Brdy 6Ed Ch15 ChemicalEquilibrium

8/9/2019 Brdy 6Ed Ch15 ChemicalEquilibrium

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Chapter 15:Chemical Equilibrium

Chemistry: The Molecular Nature

of Matter, 6E

Jespersen/Brady/Hyslop



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 2

Dynamic Equilibrium in ChemicalSystems

Chemical equilibrium exists when Rates of forward and reverse reactions are equal

Reaction appears to stop

reactants! and proucts! don"t chan#e overtime

Remain constant

Both forward and reverse reaction never cease Equilibrium si#nified by double arrows $ %

or equal si#n $&%



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E '

Dynamic Equilibrium

N!"# ! N"!

(nitially forward reaction rapid

)s some reacts ↓N!"#! so rate forward ↓

(nitially Reverse reaction slow *o products

)s N"! forms

↑ Reverse rate

(ons collide more frequently as ions! ↑

Eventually rate for$ar & rate re%erse

Equilibrium



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E +

Dynamic Equilibrium

)lmost all systems come to equilibrium

,here equilibrium lies depends on system

-ome systems. equilibrium hard to detect

Essentially no reactants or no products present

(Fig. 15.1)



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

&eaction &e%ersibilityClose system

Equilibrium can bereached from eitherdirection

(ndependent of whetherit starts with 0reactants1or 0products1

'l$ays have the samecomposition atequilibrium under same

conditionsN

2O

4 2 NO

2




3or #iven overall system composition

)lways reach same equilibrium concentrations

,hether equilibrium is approached from for$ar or

re%erse direction

N!"# ! N"!

&eactants (rouctsEquilibrium


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Equilibrium

-imple relationship amon# reactants! and

proucts! for any chemical system atequilibrium

Called & mass action e)pression

5erived from thermodynamics 3orward reaction6 ' → * &ate + , f -'.

Reverse reaction6 ' ← * &ate + , r-*.

)t equilibrium6 ' * , f -'. + , r-*. rate for$ar + rate re%erse

rearran#in#6 [ B ]

[ A ]

=k f

k r

=constant

k f

k r



E): /!01 2 3 4!01 2 !/401 2 ##C

E)p’t7 4nitial 'mts Equil’m 'mts Equil’m-M .

4 89:: mol H2

:9222 mol H2

:9:222 ; H2

1 8 89:: mol (2

:9222 mol (2

:9:222 ; (2

:9:: mol H( 89 mol H( :98 ; H(

44 :9:: mol H2

:9': mol H2

:9:': ; H2

1 8 :98:: mol (2

:9+: mol (2

:9:+: ; (2

'9: mol H( 297: mol H( :927: ; H(


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E): /!01 2 3 4!01 2 !/401 2 ##C

E)p9t7

4nitial 'mts Equil9m 'mts

Equil9m-M .

444 :9:8: mol H2

:98: mol H2

:9:8: ; H2

1 8 :9:: mol (2 :98' mol (2 :9:8' ; (2

8924 mol H( 89:: mol H( :98:: ; H(

4 :9:: mol H2

:9++2 mol H2

:9:++2 ; H2

1 8 :9:: mol (2

:9++2 mol (2

:9:++2 ; (2

+9:: mol H( '988 mol H( :9'88 ; H(


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Mass 'ction E)pression 0M'E2 =ses stoichiometric coefficients as exponent

for each reactant 3or reaction6 a' 3 b* cC 3 D

&eaction quotient *umerical value of mass action expression

Equals 0;1 at any time> an

Equals 0< 1 only when reaction is ?nown to be atequilibrium

Q=[C ] c[ D ]d

[ A ]a [ B ]b



Mass 'ction E)pression

& same for all data sets at equilibriumQ=

[HI ]2

[ H 2 ][ I 2 ]

Q= [HI ]2

[ H 2

][ I 2

]

(0.156)2

(0.0222)(0.0222)=49.4

(0.280 )2

(0.0350)(0.0450 )=49.8

(0.100)2(0.0150)(0.0135)

=49.4

(0.311)2

(0.0442)(0.0442)=49.5

Equilibrium Concentrations0M2

E)p’t -/!

. -4!

. -/4.

