## langmuir isotherm

In 1916 Langmuir proposed his theory which said that adsorption of a gas on the surface of a solid to be made up of elementary sites each of which could absorb one gas molecule.
It is assumed that all the adsorption sites are equivalent and the ability of the gas molecule to get bound to any one site is independent of whether the neighboring sites are occupied or not.
It is further assumed that a dynamic equilibrium exists between the adsorbed molecule and the free molecule.
If A is the gas molecule and M is the surface site then,

Where “Ka” and “Kd” are the rate constants for adsorption and desorption, respectively.

The rate of adsorption is proportional to the pressure of A, viz, PA and number of vacant sites on the surface, viz, N(1-θ)  where N is the total number of sites and θis the fraction of surface sites occupied by the gas molecules, i.e.

θ =  Number of adsorptions sites occupied / Number of adsorptions sites are available

Thus the rate of adsorption = kapAN(1- θ)                  ……. (1)

The rate of desorption is proportional to the number or adsorbed molecules, Nθ .

Thus the rate of desorption = kd                              ……. (2)

Since at equilibrium, the rate of adsorption is equal to the Rate of desorption, we can write from equation (1) and (2)

KapAN(1-θ) = kd                                             ……. (3)

Or,       KpA(1-θ) = θ                                                      ……. (4)

where,  K = ka / Kd

Equation (4), may thus be written as

(1-θ) / θ = 1 / KpA                                               ……. (5)

Or,      (1 / θ) – 1 = 1 / KpA                                             ……. (6)

(1 / θ) = (1 / KpA ) + 1 = (1 + KpA) / KpA           ……. (7)

Hence,   θ = KpA/ (1 + KpA )                                         ……. (8)

Equation (8) is called the “Langmuir adsorption isotherm”.

The following five assumptions are involved in derivation of the Langmuir adsorption isotherm:

1.    1. The absorbed gas behaves ideally in the vapour phase.

2.    2. Only a monolayer is formed by the adsorbed gas.

3.    3. The surface of the solid is homogeneous so that each binding site has the same affinity for the gas molecules.

4.    4. There is no lateral interaction between the adsorbate molecules.

5.    5. The adsorbed gas molecules are localized, i.e. they do not move around on the surface.

The first assumption holds at low pressure, the second assumption breaks down when the pressure of the gas is increased. The third assumption is not strictly true because the real surfaces are quite heterogeneous so that affinity for gas molecule is different at different sites. Crystal imperfections and cracks lead to the creation of different sites on the surface. The fourth and fifth assumptions, too, are not strictly valid.

## Concept of free energy

• Free energy is a thermodynamic function like enthalpy and entropy.
• It is also known as free enthalpy.
• Free energy is used to determine the changes made by a system and how much work it produced.
• It is expressed in two forms i.e. (1) Helmholtz free energy
(2) Gibb’s free energy

### Gibbs free energy:

• It is named in honor of the great American physicist J.W. Gibb’s(1839-1903).
• Gibb’s free energy(G) is used to calculate the maximum reversible work done of a thermodynamic system at constant temperature and pressure.
• Or this is the maximum amount of non expansion work, that can be extracted from a  completely reversible closed system .
• Gibb’s free energy also tells weather the process is spontaneous or not, i.e.if the change in free energy is negative the process is spontaneous, and if positive the process is non spontaneous.
• Mathematically
G = H – TS
=> ∆G = ∆H – T∆S
Where ∆G = change in Gibb’s free energy
∆H = change in enthalpy
∆S = change in entropy

### Helmholtz free energy:

• It is named in the honor of German physicist Hermann von Helmholtz(1821-1894).
• It is also termed as Work function or Helmholtz function ( A)
• It is a type of thermodynamic potential that measures the useful work obtained from a closed system at constant temperature and volume.
• If the change in Helmholtz free energy is negative, that means maximum amount of work is performed by a system at constant volume.
• And if the volume is not constant the part of this work is performed as boundary work.
• Mathematically, A = U – TS

=>∆A = ∆U – T∆S

## Le-Chatelier’s principle:

• This is based on the fundamental of a stable equilibrium.
• It states,when a system at equilibrium is subjected to a change in temperature,pressure or concentration of a reacting species,the system react in a way that partially offset the change while reaching a new state of equilibrium.

