Equilibrium is a state in which there are no observable changes as time goes by.When a chemical reaction has reached the equillibrium stae,the concentrations of reactants and products remain contant over time and there are no visible changes in the system.

In reversible reaction like-R + R’ P + P’Initially only reactants are present

R and R’ conbine to form P and P’.As soon as P and P’ are formed they start the backward reaction.As concentrations of R and R’ decrease rate of forward reaction decreasesand rate of backward reaction increases.Ultimately a stage is reached when both the rates become equal.Such a state is known as ‘CHEMICAL EQUILIBRIUM’.

Attainment of chemical equilibrium

1. Equilibrium is Dynamic in nature.

2. Equilibrium can be observed by constancy of some observable properties like colour,pressure,concentration,density,temperature etc.which may be suitable in a given reaction.

3. Equilibrium state can be affected by altering factors like pressure,volume,concentration and temperature.

At a given temperature,the product of concentrations of the reaction products raised to the rerespective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual coefficients has a constant value.This is known as the equilibrium law or law of chemical equilibrium.

For a general reaction :

the equillibrium constant can be defined as:

aA + bB <--> cC + dD

is a constant and is called the equilibrium constant in terms of concentration, where all the concentrations are at equilibrium and are expressed in moles per litre.

EXAMPLE

Q.1 The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500K.

[N2 ] = 1.5 X 10-2 M

[H2] = 3.0 X 10-2 M

[NH3] = 1.2 X 10-2 M

Calculate equilibrium constant.

SOLUTION

The equilibrium constant for the reaction3 H2 + N2 → 2 NH3 can be written as:

Kc = [NH3]2 / [H2]3 [N2]

(1.2 X 10-2)2 / (1.5 X 10-2)(3.0 X 10-2)3

= 0.106 x 104 = 0.106 x 103

Kp in homogeneous gaseous equilibria

A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase. In this case, to use Kp, everything must be a gas.

A good example of a gaseous homogeneous equilibrium is the conversion of sulphur dioxide to sulphur trioxide at the heart of the Contact Process:

Writing an expression for Kp

If you allow this reaction to reach equilibrium and then measure (or work out) the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, Kp.

RELATION BETWEEN KP AND KC

PV = nRT P=CRT where C= n/v = (moles per litre)Pc=[C] RT;

Pd=[D] RT;

PA=[A] RT;

PB=[B] RT;

KP =[C]c(RT)c [D]d(RT)d / [A ]a (RT)a[B]b(RT)b

=[C]c [D]d (RT)(c+d) - (a+b)

[A]a [B]b

Kp = Kc (RT) n

where ng = (c+d) – (a+b), calculation of n

involves only gaseous components.

n = sum of the no. of moles of gaseous products- sum of the no. of moles of gaseous reactants

1. Equilibrium constant is applicable only when concentrations of the reactants and products have attained their equilibrium state.

2. The value of equilibrium constant is independent of initial concentrations of the reactants and products.

3. Equilibrium constant is temperature dependent.

4. The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.

5. The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction,whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

LET US CONSIDER APPLICATIONS OF EQUILIBRIUM CONSTANT TO:

1.Predict the extent of a reaction on the basis of its magnitude.

2.Predict the direction of the reaction.3.Calculate equilibrium concentrations.

K = [Product]eq / [Reactant]Eq

CASE 1

If K is large (K>103) then the product concentration is very very larger than the reactant {[Product] >> [Reactant]}.Hence concentration of reactant can be neglected with respect to the product. In this case, the reaction is product favourable and equilibrium will be more in forward direction than in backward direction.

CASE 2 If K is small (K<10-3) Product << Reactant

Hence the concentration of product can be neglected as compared to the reactant.

In this case ,the reaction is reactant is favourable.

Reaction Quotient (Q) At each point in the reaction , we can write

a ratio of concentration terms having the same form as the equilibrium constant expression.This ratio is called the reaction qoutient denoted by symbol Q.

It helps in predicting the direction of a reaction.

The expression Q = [C]c [D]d / [A]a [B]b at any time during the reaction is called reaction quotient.The concentrations [C] , [D] , [A] , [B] are not necessarily at equilibrium.

