NEET  >  Revision Notes: Equilibrium

# Revision Notes: Equilibrium - Notes | Study Chemistry Class 11 - NEET

``` Page 1

? Equilibrium : It is a state in a process when two opposing processes
(forward and reverse) occur simultaneously at the same rate. The free
energy change at equilibrium state is zero i.e., ?G = 0.
? Equilibrium constant : For a general reaction :
aA + bB
???
? ??  cC + dD
K
c
=  and K
p
=
? Relationship between K
p
and K
c
:
K
p
= K
c
(RT)
?n
g
?n
g
= n
p
(g) – n
r
(g)
? Magnitude of equilibrium constant depends upon the way in which a
reaction is written :
Chemical equation Equilibrium constant
aA + bB
???
? ?? cC + dD K
cC + dD
???
? ??  aA + bB K
1
=
naA + nbB
???
? ?? ncC + ndD K
2
= K
n
1 1 11
aA bB cC dD
n n nn
+ +
K
3
= K
1/n
? Predicting the direction of reaction :
If Q
c
= K
c

? the reaction is in a state of equilibrium.
Q
c
> K
c

? the reaction proceeds in reverse direction.
Q
c
< K
c

? the reaction proceeds in forward direction.
???
? ??
Page 2

? Equilibrium : It is a state in a process when two opposing processes
(forward and reverse) occur simultaneously at the same rate. The free
energy change at equilibrium state is zero i.e., ?G = 0.
? Equilibrium constant : For a general reaction :
aA + bB
???
? ??  cC + dD
K
c
=  and K
p
=
? Relationship between K
p
and K
c
:
K
p
= K
c
(RT)
?n
g
?n
g
= n
p
(g) – n
r
(g)
? Magnitude of equilibrium constant depends upon the way in which a
reaction is written :
Chemical equation Equilibrium constant
aA + bB
???
? ?? cC + dD K
cC + dD
???
? ??  aA + bB K
1
=
naA + nbB
???
? ?? ncC + ndD K
2
= K
n
1 1 11
aA bB cC dD
n n nn
+ +
K
3
= K
1/n
? Predicting the direction of reaction :
If Q
c
= K
c

? the reaction is in a state of equilibrium.
Q
c
> K
c

? the reaction proceeds in reverse direction.
Q
c
< K
c

? the reaction proceeds in forward direction.
???
? ??
? Ostwald’s dilution law : Degree of dissociation of weak electrolyte,
a =
? Ionic Product of water (K
w
) = [H
3
O
+
] [OH
–
] = 10
–14
at 298K
? Le-Chatelier’s Principle : When a system of equilibrium is subjected to
a change in temperature, pressure or concentration, the equilibrium shifts
itself in such a way so as to undo or nullify the effect of change.
? Outcomes of Le-Chatelier’s Principle
Change at equilibrium Shift in equilibrium
Increase in temperature Endothermic direction
Decrease in temperature Exothermic direction
Increase in pressure Towards lesser gaseous moles
Decrease in pressure Towards greater gaseous moles
Increase in Conc. of reactants Forward direction
Increase in Conc. of products Reverse direction
? Conjugate Acid or Base : Acid-base pair which differ by H
+
ion.
Species – H
+
= Conjugate base
Species + H
+
= Conjugate acid
? pH of solution :
pH = –log [H
3
O
+
] or [H
+
] = 10
–pH
,

pOH = –log [OH
–
]
pH + pOH = pK
w
= 14 at 298K
? Common ion effect : The depression of ionisation of weak electrolyte by
the presence of common ion from a strong electrolyte is called common
ion effect. For example degree of dissociation of NH
4
OH decreases in the
presence of strong electrolyte NH
4
Cl.
? Hydrolysis of salts and pH of their solutions : Hydrolysis of salt is
defined as the reaction of cation or anion with water as a result of which
the pH of water changes.
1. Salts of strong and strong bases (e.g., NaCl) do not hydrolyse. The
solution pH will be 7.
2. Salts of weak acids and strong bases (e.g., CH
3
COONa) hydrolyse,
pH >7 (The anion acts as a base).
X
–
+     H
2
O
???
? ??          HX          +        OH
–
(Weak acid)         (Weak base)
Page 3

? Equilibrium : It is a state in a process when two opposing processes
(forward and reverse) occur simultaneously at the same rate. The free
energy change at equilibrium state is zero i.e., ?G = 0.
? Equilibrium constant : For a general reaction :
aA + bB
???
? ??  cC + dD
K
c
=  and K
p
=
? Relationship between K
p
and K
c
:
K
p
= K
c
(RT)
?n
g
?n
g
= n
p
(g) – n
r
(g)