### Effect of concentration:

If the concentration of a reacting component is increased,reaction shifts in a direction which tends to decrease its concentration.e.g. in the following example.
N2(g) + 3H2(g)  2NH3
[reactant] ↑         foreward shift
[product] ↑          backward shift
• If the concentration of reactant is increased at equilibrium then reaction shifts in the foreward direction.
• If concentration of product is increased then reaction shifts in backward direction.
Note: The addition of any solid component doesnot effect the concentration.

### Effect of volume:

• If volume is increased,pressure decreases hence reaction will shift in the direction in which pressure increases that is in the direction in which number of moles of gases increases and vice versa.
• If volume is increased then for,
1. ∆ng > 0 ,reaction will shift in the foreward direction
PCl5(g)  PCl3(g) + Cl2(g)  , ∆ng = 2-1 =1

2. ∆ng < 0 ,reaction will shift in the backward direction
N2(g) + 3H2(g) = 2NH3(g)  , ∆ng =2-4= -2 < 0

3. ∆ng= 0 ,reaction will not shift e.g.
H2(g) + I2(g)   2HI(g)  (no effect)

### Effect of pressure:

• On increasing pressure,equilibrium will shift in the direction in which pressure decreases,i.e. number of moles in the reaction decreases and vice versa.
• Pressure is directly proportional to number of moles.

1. For ∆ng= 0 ,no effect
2. For  ∆ng > 0
If pressure decreases equilibrium shifts to forward direction.
If pressure increases equilibrium shifts to backward direction.
3.For  ∆ng < 0
If pressure  increases equilibrium shifts to forward direction.
If pressure decreases equilibrium shifts to backward direction.

### Effect of catalyst:

• Due to catalyst ,state of equilibrium is not affected i.e.no shift will occur as catalyst lowers the activation energy of both the forward and reverse reaction by same amount.
• Thus alternating the forward and reverse rate equally and hence, the equilibrium will be attained faster i.e.time taken to reach the equilibrium is less.

### Effect of inert gas addition:

• at constant volume: Inert gas addition has no effect at constant volume
• at constant pressure: If inert gas is added at constant pressure,volume is increased and hence equilibrium will shift in that direction in which larger number of gaseous moles are formed.

1.∆ng > 0 , reaction will shift in forward direction.
2.∆ng < 0 , reaction will shift in backward direction.
3.∆ng= 0 ,  no effect

### Effect of temoerature:

• :Exothermic reaction:The reaction in which heat is evolved.

1.If temperature increases equilibrium will shift in backward direction.
2.If temperature decreases equilibrium will shift in forward direction.

• Endothermic reaction: The reaction in which heat is consumed

1.If  temperature increases equilibrium will shift in forward direction.
2..If  temperature decreases equilibrium will shift in backward direction.

## Thermodynamics of equilibrium

### Reaction Quotient(Q):

At each point in a reaction ,we can formulate a ratio of concentration terms having the same form as the equilibrium constant expression. This ratio is called the reaction quotient denoted by symbol Q.
It helps in predicting the direction of a reaction.

#### For a general reaction:

aA(g) + bB(g) ⇔ cC(g) + dD(g)

Q=[C]tc[D]td/ [A]ta[B]tb

1. If the net reaction is at equilibrium,Q= Kc
2. A the net reaction proceeding from right to left(the reverse direction) if  ,Q > Kc

### Thermodynamics of equilibrium:

For a general reaction, mA + nB ⇔ pC + qD

### ∆G =∆G⁰ + 2.303RTlog10Q

Where,    ∆G   = Gibb’s free energy change.
∆G⁰ = Standart Gibb’s free energy change.
Q     =  Reaction quotient.

