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’ combine 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 decreases and 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 is possible only in a closed

system at a given temperature.3. Both the opposing processes occur at

the same rate and there is a dynamic but stable condition.

4. All measurable properties of the system remain constant.

5. 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 respective 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 equilibrium 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

= 0.0355 x 104 = 355.0 mol/L

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:

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.

Writing an expression for 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.

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 favourable.

If K is in between 10-3 to 103

Then reactants = productsThe system is in equilibrium.

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 reaction 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 equilibrium 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 them 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. [given: Gas constant = 0.083 bar L mol-1 K-1]

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 Effect of pressure change Effect of inert gas addition Effect of temperature change Effect of catalyst

LE CHATELIER’S PRINCIPLEIt states that a change in any of the factors 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.

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.

Effect of concentration change

• CHANGES IN CONCENTRATION CONTINUED

Change

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD aA + bB cC + dD

AddAddRemove Remove

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.

C(s) + CO2(g) ---------------> 2CO(g) When pressure is increased, the

reaction goes in the reverse direction because the number of moles of gas increases in the forward direction.

• Changes in Volume and Pressure(Only a factor with gases)

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

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.

• Temperature – increasing the temperature causes the equilibrium position to shift in the direction that absorbs heat• If heat is one of the products (just like a

chemical), it is part of the equilibrium so cooling an exothermic reaction will produce more product, and heating it would shift the reaction to the reactant side of the equilibrium: C + O2(g) → CO2(g) + 393.5 kJ

CHANGES IN TEMPERATUREOnly factor that can change value of K

Change Exothermic Rx

Increase temperature K decreases

Decrease temperature K increases

Endothermic Rx

K increases

K decreases

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 equilibrium will be achieved faster.

uncatalyzed catalyzed

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

Change Shift Equilibrium

Change Equilibrium Constant

Concentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Catalyst no no

ACIDS & BASESACIDS & BASES

Acids:Acids: acids are sour tasting Arrhenius acidArrhenius acid: Any substance that, when dissolved in

water, increases the concentration of hydronium ion (H3O+)

Bronsted-Lowry acidBronsted-Lowry acid: A proton donor Lewis acidLewis acid: An electron acceptor

Bases:Bases: bases are bitter tasting and slippery Arrhenius baseArrhenius base: Any substance that, when dissolved in

water, increases the concentration of hydroxide ion (OH-)

Bronsted-Lowery baseBronsted-Lowery base: A proton acceptor Lewis acidLewis acid: An electron donor

ARRHENIUS (OR CLASSICAL) ACID-BASE DEFINITION

Neutralization is the reaction of an H+ (H3O+) ion from the acid and the OH - ion from the base to form water, H2O.

H+(aq) + OH-

(aq) H2O(l) H0rxn = -55.9 kJ

An acid is a substance that contains hydrogen and dissociates in water to yield a hydronium ion : H3O+

A base is a substance that contains the hydroxyl group and dissociates in water to yield : OH -

The neutralization reaction is exothermic and releases approximately56 kJ per mole of acid and base.

BRØNSTED-LOWRY ACID-BASE DEFINITION

An acid is a proton donor, any species that donates an H+ ion. An acid must contain H in its formula; HNO3 and H2PO4

- are two examples, all Arrhenius acids are Brønsted-Lowry acids.

Therefore in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process.

Acids donate a proton to water

Bases accept a proton from water

A base is a proton acceptor, any species that accepts an H+ ion. A base must contain a lone pair of electrons to bind the H+ ion; a few examples are NH3, CO3

2-, F -, as well as OH -. Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases contain the Brønsted-Lowry base OH-.

MOLECULAR MODEL: THE REACTION OF AN ACID HA WITH WATER TO FORM H3O+ AND A CONJUGATE BASE.

Acid Base Conjugate Conjugate acid base

AUTOIONIZATION OF WATER

H2O(l) + H2O(l) H3O+ + OH-

Kc =[H3O+][OH-]

[H2O]2

The ion-product for water, Kw:

Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C)

For pure water the concentration of hydroxyl and hydronium ions must be equal:

[H3O+] = [OH-] = 1.0 x 10-14 = 1.0 x 10 -7 M (at 25°C)

ACID AND BASE CHARACTER AND THE PH SCALEIn acidic solutions, the protons that are released into solution will notremain alone due to their large positive charge density and small size.They are attracted to the negatively charged electrons on the oxygen atoms in water, and form hydronium ions.

H+(aq) + H2O(l) = H3O+

(l)

To handle the very large variations in the concentrations of the hydrogenion in aqueous solutions, a scale called the pH scale is used which is:

pH = - log[H3O+]

What is the pH of a solution that is 10-12 M in hydronium ion ?pH = -log[H3O+] = (-1)log 10-12 = (-1)(-12) = 12

What is the pH of a solution that is 7.3 x 10-9 M in H3O+ ?

pH = -log(7.3 x 10-9) = -1(log 7.3 + log 10-9) = -1[(0.863)+(-9)] = 8.14pH of a neutral solution = 7.00pH of an acidic solution < 7.00pH of a basic solution > 7.00