? Magnitude of equilibrium constant depends upon the way in which a
reaction is written :
Chemical equation Equilibrium constant
aA + bB
???
? ?? cC + dD K
cC + dD
???
? ??  aA + bB K
1
=
naA + nbB
???
? ?? ncC + ndD K
2
= K
n
1 1 11
aA bB cC dD
n n nn
+ +
K
3
= K
1/n
? Predicting the direction of reaction :
If Q
c
= K
c

? the reaction is in a state of equilibrium.
Q
c
> K
c

? the reaction proceeds in reverse direction.
Q
c
< K
c

? the reaction proceeds in forward direction.
???
? ??
? Ostwald’s dilution law : Degree of dissociation of weak electrolyte,
a =
? Ionic Product of water (K
w
) = [H
3
O
+
] [OH
–
] = 10
–14
at 298K
? Le-Chatelier’s Principle : When a system of equilibrium is subjected to
a change in temperature, pressure or concentration, the equilibrium shifts
itself in such a way so as to undo or nullify the effect of change.
? Outcomes of Le-Chatelier’s Principle
Change at equilibrium Shift in equilibrium
Increase in temperature Endothermic direction
Decrease in temperature Exothermic direction
Increase in pressure Towards lesser gaseous moles
Decrease in pressure Towards greater gaseous moles
Increase in Conc. of reactants Forward direction
Increase in Conc. of products Reverse direction
? Conjugate Acid or Base : Acid-base pair which differ by H
+
ion.
Species – H
+
= Conjugate base
Species + H
+
= Conjugate acid
? pH of solution :
pH = –log [H
3
O
+
] or [H
+
] = 10
–pH
,

pOH = –log [OH
–
]
pH + pOH = pK
w
= 14 at 298K
? Common ion effect : The depression of ionisation of weak electrolyte by
the presence of common ion from a strong electrolyte is called common
ion effect. For example degree of dissociation of NH
4
OH decreases in the
presence of strong electrolyte NH
4
Cl.
? Hydrolysis of salts and pH of their solutions : Hydrolysis of salt is
defined as the reaction of cation or anion with water as a result of which
the pH of water changes.
1. Salts of strong and strong bases (e.g., NaCl) do not hydrolyse. The
solution pH will be 7.
2. Salts of weak acids and strong bases (e.g., CH
3
COONa) hydrolyse,
pH >7 (The anion acts as a base).
X
–
+     H
2
O
???
? ??          HX          +        OH
–
(Weak acid)         (Weak base)
pH  =    (pK
a
+ log C)
3. Salt of strong acids and weak bases (e.g., NH
4
Cl) hydrolyse, pH < 7.
(The cation acts as an acid).
M
+
+ H
2
O
???
? ??  MOH + H
+
pH  =  (pK
b
+ logC)
4. Salt of weak acids and weak base (e.g., CH
3
COONH
4
) hydrolyse.
The cation acts as an acid and anion as a base but whether the solution is
acidic or basic depends upon the relative values of K
a
and K
b
for these
ions.
M
+
+ X
–
+ H
2
O
???
? ??  MOH + HX
pH =  (pK
a
– pK
b
)
? Buffer solutions : The solutions, which resist the change in pH on dilution
or addition of small amounts of acid or base, are called buffer solutions.
? Basic buffer : Solution of weak base and its salt with strong acid, For
e.g., NH
4
OH + NH
4
Cl
? Acidic buffer : Solution of weak acid and its salt with strong base, For
e.g., CH
3
COOH + CH
3
COONa.
? Henderson Hasselbalch Equation for the pH of Buffer solution—
pH = pK
a
+ log (for acidic buffer)
pOH = pK
a
+ log (for basic buffer)
? Solubility Product (K
sp
) : The equilibrium constant that represent the
equilibrium between undissolved salt (solute) and its ions in a saturated
solution is called solubility product constant (K
sp
).
For  A
x
B
y
xA
y+
+ yB
x–
K
sp
= [A
y+
]
x
[B
x–
]
y
= (xs)
x
(ys)
y
= x
x
. y
x
. s
(x+y)
where s = Molar solubility
If ionic product < K
sp
; salt remain dissolve.
If ionic product > K
sp
; salt will be precipitated.
+
Page 4

? Equilibrium : It is a state in a process when two opposing processes
(forward and reverse) occur simultaneously at the same rate. The free
energy change at equilibrium state is zero i.e., ?G = 0.
? Equilibrium constant : For a general reaction :
aA + bB
???
? ??  cC + dD
K
c
=  and K
p
=
? Relationship between K
p
and K
c
:
K
p
= K
c
(RT)
?n
g
?n
g
= n
p
(g) – n
