Conver%ng)Between)Kc)and)Kp))• Because'concentration'and'pressures'of'gases'are'related'via'the'ideal'gas'law,'we'can'obtain'convert'between'Kc'and'Kp.'
'PV'='nRT'P'='(n/V)RT'P'='cRT'
• In'order'to'obtain'P'from'['],'we'x'the'[']'by'RT'• Since'we'have'multiple'concentrations'in'an'equilibrium)expression,'we'have'to'do'this'multiple'times'
• Use'R'='0.08206'(P'is'in'atm)'and'T'in'Kelvin'115'
Conver%ng)Between)Kc)and)Kp))• Write'both'Kc'and'Kp'expressions'for'the'following'reactions.'Then,'relate'the'two'constants'using'the'ideal'gas'law'
2NO2(g)'⇄'N2(g)'+'2O2(g)''''3H2(g)'+'N2(g)'⇄'2NH3(g)'''
116'
Kp = Kc - - - - .
PV=nRTp=grtfP=CRT
Kc=[Ny?÷fg?" kp=(PoI¥=(Eoz3RT)YIRT)
(PNOZ)" ([Nor] RT)'
= TIE KIRTII. Kp -- Kc (RT)
't
ke Kp =
Conver%ng)Between)Kc)and)Kp))• Write'both'Kc'and'Kp'expressions'for'the'following'reactions.'Then,'relate'the'two'constants'using'the'ideal'gas'law'
H2(g)'+'F2(g)'⇄'2HF(g)''''2C6H6(l)'+'15O2(g)'⇄'12CO2(g)'+'6H2O(g)''
117'
Kc-_[H⇒ Kp=(PH=(EHFTR =Kc (RT)
°
.
.
. Kp -- KCLRT)°
THz] [Fz] Pttz PFZ (EHZTRTHEFZJRT)
Kc- [C0z]REHz0# Kp=(Roz)t){ ([coz]RT)%LHzDRTY[02315 ( Poz) 's (EOz]RT
= Kc (RT)' ! . Kp - Kcc RTP
Conver%ng)Between)Kc)and)Kp))• Since'the'number'of'‘RT’'terms'depend'on'the'balance'between'the'#'of'gases'in'the'reactant'and'product'sides,'we'de_ine'
Δn'='nproducts(g)'^'nreactants(g)''• If'we'know'Δn,'then'Kp'and'Kc'are'related'as'such:'
Kp'='Kc(RT)Δn''
118'
Conver%ng)Between)Kc)and)Kp))• If'the'value'of'Kc'is'55.0'at'25oC,'what'is'the'value'of'Kp'for:'
N2(g)'+'2O2(g)'⇄'N2O4(g)'
Kp'='Kc(RT)Δn'''''''
Kp'=0.0919'
119'
③ ⑦1 - 3
= 55.0 ( 0.08206 ° 298 .15)
= 55.0 (0.08206 . 298.15)-2
= 9.19×10-Z
Kc → Kp
Conver%ng)Between)Kc)and)Kp))• For'the'reaction'A(g)'+'B(g)'⇄2C(g)'+'2D(g),'the'value'of'Kp'is'24.1'at'0oC.'What'is'the'value'of'Kc?'
Kp'='Kc(RT)Δn'''''''''''
Kc'='0.0480'
120'
P n④
Kc = Kp 24. I
⇒n
=
T.ogzog.zpz.is#=0.0480T
Conver%ng)Between)Kc)and)Kp))• Example:'For'the'reaction'2'O3(g)'⇄'3'O2(g),'the'equilibrium'concentrations'are'0.0420'M'for'O3'and'0.1402'M'for'O2'at'300.'K.'! 'Calculate'Kc'! 'Calculate'Kp''
121'
Kc = {0023-3}={0*43}--1.56223 - --
= 1.56
Kp = Kc (RT)3-2=1.56223 - - - (O -08206.300. K)
= 38.459 - - . = 38.5
Conver%ng)Between)Kc)and)Kp))• Example:'For'the'reaction'2'HF(g)'⇄'H2(g)'+'F2(g),'the'value'of'Kp'is'1.50'*'10^2'at'30oC.'What'is'the'value'of'Kc'at'30oC?'
