Equilibrium & Further Aspects of Equilibria
.
☒Pine
Equilibrium GhÑvihF⑨£psFs⑧B Gonville
~
Reversible Reaction
Equilibrium
A
:
A reaction in which the products can react to
:
stage where the rate of forward reaction is equal to the rate of backward reaction
① Closed system
is
none of the
in which
one
reactants
or
products escape from the system
④ Concentration of Products d. Reactants remain constant
It
is
dynamic i e
•
-
•
•
•
Eg
H2
.
i
+
is called a reversible reaction
reform the original reactants
Is
=
In
a
closed system
.
(Reaction mixture )
.
-
Reactants & products are continoosly reacting
Where
of forward reaction
rate
is
rate of backward
equal to
& concentration of products { reactants
in
a
closed
system
reaction
constant
remain
.
.
.
2 HI
:-| (
.
.
lntial Cone
§ I
Hz
Iz
HI
5
5
0
-4.32
-432
☐
7
=
-
6-
jѧ
S
kami "
Gmt " " "
"
° "
"
432×2=8 "
"8
'
'
-
÷: :
3
O
-
j
Time 1sec
{
g.
21-1--1
2
10
=
1-12
+
Iz
A
ÉÉ$ ?
|
^
G
ro
=
s
EEEof
a
3
-
-
-
§É÷ ; : #
he
•
If
-
Chatelier 's
Principle
more
factors that
one
or
affect
"
""
" "" " ""
Equilibrium
-
-
o
-
8.64
°
°
" °
[-1-36]
I -3612
1.3612
=
0-68
=
0-68
•
equilibrium
Equilibrium Shifts to Right
field of Products Increase
are
changed the system will
ohhose
the
Equilibrium Shifts to Left
Cone / Yield of Reactants increases
.
change to
restore
equilibrium
FACTORS
AFFECTING EQUILIBRIUM
②
① Temperature
Pressure
Exothermic
9 Temp
to Product
Rmde
>
Pmoie
+ Pressure
Product
( doesn't like heat
b. Temp
9 Product
Rmde
-
Pmole
$ Pressure
↳ Product
R mole
=
Pmole
NO EFFECT
Endothermic
( likes
heat)
* Temp
- Product
↳ Temh
↳ Product
③ Concentration
④ Catalyst
Product
eg
.
to concentration
9 Yield
Helps active
T Concentration
Ha Yield
affect equilibrium
A Concentration
T Yield
b- Concentration
* Yield
equilibrium faster
,
because it
forward reaction
affect
Doesn't
PRESSURE AT EQUILIBRIUM
Fez , Vzos
backward &
Reactant
OF
yield
but
doesn't
sheds uh both
equaby
.
.
Equilibrium 9 Industry
→
Haters Process
CNH })
☐ H=
-92 KJMOI
N2 (9)
31-12
+
(9)
'
-
-
l
2MHz
→
Contact Process
①
S -102
②
→
(1-12504)
502
exo
802+02
÷
[ V20 5
(g)
2503
at
1-2 atm ]
4502
450C
200 atm
Fe
Increasing field
decrease temperature
2)
Increase
3)
Condense }
it
is formed
D
b.
3
Pressure
Remove
To Increase field
NHz
as soon
:
temperature
2) F Pressure
;
1)
as
Catalyst
503
,
although unnecessary as yield is
+
④ Hasaoz
Hasoa
+
Hao
>
→
enough at 2 atm (expensive
1-125207
( oleum
21-12504
( 2 moles)
EQUILIBRIUM CONSTANT
A relationship
that
Equilibrium expression
Eg
Nz + 31-12
.
KC
links
to
equilibrium
Kc
[NH }]
=
"
"
eg
3
( Molden-3 )
solid
Kc
Fe
+
,
"
included weather
"
Fe
( magma , ,
Kc
concentration remains
constant
( aq )
or
product
Agtcaq)
Fe
of
in
change
caqi
"
concentration
[ HIT
Kc
=
no
-
pressure
N2
+
lvtemh
0
=
& temperature
Temperature
2 HI
=
,
Kc
.