Since at equilibrium,Q=K

Here K is the thermodynamic equilibrium constant replacing  Kc or  Kp
Here aX   denotes the activity of x.
In fact,is the ratio of activity of substance at equilibrium and its activity in standard condition.That is why it is unitless and K is also unitless.

#### Note:

1. Thermodynamics equilibrium constant is unitless since activity is unitless.
2. For pure solids and pure liquids activityis unity.
3. For gases (ideal behavior),the activity is partial pressure(in atm)
4. For components in solution, activity is molar concentration
5. For endothermic reaction()value of equilibrium constant increases with the rise in temperature.
6. For exothermic reaction, value of equilibrium constant decreases with increase in temperature.

#### Condition of spontaneity:

•    ∆G<0 for a spontaneous process or reaction.
Since, ∆G = ∆H – T∆S
= > ∆H – T∆S < 0
= >  T >  ∆H/T∆S
• ∆G > 0 for a non spontaneous process or reaction

## Important relationship involving equilibrium constant

1. If we reverse an equation  KP or  KC  is reversed.

If              A + B ⇔ C + D         KC = 10
then for    C + D A + B          K= 10-1

2. If we multiply each of the coefficient in a balanced equation by a factor n,then equilibrium constant is raised to the same factor.

½ N2 + ½O2 ⇔ NO            KC = 5

N2  +  O2 ⇔ 2NO               KC = 52 = 25

3. If we divide each of the coefficients in a balanced equation by a factor n, then new equilibrium constant is the nth root of the previous value.
2SO2 + O2      2SO3              KC =25

SO2 + 1/2 O2   SO3B     KC = (25)1/2

4. When we combine individual equation, we have to multiply their equilibrium constant for net reaction.
If K1 , K2 and K are stepwise equilibrium constant for  A⇔B , B⇔C , C⇔D   .
Then for A⇔D , the equilibrium constant is K= K1K2K

#### Significance of the magnitude of equilibrium constant:

1. If a large value of KP or  KC signifies that the forward reaction goes to completion or very nearly so.
2. A large value of KP or KC signifies that the forward reaction doesn’t occur to any significant extent.
3. A reaction is most likely to reach a state of equilibrium in which both reactant and product are present if the numerical value of KP or  KC is neither very large nor very small

#### Units of the equilibrium constant:

For the sake of simplicity in the units of Kc or Kp the relative molarity of pressure of reactants and products are used with respect to standard condition. (For solution standard state 1 mol/liter and for gas standard pressure=1 atm).Now resulting equilibrium constant becomes unitless by using relative molarity and pressure. But the numerical value may change in the value of standard states molarity and pressure.

## Relationship between Kp ,Kc, Kx and Kn

• Kp=Equilibrium constant in terms of partial pressure.
• Kc=Equilibrium constant in terms of concentration.
• Kx=Equilibrium constant in terms of  mole fraction.
• Kn=Equilibrium constant in terms of  number of moles.
•  Kp and Kc are related as  KP = KC (RT)∆n ………………………(eq 1)

#### Relationship between KP and Kn ………………………(eq 2)
From ideal gas equation,PV=nRT
=>P=n(RT/V)
Where, n is the number of moles

So,  PA =nA(RT/v), PB = nB(RT/V), PC =nC(RT/V) and PD =nD(RT/V)

Replacing equation 2 by the above value we get that,
[nC(RT/V)]c  . [nD(RT/V)]d
K=  ————————————-
[nA(RT/V)]a  . [nB(RT/V)]b
nCc . nDd
= > Kp =  ————– .  ( RT/V ) (c + d) – (a + b)
nAa. nBb
= >  Kp = Kn . (RT/V) ∆n

## = >  Kp = Kn . (PT/nT) ∆n

∆n=number of gaseous moles of product – number of gaseous moles of reactant

#### Relation between  Kp and Kx

From  above (eq 2)

Partial pressure(P) = Mole fraction(x) . Total pressure(PT)

So,   PA =  xA .  PT

PB =  xB .  P

PC =  xC .  PT
PD =  xD .  PT

Putting the values in whole equation:
( xC .  P)c . ( xD .  PT )d
K= ———————————
( xA .  PT )a . ( xB .  PT )b
xCc .  xDd
=  ————–   PT(c+d) – (a+b)
xAa .  xBb

