The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. law of mass action- jA + kB  lC + mD where A, B, C, and D represents chemical species and j, k, l, and m are their coefficient in the balanced equation. The law of mass action is represented by the equilibrium expression: The square brackets indicate the concentrations of the chemical species at equilibrium, and K is a constant called the equilibrium constant.

1. CHEMICAL
EQUILIBRIUM
BY- SANCHIT DHANKHAR

2. CHEMICAL EQUILIBRIUM
The state where the concentrations of all reactants and
products remain constant with time.
On the molecular level, there is frantic activity. Equilibrium
is not static, but is a highly dynamic situation.

3. Figure 13.1 A Molecular Representation of the Reaction
2NO2(g) 2O4(g) Over Time in a Closed Vessel

4. Figure 13.2 Changes in Concentrations
H2O(g) + CO(g)  H2(g) + CO2(g)

5. Figure 13.4 The Changes with Time in the Rates of Forward and Reverse Reactions
H2O(g) + CO(g)  H2(g) + CO2(g)

6. Figure 13.5 The Ammonia Synthesis Equilibrium
N2(g) + 3H2(g)  2NH3(g)

7. THE LAW OF MASS ACTION
jA + kB  lC + mD
where A, B, C, and D represents chemical species and j, k, l, and m are
their coefficient in the balanced equation.
The law of mass action is represented by the equilibrium expression:
The square brackets indicate the concentrations of the chemical
species at equilibrium, and K is a constant called the equilibrium
constant.
K
l m
j k

C D
A B

8. EQUILIBRIUM EXPRESSION
Write the equilibrium expression for the following
reaction:
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)
Applying the law of mass action gives,
Superscript 4, 6, 4, and 7 are the coefficients of
NO2, H2O, NH3, and O2 respectively. The value of
the equilibrium constant at a given temperature
can be calculated if we know the equilibrium
concentrations of the reaction components.
K 
NO H O
NH O
2
2
2
4 6
3
4 7

9. NOTES ON EQUILIBRIUM
EXPRESSIONS (EE)
4 The Equilibrium Expression for a reaction is the reciprocal of
that for the reaction written in reverse.
4 When the equation for a reaction is multiplied by n, EEnew =
(EEoriginal)n
4 The units for K depend on the reaction being considered. K
values are customarily written without units.

10. EQUILIBRIUM EXPRESSIONS
INVOLVING PRESSURES:
Equilibria involving gases can be described in terms of
pressures.
Ideal gas equation:
where, C equals n/v or number of moles n per unit
volume V. Thus C represents the molar concentration
of the gas. N2(g) + 3H2(g) 2NH3(g)
In terms of the equilibrium partial pressures of the
gasses
PV = nRT P =
n
v
RT = CRT








K =
[NH ]
[N ][H ]
C
(C )(C )
K
3
2 2
NH
N H
c
3
2 2
 
K =
P
(P )(P )
p
NH
N H
3
2 2
2
2
3
2
3
3

11. K V. KP
For
jA + kB  lC + mD
Kp = K(RT)n
n = sum of coefficients of gaseous products minus
sum of coefficients of gaseous reactants.
K
P P
P P
C RT C RT
C RT C RT
C C
C C
RT
RT
K RT K RT
p
C D
A B
c D
A B
c D
A B
 
 
 
   
( )( )
( )( )
( ) ( )
( ) ( )
)( )
( )( )
( )
( )
( ) ( )
(
l l
l
m m
m
j j
j
k k
k
l m

j k

( ) ( )
l m j k
   n

12. HETEROGENEOUS EQUILIBRIA
. . . are equilibria that involve more than one phase.
CaCO3(s)  CaO(s) + CO2(g)
K = [CO2]
The position of a heterogeneous equilibrium does not depend on
the amounts of pure solids or liquids present.

13. Figure 13.6 CaCO3(s)  CaO(s) + CO2(g)

14. EXAMPLES
The decomposition of liquid water to gaseous
hydrogen and oxygen,
2H2O(l) 2H2(g) + O2(g)
K = [H2]2[O2] and Kp=(P2
H2)(PO2)
Water is not included in either equilibrium
expression because it is a pure liquid. However, if
water is a gas rather than a liquid,
2H2O(g) 2H2(g) + O2(g)
because the concentration or pressure of water
vapor can change.
K
H O
H O
Kp
P P
P
H O
H O
 
[ ] [ ]
[ ]
( )( )
2 2
2
2 2
2
and
2
2
2
2

15. REACTION QUOTIENT
. . . helps to determine the direction of the move
toward equilibrium.
The reaction quotient is obtained by applying the
law of mass action using initial concentrations
instead of equilibrium concentrations.
H2(g) + F2(g)  2HF(g)
where the subscript zeros indicate initial
concentrations.
Q 
HF
H F
2 2
0
2
0 0

16. DIRECTION OF REACTION
To determine in which direction a system will shift to reach
equilibrium, we compare the values of Q and K. There are three
possible cases:
1.Q is equal to K. The system is at equilibrium; no shift will occur.
2.Q is greater than K. The system shifts to the left, consuming products
and forming reactants, until equilibrium is achieved.
3.Q is less than K. The system shifts to the right, consuming reactants
and forming products, to attain equilibrium.

17. SOLVING EQUILIBRIUM
PROBLEMS
1. Balance the equation.
2. Write the equilibrium expression.
3. List the initial concentrations.
4. Calculate Q and determine the shift to equilibrium.
5. Define equilibrium concentrations.
6. Substitute equilibrium concentrations into equilibrium
expression and solve.
7. Check calculated concentrations by calculating K.

18. LE CHÂTELIER’S
PRINCIPLE
. . . if a change is imposed on a
system at equilibrium, the
position of the equilibrium
will shift in a direction that
tends to reduce that change.

19. EFFECTS OF CHANGES ON THE
SYSTEM
1. Concentration: The system will shift away from the added
component.
2. Temperature: K will change depending upon the temperature
(treat the energy change as a reactant).
3. Pressure:
a. Addition of inert gas does not affect the equilibrium position.
b. Decreasing the volume shifts the equilibrium toward the side
with fewer moles.

20. Figure 13.8 A Mixture of N2, H2, and NH3

21. Figure 13.9 The Effect of Decreased Volume on the Ammonia Synthesis Equilibrium

22. THANKYOU