This document discusses chemical equilibrium, including definitions, characteristics, and factors that affect equilibrium. It defines chemical equilibrium as a state where the forward and reverse reaction rates are equal. Characteristics include the dynamic nature of equilibrium and constant concentrations of reactants and products at equilibrium. Factors that affect equilibrium position include concentration, pressure, temperature, and catalyst additions according to Le Chatelier's principle. The relationship between the equilibrium constant K and standard Gibbs free energy change ΔG° is also described.

1. By ARUNESH GUPTA
PGT (CHEMISTRY), KV Barrackpore (AFS)
*

2. Irreversible reactions:
A reaction that takes in forward direction reactants forming products but the
reverse reaction does not take place is called an irreversible reaction.
Example:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq)
2Mg(s) + O2 (g) → 2MgO (s)
 Reversible reactions
 A reaction in which reactants react to form
products & at the same time the products react
to form reactants under the same conditions is
called reversible reactions.
 In general, A + B ⇄ C + D
 E.g. N2(g) + 3H2(g)⇄ 2NH3(g)

3. Equilibrium: A reaction (or a process) is said to be in
equilibrium when the rate of forward reaction (or process) is
equal to the rate of backward reaction (or a process).
Types of Equilibrium:
(1)Physical equilibrium
(2) Chemical equilibrium
(3) Ionic Equilibrium

5. Chemical Equilibrium:
Definition ?
General Characteristics of physical & chemical equilibrium:
• Dynamic nature of equilibrium.
• At equilibrium, concentration terms of each reactant & product
becomes constant
• In a closed system
• Chemical equilibrium can be attained from either direction.
• A catalyst does not alter the state of equilibrium. It increases
the rates of forward & backward reactions to an equal extent.

6. Therefore, Kf[A][B]b = Kb[C]c[D]d
Kf/Kb = K = Here K = equilibrium constant for the reaction

7. Types of Equilibrium constants:
(a) In terms of molar concentration Kc =
(b) In terms of partial pressure: Kp =
Where [A] denotes molar concentration & PA denotes partial pressure of A(g)
Try: Derive a relationship between KP & KC for a reaction in equilibrium.
Kp = Kc (RT)∆n

8. For any reaction: aA + bB cC + dD
At any stage, may not be in equilibrium,
Qc =
where QC is the reaction quotient in terms of molar concentration.
 If QC = KC the reaction is in equilibrium.
 If QC < KC the reaction tends to move forward in order to attain
equilibrium.
If QC > KC the reaction tends to move backward in order to attain
equilibrium.

9. Characteristic of Equilibrium constant (K) :
• K is constant for a reaction & it alters only with the temperature change.
•K is not affecte4d by adding a catalyst.
•If the equilibrium reaction is reversed, the new equilibrium constant value
becomes the reciprocal. [K.K´ = 1]
•If the reaction is written in two steps, the equilibrium constant is equal to
the product of the equilibrium constant values of the two step reactions.
• K value helps to predict the extent of a reaction.
If K > 1, the is more to the forward direction & vice versa.

10. Effect of Temperature on Equilibrium Constant.
For a reaction in equilibrium: aA + bB cC +dD
K = =
For an exothermic reaction ∆H = -ve, Kf decreases with the
increase in temperature, so that K decreases.
For an endothermic reaction ∆H = +ve, Kf increases with the
increase in temperature, so that K increases.
Kf
Kb

11. Mathematically the effect of temperature can be derived as:
∆G° = -RT ln K
or
or = - R …………..(i)
Also, ∆G° = ∆H°- T ∆S°
or = - ∆S°
or = - …………………………(ii)
Comparing (i) & (ii),
= ……………………………………….(iii)
Eqn. (iii) is names as van’t Hoff equation.

12. Integrated form of van’t Hoff equation:
Integrating equation(iii)
= or ln Kp = –
The plot of ln K vs 1/T gives a straight line with slope = - ∆H°/T,
If K1 & K2 are equilibrium constants at T1 & T2 respectively, then,
where ∆H° = enthalpy change of the reaction (assuming it to be constant in the
temperature range of T1 & T2.
 We conclude: If ∆H = 0, no heat is evolved or absorbed in the reaction,
= 0 , i.e. = 1 or K1 = K2
i.e. the equilibrium constant does not change with temperature.
 If ∆H = +ve, (endothermic reaction) = +ve, or log K2 > log K1
or K2 > K1increases with the increase in temperature,
If ∆H = -ve, (exothermic reaction) == -ve, or log K2 < log K1
or K2 < K1 decreases with the increase in temperature,

13. Relationship between equilibrium constants of different reactions :
For a reaction in equilibrium: 3A + B ⇌ 2C
where equilibrium constant = K1
(a)Equilibrium constant of 6A + 2B ⇌ 4C : K2 = (K1)2
(b)Equilibrium constant of 2C ⇌ 3A + B : K3 = 1/K1 = (K1) -1
(c)Equilibrium constant of 3nA + nB ⇌ 2nC : K4 = (K1)n
(d)Equilibrium constant of 2nC ⇌ 3nA + n B K5 = (K1)-n
In general, Kn = (K1)±n
Try: Equilibrium constant for the reaction: N2O4 ⇌ 2NO2 is 64.
Find the equilibrium constant for the reaction ⅓ NO2 ⇌ ⅔ N2O4.

