EQUILIBRIUM CONSTANT Kp ,Kc AND Kx (LAW OF MASS ACTION)

Law of mass action is applicable for only reversible chemical reactions and it is an imperial law.

The law state that “ At a fixed temperature the rate of a chemical reaction is directly proportional to the product of reactive mass of reactants raised to the their respective Stoichiometric cofficients ” The law of mass action is by Guldberg and Waage.

(1) Equilibrium Constants in term of concentration ( K_C):

At the constant temperature, let us consider the following reversible reaction

According to law of mass action - The rate of forward reaction

The rate of reverse reaction-

Where K_f and K_b  is the rate constant of the forward reaction and backward reaction respectively

We know at equilibrium, the two rates of forward as well as backward are equal. ie

Rate of reaction = Rate of forward reaction – Rate of backward reaction = 0

Kc=Kf/Kb

Unit of Kc= (Conc)ng

(2) Equilibrium Constants in term of Pressure( Kp):

Consider the general chemical  reaction taking place at constant temperature.

From law of mass action- rate forward reaction is directly proportional  to product of active mass of reactants and rate backward reaction is directly proportional  to product of active mass of products.

For an ideal gas PV=nRT

Where 

P= Pressure in atm

V=Volume in liters

n=Number of gaseous moles

R=Gas constant  

  =  0.0821 L atm/mol/K or 1/12 L atm /mole/K

T=Temperature in kelvin

                                      = total number of moles of gaseous products -total number of moles of gaseous reactants

EXAMPLE (1): At 700 K, the equilibrium constant Kp, for the reaction

is 1.8 × 10³  kPa What is the numerical value of Kc for this reaction at the same temperature

SOLUTION:

EXAMPLE (2) : Methanol (CH₃OH) is manufactured industrially by the reaction

 CO(g) + 2H₂(g) ⇌ CH₃OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of Kp at this temperature ?

SOLUTION: Give Data ,Kc = 10.5 ,T= 220oC = (220 + 273)K = 493 K

EXAMPLE(3): For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) The Kp is 4.3 × 10^-4 at 375°C. Calculate Kc for the reaction. ?

SOLUTION:

EXAMPLE(4):   The value of Kc for the reaction is 0.50 at 400º C.

What will be the value of Kp at 400ºC when concentration are expressed in mole litre.1 and pressure in atmosphere ?

SOLUTION: 

EXAMPLE(5):

EXAMPLE(6):

SOLUTION:

PH CALCULATION OF WEAK ACIDS AND WEAK BASES

PH AND POH CALCULATION OF STRONG ACIDS AND BASES

PH AND POH AND SCALE

IONISATION AND IONIC PRODUCT OF WATER (Kw)

Water is a polar ,protic inorganic compound that is at room temperature a tasteless and odourless liquid nearly colourless with hint of blue. It covers about 71.4% of earth. Water is universal solvent due to its ability to dissolved many substance .The IUPAC name of is Water  and  Oxidane.

Pure water is a very weak electrolyte -

On applying the law of mass action at equilibrium,

              Then

Since, ionization takes places to a very small extent, so the concentration of unionized water molecule is regarded as constant. Thus the product of K[H2O] gives another constant Kw.

                       So    

The product of concentration of H+ and OH. ion in water at a particular temperature is known as ionic  product of water.

IMPORTANT POINT ABOUT WATER:

(1) Mass of 1 litre of water = 997 gm.

(2) Molar concentration of water = 55.5 gm-mole / litre.

(3) Number of water molecule in 1 litre of water = 55.5×6.023×10²³ = 3.34×10²⁵.

(4) Concentration of H⁺ ion in one litre of neutral water = 10^-7 moles / litre.

(5) Concentration of OH^- ion in one litre of neutral water = 10^-7 moles / litre.

(6) Number of H+ ion in one litre of neutral water = 6.023 x 10¹⁶.

(6) Number of OH.- ion in one litre of neutral water = 6.023 x 10¹⁶.  

IONIZATION CONSTANT (Ki) OR DESSOCIATION CONSTANT (Kd) OF WATER:

At 25^c  Kw=1x10^-14  and conc^. of water 1000/18=55.55 mole /liter

Ki or Kd are also called acid or base constant of water.

Hence Pka and Pkb of water is equal (Pka=Pkb)

DISSOCIATION CONSTANT OF WATER (DOD):

Pure water is weak electrolyte and dissociates as

EFFECT OF TEMPERATURE ON IONIC PRODUCT OF WATER ( Kw):

(1) Ionisation constant of water is endothermic process, so on increasing temperature Keq increase. Kw increases with increase in temperature.

(2) During self ionisation of water dissociation concentration of water is remain constant=55.55 mole /liter.

(3) Kw is a thermodynamics equilibrium constant for water it depend upon only temperature, if temperature is increases then value of Kw is increases. and temperature is decreases then Kw decreases

(4) The numerical value of Kw increases considerably with temperature from 0.11 × 10^–14 at 0°C to 50 × 10^–14 at 100°C. It is 1.0 × 10^–14 at 25°C which we will use frequently the variation of ionic product of water with temperature is given by

On the basis of above equation we calculate the value of kw at different temperature which are given below .

We know ionic product of water at 25 ^c is

The pH scale was marked from 0 to 14 with central point at 7 at 25ºC, taking water as solvent.

If the temperature and the solvent are changed, the pH range of the scale will also change. For example

 at 25 ºC (Kw = 10^–14) Neutral point, pH= 7

 at 80 ºC (Kw = 10^–13) Neutral point, pH = 6.5

ILLUSTRATIVE EXAMPLE (1): H₂O has its pH 6.5 predict the nature of solution when

(a) Kw = 10^–14 at 25 ºC

(b) Kw = 10^–15 at 70 ºC

(c) Kw = 10^–12 at 90 ºC

SOLUTION:

(a) acidic  (b) acidic (c) basic

ILLUSTRATIVE EXAMPLE (2): The ionization constant for water is 1x10^-13.6 at 37ºC.What will be H₃O⁺ and OH^-Concentration at that temperature.

(1) 3.75 x10^-8          (2) 1.75 x 10^-8                      (3) 1.58 x 10-7            (4) 1.85 x 10^-8

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): Calculate the ionic product of water at 25^c if Kw is 1.0 X10^-14 m² at 25^c

(Ans - 5.5x10-14)

ILLUSTRATIVE EXAMPLE (4): At 25 ^c the degree of dissociation of water is 1.8 x 10^-9 calculate the dissociation constant and ionic product of water.

(Ans-3.24x10^-18)

ILLUSTRATIVE EXAMPLE (5): At 30 ^c the self ionization constant of liquid ammonia (NH₃) is 10^-30.the density of NH3 at -30 ^c is 0.85 gm /liter. 

(1) Calculate the ionic product of liquid ammonia?

(2) Calculate % extent of self ionization of NH₃ molecules?

(3) How many NH₄⁺ ions are present in 5ml of liquid NH₃?

(1- 2.5x10-27 m², 2- 10^-13% 3- 25x10^-6x6.022x10²³ )

ILLUSTRATIVE EXAMPLE (6):  At 25^c the ionic product of heavy water (D₂O) is 1.44x 10^-15 m² Calculate it's dissociation constant and degree of dissociation (density of D₂O=1.02) gm/ml)

(Ans-7.44x10^-10)                                               

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