2MHz
tmtemh
equilibrium to RHS
-
Nz -131-12
unit
÷÷
31-12
on
>
conc to
:
•
•
.
concentration does not affect Kc
Pressure
doesn't
affect
>
Nat 31-12
conc to
conctn
Kc
.
equilibrium to LHS
-
2MHz
2MHz
To Restore temp
•
.
Cag ,
[ Hz ] [ Is ]
.
Zdmb
,
Hat Iz
•
>+
3)
,
reactant
in
Concentration } Pressure
•
Fe
+
-
solid because its
ignore
we
MOI
=
concentration
equilibrium
Cmd dm-
I
4
"
]
}
=
Ag Cs)
on
caq)
is not
=
Effect
•
expression
"
Agtcaq)
.
( MddM→ )
=
"
[
[Not [Hz]
In equilibrium
of reactants d. products is called
stoichiometry
2MHz
Equilibrium expression
[Nz] [1-12]
concentration
c-
.
=
[ Ntts]
&
concentration
( KC)
To
restore
Kc
Increases
conct
temp
Decreases
Kc
concentration of [ HI] 9-
equilibrium shifts to LHS
.
!
ONLY
0
TEMPRATURE
As
change in
to
ooieerse
change
.
its
AFFECTS
conc
Kc
& Pressure
forces
the reaction to
original state , only temperature causes
EQUILIBRIUM CONSTANT ( KP)
Partial :
The pressure
Pressure
The total
Ptotal
Equilibrium
EXAMPLE
exerted
by
one
any
pressure of a gas equals
Pa
=
+
expression
PB
P
in the
gas
to the
sum
is
mixture
of the
Pressure
-
partial pressures of the individual gases
Pc
+
partial pressures
involving
.
N2 (9)
+
31-12
=
(9)
(21-1419)
EXAMPLE
2MHz
'
H2O (9)
+
Kh
(9)
O_0
Equilibrium md
?
?
1-0
Initial md
20
Equilibrium md
, #mafia
Total pressure
( 214g OH (9)
-
2022am:*
o_0
to
1-0
1.0
Pot
equilibrium
at
\
y Eq
is
4014Pa
formed
.
moles
calculate Mole Fraction ( X)
✗
✗
"
"
"
Hao
( 21-14
=
( 21-14
=
^
total
Hao
=
total
£9M" brim moles
( 21-14
( of
)
I
=
( zgiiiiiln.ua?0ksat)
=
1+1+1
\
§
1-3
/
n
CZHSOH
CZHSOH
=
n
§
=
total
Calculate Partial Pressures
Pcztlq
✗ ↳ Hq
✗
Ptotal
=
113
P Hao
=
✗
✗
Ptotal
=
V3
pgµ , ,
=
ygµµ, xp,, ,
,
,
Kp
=
Hao
Pc2Hs0H
Pcztlq ✗ PHD
=
Use moles to
Calculate
mole
fraction
.
=
13-33
13-33×13-33
y,
Paid
=
pa×pa3
Calculate
mol left
left
Mol
'
H2O (9)
+
°Fkh
( 21-1,01-1
'
20
↳ Ha (9)
""
Pnyggx Piz
20
SOLUTION
ÑNHZ
=
Initial mol
✗
called it 's partial pressure
✗
40
✗ 40
✗
=
go
¥3
13.33
=
13.33 KPa
,
y.gg
}
KPa
=
ypa
-1
=
0.0750 Pa
Calculate
payygygyeggwy
} Find Kp
.
Calculate
KP
.
.
.
Acids & Bases
Strong Acid
into
HCl
+
Hao
+
A Base
-
NaOH
which
21-120
•
+
+
21-130
H2O
"
said
+
Nat Caq)
>
Base : Is
a
is
proton
a
donor
.
'
H2O
+
CH , COO
-
+
.
1-130-1
-
disassociates into ions
OH
+
-
.
Caq)
-
An acid
partially disassociates into ions
Acid which
CHZCOOH
Bronsted dowry Acids and Bases
Acid :
An
-
Ci
completely
which
Acid
.