r
(g)
? Magnitude of equilibrium constant depends upon the way in which a
reaction is written :
Chemical equation Equilibrium constant
aA + bB
???
? ?? cC + dD K
cC + dD
???
? ??  aA + bB K
1
=
naA + nbB
???
? ?? ncC + ndD K
2
= K
n
1 1 11
aA bB cC dD
n n nn
+ +
K
3
= K
1/n
? Predicting the direction of reaction :
If Q
c
= K
c

? the reaction is in a state of equilibrium.
Q
c
> K
c

? the reaction proceeds in reverse direction.
Q
c
< K
c

? the reaction proceeds in forward direction.
???
? ??
? Ostwald’s dilution law : Degree of dissociation of weak electrolyte,
a =
? Ionic Product of water (K
w
) = [H
3
O
+
] [OH
–
] = 10
–14
at 298K
? Le-Chatelier’s Principle : When a system of equilibrium is subjected to
a change in temperature, pressure or concentration, the equilibrium shifts
itself in such a way so as to undo or nullify the effect of change.
? Outcomes of Le-Chatelier’s Principle
Change at equilibrium Shift in equilibrium
Increase in temperature Endothermic direction
Decrease in temperature Exothermic direction
Increase in pressure Towards lesser gaseous moles
Decrease in pressure Towards greater gaseous moles
Increase in Conc. of reactants Forward direction
Increase in Conc. of products Reverse direction
? Conjugate Acid or Base : Acid-base pair which differ by H
+
ion.
Species – H
+
= Conjugate base
Species + H
+
= Conjugate acid
? pH of solution :
pH = –log [H
3
O
+
] or [H
+
] = 10
–pH
,

pOH = –log [OH
–
]
pH + pOH = pK
w
= 14 at 298K
? Common ion effect : The depression of ionisation of weak electrolyte by
the presence of common ion from a strong electrolyte is called common
ion effect. For example degree of dissociation of NH
4
OH decreases in the
presence of strong electrolyte NH
4
Cl.
? Hydrolysis of salts and pH of their solutions : Hydrolysis of salt is
defined as the reaction of cation or anion with water as a result of which
the pH of water changes.
1. Salts of strong and strong bases (e.g., NaCl) do not hydrolyse. The
solution pH will be 7.
2. Salts of weak acids and strong bases (e.g., CH
3
COONa) hydrolyse,
pH >7 (The anion acts as a base).
X
–
+     H
2
O
???
? ??          HX          +        OH
–
(Weak acid)         (Weak base)
pH  =    (pK
a
+ log C)
3. Salt of strong acids and weak bases (e.g., NH
4
Cl) hydrolyse, pH < 7.
(The cation acts as an acid).
M
+
+ H
2
O
???
? ??  MOH + H
+
pH  =  (pK
b
+ logC)
4. Salt of weak acids and weak base (e.g., CH
3
COONH
4
) hydrolyse.
The cation acts as an acid and anion as a base but whether the solution is
acidic or basic depends upon the relative values of K
a
and K
b
for these
ions.
M
+
+ X
–
+ H
2
O
???
? ??  MOH + HX
pH =  (pK
a
– pK
b
)
? Buffer solutions : The solutions, which resist the change in pH on dilution
or addition of small amounts of acid or base, are called buffer solutions.
? Basic buffer : Solution of weak base and its salt with strong acid, For
e.g., NH
4
OH + NH
4
Cl
? Acidic buffer : Solution of weak acid and its salt with strong base, For
e.g., CH
3
COOH + CH
3
COONa.
? Henderson Hasselbalch Equation for the pH of Buffer solution—
pH = pK
a
+ log (for acidic buffer)
pOH = pK
a
+ log (for basic buffer)
? Solubility Product (K
sp
) : The equilibrium constant that represent the
equilibrium between undissolved salt (solute) and its ions in a saturated
solution is called solubility product constant (K
sp
).
For  A
x
B
y
xA
y+
+ yB
x–
K
sp
= [A
y+
]
x
[B
x–
]
y
= (xs)
x
(ys)
y
= x
x
. y
x
. s
(x+y)
where s = Molar solubility
If ionic product < K
sp
; salt remain dissolve.
If ionic product > K
sp
; salt will be precipitated.
+
88
? Relationship between solubility (s) and solubility product (K
sp
).
K
sp
= x
x
.y
y
. s
x + y
For binary salts (e.g., AgCl, AgBr, AgI) K
sp
= s
2
For Ternary salts (e.g., PbI
2
) K
sp
= 4s
3
```

## Chemistry Class 11

204 videos|331 docs|229 tests

## Chemistry Class 11

204 videos|331 docs|229 tests

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