4 :9:222 :9:222 :98

44 :9:': :9:+: :927:

444 :9:8: :9:8' :98::

4 :9:++2 :9:++2 :9'88

Average = 49.5



Equilibrium 8a$

3or reaction at equilibrium write the followin#

Equilibrium 8a$ $at ++: @C%

Equilibrium constant & < c & constant at #iven A

=se < c since usually wor?in# with concentrations in

mol=8

3or chemical equilibrium to exist in reactionmixture> reaction quotient ; must be equal toequilibrium constant , < c

K c=

[HI ]2

[ H 2

][ I 2

]=49.5


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(reictin Equilibrium 8a$

3or #eneral chemical reaction6

D 3 eE f> 3 ? ,here D > E > > > and ? represent chemical formulas

> e > f > and 1 are coefficients

;ass action expression &

Note: E)ponents in mass action expression

are stoichiometric coefficients in balancedequation9

Equilibrium la$ is6

[ F ] f [G ] g

[ D ]d [ E ]e

K c=[ F ] f [G ] g

[ D ]d [ E ]e


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(reictin Equilibrium 8a$

,here only concentrations that satisfy this

equation are equilibrium concentrations Numerator

;ultiply proucts! raised to their

stoichiometric coefficients Denominator

;ultiply reactants! raised to their

stoichiometric coefficients

is scientist.s convention K c=

[ products ] f

[reactants

]

d



E)@ Equilibrium 8a$

A /!01 2 3 N!01 2 ! N/A01 2

< c & +92 x 8:7 at 2 @C

,hat is equilibrium law

K c= [ NH3 ]2

[ H 2 ]

3 [ N 2 ]=4.26×10

8



8earnin Chec

,rite mass action expressions for the followin#6

! N"! 01 2 N!"# 01 2

!C" 01 2 3 "! 01 2 ! C"! 01 2

Q=[ N 2O 4 ]

[ NO2]2

Q= [CO 2 ]2

[CO ]2 [O 2 ]


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Bour Turn,hich of the followin# is the correct mass

action expression $;)E% for the reaction6Cu2$aq % +*H'$aq % Cu$*H'%+

2!$aq %

A. Q =

[Cu ( NH3)42+ ]

[Cu2+ ][ NH3]4

B . Q=[Cu( NH

3)42+ ]

[Cu

2+

][ NH3 ]

C . Q =[Cu2+ ] [ NH

3]4

[Cu( NH3 )42+ ]

D . none of these 84



Manipulatin Equations for ChemicalEquilibria

Darious operations can be performed onequilibrium expressions

1@ ,hen irection of equation is re%erse>

new equilibrium constant is reciprocal ofori#inal

' 3 * C 3 D

C 3D ' 3 * K c

' =[ A ][ B ]

[C ][ D ]

=1

K c

K c=[C ][ D ]

[ A][ B ]


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E)@ Manipulatin Equilibria 1

1@ ,hen irection of equation is re%erse> new

equilibrium constant is reciprocal of ori#inal

A /!01 2 3 N!01 2 ! N/A01 2 at 2@C

! N/A01 2 A /!01 2 3 N!01 2 at 2@C

K c

' =[ H

2]3 [ N

2]

[ NH3]2 =

1

K c

= 1

4.26×108=2.35×10

−9

K c= [ NH3 ]2

[ H 2 ]3 [ N 2 ]

=4.26×108


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Manipulatin Equilibria !

!@ ,hen coefficients in equation are

multiplie by a factor> equilibriumconstant is raise to a po$er equal to thatfactor9

' 3 * C 3 D

A' 3 A* AC 3 AD

K c=[C ][ D ][ A][ B ]

K c

' ' =[C ]3 [ D ]3

[ A ]

3

[ B ]

3=[C ] [ D ][ A ][ B ]

×[C ][ D ][ A] [ B ]

×[C ][ D ][ A] [ B ]

= K c

3





Manipulatin Equilibria A

A@ ,hen chemical equilibria are ae>

their equilibrium constants aremultiplie

' 3 * C 3 D

C 3 E > 3 ?

' 3 * 3 E D 3 > 3 ?

K c1

=[C ][ D ]

[ A] [ B ]

K c

2

=[ F ] [G ][C ][ E ]

K c

3

=[C ][ D ]

[ A ] [ B ]

×[ F ] [G ]

[C ] [ E ]

=[ D ][ F ][G ]

[ A ][ B ][ E ]

= K c1

× K c2



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 2'

A@ ,hen chemical equilibria are ae> their

equilibrium constants are multiplie

[ NO ][ NO3 ]

[ NO2]2

×[ NO

2 ] [CO2 ]

[ NO3][CO ]

2 *2$# % *'$# % *$# %

*'$# % C$# % *2$# % C2$# %

*2$# % C$# % *$# % C2$# %

K c

1=[ NO ][ NO3 ]

[ NO2]2

K c2=

[ NO2][CO

2]

[ NO3 ][CO ]

K c

3=[ NO ][CO2]

[ NO2 ][ CO ]

Therefore K

c1

× K c2

= K c3

Manipulatin Equilibria A

=[ NO ][CO

2 ][ NO

2][CO ]



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 2+

8earnin Chec 3or6 N!01 2 3 A /!01 2 ! N/A01 2

< c + 5 at a particular temperature9

,hat would be < c for followin#

2 *H'$# %

*

2$# %

' H

2$# %

F *2$# % '/2 H2$# % *H'$# %

K c

' = 1

K c

= 1

500=

22.4

0.002

K c

' ' = K

c1/2=√ 500=




Equilibrium Constant, < c Constant value equal to ratio of product

concentrations to reactant concentrationsraised to their respective exponents

Chan#es with temperature $van.t HoffEquation%

5epends on solution concentrations )ssumes reactants and products are in

solution

K c=

[ products ] f

[ reactants ]d




Equilibrium Constant, < p Based on reactions in which substances are

#aseous

)ssumes #as quantities are expressed inatmospheres in mass action expression

=se partial pressures for each #as in place ofconcentrations

Ex9 N!

