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