122'
↳ 303.15K
Kp = Kc (RT)M
Kc = Kp¥on
= 1.50×210-2(0.08206 . 303.15×72-2
= 1-50×10-2
Posi%on)of)the)Equilibrium)• The'magnitude)of)the)Keq)tells'us'which'side'of'the'reaction'is'favoured'at'equilibrium'(‘position’)'• Since'in'general,'Keq'=''
• When'K'>>'1'(usually'>'102),'products'are'favoured'at'eq’m'([P]1[P]2…'>'[R]1[R]2…)'
• When'K'~'1,'products'and'reactants'are'equally'favoured'at'eq’m'([P]1[P]2…'~'[R]1[R]2…)'
• When'K'<<'1,'(usually'<'10^2)'reactants'are'favoured'at'eq’m'([P]1[P]2…'<'[R]1[R]2…)'
124'
eq > 1 = 1 LI
[Reactant ]eq EP3eq5LR3egEPI eq 7 ER] eq EP]eq=ER]eq
K’s)Dependency)
• Keq'depends'on'several'facts:'1. Coef?icients'used'in'chemical'equation'2. How'reversible'reaction'is'written'
(direction)'3. Multiple'steps'or'sequential'reactions'4. Temperature'at'which'reaction'takes'place'
125'
1.)K’s)Dependence)on)Coefficients)• The'equilibrium'constant'value'depends'on'how'the'reaction'is'written'
• For'the'Haber^Bosch'process,'the'following'concentrations'were'obtained'at'eq’m:'
N2(g)'+'2H2(g)'⇄'2NH3(g)'[N2]eq'='1.0'x'10^6'[H2]'eq'='1.0'x'10^2'M'[NH3]'eq'='3.0'x'10^5'M'
126'
Calculate)Kc)[N2]eq'='1.0'x'10^6'[H2]'eq'='1.0'x'10^2'M'[NH3]'eq'='3.0'x'10^5'M'
127'
N2(g)'+'2H2(g)'⇄'2NH3(g)'' 2N2(g)'+'4H2(g)'⇄'4NH3(g)' 1/2N2(g)'+4H2(g)'⇄'NH3(g)'Original'⇐÷±
to"i:÷÷÷÷÷÷÷i÷. of " ÷ f . ⇒
K’s)Dependence)on)Coefficients)
• In'general,'when'we'multiply'the'coef_icients'by'a'constant,'n,'then'the'new'equilibrium'constant'is'equal'to'the'original'raised'by'n'
Knew'='Koriginaln'
128'
K’s)Dependence)on)direc%on)[N2]eq'='1.0'x'10^6'[H2]'eq'='1.0'x'10^2'M'[NH3]'eq'='3.0'x'10^5'M'
'
129'
N2(g)'+'2H2(g)'⇄'2NH3(g)''Original'
2NH3(g)'⇄'N2(g)'+'2H2(g)'''
②
Terse.
" ao
:c::÷i. I "÷÷÷÷÷÷÷¥÷⇒
2.)K’s)Dependence)on)direc%on)'• In'general,'if'we'reverse'the'reaction,'then'the'new'equilibrium'constant'is'equal'to'the'reciprocal'of'the'original'
Kreverse'='1/Koriginal''
130'
3.)K’s)dependence)on)sequen%al)reac%ons)
• Consider'the'following'sequential'reactions:'K1 '' 'A+B'⇄'C''K2 '' 'C'⇄'2E'Koverall 'A+B'⇄'2E''What'is'the'relationship'between'Koverall'and'K1'&''K2?'