01 2

3 A /!

01 2

! N/A

01 2

K P =

P NH3

2

P N 2 P

H 2

3




/o$ are < p an < c &elate -tart with (deal Gas aw

(+n&T

Rearran#in# #ives

-ubstitutin# (=&T for molar concentrationinto I c results in pressurebased formula

Fn & moles of #as in product K moles of#as in reactant

K p= K c( RT )

Δn

P =

n

V RT = MRT




8earnin Chec

Consider the reaction6 !N"!01 2 N!"#01 2

(f < p & :9+7: for the reaction at 2@C> what is

value of < c at same temperature

n + nproucts G nreactants

& 8 K 2 & G1

K p= K c(R)!n

K c= K p

(R ) Δn= 0.480(0.0821×298 K )−1

Kc = 11.7




Bour TurnConsider the reaction '01 2 3 !*01 2 #C01 2

(f the I c for the reaction is :9<< at 2LC> whatwould be the I p

)9 :9<<

B9 29:C9 2+9

59 2+::

E9 *one of these

!=(4 " #)=1K $ = K c(%&)!

K $= 0.99*(0.082057*298.15)1

K $ = 24

2<



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E ':

/omoeneous reaction=equilibrium )ll reactants and products in same phase

Can mix freely

/eteroeneous reaction=equilibrium Reactants and products in different phases

Can.t mix freely

-olutions are expressed in ;

Gases are expressed in ;

Governed by I c



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E '8

/eteroeneous Equilibria

2*aHC'$s % *a2C'$s % H2$# % C2$# %

Equilibrium aw &

Can write in simpler form

3or any pure liquid or solid> ratio of moles tovolume of substance $;% is constant

Ex9 8 mol *aHC' occupies '79< cm'

2 mol *aHC' occupies 4497 cm'

K =[ NaCO!

( " ) ][ H 2O ( g ) ] [CO( g ) ]

[ NaHCO! ( " ) ]2

M =

1 "o# NaHCO3

0. 0389 $ =25.%&

M =2 "o# NaHCO

3

0 .0%%8 $ =25.%&




/eteroeneous Equilibria2*aHC'$s % *a2C'$s % H2$# % C2$# %

Ratio $n/D% or ; of *aHC' is constant $294 mol/%re#ardless of sample siMe

i?ewise can show that molar concentration of

*a2C' solid is constant re#ardless of sample siMe -o concentrations of pure solids and liquids

can be incorporated into equilibrium constant>

I c

Equilibrium law for hetero#eneous system written

without concentrations of pure solids or liquids

K c= K [

Na2

CO3 ( ")]

[ NaHCO3( " )]2

=[ H 2O ( g )][CO2( g )]



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E ''

8earnin Chec

,rite equilibrium laws for the followin#6

)#$aq % Cl K$aq % )#Cl$s %

H'N+$aq % H2$O% H'$aq % H2N+

K$aq %

K c=1

[ Ag )

][C# *

]

K c=[ H

3O

)][ H

2 PO

4

*]

[ H 3 PO

4]




Bour TurnGiven the reaction6

'Ca2$aq % 2N+'K$aq % Ca'$N+%2$s %

,hat is the mass action expression

A . Q =

[Ca2+ ]3 [+O43− ]2

[Ca3(+O4 )2 ]

B . Q =[Ca

2+ ]3 [+O43− ]2

[1 ]

C . Q =[Ca3(+O4 )2 ][Ca2+ ]3[+O

43− ]2

D . Q =[1 ]

[Ca

2+

]

3

[+O4

3−

]

2

'+




Bour TurnGiven the reaction6

'Ca2$aq % 2N+'K$aq % Ca'$N+%2$s %

,hat is mass action expression for the reverse

reaction A . Q =

[Ca2+ ]3 [+O43− ]2

[Ca3(+O

4)2 ]

B . Q =[Ca

2+ ]3 [+O43− ]2

[1 ]

C . Q =[Ca3(+O4)2 ][Ca2+ ]3[+O

43− ]2

D . Q=[1 ]

[Ca2+

]3

[+O43−

]2

'




4nterpretin < C 8are < 0<HH12

;eans product rich mixture

Reaction #oes far towardcompletion

E)@

2-2$# % 2$# % 2-'$# %

I c & 49: × 8:2 at 2 @ C

K c=

[,O3]2

[,O2]2 [O

2]=

% .0×1025

1




4nterpretin < C Small < 0<II12

;eans reactant richmixture

nly very small amounts of

product formed

E)@

H2$# % Br2$# % 2HBr$# % I c & 89+ × 8: K28 at 2 @C

K c=

[H-r ]2

[ H 2 ][-r 2 ]

=1 .4×10

−21

1




4nterpretin < C

<≈

1 ;eans product and

reactant concentrations

close to equal Reaction #oes only P

halfway



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E '<

SiJe of < i%es measure of ho$reaction procees

< HH 1 products! QQ reactants!

