Chapter 7 Equilibrium chemistry class 11 -ppt.pdf

this chapter is based on physical chemistry of class 11

1. Chemical Equilibrium

2. Copyright McGraw-Hill 2009
Double arrows ( ) denote an equilibrium reaction.

15.1 The Concept of Equilibrium
Most chemical reactions are reversible.
reversible reaction= a reaction that proceeds
simultaneously in both directions
Examples:)(NO 2 )(ON
242 gg
 )(NH 2 )(H 3 )(N
322 ggg 

 )OH(CH )(H 2 )CO(
32 ggg 



3. Copyright McGraw-Hill 2009
Equilibrium)(NO 2 )(ON
242 gg

Consider the reaction
At equilibrium,
the forward reaction: N
2
O
4
(g) 2 NO
2
(g), and
the reverse reaction: 2 NO
2
(g) N
2
O
4
(g)
proceed at equal rates.
Chemical equilibria are dynamic, not static –the
reactions do not stop.

4. Copyright McGraw-Hill 2009
Equilibrium
Let’s use 2 experiments to study the reaction
each starting with a different reactant(s).)(NO 2 )(ON
242 gg

Exp #2
pure NO
2
Exp #1
pure N
2
O
4

5. Copyright McGraw-Hill 2009
Equilibrium
Experiment #1)(NO 2 )(ON
242 gg


6. Copyright McGraw-Hill 2009
Equilibrium
Experiment #2)(NO 2 )(ON
242 gg


7. Copyright McGraw-Hill 2009
Equilibrium)(NO 2 )(ON
242 gg

Are the equilibrium pressures of NO
2
and N
2
O
4
related? Are they predictable?

8. Copyright McGraw-Hill 2009
15.2 The Equilibrium Constant

rate
forwardrate
reverse
At equilibrium,2
f 2 4 eq r 2 eq
[N O ] [NO ]kk  )(NO 2 )(ON
242 gg

or2
2 eqf
c
r 2 4 eq
[NO ]
[N O ]
k
K
k

where K
cis the equilibrium constant

9. Copyright McGraw-Hill 2009
The Equilibrium Constant
This constant value is termed
the equilibrium constant,
K
c, for this reaction at 25°C.

10. Copyright McGraw-Hill 2009
The Equilibrium Constant
For the NO
2
/ N
2
O
4
system:
equilibrium constant expression
equilibrium constant2
4
[NO ]
O]
K  
2
2
0.143 at 25 C
[N )(NO 2 )(ON
242 gg

Note: at100°C, K=
6.45

11. Copyright McGraw-Hill 2009
The Equilibrium Constant
reaction quotient= Q
c= the value of the
“equilibrium constant expression” under
anyconditions.)(COCl )(Cl )CO(
22 ggg 


For,2 eq
c
eq 2 eq
[COCl ]

[CO] [Cl ]
K 2
c
2
[COCl ]

[CO][Cl ]
Q
Q> Kreverse reaction favored
Q= Kequilibrium present
Q< Kforward reaction favored

12. Copyright McGraw-Hill 2009
The Equilibrium Constant
The Law of Mass Action:
Cato Maximilian Guldberg & Peter Waage,
Forhandlinger: Videnskabs-Selskabet i
Christiana1864,35.K
cd
c ab
[C] [D]

[A] [B]
For a reaction:D C B A dcba 

 CD
AB
P
PP
K
PP



cd
ab

For gases:
For solutions: [ ] = mol/L
Pin atm

13. Copyright McGraw-Hill 2009
The Equilibrium Constant
Note:
•The equilibrium constant expression has products
in the numerator, reactantsin the denominator.
•Reaction coefficients become exponents.
•Equilibrium constants are temperature dependent.
•Equilibrium constants do nothave units. (pg. 622)
•If K>>> 1, productsfavored (reaction goes
nearly to completion).
•If K<<< 1, reactantsfavored (reaction hardly
proceeds).