1-130-1
•
Weak
completely disassociates
added to Water
when
ions
1-12504
Strong Base
Acid
An
-
Aap
( Ht)
Ctlz COOH Cag)
Acid
Base
H2O A)
-1
Conjugate
conjugate
Base
1-130-1 Cag)
-
→
cuz
Caql +
COO
•
proton accentor ( Ht)
I
1
conjugate
H+ donated
Acid Base
-
Pair
Measuring PH
The
PH scale
lower than
We
can
7
use
be used to
can
acidic
is
pH
a
indicate weather
a
,
meter
solution
a
with
solution
7
is
neutral
pH
is
the
fit
universal indicator
or
is acidic, alkali ,
,
a
or
neutral
solution
.
At
25C
pH
with
(or 298 K )
-
greater
a
than
solution
7
is
with
alkali
hH
.
.
-
3IV
0
I
2345
6
7 8 9 10 11
13 '
12
4-
concentration
pH
1-120
H+ caq)
A)
Equilibrium 1kW )
[ H+
=
can
log ,o[
an
aqeous
base
solution
Ht Caa!
it
has
] [ OH
OH
+
-
,
-
>
can ]
a
Degree
very
of
small
has
dissociation is
so
[1-12014]
remains constant
=
[Htcaa ] [ OH can
of
value
1-0
✗
10
"
at
298k
pure water the concentrations of H+ & OH
[ Htcaq) ] =/ OH law]
1-0×10-14 mddm→
-
=
-
ions
are
equal so
,
.
of the
10
hydrogen
.
Caq )
-
Kw ( Ionic Product)
kw
In
-
[1-12014]
Constant
:
=
in
to
Dissociation of water
€1B Yhe
☐
negative logarithm
no
units
.
ion
Calculating the pH of strong Bases Strong Acids Weak Bases } Weak Acids
,
Strong
acids
of acid
concentration
molecule
eg
ionised
( For
.
/ monohrotic
monobasic acids
,
of
concentration
in solution so
H+ ions
is
approximately the same
which contain only
acids
replaceable H+
one
as
the
ion
her
like HU )
pH
.
completing
are
of
(o / mddm- 3)
HCl
.
Htlaql + a- caq)
HCl lag)→
[1-14]=[1-1]
mddm-3.i.PH
-109,010.17--1
=
0.1
=
logo [ 1-17
-
strong bases
:
eg Calculate
NaOH (ag)
Kw
=
0.05
100×10+4
=
[ OH ]
pH
OH Cag)
+
0-05
[1-1+1101-5]
=
[1-1-1]
-
Nat Caq)
→
0.05
0.0500 mddm?
solution of NaOH of concentration
fit of a
the
.
Kw
=
=
-
=
-
log
-
2×10
=
"
-3
moldm
0.05
[1-1-1]
,o
log ,o [2×10-3]
12-7
=
Weak Acids
Weak
Acids
dissociate
( Hz COOH Cag )
Dissociation
of
HA 1991
only
'
-
a
partially in aq
CH , COO
weak Acid
=
CHZCOO
-
caqj
is
law
.
solution
+
:
Carbonic Acid (1-12103)
pKa
:
Ht can
=
[ Ht can ] [ A- can ]
(Acid Dissociation )
[ HA law
constant
=
(1-1+49)
The
Value
[ " A ""]
.
of the acid
.
acid
of
ka
indicates
the extent
OF
Ka
form
constants
by taking
the
can
be
expressed
negative logarithm
of dissociation
hka
=
-
.
tog oka
,
The
higher the value of Ka
The
lower
the
value
,
of pKa
the
,
the
in a
to base
"
(Equilibrium Expression for )
monroprotic
(CH} COOH)
Ythanoic Acid
Acid dissociation
→
convienent
more
10
Ka
,
Ht Caq)
represented by
+
Examples
.
stronger the acid
.
stronger the acid
.
Eg
50M
.
.
Calculate
pH
=
-
2. 9
=
[1-1-1]
=
.
Sol ?
of 0.010 motdm
}
ethanol
log ,o[Htcaq )]
1-26×10-3 molding }
[1-1-1]
?
[1-1-1]
=
[ HA ]
Eg
A solution
.
10-2-9
=
i.
acid
log,o[Htcaql]
-
=
Ka
for ethanol
of Ka
the value
Ka
=
Calculate
2
=
[CHZCOOH ]
I -59
✗
Ka
=
=
of
=
0.100
0.100 molars
=
[1-1-1]
( Ht )
}
mo1dm→
acid
methanol 'c
[1-1-1]
=
.