< + 1 products! & reactants!

< II 1 products! reactants!

8 i Ch



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E +:

8earnin Chec

Consider the reaction of !N"!01 2 N!"#01 2

(f I p & :9+7: at 2@C> does the reaction favor

product or reactant

K is small (K ' 1)

Reaction favors reaca!Since K is close to 1 si!nificant amo"nts of

both reactant an# $ro#"ct are $resent

ilib i i i KShif L



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E +8

Equilibrium (ositions an KShiftsL

Equilibrium positions Combination of concentrations that allow ; + <

(nfinite number of possible equilibrium positions

8e Chtelier9s principle -ystem at equilibrium $; + < % when upset by

disturbance $; < % will shift to offset stress

-ystem said to 0shift to riht1 when

for$ar reaction is dominant $; I < % -ystem said to Kshift to leftL when re%erse

direction is dominant $; H < %

& l ti hi * t ; <




&elationship *et$een ; an <

; + < reaction at equilibrium

; I < reactants→

proucts

Aoo many reactants

;ust convert some reactant to product to movereaction toward equilibrium

; H < reactants ← proucts

Aoo many products ;ust convert some product to reactant to move

reaction toward equilibrium

E l f 8 Cht li 9 ( i i l



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E +'

E)amples of 8e Chtelier9s (rinciple

et.s see how this wor?s with chan#es in

89 Concentration

29 Nressure and volume

'9 Aemperature

+9 Catalysts

9 )ddin# inert #as to system at constant

volume

1 Eff t f Ch i C t ti



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E ++

1@ Effect of Chane in ConcentrationCu0/!"2#

!30aq 2 3 #Cl G0aq 2 CuCl#!G0aq 2 3 #/!"

blue yello$ Equilibrium mixture is blue#reen

)dd excess Cl K $conc HCl% Equilibrium shifts to products

;a?es more yellow CuCl+2K

-olution becomes #reen

K c= [CuC#4

2−(a$ ) ][ H 2O ]4

[Cu ( H 2

O )4

2+( a$ ) ][C#−( a$ ) ]4

K c

' = K

c

[ H 2O ]4

= [CuC#

4

2−(a$)]

[Cu( H 2O )

4

2+(a$)][C#

−(a$)]4

1 Eff t f Ch i C t ti




1@ Effect of Chane in ConcentrationCu0/!"2#

!30aq 2 3 #Cl G0aq 2 CuCl#!G0aq 2 3 #/!"

blue yello$

)dd )# Removes Cl K6 '30aq 2 3 Cl G0aq 2 → 'Cl0s 2

Equilibrium shifts to reactants

;a?es more blue Cu$H2%+2

-olution becomes bluer

)dd H2

K c= [CuC#4

2−( a$ ) ]

[Cu ( H 2O )42+

( a$ ) ][C#−( a$ ) ]4

Effect of Chane in Concentration




Effect of Chane in Concentration 2-2$# % 2$# % → 2-'$# %

I c & 29+ x 8:' at 4:: oC ,hich direction will the reaction move if

:982 moles of 2 is added to an equilibrium

mixture )9 Aowards the products

B9 Aowards the reactants

C9 *o chan#e will occur

+

Eff t f Ch i C t ti




Effect of Chane in Concentration

,hen chan#in# concentrations of reactants

or products Equilibrium shifts to remo%e reactants or

products that have been ae

Equilibrium shifts to replace reactants orproducts that have been remo%e

Eff t f ( l Ch




Effect of (ressure an olume Chanes

Consider #aseous system at constant T and n

A/!01 2 3 N!01 2 !N/A01 2

(f reduce volume $ % Expect Nressure to increase $(%

Ao reduce pressure> loo? at each side of reaction ,hich has less moles of #as

Reactants & ' 8 & + moles #as

Nroducts & 2 moles #as

Reaction favors products $shifts to ri#ht%

K P =

P NH3

2

P N 2 P H

2

3

Eff t f ( Ch



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E +<

Effect of ( an Chanes

Consider #aseous system at constant T and n

Ex9 /!01 2 3 4!01 2 ! /401 2

(f pressure is increased> what is the effect onequilibrium

nreactant

& 8 8 & 2

nproduct & 2

Nredict no chan#e or shift in equilibrium

K P =

P HI

2

P H 2 P

I 2

Eff t f ( Ch



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E :