14. . .
Relationship between K
cand K
p
•Relationship between concentration and pressure obtained from the
ideal gas law.
–Assuming all the species behaving like ideal gasses
–PV = nRTor P=(n/V)RT=[Conc]RT
–Substitute for P in equilibrium expression.
–Consider the reaction:
aA(g) + bB(g) cC(g) + dD(g)RT]A[
RT
V
n
P
A
A

   
  


 
n
c
)ba()dc(
cP
baba
dcdc
ba
dc
b
B
a
A
d
D
c
C
P
RTKRTKK
RT]B[]A[
RT]D[]C[
RT]B[RT]A[
RT]D[RT]C[
PP
PP
K







15. Copyright McGraw-Hill 2009
15.3 Equilibrium Expressions
homogeneous equilibria= equilibria in which all
reactants and products are in the same phase.
•[CaO] and [CaCO
3
] are solids.
•Pure solids and liquids are omittedfrom equilibrium
constant expressions.)(CO )CaO( )(CaCO
23 gss 

Ex:
The equilibrium constant expression is,
K= [CO
2]
heterogeneous equilibria= equilibria in which all
reactants and products are notin the same phase.

16. .
.
Applications of the
Equilibrium Constant
•Extent of reaction: The magnitude of the equilibrium constant
allows us to predict the extent of the reaction.
–Very large K (e.g. 10
10
) mostly products.
–Very small K (e.g. 10
10
) mostly
reactants.
–When K is around 1, a significant amount
of reactant and product present in the
equilibrium mixture.
E.g. For each of the following decide which species will
predominate at equilibrium.AgCl(s)  Ag
+
(aq) + Cl

(aq) Ksp = 1.8x10
10

Ag
+
(aq) + 2NH3(aq)  
23)Ag(NH
7
f 10x7.1K
H2CrO4(aq) + H2O(l)  
4HCrO
(aq) +
H3O
+
(aq)
Ka1 = 0.15

17. . .
Using the Equilibrium
Constant
•Direction of Reaction: the reaction quotient can be used to determine
the direction of a reaction with certain initial concentrations.
•Reaction Quotient: For the general reaction,
aA + bB cC + dD,.
where the t refers to the time that concentrations of the mixture are
measured; not necessarily at equilibrium.
•Comparison of Q
cwith K
creveals direction of reaction.
•When only reactants Q
c= 0; leads to
–IfQ
c< K
c, products will form.
•When only products present, Q
c.
–If Q
c> K
c, reactants will form.
•When Q
c= K
c, no net reaction.
E.g. Determine direction of reaction: H
2(g) + I
2(g) 2HI(g). Assume
that [H
2]
o= [I
2]
o= [HI]
o= 0.0020M at 490
o
C for which K
c= 46. ba
c
t
d
t
c
[B][A]
[C][D]
Q
tt


18. Copyright McGraw-Hill 2009
Exercise:Write the expressions for K
p
for the
following reactions:)O(H 2 )O(N )(NONH (a)
2234 ggs 
 )(CuCl )(Cl )Cu( (b)
22 sgs 


Solution: 
22
2
N O H OP
K P P

(a) 2
Cl
1
P
K
P
(b)

19. Copyright McGraw-Hill 2009
Equilibrium Expressions
A. Reverse Equations[1] )(NO 2 )(ON
242 gg

For,
Conclusion:2
1
1
K
K
 C
1
6.99 at 25
0.143 
C
P
K
P

2
24
NO
1
NO
2
0.143 at 25 [2] )(ON )(NO 2
422 gg

For,
24
2
NO
NO
P
K
P

2 2

20. Copyright McGraw-Hill 2009
Equilibrium Expressions
B. Coefficient Changes[1] )(NO 2 )(ON
242 gg

For,
Conclusion:31
KK C 0.143 0.378 at 25
For,[3] )(NO )(ON
2
1
242 gg
  
2
24
NO
1/2
NO
P
K
P

3

C
P
K
P

2
24
NO
1
NO
2
0.143 at 25

21. Copyright McGraw-Hill 2009
Equilibrium Expressions
C. Reaction Sum (related to Hess’ Law) [1] )(NO 2 )(ON
242 gg

For,
For,[4] )(O )NO( 2 )(NO 2
22 ggg 
 

2
2
NO O
NO
PP
K
P

2
4 2

Add [1] + [4], [5] )(O )NO( 2 )(ON
242 ggg 
 
2
24
NO O
NO
PP
K
P

2
5
42
2
ON
NO
1
P
P
K
2
 KK
14

22. Copyright McGraw-Hill 2009
Equilibrium Expressions

23. Copyright McGraw-Hill 2009
Exercise:At 500ºC, K
P
= 2.5 10
10
for,)(SO 2 )(O )(SO 2
322 ggg 


Compute K
P
for each of the following:
(a) At 500ºC, which is more stable, SO
2
or SO
3
?
(g)O
2
1
(g)SO(g)SO(d)
223