If Q > Kc reaction will proceed in backward direction until equilibrium is reached.

If Q < Kc reaction will proceed in forward direction until equilibrium is established.

If Q = Kc reaction is at equilibrium.

Q. The value of Kc for a reaction

2A B + C is 2 x 10 -3. At a given time,the composition of reaction mixture is [A]=[B]=[C] = 3x10-4 M. In which direction the reaaction will proceed?

ANS : For the reaction the reaction quotient Q is given by,

Qc = [B][C]/[A]2

As [B] = [C] = [A] = 3x10-4 M Qc = (3x10-4)(3x10-4) / (3x10-4)2 = 1 As Qc > Kc so the reaction will proceed in the reverse

direction

The concentration of various reactants and products can be calculated using the equilibrium constant and the initial concentrations.In case of a problem in which we know the

initial concentrations but do not know any of the equilibrium concentrations,the following three steps shall be followed :1.Write the balanced chemical equation for the reaction.2.Under the balanced equation,make a table that lists for each substance involved in the reaction:(a)The initial concentration(b) the change in concentration on going to equilibrium(c) the equlibrium concentration

3. Substitute the equlibrium concentrations into the equilibrium equation for the reaction and solve for x.If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.

4. Calculate the equilibrium concentrations from the calculated value of x.

5. Check your results by substituting then into the equilibrium equation.

Q.13.8g of N2 O4 was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium.

N2 O4 2NO2

The total pressure at equilibrium was found to be 9.15 bar.Calculate KC , Kp and Partial pressure at equilibrium.

We know pV = nRT

Total volume = 1LMolecular mass of N2O4 = 92gNumber of moles = 13.8 / 92 = 0.15Gas constant = 0.083 bar L mol-1 K-1

Temperature = 400K pV=nRT

P = 4.98

N2O4 2NO2

Initial pressure 4.98 0At equilibrium 4.98 – x 2xHence,Ptotal at equilibrium = P (N2O4) + P(NO2)9.15 = 4.98-x + 2xX = 9.15 – 4.98 = 4.17.Partial pressures at equilibrium are,P(N2O4) = 4.98 – 4.17 = 0.81 barP(NO2) = 2x = 2 x 4.17 = 8.34 barKp = [P(NO2)]2 / P(N2O4) = (8.34)2 / 0.81 = 85.87KP = Kc (RT) n

85.87 = Kc (0.083 X 400) Kc = 2.586 = 2.6

Effect of concentration change If the concentration of a component is

increased,reaction shifts in a direction which tends to decrease its concentration.

Eg : N2 + 3H2 2NH3

[Reactant] Forward shift

[Product] Backward shift

If concentration of reactant is increased at equilibrium then reaction shifts in the forward direction.

If concentration of product is increased then reaction shifts in the backward direction.

On increasing pressure,equilibrium will shift in the direction in which pressure decreases i.e no. of moles in the reaction decreases and vice versa.

For n = 0 (No effect) For n > 0

If P ; QP (Forward shift)

If P ; Qp (Backward shift)

For n < 0

P QP (Forward shift)

P QP (Backward shift)

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 then to maintain the pressure

constant,volume is increased.Hence equilibrium will shift in the direction in which larger no. of moles of gas is formed.

I. ng > 0 reaction will shift in forward direction

II. ng < 0 reaction will shift in backward direction

III. ng = 0 no effect

EFFECT OF TEMPERATURE CHANGE

Whenever an equilibrium is disturbed by a change in concentration , pressure or volume, the composition of the equilibrium mixture changes because the reaction quotient Qc no longer equals the equilibrium constant Kc. However,when a change in temperature occurs the value of equilibrium constant is changed.

In general the temperature dependence of the equilibrium constant depends on the sign of H for the reaction.

The equilibrium constant for an exothermic reaction decreases as the temperature increases.

The equilibrium constant for an endothermic reaction increases as temperature increases.

EFFECT OF CATALYST

Due to catalyst,the 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 altering the forward and reverse rate equally and hence,the equlibrium will be altered faster.

LE CHATELIER’S PRINCIPLE

It states that a change in any of the factors that determine the eqiulibrium 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.