( Ka
pH
=
-
?
=
-
1-32×10-3 motocross
log ,o[ Ht ]
log ,o[ 1.32×10-3]
0.100
=
1-74×10 smokers } )
-
=
2
[ HA ]
1.74×10-5
1-59×10-4 mddm→
'
0.010
10-4 moldrri }
the pH
[1-11001-1]
(1-26×10-3)
2-88
acid ,
has
a
pH of 2^90
.
Buffer Solution
A
Buffer solution
is one
types of Buffer solutions
There are two
:
small amounts of acid d. alkali
changes in pH when
that resists
are added
.
-
1) Acidic Buffer Solution
It
from
commonly made
is
It has anti
Grande
weak acid & it's
a
conjugate base (salt)
7
-
Ethan oic acid of Sodium Ethanoate
CH } COOK
G- 1300 Nat
I
can
↳
( 143 COOH
2) Alkaline Buffer
It is
&
Cag )
4-13 COO
-
Caq )
.
Conjugate base lodnanoate Ion)
caq )
Solution
commonly made from
a
&
weak base
conjugate
it's
(its salt)
acid
It has a pH > 7
Example
NH } Cag )
Bd
&
HOW A Buffer solution Works
Consider
The
of Ammonia solution
Ammonium chloride solution
.
N
buffer
equilibrium that exsists
HA lag )
Ht Caq)
'
-
If some
HCl
+
containing
this
in
solution is
That means equilibrium shifts
IF some
HA Caq)
a
The OH
is
NaOH
+
OH
-
Equilibrium shifts
-
weak acid
added is
Can
to
'
and its
conjugate base
A-
.
-
to LHS
,
the
.
•
from acid
The
H+ added
with
A- can
+
in
the solution
is
"
mopped up
"
the
OH
the weak acid &
once
-
by reaction with the conjugate
'
by
reaction
reacts with
tweak acid)
HA
Hao CD
with
base
and
also because
in
the
solution
.
Pitts
mopped up
.
weak acid
extra
,
A- (conjugate base)
HA Can
pH doesn't change it remains stable
added to the solution
→
Ht reacts
H+ lag)
+
conjugate base
is
tweak acid )
HA
solution , extra
A- (aq)
HA formed
added *
are
A- Cag)
added to this
is
Equal Molar Proportions
:
general acidic
a
*
Hata lag)
-
again pH
doesn't change
,
remains stable
.
A common buffer used in the study
of biochemical
( Hzpoci )
the
acid
And
Hpoa
1-12 POI
caqj
.
,
"
acts
ions
as
a
prostrate buffer containing dihydrogen phosphate
reactions is a
conjugate base
ions
.
HPOÉ lag) + Ht Cag,
Blood hH
In order for the
between
&
7.35
body to function properly
human
7.45
.
,
7-8
greater than
Values
the
human
or
less than
body needs to
maintain
a
often result in death
6.8
tAHADmAAEA
Nor
I
6- 6
6-8
I
1
¥ 7.2
1
74
I
7.678
To prevent acidosis or alkalosis the body relies on the
The cells in our
body produce CO2
interaction
of
a
chemical
buffer system
.
CGH 206
,
602
+
>
602
61-120
+
glucose oxidation
via
?⃝
*
help of carbonic
anhydrase enzyme
Hzcozcaq)
In
blood
carbonic acid
[weak acid
Ht caq)
Hcoj
,
Hzcoz lag)
Now if
[1-1-1]
increases
if
[ Ht ]
decreases
,
,
'
-
equilibrium
equilibrium
shifts
shifts
to
+
Cag )
to
LHS
and
hit remains
RHS
and
keeps
the
pH
constant
constant
.
.
in the
blood
.
.
fit
Calculating the hit of a Buffer solution
6cg I
calculate
.
and
moldnri
0-250
solution
hit of a
the
}
containing
ethanoate
sodium
.
0-200 moldrñ}
ethanoic
acid
.
Solution
CH
199)
COOH
}
Ctlzcoo ( aq)
'
-
0.200
Mol
Ka
[ CH , COO ] [1-1-1]
=
0
-
+
H+
faq )
250m01
[ CH } COOH]
[1-1-1]
Ka [ CH > COOH]
=
=
1.74×10-5 ✗ 0250
[ CH } COO ]
-
pH
=
=
.
