? Is this system at equilibrium?

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


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2(g) 2(g) (g)

At time 32 s, do we have more products or more reactantsin the mixture? Which is now faster the forward reaction or

the reverse reaction? Is this system at equilibrium?

[H2] = 4, [I2] = 4, [HI] = 8

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


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2(g) 2(g) (g)

As the concentration of product increases and theconcentrations of reactants decrease, the rate of the forward

reaction slows down, and the rate of the reverse reaction speedsup.

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


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2(g) 2(g) (g)

At time 48 s, the amounts of products and reactants in themixture haven’t changed. Are the forward and reverse

reactions are proceeding at the same rate? Has the system

reached equilibrium?

[H2] = 4, [I2] = 4, [HI] = 8

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


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2(g) 2(g) (g)

Calculate Keq

[H2] = 4, [I2] = 4, [HI] = 8

@//3-' [H2

] = P ?6 '0+L

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


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2(g) 2(g) (g)

Keq =

[H2] = 4, [I2] = 4, [HI] = 8

p


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2(g) 2(g) (g) At dynamic equilibrium, the rate of the forward reaction is equalto the rate of the reverse reaction.

The concentrations of reactants and products no longer change.

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


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2(g) 2(g) (g) At dynamic equilibrium, the rate of the forward reaction is equalto the rate of the reverse reaction.

The concentrations of reactants and products no longer change.

"34 63 4' 6'('E&97' 4V'7 @ )Y)('& 9) T8+ @( 'Q5929DE95&X

The Reaction Quotient, Q


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, Q

For aqueous phase or gas phase reactions NOT at

equilibrium

a A + bB! c C + d D

the reaction quotient is Q



70/75

Q

For the aqueous phase or gas phase reactions

NOT at equilibrium

a A + bB! c C + d D

the reaction quotient is QDefine quotient



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Q


NOT at equilibrium

a A + bB! c C + d D

the reaction quotient is QDefine quotient

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72/75

Q


NOT at equilibrium

a A + bB! c C + d D

the reaction quotient is QWho is Q?

What is Q?



73/75

Q


NOT at equilibrium

a A + bB! c C + d D

the reaction quotient is QWho is Q?

What is Q?What does Q have to do with Space &

Time?



74/75


a A + bB! c C + d D

the reaction quotient is:

J,' R4&3' $* Z6 %,'# +$-941'( 0$ "'X6 91'(.+0/ .* 0,' /2/0'- ./

4991$4+,.#) 'X3.&.71.3- $1 ./ 94/0 'X3.&.71.3-L

_# -40,'-48+/6 4 X3$8'#0 ]*1$- E48#F X3$8'#/^

./ 0,' 1'/3&0 $* (.R./.$#L

Reaction direction and the relative sizes of Q and K


75/75

Reaction direction and the relative sizes of Q and K .

Q < K

Q > K

Q = K

Chemistry 223 Equilibrium Lecture

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Transcript of Chemistry 223 Equilibrium Lecture