!Na/S"A0s 2 NaS"A0s 2 3 /!"01 2 3 S"!01 2

(f you decrease volume of reaction> what is

the effect on equilibrium Reactants6 no moles #as & all solids

Nroducts6 2 moles #as

D> causes N

Reaction shifts to left $reactants%> as this hasfewer moles of #as


K P = P

H 2O P ,O

2

Eff t f ( Ch




Reducin# volume of #aseous reaction

mixture causes reaction to decrease numberof molecules of #as> if it can (ncreasin# pressure

;oderate pressure chan#es have ne#li#ibleeffect on reactions involvin# only liquidsand/or solids

-ubstances are already almost incompressible Chan#es in D> N and S! effect position of

equilibrium $T%> but not I


Effect of Chane in Temperature




Effect of Chane in Temperature

Cu0/!"2#!30aq 2 3 #Cl G0aq 2 CuCl#

!G0aq 2 3 #/!"

blue yello$ Reaction endothermic

)ddin# heat shifts equilibrium toward products

Coolin# shifts equilibrium toward reactants 2

%ce

&ater

'oilin!

&ater

Effect of Temperature Chanes




Effect of Temperature Chanes/!"0s 2 /!"0O2 ∆H@ & ?J $at : @C%

Ener#y H2$s % H2$O%

Ener#y is reactant

)dd heat> shift reaction ri#htA/!01 2 3 N!01 2 !N/A01 2 ∆Hf @& K+498< ?J

' H2$# % *2$# % 2 *H'$# % ener#y

Ener#y is product

)dd heat> shift reaction left

Effect of Temperature Chanes




Effect of Temperature Chanes↑ T shifts reaction in direction that produces

enothermic $heat absorbin#% chan#e↓ T shifts reaction in direction that producese)othermic $heat releasin#% chan#e

Chan#es in A chan#e value of mass actionexpression at equilibrium> so I chan#ed I depends on A

↑A of exothermic reaction ma?es I smaller

;ore heat $product% forces equilibrium to reactants

↑A of endothermic reaction ma?es I lar#er

;ore heat $reactant% forces equilibrium to products

Catalysts 'n Equilibrium




Catalysts 'n Equilibrium Catalyst lowers Ea

for both forwardand reversereaction

Chan#e in Ea affects rates ? r

and ? f equally

Catalysts have noeffect on

equilibrium

Effect of 'in 4nert ?as




Effect of 'in 4nert ?as

4nert as ne that does not react with components of

reaction

E)@ )r#on> Helium> *eon> usually *2

)ddin# inert #as to reaction at fixed $n andA%> ↑ ( of all reactants and products

-ince it doesn.t react with anythin# *o chan#e in concentrations of reactants or

products

*o net effect on reaction

/o$ to Pse 8e Chtelier9s (rinciple




/o$ to Pse 8e Chtelier s (rinciple

89 ,rite mass action expression for reaction

29 Examine relationship between affectedconcentration and T $direct or indirect%

'9 Compare T to I (f chan#e ma?es T Q I> shifts eft

(f chan#e ma?es T I> shifts Ri#ht

(f chan#e has no effect on T> no shift expected

8earnin Chec:




8earnin Chec:

onsi#er

+,-(aq) / +- (aq) +2-() / ,-+(aq)

hat &ill ha$$en if ,-+ is remove#3

4 is $ro$ortional to ,-+ 6 ↓ ,-

+ 6 ↓ 4

4 K e"iliri"m shifts to ri!ht

Q=

[ PO%

!− ]

[OH − ]3 [ H 3 PO% ]

8earnin Chec:



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E <

8earnin Chec:The reaction

+,-(aq)

/ +-

(aq)

+2-(aq)

/ ,-

+(aq)

is e:othermic.

hat &ill ha$$en if s;stem is coole#3

Since reaction is e:othermic heat is $ro#"ct

eat is #irectl; $ro$ortional to 4

↓T ↓ 4

4 K e"iliri"m shifts to ri!ht

heat

Q= [ PO%

!−

][OH − ]3 [ H

3 PO

%]

Bour Turn




Bour TurnAhe equilibrium between aqueous cobalt ion

and the chlorine ion is shown6Co$H2%!