*
The Henderson
htl
So
=
-
pKa
:
-
log ,o[ Ht]
-
logo [1-39×10-5]
pH of this
Hasselbach
+
109,0
(
=
-
-
log
,,
Ka
+
=
Equation
10-5 mddrri }
200
4.86
hka
+
logo
[ conjugate base or salt]
[ weak acid]
[ HA ]
above we can use this formula
logo (0-2-50)
(0-200)
+
1-39 ✗
.
[A ]
logic (1-74×10-5)
=
4.86
buffer system is
for the same question
hH=
0
-
logo (0-250)
(0-200)
=
4-86
as
follows
:
( Ka
-
1-74×10-5 moidms )
Titration curves
① Calculate initial pH using
strong Acid
a
"
-
concentration whenever possible :eg
?⃝
a
13
"
HH
"
10-5
equivalence
at
7.0
-
o,;oÉ
i
-
•
sharp fall at
•
Graf
is
3- 5-
inverted
Weak Acid
HH equivalence
-
4
Curve flattens at ñ 3. g
Volume of NaOH
%
s
10
's
Y
's:
'
so
as
added 1cm
-
pH
-
÷
S
4
-
2
"
'
s
to
i
do
's
?
z
's
Jo
3
56
"
as
added 1cm
Volume of NaOH
•
Titration
Sharp Fall
No
Strong Acid
Curve
Final
equivalence
pH approaches
s
}
to
*
?
,
gradual Change No Indicator
,
suitable
-
Titration Curve
Weak Base
Y
"
ios
,
"
'
s
to
i
's
do
z
9
h"
-
7-
70
-
-
8
- -
-
4
-
-
z
Volume of NaOH
→
to
3
-
10.5
's
at
a
56
}
added 1cm
I
Ss
1
Dt
equivalence at
7.0
5.0
ñ
curve flattens at ñ 3. g
I
-
,
so
,
-
2
Required
fÉÉ
f
h"
-
g
's
70
-
-
6
Volume of NaOH
Shark fall from
,
10
l
i
-
,,
at a g. o
Curve flattens at ñ
13
13
:
-
-
o
-
o
.
Éu
- -
-
do
~
-
,
¥
-
7-
's
point
"
-
9
7. I
5
-
Strong Base
-
do
-
2
Required
→
at
-
3
vertically when strong acid is
-
BAM
-
b
I
12
i
-
g
?
"
13
g
curve
-
?⃝
-
added to base instead
pH
logo (0-1)=1
-
"
-
Volume of NaOH
o
-
-
a
,
11
hH=
Titration
Weak base
-
,
"
h"
- -
-
i
ñ
till
"" " "" " "
-
4
"
Weak acid
-
6-
z
""
/
"
7-
s
0-100 mddrris
Ht is
FG
Iat
g
.
Concentration
.
Titration Curve
Strong base
"
pH
{ Indicators
l
,
-
"
0
'
s
to
i
's
?
do
to
Volume of NaOH
ME
Sharp Fall
3.5
-
7-0
3
's
at
a
56
}
added 1cm
?
s
's
,
so
Bt
Indicators
Equivalence Point
:
The
acid & alkali
colour
pH Range
because
Indicator
methyl
3. I
bromo phenol
-
not
of an indicator
Yellow
4- 6
Yellow
Blue
6.3
Red
Yellow
phenol
6- 8- 8. a
Yellow
Red
82
Colorless
Pink
Example
:
-
10.0
weak Acid
-
-
same
pH values
ionised &
ionised
Strong Base titration
since Addition of a
base will
single dsoh of
change pH it is
important to use
suitable
indicator
r
.
point
at which
equivalent
of a titration is the
point
between
indicator
numbers
of
at which the indicator
moles
forms
which the
are
present
.
has
intermediate
of
changes
.
Alkali Color
Acid Color
4.2
Phenol hhtalein
un
Red
-
the
is the
of
the
is
The end point
necessarily
mothy red
red
.
4.4
2- 9-
blue
are
compare able amounts
pH Range
orange
have been added
these
-
The pH range
:
point of a titration
equivalence
colours,