131'
[C ]KI = -EA] EB ]
Kz = EELEC ]
kwaei-EIF.j.at#.gEE3Ik""
K overall = K , × Kz
K’s)dependence)When)Combining)Reac%ons)
• In'general,'when'we'add'reactions'together'to'get'an'overall'reaction,'the'equilibrium'constant'of'the'overall'reaction'is'the'product'of'the'individual'equilibrium'constants'
'K1"K2"'K3'…'='Koverall'
132'
K’s)dependence)When)Combining)Reac%ons)
• Consider'the'following'sequential'reactions:'(assume'all'species'are'in'aq)'
'1. 2HCl'⇄'H2'+'Cl2'2. 2NaF'⇄'2Na'+'F2'3. 2Na'+'Cl2'⇄'2NaCl'4. H2'+'F2''⇄'2HF'
'• Determine'the'overall'reaction'
133'
% -win
mm -
Xoverall : 2 HCl t 2Naf 5 2Nacl t 2HF
K’s)dependence)When)Combining)Reac%ons)
• Consider'the'following'sequential'reactions:'(assume'all'species'are'in'aq)'
'1. 2HCl'⇄'H2'+'Cl2'2. 2NaF'⇄'2Na'+'F2'3. 2Na'+'Cl2'⇄'2NaCl'4. H2'+'F2''⇄'2HF'
'If'K1'='1'K2'='2'
K3'='0.5,'and''K4'='0.25'
• What'is'the'equilibrium'constant'for'overall'reaction'where'all'coef_icients'are'simpli_ied?'
'
134'
Overall : 2 HCl t2NaF 5 2Nacl t 2HF
XI or ÷ 2 t simplifyoverall
simp: HH t Na't 5 NaCl t HF
Kor-K4= l - 2. 0.5 . 0.25 = 2 . I . If = 4=0.25Kor . simp = (Kou)÷ = (F)
±= I = 0.5J
4.)K’s)Dependence)on)Temperature)• As'mentioned'before,'the)value)of)K)can)change,)but)only)with)temperature)changes)• It'is'tempting'to'think'that'any'shift'in'equilibrium'should'cause'a'change'in'the'value'of'K.)However,'this'is'not'true.''
• We'can'think'of'shifts'occurring'to'counteract'initial'stresses,'“balancing”'out'changes'in'concentration'
135'
K’s)Dependence)on)Temperature)• For'example,'if'you'increase'the'product'concentration'(stress),'the'shifts'causes'a'decrease'in'[P]'and'increase'in'[R].'• This'shift'effect'balances'out'the'initial'increase'of'product,'and'K'remains'the'same'
136'
K’s)Dependence)on)Temperature)• How'K'changes'with'temperature'depends'on'whether'or'not'the'reaction'is'endothermic'or'exothermic'• We'can'use'Le'Chatelier’s'Principle'to'help'us'determine'if'it'is'an'increase'or'decrease'
137'
K’s)Dependence)on)Temperature)• Consider'the'following'endothermic'reaction'
Heat'+'A'⇄'B'• The'equilibrium'expression'is:'
'Keq'=''• Which'way'does'the'equilibrium'shift'when'the'temperature'is'increased?''
• What'happens'to'[B]'and'[A]?'• What'happens'to'the'value'of'Keq?''
138'
EB ] T=
LATH
Right4 Hr
increase
K’s)Dependence)on)Temperature)• Consider'the'following'endothermic'reaction'
Heat'+'A'⇄'B'• The'equilibrium'expression'is:'
'Keq'=''• Which'way'does'the'equilibrium'shift'when'the'temperature'is'decreased?''
• What'happens'to'[B]'and'[A]?'• What'happens'to'the'value'of'Keq?''
139'
EBLAEAST
Left
It Idecrease
K’s)Dependence)on)Temperature)Endothermic)Reaction)
Exothermic)Reaction)
Increase)Temperature) K'increases' K'decreases'
Decrease)Temperature)
140'
Heat + A E B A E Bt heat
shift left → EB Td
[A] T
shift right → EBJTEAT d
k decreases K increases
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