2$aq % +Cl K$aq % Co$Cl%+!2K$aq % H2$O%

pin blue

(t is noted that heatin# a pin? sample causes itto turn violet9

Ahe reaction is6

)9 endothermicB9 exothermic

C9 cannot tell from the #iven

information :

Bour Turn




Bour Turn

Ahe followin# are equilibrium constants for the

reaction of acids in water> I a9 ,hich is themost efficient reaction

)9 I a & 292.10 K3

B9 I a & 897.10 K5

C9 I a & +9:.10 K10

59 I a & 9'.10 K3

8

Equilibrium Calculations




Equilibrium Calculations

3or #aseous reactions> use either < ( or < C

3or solution reactions> must use < C

Either way> two basic cate#ories of

calculations89 Calculate < from ?nown equilibriumconcentrations or partial pressures

29 Calculate one or more equilibrium

concentrations or partial pressures usin#?nown < ( or < C

Calculatin <C from Equilibrium




Calculatin < C from Equilibrium

Concentrations

,hen all concentrations at equilibrium are?nown =se mass action expression to relate

concentrations to < C Awo common types of calculations

)9 Given equilibrium concentrations> calculate I

B9 Given initial concentrations and one final

concentration Calculate equilibrium concentration of

all other species

Ahen calculate I

Calculatin <C ?i%en Equilibrium




Calculatin < C ?i%en Equilibrium

Concentrations

E)@ 1 N!"#01 2 !N"!01 2 (f you place :9:': mol *2+ in 8 flas? at

equilibrium> what is < C

*2+!eq & :9:2<2 ;

*2!eq & :9:88 ;

K c= [ NO

]2

[ N 2O 4 ] K c=[ 0.0116 ]

2

[ 0.0292]

K = 4.*1 × 10 "#



Calculatin < C ?i%en 4nitial Concentrations




C

an "ne >inal Concentration

E)@ ! !S"!01 2 3 "!01 2 !S"A01 2 89::: mol -2 and 89::: mol 2 are placed in

a 89::: flas? at 8::: I9 )t equilibrium

:9<2 mol -' has formed9 Calculate I C forthis reaction9

8st calculate concentrations of each

(nitial

Equilibrium

[ &O ]=[O2 ]=1.00 (#

1.00 ) =1.00 M

[ &O!]=

0.925 (#

1.00 )

=0.925 M

/o$ to Sol%e:




/o$ to Sol%e: -et up Concentration Aable

Based on the followin#6 Chanes in concentration must be in same ratio

as coefficients of balance equation

-et up table under balanced chemical equation

4nitial concentrations Controlled by person runnin# experiment

Chanes in concentrations

Controlled by stoichiometry of reaction Equilibrium concentrations

EquilibriumConcentration

&(nitial

Concentratio

n

KChan#e in

Concentration

Ne)t Set up Concentration Table




Ne)t Set up Concentration Table

2-2$# % 2$# % 2-'$# %

4nitial Conc9 $;% 89::: 89::: :9:::

Chan#es in Conc9 $;%

Equilibrium Conc9 $;%

-2! consumed & )mt of -' formed

& -'! at equilibrium & :9<2 ;

2! consumed & F amt -' formed

& :9<2/2 & :9+2 ;

-2! at equilibrium & 89::: K :9<4 & :9:4

2! at equilibrium & 89:: K :9+2 & :9'7 ;

"0.925 "0.4*2 +0.925

0.075 0.5#, 0.925

E) !



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E <

E)@ !

3inally calculate I C at 8::: I

K c=

[,O3]2

[,O2]2 [O

2]

K c= [ 0.925 ]2

[ 0.0%5]2 [0.538 ]

< c + !@Q R 1! + !Q

Summary of Concentration Table



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 4:


=sed for most equilibrium calculations $Ch

8> 84> and 87%89Equilibrium concentrations are only values

used in mass action expression

Dalues in last row of table89(nitial value in table must be in units of

mol/ $; %

S!initial & those present when reaction prepared *o reaction occurs until everythin# is

mixed





Summary of Concentration Table'9 Chan#es in concentrations always occur in

same ratio as coefficients in balancedequation

+9 (n 0chan#e1 row be sure all reactants!chan#e in same directions and all products!chan#e in opposite direction9

(f reactant!initial & :> its chan#e must be $↑%

because reactant!final ≠ ne#ative

(f reactants! ↓> all entries for reactants in chan#erow should have minus si#n and all entries forproducts should be positive

Calculate -.equilibrium from < c an -.initial




- .equilibrium c - .initial

,hen all concentrations but one are ?nown

=se mass action expression to relate I c and ?nown concentrations to obtainmissin# concentrations

E)@ A C/#01 2 3 /!"01 2 C"01 2 3 A/!01 2 )t 8:: @C> I c & 949 )n equilibrium

mixture of #ases had the followin#

concentrations6 CH+! & :9+:: ; andH2! & :97::; and C! &:9'::;9

,hat is H2! at equilibrium

Calculate -.equilibrium from < c an -.initial



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 4'

q

E)@ A C/#01 2 3 /!"01 2 C"01 2 3 A/!01 2 I c & 94

CH+! & :9+:: ;U H2! & :97::;U C! &:9'::; ,hat is H2! at equilibrium

3irst> set up equilibrium

*ext> plu# in equilibrium concentrations and < c

K c= [CO ] [ H 2 ]

3

[CH4][ H

2O ]