(g)SO(g)O(g)SO(b)
322 


2
1
(g)SO3(g)O(g)SO3(c)
322 


2
3

24. Copyright McGraw-Hill 2009
15.4 Using Equilibrium
Expressions to Solve Problems
Q> Kreverse reaction favored
Q= Kequilibrium present
Q< Kforward reaction favored
Predicting the direction of a reaction
Compare the computed value of Qto K

25. Copyright McGraw-Hill 2009
Exercise #1:At 448°C, K= 51 for the reaction,)HI( 2 )(I )(H
22 ggg 


Predict the directionthe reaction will proceed, if at
448°C the pressures of HI, H
2
, and I
2
are 1.3,
2.1 and 1.7 atm, respectively.
Solution:
22
2
HI
HI

P
Q
PP

 0.47
)7.1()1.2(
)3.1(

2



0.47 < 51system not at equilibrium
Numerator must increase and denominator must
decrease.
Consequently the reaction must shift to the right.

26. Copyright McGraw-Hill 2009
Exercise #2:At 1130°C, K= 2.59 10
2
for)(S )(H 2 )S(H 2
222 ggg 

At equilibrium, P
H
2
S
= 0.557 atmand P
H
2
= 0.173 atm,
calculate P
S
2
at 1130°C.
Solution:

22
2
2
HS 2
2
HS
2.59 10
PP
K
P


  
P
S
2
= 0.268 atm2
2
S 2
2
(0.173)
2.59 10
(0.557)
P


  

27. Copyright McGraw-Hill 2009
Exercise #3:K= 82.2 at 25°Cfor,
Initially, P
I2
= P
Cl2
= 2.00 atm and P
ICl
= 0.00 atm.
What are the equilibrium pressures of I
2
, Cl
2
, and ICl?
Solution:)ICl( 2 )(Cl )(I
22 ggg 


Initial 2.00atm2.00atm 0.00atm
Change x x +2x
Equilibrium(2.00 –x)(2.00 –x) 2x22
2
ICl
I Cl

P
K
PP


perfect square2
(2 )
82.2
(2.00 )(2.00 )
x
xx

 )ICl( 2 )(Cl )(I
22 ggg 



28. Copyright McGraw-Hill 2009(2 )
9.066
(2.00 )
x
x


square root 
2x= 18.132 –9.066x
11.066x= 18.132
x= 18.132 / 11.066 = 1.639
P
I2
= P
Cl2
= 2.00 –x= 2.00 –1.639 = 0.36 atm
P
ICl
= 2x= (2)(1.639) = 3.28 atm2
(2 )
82.2
(2.00 )(2.00 )
x
xx


Exercise #3:(cont.)

29. Copyright McGraw-Hill 2009
Exercise #4:At 1280°C, K
c
= 1.1 10
3
for)Br( 2 )(Br
2 gg

Initially, [Br
2
] = 6.3 10
2
Mand [Br] = 1.2 10
2
M. What are the equilibrium concentrations of Br
2
and Br at 1280°C?
Initial 6.3 10
2
M1.2 10
2
M
Change -x +2x
Equilibrium (6.3 10
2
) -x(1.2 10
2
) + 2x
Solution:)Br( 2 )(Br
2 gg
 2 2 2
3
c 2
2
[Br] [(1.2 10 ) 2 ]
1.1 10
[Br ] (6.3 10 )
x
K
x




   