## = >     KP  = Kx  (PT) ∆n

N2O4(g)=2NO2(g) ,Calculate Kand Kfor the given reaction for which Kp=0.157

## Equilibrium constant in terms of partial pressure.

Both KP and Kc  are equilibrium constants,but  Kc  is defined by molar concentration

#### Consider the following reversible reaction

aA(g) + bB(g) ⇔ cC(g) + dD(g)

• If equilibrium constant for the reaction  is expressed in terms of the molar concentration , • If the equilibrium involves gaseous species, then concentration may be expressed in terms of partial pressure of the reactants and products and it is represented as ……………………. (eq1)

PA  ,  PB  ,  PC and PD  are  the partial pressure of A,B,C and D respectively.

#### Relationship between  KP and Kc  :

We know that ,for a ideal gas pressure is directly proportional to the concentration assuming volume and temperature constant.
So,    PV = nRT
Divide both sides by V,
P=(n/V)RT
P=CRT
Where,C is concentration
n is the number of moles
V is the volume
R is ideal gas constant,and
n/V is the molar concentration.
Thus the partial pressures of individual gases A, B, C and D are:
PA = [A]RT; PB = [B]RT; PC = [C]RT; PD = [D]RT
Substituting these value in equilibrium constant expression equation 1, we have  KP = KC (RT)(c+d)-(a+b)

KP = KC (RT)∆n
where ∆n = (c+d) – (a+b)
=number or gaseous moles of the product-number of gaseous moles of the reactant

## Law of mass action/Law of chemical equilibrium

### Principle of Chemical equilibrium:

• This theory was proposed by Goldberg and Waage in 1884.
• According to this law ,rate of reaction at a given temperature is directly proportional to the product of active masses of the reactants, raised to the power equal to stoichiometric co- efficient of the balanced chemical equation at a particular instant of reaction.
• It explains and predicts the behaviour of solution in dynamic equilibrium.
• Specifically it implies that for a chemical reaction reaction mixture in equilibrium, the ratio between concentration of reactants and products is constant.

#### Derivation of equilibrium constant in terms of concentration:

Consider a hypothetical reversible reaction i.e.

aA(g) + bB(g) ⇔ cC(g) + dD(g)
Applying law of mass action here,
• Rate of forward reaction would be k[A]a[B]b
• Rate of backward reaction would be k[C]c[D]d

where,
[A], [B], [C] and [D] being the active masses
kf and kb are rate constants of forward and backward reactions,
a, b, c, d are the stoichiometric coefficients related to A, B, C and D respectively. However, at the equilibrium ,
Rate of forward reaction = Rate of backward or, or, or, or,

Keq  = Kf / K

Where K is known as equilibrium constant(if concentration is given).
Session Quiz:
Q. For the reaction  Br2(g)  + BF2(g) → 2BrF2(g) the equilibrium constant at 2000K and 0.1 bar is 5.25.When the pressure is increased by 8 fold, what is the change in equilibrium constant?

## CHEMICAL EQUILIBRIUM

### EQUILIBRIUM:

• Equilibrium is a state of reversible reaction in which rate of forward reaction is same as that of reverse reaction.
• So we can say that all the equilibrium reaction is reversible reversible in nature but the reverse is not true.
• In a chemical reaction, chemical equilibrium is a state in which reactants and products are present in concentration which have no further tendency to change with time.
• Chemical equilibrium is also known as dynamic equilibrium because there is no net changes in the concentration of reactant and product.
• The equilibrium between different chemical species present in the same or different phases is called chemical  equilibrium.
• The entire process can be understood by considering a reversible reaction i.e.
• There are two types of chemical equilibrium i.e.
• Homogeneous Equilibrium:The equilibrium reaction in which all the reactants and products are present in the same phase are called homogeneous equilibrium.

• Heterogeneous equilibrium:The equilibrium reaction in which all the reactants and products are present in different phase are called heterogeneous equilibrium.