14. Expressions for Kc & Kp in terms of the amount of a substance reacted or formed
N2(g) + 3 H2(g) ⇌ 2NH3(g)
Initially, t = 0 1 mol/L 3M 0
At eqlbm, x mol/L (1-x)M (3 – 3x)M 2x M
In terms of pressure,
Total no. of moles at equilibrium = (1-x) + (3-3x) + 2x = 4-2x = 2(2-x)
If P is the total pressure of the of equilibrium reaction mixture,
Substituting in the expression,
(a)

15. (b) Expression for Kc & Kp in terms of degree of dissociation (∝)
2HI(g) ⇌ H2(g) + I2 (g)
Initially, no. of moles a 0 0
At eqlbm. No. of moles a(1- ∝) a∝/2 a∝/2
Molar conc. in V L
Expression for Kp : no. of moles of reactants & products are equal , nr = np

16. (c) Expression for Kc & Kp in terms of degree of dissociation (∝) for
“ dissociation of PCl5”
Expression for Kc:
PCl5 (g) ⇌ PCl3(g) + Cl2 (g)
Initial no. of moles a 0 0
Moles at equilibrium: a(1- ∝ ) a∝ a∝
Molar conc.
=
Kc =
Kc = a∝2/(1- ∝)V

17. Expression for Kp for : PCl5 (g) ⇌ PCl3(g) + Cl2 (g)
Total no. of moles at equilibrium= a(1- ⇌) + a⇌ + a ⇌ = a(1+ ⇌)
Let P is the total pressure at equilibrium, then.
Kp =

18. (iii) Dissociation of N2O4 (a) Expression for Kc
N2O4(g) ⇌ 2NO2(g)
Initial moles : a 0
Moles at eqlbm. a(1-α) 2aα mol
Molar conc. in V L a(1-α)/V 2aα/V mol/L

19. Expression for Kp : N2O4(g) ⇌ 2NO2(g)
Total no. of moles at equilibrium= a(1- ∝) + 2a∝ = a(1+ ∝)
Let P is the total pressure at equilibrium, then,

20. Some useful points in solving numericals on equilibrium
To calculate KP , partial pressure of a reactant or product at equilibrium is
Using equation, KP = KC (RT)∆n, take ∆n = np – nr
If np = nr , KP = KR where ∆n = np – nr
Take, Kp & KC as dimensionless.
If pressure is in bar, use R = 0.0831 L bar / mole/ K
If pressure is in pascal,(Pa or Nm-2), use R = 8.314 J/mol/K
If pressure is in atm. Use R = 0.0821 L atm / mol / K
For Kc , molarity (mol / L) should be used in KC terms:,
Remember:
1atm. = 1.01 bar = 760 torr = 760mm = 76 cm mercury = 1.01x 105 Pa
For easy calculations: take, 1 atm =1.013bar ≈ 1bar
Partial pressure of reactant or product =
P(reactant or product) = Ptotal
χ
(reactant or product)

21. 1. Equilibrium constant in terms of mole fraction ( χ )
For a reaction: aA + bB ⇌ cC + dD
where χA , χB, χC & χ D are mole fractions
of A, B, C & D respectively.
where P is the total pressure of the reaction mixture,
and (∆ng) = np - nr

22. 2. Calculation of degree of dissociation (∝) from vapour density measurements:
(a) Dissociation of PCl5 (g) : PCl5 (g) ⇌ PCl3 (g)+ Cl2 (g)
No. of moles before dissociation 1 0 0
No. of moles after dissociation: 1-∝ ∝ ∝ (in V litre)
Before dissociation: No. of moles in V litres = 1 mol
If D is the vapour density (VD) of PCl5 before dissociation (called theoretical vap. density),
then
D ∝ 1/V (∵ Density ∝ 1/Volume ).
After dissociation, total no. of moles = (1- ∝) + ∝ + ∝ = 1+ ∝
Total volume occupied by the reaction mixtures = (1+ ∝) V litres.
Now, if d is the density of the vapour (called observed density)
Then,
Dividing D by d, we get

23. As mol. mass = 2 x vapour density
Where Mcal = calculated (theoretical) mol. mass
Mobs = observed (experimental) mol. mass
Alternately, observed mol. Mass is calculated from the mass of a definite volume of
the vapour at particular temperature using relation:

24. In case of dissociation of N2O4,
N2O4 (g) ⇌ 2NO2 (g)
No. of moles before dissociation = 1 mol 0 mol
No. of moles after disso. 1- α 2α
Total no. of moles = 1 – α + 2 α = 1+ α
&
Where Mcal = calculated (theoretical) mol. mass
Mobs = observed (experimental) mol. mass

25. In general, if 1 mole of reactant A dissociates to give n moles of gaseous products, i.e.
A ⇌ n1B + n2C + ……… (where n = n1 + n2 + …….)
Initial moles 1 mol 0 mol
Moles after disso, 1- α nα
Total no. of moles = 1 – α + nα = 1 + (n -1)α
Then, D/d = 1 + (n -1)α so,
Example: NH2COONH4 ⇌ 2 NH3 + CO2, n = 2+1 = 3
So,

26. Relationship between standard free energy change (∆G°) & equilibrium constant (KP)
▬► van’t Hoff reaction isotherm,
∆G° = - RT ln KP = - 2.303 RT ln KP.
Derivation: Let ∆G of a reaction under any set of conditions is related to ∆G°
∆G = ∆G° + RT ln Q
where Q is the reaction quotient under a given set of conditions.
At equilibrium, ∆G = 0 and Q becomes K (Q = K)
Hence, 0 = ∆G° + RT ln K
Or, ∆G° = - RT ln K = - 2.303 RT ln K.
We have:
If ∆G°is negative, K>1, so, forward reaction is favoured.
If ∆G°is positive, K<1, so, backward reaction is favoured.

27. To discuss:
Qu. 1) Why do some solids sublime ?
Ans. Some solids like naphthalene, camphor etc donot melt but forms vapour
directly. This is because the heat absorbed from surrounding cuts off their inter
– molecular forces so significantly small that the liquid state is not observed
normally.
Why is dissolution of gas exothermic?
Ans. Dissolution of a gas in a liquid is a spontaneous process. (i.e. ∆G = -ve).
As it is accompanied by the degree of randomness (i.e. ∆S = -ve )
Hence , ∆G = ∆H - T∆S, ∆G can be –ve only when ∆H is –ve.
So,…….

28. A mathematical expression of this thermodynamic view
can be written as:-
∆G = ∆G°+ RT lnQ
At equilibrium ∆G =0 and Q = Kc
Then the equation becomes:-
∆G°= - RT ln K
Which is equal to : ln K = ∆G°/RT
Therefore:-
Taking antilog on both sides we get :-
K = e - ∆G°/RT

29. If ∆GO < 0 then ∆GO/RT will be positive and K>1
which implies a spontaneous reaction or the
reaction which proceeds in the forward direction to
such an extent that the products are present
predominantly.
If ∆GO > 0 then ∆GO/RT will be negative that is
K<1 which means a non spontaneous reaction or a
reaction which proceeds in the forward direction to
such a small degree that only a very minute quantity
of product is formed.

30. Factors affecting equilibria
Le Chateliers’s principle:-
It states that a change in any of the factors in
equilibrium that determine the equilibrium conditions
of a system will cause the system to change in such
a manner so as to reduce or to counteract the effect
of the change.

32. Effect of concentration change
The Le Chatelier’s principle predicts that:-
In an equilibrium mixture the concentration stress of an added reactant /
product is relieved by net reaction in that direction that consumes the added
substance.
The concentration stress of a removed reactant / product is relieved by net
reaction in the direction that replenishes the removed substance.
Or in other words
“ When the concentration of any reaction or products in a reaction at
equilibrium is changed , the composition of the equilibrium mixture changes
so as to minimize the effect of concentration changes”.

33. Effect of pressure change on equilibrium mixture:
On increasing the pressure of an equilibrium
mixture, the volume of the equilibrium mixture
decreases. So equilibrium shifts t that side where
no. of moles of reactants or products decreases &
vice versa.
Try : If ∆g(n) = 0, what is the effect of increasing
pressure on equilibrium?

34. Effect of temperature changes
In general temperature depends on the ∆H for the reaction:-
Increasing the temperature shifts the equilibrium to left for an
exothermic process & vice versa.
Effect of adding a catalyst to an equilibrium mixture:
A catalyst does not change the state of equilibrium. It increases
the rates of forward & backward reactions to an equal extent. So
it helps to attain equilibrium faster.

35. EFFECT OF ADDING AN INERT GAS TO AN EQUILIBRIUM MIXTURE
If the volume is kept constant and an inert gas such
as argon is added which does not take part in the
reaction, the equilibrium remains undisturbed. It is
because the addition of an inert gas at constant
volume does not change the partial pressures or
the molar concentrations of the substance involved
in the reaction.
The reaction quotient changes only if the added
gas is a reactant or product involved in the reaction.

36. THANK YOU