4x
2
+ 0.0491x+ (7.47 10
5
) = 0

30. Copyright McGraw-Hill 2009
4x
2
+ 0.0491x+ (7.47 10
-5
) =
0
quadratic equation:ax
2
+ bx+ c= 02
4

2
b b ac
x
a
  

solution:
x= 1.779 10
3
and 1.050 10
2
Q:Two answers? Both negative? What’s happening?
Equilibrium Conc.x=1.779 10
3
1.050 10
2
[Br
2
]= (6.3 10
2
) –x= 0.0648 M 0.0735 M
[Br]= (1.2 10
2
) + 2x=0.00844 M0.00900 M
impossible
[Br
2
]= 6.5 10
2
M
[Br]= 8.4 10
3
M

31. Copyright McGraw-Hill 2009
Exercise #5:A pure NO
2
sample reacts at 1000 K,
K
P
is 158. If at 1000 K the equilibrium partial
pressure of O
2
is 0.25 atm, what are the equilibrium
partial pressures of NO and NO
2
.)(O )NO( 2 )(NO 2
22 ggg 

Solution:)g(O )NO( 2 )(NO 2
22 

gg
Initial ? 0 atm 0 atm
Change
Equilibrium 0.25 atm
+0.25+0.50
+0.50 atm
0.50
 
2
22
2
2
NO O
22
NO NO
(0.50) (0.25)
158
P
PP
K
PP
  

rearrange and solve
P
NO2

32. Copyright McGraw-Hill 2009
2
2
NO
(0.50) (0.25)
158
P

2 
2
P 
2
2
NO
(0.50) (0.25)

158 P

  
2
4
NO
3.956 10 0.01989
Exercise #5:(cont.)
= 3.956 10
4
P
NO2
= 0.020 atm
P
NO
= 0.50 atm
see ICE table

33. Copyright McGraw-Hill 2009
Exercise #6:The total pressure of an equilibrium
mixture of N
2
O
4
and NO
2
at 25°C is 1.30 atm.
For the reaction:
K
P
= 0.143 at 25°C. Calculate the equilibrium
partial pressures of N
2
O
4
and NO
2
.)(NO 2 )(ON
242 gg
 
2
24
2
NO
NO
0.143
P
P
K
P

P
NO2
+ P
N
2
O
4
= 1.30 atm
two equations and two unknowns –BINGO!

34. Copyright McGraw-Hill 2009
P
NO2
2
+0.143P
NO2
0.1859=
0
P
N
2
O
4
= 1.30 atm -P
NO2
2
2
2
NO
NO
0.143
(1.30 )
P
P

 
2
24
2
NO
NO
0.143
P
P
K
P

P
NO2
+ P
N
2
O
4
= 1.30 atm
Exercise #6:(cont.)
Use the quadratic formula,
P
NO2
= +0.366atm and 0.509atm
P
N
2
O
4
=1.30atm-P
NO2
=1.300.366=0.934atm
P
N
2
O
4
=0.93 atm

35. Le Châtelier’s principle:
restoring balance

36. Factors Affecting an Equilibrium System
Equilibrium represents a balance between the
reactants and the products of a chemical reaction.
Changes to the conditions of the system can
disrupt that equilibrium.
When this occurs, the system reacts in such a
way as to restore the equilibrium.

37. Stresses to a chemical system include:
•changes in the concentrations of reactants
•changes in the concentrations of products
•changes in the temperature of the system
•changes in the pressure of the system
Factors Affecting an Equilibrium System

38. Change in Concentration
A change in the concentration of one of the substances
in an equilibrium system typically involves either the
additionor the removalof one of the reactants or
products.
If the concentration of one substance in a system is
increased by additionor decreased by removal, the
system will respond by favoring the reaction that
consume or produce that substance, reestablishing
equilibrium.

40. Change in Temperature
Increasing or decreasing the temperature of a
system at equilibrium is also a stress to the
system.
An increase in the temperature of a system favors
the reaction that absorbs heat, the endothermic
reaction.
A decrease in the temperature of a system favors
the reaction that releases heat: the exothermic
reaction.
exothermic reaction

41. Change in Pressure
Changing the pressure of an equilibrium system in which
gases are involved is also a stress to the system.
The reaction system contains primarily
N
2and H
2, with only one molecule of
NH
3present.

