1. The document introduces the concept of chemical equilibrium, including that it is a state where the concentrations of reactants and products do not change over time, and that forward and backward reactions continue at the microscopic level. 2. It describes the law of mass action, which states that the rate of a reaction is directly proportional to the product of the molar concentrations of reactants raised to their stoichiometric coefficients. 3. It provides expressions for calculating equilibrium constants (Kc) in both gaseous and liquid/solid systems based on the law of mass action and reaction stoichiometry.

1. Chemical Equilibrium
11th
Law of Mass ACTION

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18. EXCLUSIVE

19. Equilibrium is the state at which the concentrations of reactants and
products do not change with time.
This state can be recognized by the constancy of certain measurable
properties such as pressure, density, colour, concentration, etc.
What is Equilibrium

20. ..these viral ones

21. Dynamic equilibrium. Dynamic means moving and at a microscopic level,
the system is in motion.
There is no apparent change at equilibrium yet both forward and backward
reactions continue to take place.
Lets understand

22. Lets understand

23. Rate of forward reaction = Rate of backward reaction
Lets understand

24. CONCEPT OF EQUILIBRIUM STATE
Rate of forward reaction = Rate of backward reaction

25. 1. The observable properties of the system, such as pressure, colour,
concentration, etc., become constant at equilibrium and remain unchanged
thereafter.
2. The equilibrium can be approached from either direction.
3. The equilibrium can be attained only if the system is closed.
4. A catalyst does not alter the equilibrium point.
CHARACTERISTICS OF CHEMICAL EQUILIBRIUM

26. LAW OF MASS ACTION
At constant temperature the rate of a chemical reaction is directly
proportional to the product of the molar concentrations of reacting species
with each concentration term raised to the power equal to the numerical
co-efficient of that species in the chemical equation.
LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

27. For any general reaction,
aA + bB-------> Products
The law of mass action may be written as
Rate = k [A]a[B]b
LAW OF MASS ACTION

28. LAW OF MASS ACTION
Let us consider a simple reversible reaction,

29. LAW OF MASS ACTION

30. LAW OF MASS ACTION

31. LAW OF MASS ACTION

32. EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

33. EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

34. EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

35. EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

36. EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

37. (i) N2(g) + 3H2 (g) ⇌ 2NH3 (g)
Kp = Kc (RT)Dn
(ii) H2 (g) + I2 (g) ⇌ 2HI (g)
Dng = 2 – (1 + 1) = 0
(iii) 2 NOCI (g) ⇌ 2NO (g) + Cl2 (g)
Dng = 2 – (1 + 1) = 0
g
Lets Practice

38. By applying law of mass action to a reversible reaction, at equilibrium, it is
possible to derive a simple mathematical expression known as law of
chemical equilibrium.
Let us consider a simple reversible reaction,
A + B  C + D
Rate of forward reaction  [A][B]
= kf [A][B]
where kf is the rate constant for the forward reaction.
Similarly,
Rate of backward reaction  [C][D]
= kb[C][D]
where kb is the rate constant for the backward.
LAW OF MASS ACTION: Application

39. Rate of forward reaction = Rate of backward reaction.
kf[A][B] = kb[C][D]
or
Since kf and kb are constants, therefore, the ratio kf/kb is also constant and
is represented by Kc.
  
  
f
b
k C D
k A B

  
  
f
c
b
k C D
K
k A B
 
LAW OF MASS ACTION: Application

40. The constant, Kc is called equilibrium constant.
For a general reaction of the type :
aA + bB  cC + dD
The equilibrium constant, Kc, may be defined as the ratio of product of the
equilibrium concentrations of the products to that of the reactants with
each concentration term raised to the power equal to the stoichio-metric
co-efficient of the substance in the balanced chemical equation.
   
   
c d
c a b
C D
K
A B

Equilibrium Constant

41. Write the equilibrium constant expressions for the following reactions:
(a) BaCO3(s)  BaO(s) + CO2(g)
(b) AgBr(s)  Ag+ (aq) + Br– (aq)
(c) CH3COCH3(l)  CH3COCH3(g)
(d) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Lets Practice

42. Law of chemical equilibrium may be stated as :
At a given temperature, the ratio of product of equilibrium
concentrations of the products to that of the reactants with each
concentration term raised to the power equal to the respective
stoichiometric co-efficients in the balanced chemical has a constant
value.
The concentration ratio =
   
   
c d
a b
C D
A B
Equilibrium Constant

43. If an equilibrium involves gaseous species, then the concentrations in the
concentration quotient may be replaced by partial pressures because at a
given temperature the partial pressure of a gaseous component is
proportional to its concentration. If the above mentioned reaction has all
the gaseous species, then
   
   
c d
C D
p p
a b
A B
p p
Q K
p p
 
EQUILIBRIUM CONSTANT IN GASEOUS SYSTEMS

44. For an ideal gas, PV = nRT
p = n/V RT = CRT
pa = CART
pB = CBRT
pC = CCRT
pD = CDRT
Kp = Kc(RT)Dn
Dn = (sum of the exponents in the numerator of Qc)
– (sum of the exponents in the denominator of Qc)
   
   
c d
C D
P a b
A B
C RT C RT
K
C RT C RT

   
   
    
.
c d
c d a b
C D
a b
A B
C C
RT
C C
  
RELATIONSHIP BETWEEN Kp AND Kc

45. 1. At equilibrium, the concentrations of N2, O2 and NO in a sealed vessel at
800 K are :
N2 = 3.0 x 10–3 M, O2 = 4.2 x 10–3 M and NO = 2.8 x 10–3 M
Calculate the equilibrium constant Kc for the reaction:
N2(g) + O2(g) ⇌ 2NO(g)
Lets Practice

46. 2. What is Kc for the following equilibrium, when the equilibrium
concentration of each substance is
[SO2] = 0.60 M, (O2) = 0.82 M and (SO3] = 1.90 M
2SO2(g) + O2(g) ⇌ 2SO3(g)
Lets Practice

47. 3. PCl5, PCl3 and Cl2 are at equilibrium at 500 K and having
concentration 1.59 M PCl3, 1.59 M Cl2 and 1.41 M PCl5. Calculate Kc
for the reaction :
PCl5 ⇌ PCl3 + Cl2
Lets Practice

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62. EXCLUSIVE

63. Step 2
Step 1 Step 2 Step 3 Step 4 Step 5

64. EXCLUSIVE