-/!". + @6Q M

[ H 2O ]=[0.300 ] [0.800 ]3

[0.400 ](5.6%) =0.1542.2%

[ H 2O ]=[CO ][ H 2 ]3

[CH4] K

c

Calculatin -.E ilib i from <




Calculatin -.Equilibrium from < c

Uhen 4nitial Concentrations 're ?i%en

,rite equilibrium law/mass action expression

-et up Concentration table

)llow reaction to proceed as expected>usin# 0x1 to represent chan#e inconcentration

-ubstitute equilibrium terms from table into

mass action expression and solve

Calculate -.equilibrium from -.initial an <C




Calculate -.equilibrium from -.initial an < C

E)@ # /!01 2 3 4!01 2 !/401 2 at #!5 C

< C & 9+

(f one mole each of /! an 4! are placed in a

:9:: flas? at +2 @C> what are theequilibrium concentrations of H2> (2 and H(

Step 1@ Urite Equilibrium 8a$

K c= [ HI ]

2

[ H 2][ I

2]=

55.64

E)@ # Step !@ Concentration Table




E)@ # Step !@ Concentration Table

Conc $;% /!01 2 3 4!01 2 !/4 01 2

(nitial 29:: 29:: :9:::

Chan#e

Equil.m

(nitial H2! & (2! & 89:: mol/:9:: &29::;

)mt of H2 consumed & )mt of (2 consumed & x

)mt of H( formed & 2x

G ) 3!) G )

3!)!@ G )!@ G )

55.64= ( 2/ )2

(2.00− * )(2.00− * )=

( 2/ )2

(2.00− * )2

E)@ # Step A@ Sol%e for )




E)@ # Step A@ Sol%e for ) Both sides are squared so we can ta?e

square root of both sides to simplify

√ K =√ 55.64=√ (2/ )2

(2.00− * )2

%.459=2/

(2.00− * )%.459(2.00− * )=2/

14 .918−%.459 *=2/

*=14.918

9.459 =1.58

14 .918=9. 459 *






Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 4<

- .equilibrium - .initial C

E)@ 5 /!01 2 3 4!01 2 !/401 2 at #!5 C

< C & 9+

(f one mole each of /! , 4! and /4 are placed

in a :9:: flas? at +2 @C> what are theequilibrium concentrations of H2> (2 and H(

*ow have product as well as reactants initially

Step 1@ Urite Equilibrium 8a$

K c=

[ HI ]2

[ H 2][ I

2]=55.64



E)@ 5 Step A@ Sol%e for )





%.459=2.00+2/

(2.00− * )%.459(2.00− * )=2.00+2/

14.918−%.459 *=2.00+2/

*=

12.918

9.459 =1.3%

12. 918=9. 459 * 26e"il = %26e"il = 2.00 : =

2.00 1.+7 = 0.<+

%6e"il = 2.00 / 2:= 2.00 / 2(1.+7)

= 2.00 / 2.7

= .7

Bour Turn




*2$#% 2$#% 2*$#%→

I c & :9:82' at '<:: oC(f :92 moles of *2 and 2 are placed in a 2:

m container> what are the equilibrium

concentrations of all species )9 :9:2 ;> :9<+4 ;> :98: ;

B9 :9<+4 ;> :9<+4 ;> :98: ;

C9 :9<+4 ; :98: ;> :9:2 ; 59 :98: ;> :98: ;> :9<+4 ;

72

Bour Turn V Solution




Conc $;% N!01 2 3"!01 2 !N" 01 2 (nitial 89:: 89:: :9::

Chan#e G ) G ) 3 !)

Equil 89:: G ) 89:: G ) 3 !)

7'

2 2

2

2

:92/: mol* ! E ! 89::

:92/: H

$2 % 2:9:82' :9:82'8$8 %

:9:/2 *E! & 2 & :98:/

;

x x x x

x ; x ;

= = =

= =

−−

=





- .equilibrium - .initial C

E)@ 6

C/AC"!/0aq 2 3 C!/5"/0aq 2 C/AC"!C!/50aq 2 3

acetic aci ethanol ethyl acetate /!"0

2

< C & :988

)n aqueous solution of ethanol and aceticacid> each with initial concentration of :978:;> is heated at 8:: @C9 ,hat are the

concentrations of acetic acid> ethanol andethyl acetate at equilibrium

E)@ 6 Step 1@ Urite Equilibrium 8a$




p q

*eed to find equilibrium values that satisfy

thisStep !: -et up concentration table usin# 0x1

for un?nown

4nitial concentrations Chan#e in concentrations

Equilibrium concentrations

K c=

[CH3CO

2C

2 H

5]