42. Change in Pressure
As the piston is pushed inward, the
same number of molecules are confined
to a smaller space, so the pressure of
the system increases.
According to Le Châtelier’s principle, the
system responds in order to relieve the
stress.

43. Change in Pressure
On the contrary, if the pressure of the system is decreased, the
equilibrium would respond by favoring the reverse reaction, in
which NH
3decomposes to N
2and H
2. This is because the
overall number of gas molecules would increase and so would
the pressure.
The forward reaction has been favored,
in which three molecules of N
2combine
with nine molecules of H
2to form six
molecules of NH
3. The overall result is
a decrease in the number of gas
molecules in the entire system.
This decreases the pressure and
counteracts the original stress of a
pressure increase.

44. Use of Catalyst
A catalystis a substance (like iridium) that speeds up
the rateof a reaction.
It has equal effects on the rates of the forward and
reverse reactions, so for a system at equilibrium, these
two rates remain equal.
A system will reach equilibrium more quickly in the
presence of a catalyst, but the equilibrium position
itself (the equilibrium constant) is unaffected.

45. Lesson Summary
• A system at equilibrium can be disrupted by a change in concentration of one of the substances or by a
change in temperature or pressure.
Le Châtelier’s principle states that such a system will respond by attempting to counteract the stress.
Either the forward or reverse reaction will temporarily be favored until equilibrium is reestablished.
•The effects of various stresses on a system at equilibrium are summarized in the table below:
Stress Response
additionofreactant forwardreactionfavored
additionofproduct reverse reactionfavored
removalofreactant reverse reactionfavored
removalofproduct forwardreactionfavored
temperature increase endothermicreactionfavored
temperature decrease exothermicreactionfavored
pressureincrease reactionthatproducesfewergas moleculesfavored
pressuredecrease reactionthatproducesmore gas moleculesfavored
presenceofcatalyst reactionratesincreaseswithno variationofKeq

46. Le Chatelier’s principle in pratice:
In the reaction 2SO3 + heat 2SO2 + O2, an increase
in the concentration of O2will produce
1. an increase in the forward reaction
2. an increase in the reverse reaction
3. an increase in the formation of SO3
4. no effect on Keq

47. Ionic Equilibria(I)
Acids andbases

48. Ionic Equilibria(I)
1. How do we identify acids andbases?

49. Ionic
Equilibria(I)
An acidis a proton (H
+)donor.•
•
•
A baseis proton acceptor.
In an acid-base reaction, the transfer of protons occurs
from an acid to a base.
1. How do we identify acids andbases?
Bronsted-Lowry Definition
Tip:It may be helpful for beginners to memorise some
common acids and bases.

50. Ionic Equilibria(I)
2. How do we identify conjugate acids andbases?

51. Ionic
Equilibria(I)
2. How do we identify conjugate acids andbases?
Acid ConjugateBase
–H
+
Base ConjugateAcid
+H
+
+H
+
–H
+
Example:
HCl Cl
–
NH
3 NH
4
+

52. Ionic Equilibria(I)
3. How do we know whether an acid/ base is
strong or weak?

53. Ionic
Equilibria(I)
3.How do we know whether an acid/ base is
strong or weak?
Strong acids/ bases ionise completelyin aqueous solution.
Weak acids/ bases ionise partiallyin aqueous solution.
Tip: An acid/ base can be predicted to be weakwhen:
•It is stated so in the question
•K
a or K
b value is given
•Degree/ percent of ionisation is given
•[H
3O
+
] < [HA]; [OH
–
] < [B]

54. Ionic Equilibria(I)
4. How do we calculate pH of an acid/base?

55. Ionic
Equilibria(I)
4.How do we calculate pH of an acid/base?
(I)Learn the important terms/relations
•pH,pOH
•K
a,pK
a
•K
b,pK
b
•K
w
pH + pOH =14
pK
a + pK
b =14
[H
+][OH
–] = 10
-14
K
a . K
b = K
w = 10
-14

56. Ionic
Equilibria(I)
4. How do we calculate pH of an acid/base?
(II) Determine strong orweak
Strong acids ionise completelyin aqueous solution.
[H
3O
+] = [HA] pH = -log [HA]
Strong bases ionise completelyin aqueous solution.
[OH
-] = [B] pOH = -log [B]
StrongAcid
StrongBase