[C 2 H 5 OH ] [CH3 CO2 H ]=0.11

E)@ 6 Step !@ Concentration Table




p

$;% CH'C2H$aq % C2HH$aq % CH'C2C2H$aq % H2$ l %

4 :978: :978: :9:::C

E

)mt of CH'C2H consumed & )mt of C2HH

consumed & K x

)mt of CH'C2C2H formed & x

CH'C2H!eq and C2HH ! & :978: K x

CH'C2C2H! & x

"- +- " -

+-0.,10 " -0.,10 " -

0.11= *

(0.810−* )(0.810−* )





p

Rearran#in# #ives

Ahen put in form of quadratic equation

ax2 bx c & :

-olution for quadratic equation #iven by

0.11×(0.6561−1.62 *+ *2)= *

0.0%21%−0. 1%82 *+0.11 *2− *=0

0.11 *2−1.1%82 *+0.0%21%=0

*=−b±√ b2−4 ac

2a





p

Ahis #ives two roots6 x & 8:9 and x &:9:+

nly x & :9:+ is possible x & 8:9 is QQ :978: initial concentrations

:978: K 8:9 & ne#ative concentration>which is impossible

*=−(−1.1%82)±√(−1.1%82)2−4(0.11 )(0.0%21% )

2(0.11)

*=1.1%82±√ (1.388)−(0.032)

0.22

=1.1%82±1.164

0.22

E)@ 6 Step #@ Equilibrium Concentrations



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 7<

$;% CH'C2H$aq % C2HH$aq % CH'C2C2H$aq % H2$ l %

4 :978: :978: :9:::

C

E

CH'C2C2H!equil & x & :9:+ ;

CH'C2H!equil & C2HH!equil & :978: ; K x

& :978: ; K :9:+ ;& :94+ ;

G@6# 3@6# G @6#

3@6#@#6@#6




Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E <:

,hen < C is very small

E)@ !/!"01 2 !/!01 2 3 "!01 2

)t 8::: @C> I C & 49' × 8: K87

(f the initial H2 concentration is :98::;>what will the H2 concentration be at

equilibrium

Step 1@ Urite Equilibrium 8a$ K

c=[ H

2]2[O

2]

[ H 2O ]2

=%.3×10−18

E)@ Step !@ Concentration Table



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E <8

Conc $;% !/!"01 2 !/!01 2 3 "!01 2

(nitial :98:: :9:: :9::Chan#e

Equil.m

G !) 3)3!)

3)3!)@1 G !)

% .3×10−18= (2/ )2 *

(0.100−2/ )2= 4/3

(0.100−2/ )2

Cubic equation K tou#h to solve

;a?e approximation < C very small> so ) will be very small

)ssume we can ne#lect )

must prove valid later

E)@ Step A@ Sol%e for )



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E <2

)ssume 0@1 G !)2 ≈ @1

*ow our equilibrium expression simplifies to

Conc $;% !/!" 01 2 !/! 01 2 3 "! 01 2

(nitial :98:: :9:: :9::

Chan#e

Equil.m

G !) 3)3!)

3)3!)@1

% .3×10−18

= (2/ )2 *

(0.100 )2=4/

3

0.010

4/3=0.010(% .3×10

−18) & 49' V 8: K2:



Simplifications: Uhen Can Boui i i l 0C 2



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E <+

4nore ) in *inomial 0Ci G )2 (f equilibrium law #ives very complicated

mathematical problems

)nd if < is small

Chan#e to reach equilibrium $ ) term% is also small

Compare initial concentration Ci in binomial to

value of <

=se proof to show that dropped ) term wassufficiently small

dropped / ter"

C +

¿,

0.05

C +

K >400

8earnin Chec




3or the reaction !'01 2 *01 2

#iven that I p & '9V8: K8 at 2@C> and we place :92

atm ) into the container> what will be the pressure of Bat equilibrium

!' ↔ *

( :92 : atm

C K2x x

E :92 K 2x x

Q= P N 2O4

P NO

2

3 .5×10

−16

=

*

(0 .2)2x & 89+V8: K84

-*.+ 1@#R1 G1 M

proof6 89+V8: K84 /:92:9:

Q=0

(0 . 2)2=0

Q< shfts r(ht

W@!

<

Bour Turn




(n the reaction shown> I & 897 V 8: K

HC2H'2$aq % H2$l % H↔ '$aq % C2H'2 K$aq %

(f we start with :9'; HC2H'2> what will be the

equilibrium concentration of C2H'2 K

)9 :9' ;

B9 :9::2 ;

C9 :9:+ ;59 :9 ;

<

Calculatin < C ?i%en 4nitial Concentrations



an "ne >inal Concentration

E)@ !a/!01 2 3 4!01 2 !/401 2 X #5 C

(nitially H2 and (2 concentrations are :92::

mol each in 29:: $& @1M%U no H( ispresent )t equilibrium> H( concentration is @16 ;

Calculate < C

Ao do this we need to ?now ' sets ofconcentrations6 initial> chan#e and

Documents

Brdy 6Ed Ch15 ChemicalEquilibrium