57. Ionic
Equilibria(I)
4. How do we calculate pH of an acid/base?
(II) Determine strong orweak
Weak bases ionise partiallyin aqueoussolution.
[OH
-
]<[B]  pOH = -log[OH
–
]
WeakAcid
WeakBase
Weak acids ionise partiallyin aqueoussolution.
[H
3O
+
]<[HA] pH = -log[H
3O
+
][H
3O
+
]=K
a ×C
[HA]
[OH
–
]=K
b×C
[B]

58. Ionic Equilibria(II)
Salts

59. Ionic Equilibria(II)
5.How do we know if a salt is neutral, acidic or
basic?

60. Ionic Equilibria
(II)
5.How do we know if a salt is neutral, acidic or
basic?
•
•
•
•
Split the salt into its cation/ anion
Identify acid that produced anion (add H
+) Identify
base that produced cation
If the acid or base are both strong, the salt is neutral

61. Ionic Equilibria
(II)
5.How do we know if a salt is neutral, acidic or
basic?
•
•
•
•
Split the salt into its cation/ anion
Identify acid that produced anion (add H
+) Identify
base that produced cation
If the acid is weak, the anion hydrolyses water and
acts as a weak base basic salt

62. Ionic Equilibria
(II)
5.How do we know if a salt is neutral, acidic or
basic?
•
•
•
•
Split the salt into its cation/ anion
Identify acid that produced anion (add H
+) Identify
base that produced cation
If the base is weak, the cation hydrolyses water and
acts as a weak acid acidic salt

63. Ionic Equilibria(II)
6.How do we calculate the pH of a saltsolution?

64. Ionic Equilibria
(II)
6.How do we calculate the pH of a saltsolution?
•
•Identify ion that hydrolyses H
2O
•Write equation of the ion with H
2O
•Find K
a/ K
b of the ion from K
b/ K
a of the parent acid/
base
Find pH by treating ion as weak acidor weak base
Recall:
[H
3O
+]=K
a ×
[HA]
[OH
–]=K
b ×
[B]

65. Ionic Equilibria
(III)
SolubilityEquilibria
-sparingly solublesalts

66. Ionic Equilibria(III)
7.How do know if a salt is sparinglysoluble?

67. Ionic Equilibria
(III)
7.How do know if a salt is sparinglysoluble?
•From O level QAknowledge.
•When K
sp isgiven.
•When it is stated so in thequestion.

68. Ionic Equilibria
(III)
General Tips
There are mainly 2 types of question:
3.A sparingly soluble salt is dissolved in water to
give a saturated solution
e.g.CaSO
4 (s) ⇌Ca
2+ (aq) + SO
4
2–(aq)
2.2 ions are mixed to form a sparingly soluble salt.
e.g.
Ca
2+ (aq) + SO
4(aq) CaSO
4 (s)
2–

69. Ionic Equilibria
(III)
Type 1Questions
K
sp solubility conc. ofions
Given the value of one of the above (e.g. K
sp), find the
other 2 values (solubility and conc. of ions)

70. Ionic Equilibria
(III)
Type 1 Questions(Strategy)
I
CaF
2 (s) ⇌Ca
2+ (aq) + 2F
–(aq)
? 0 0
C -x +x +2x
E ? x 2x
K
sp solubility conc. ofions
Expressing each term in x can help in theinterconversion.
solubility = x
K
sp = 4x
3
[Ca
2+] = x
[F
–] = 2x

71. Ionic Equilibria
(III)
Type 2 Questions
Ca
2+ (aq) + 2F
–(aq) ppt? Ksp = ……
Given the conc. of Ca
2+ and F
–and K
sp, predict whether
ppt is formed
or
Given K
sp and conc. of one ion, predict the min. conc of
the other ion that will cause precipitation.

72. Ionic Equilibria
(III)
Type 2 Questions(Strategy)
•
•Identify sparingly soluble salt (look for K
sp)
•Write ionic product for salt (same expression as for
K
sp)
•Calculate ionic product and compare:
Q ≤ Kno ppt
Q > Kppt