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Dec 23, 2018

Law of Mass Action

Low of Mass Action to the Chemical Reaction:

The Low of Mass Action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864. The basis of their formulation is the observation of huge deposit of Sodium Carbonate on Egyptian take shore. Large amount of NaCl in take water and CaCO3 on the take shore made the reverse reaction possible.

CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2NaCl(aq)

The foreword reaction occurs spontaneously in the laboratory.
They States the Low as, 
The rate of chemical reaction at a constant temperature, is directly proportional to the active mass of the reactants.
The active mass is thermodynamic quantity. 
We assumed active mass as, 
(a) Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally. 
(b) Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low. 
(c) For pure solid and pure liquid, active mass is assumed to be unity since their mass does not effect the rate of reaction.

Application of Low of Mass Action to the Chemical Reaction: 

Let us Consider a reaction 
A + B C + D
Let the reacting system contains reactants only and CA and CB are their Concentrations in molar units. According to the mass action low, the rate of the foreword reaction, 
Rf  ∝ CA × CB
Rf = Kf × CA × CB
Though the rate equation is an experimental quantity and powers of concentration terms are determined experimentally, here we consider a special type of rate equation where the stoichometric coefficients are raised in the of the conc. term(this is true for the elementary or one step reaction).
Law of Mass Action and derivation of Equilibrium Constant
Low of Mass Action
As the reaction proceeds in the foreword direction, conc. of A and B decreases and Rf also decreases. When the products are getting accumulated in the system, backwards reaction also starts and the rate of the backward reaction,
Rb  ∝ CC × CD
Rb = Kb × CC × CD
Here Kf and Kb are the rate constants of the foreword and backward reaction and they do not depends on Conc. at a given temperature.
As the reactions proceeds in the foreword reactions Rf  is decreasing but Rb is increasing. A state is then attain when they are equal. This state is called the chemical equilibrium. There will be no further change in the composition of the system.
Thus at equilibrium, Rf  = Rb
or, Kf × CA × CB  = Kb × CC × CD
or, Kf/Kb = CC × CD/CA × CB
CACBCC and CD are the equilibrium concentration of ABC and D
Again, Kf/Kb  = KC, called concentration equilibrium constant of the reaction. At a given temperature for a reaction, KC is constant does not depend on the concentration of the reacting components. 

KC  = CC × CD/CA × CB

Equilibrium Constant : 

Concentration Equilibrium Constant: 

If we write the equation as,
Ɣ1A1 + Ɣ2A2  Ɣ3A3 + Ɣ4A4
Where Ɣ1Ɣ2Ɣ3 and Ɣ4 are stoichiometric coefficient.

KC = (C3Ɣ3 × C4Ɣ4)/(C1Ɣ1 × C2Ɣ2)

Where KC is called concentration equilibrium constant of the reaction and C1 ,C2 ,C3 and C4  are the equilibrium concentration of A, B, C and D.
However the values of equilibrium constant of a chemical reaction depends on the mode of writing its Stoichiometric (balanced) equation.
Thus, for the reaction of formation of NH₃ from N₂ and H₂, we can write the equation as,
N₂ + 3 H₂  2 NH₃
The equilibrium constant can be written as, 
KC (CNH₃)2/(CN₂) (CH₂)2
But if the equation written as,
1/2 N₂ + 3/2 H₂  NH₃
Then, C = (CNH₃)/(CN₂)3/2(CH₂)1/2 

It is clear then KC and C are not Equal in magnitude.
Thus, KC = (C
The general rule is that if the equation is multiplied by n then the relation of Equilibrium concentration n will be raised to the power.
That is, KC = (C)ⁿ

Pressure Equilibrium Constant : 

When all the reactants and products are gases (that is, gas - phase reacting system), the expression of equilibrium constant for the equation,
Ɣ1A1 + Ɣ2A2  Ɣ3A3 + Ɣ4A4
Where Ɣ1Ɣ2, Ɣ3 and Ɣ4 are stoichiometric coefficient.

∴ KP = (P3Ɣ3 × P4Ɣ4/(P1Ɣ1 × P2Ɣ2)

Where KP is called Pressure equilibrium constant of the reaction and P1 ,P2 ,P3 and P4  are the equilibrium partial pressure of reacting components.

For the dissociation N2O4  2NO2, obtain an expression for the fraction of original N2O4 dissociated at equilibrium in terms of KP and total pressure.

The reaction is N2O4  2NO2 
Let a mole of N2O4 is taken initially and x moles of N2O4 is dissociated at equilibrium then mole number of N2O4 and NO2 at equilibrium is (a - x) and 2x.
Total moles number at equilibrium = a -x + 2x = a + x
Equilibrium partial pressure of the components is {(a - x)/(a + x)}P and {2x/(a + x)}P where P = Total pressure of the reacting system.
The expression of the equilibrium constant,
KP = (PNO2)2/PN2O4 = {2x/(a + x)}P/{(a - x)/(a + x)}P 
or, KP = (4x2P)/(a2 - x2
The fraction of the original N2O4  dissociated at equilibrium ɑ = x/a.
Replacing (x/a) by ɑ, we have
KP = (2P)/(1 - ɑ2)

Mole Fraction Equilibrium Constant:

Kx =  (x3Ɣ3 × x4Ɣ4)/(x1Ɣ1 × x2Ɣ2) 

Where Kx  is called Mole fraction Equilibrium Constant of the reaction and x1x2x3, and x4 are the equilibrium mole fraction of reacting components.

Relation Between KP and KC : 

The interrelations of these Equilibrium constants are as follows,
KP = (P3Ɣ3 × P4Ɣ4)/(P1Ɣ1 × P2Ɣ2)
= (C3RT)Ɣ3 × (C4RT)Ɣ4/(C1RT)Ɣ1 × (C2RT)Ɣ2

Thus, KP = KC(RT)ΔƔ
ΔƔ = (Ɣ3 + Ɣ4) - (Ɣ1 + Ɣ2)


H2 (g) + I2 (g) ⇆ 2HI
KP = (PHI)2/{(PH₂)(PI₂)}
= [{(CHI)2}/{(CH₂) (CI₂)}] × [(RT)2/(RT)(RT)]
Thus when (c+d) = (a+b), KP = KC
Example of the some reaction where KP ≠ KC :
2SO2(g) + O2(g)  SO3(g)
Here, KP = KC×(RT){1 - (2+1)}KC×(RT) - 2
PCl5(g)  PCl3(g) + Cl2(g)
KK× (RT)
2H2(g) + O2(g)  2H2O(g) 
KK× (RT) - 1
4HCl(g) + O2(g)  2H2O(g) + Cl2(g)  
KK× (RT) - 2
2CO(g) + O2(g)  2CO2(g) 
KK× (RT) - 1

Calculate the Kc value of the reaction N₂ + 3H₂ ⇆ 2NH₃ at 400°C, 
Given Kp at the same temperature 1.64 × 10⁻⁴.

We know that Kp = Kc × (RT)Δn 
For the above equation,Kp = Kc × (RT) - 2 = 1.64 × 10⁻⁴ × (0.082 × 673)
= 0.5

At 1000C the vapour density of N2O4 is 25 at 1 atm. Show that KP = 9.6.

Let 1 moles of a is taken initially (t = 0) and x mole of N2O4 has reacted at equilibrium.
So the mole number of each component is (1-x) and 2x and total moles at equilibrium (1-x+2x) = (1+x).
So total moles has increased from 1 to (1+x).
Let Volume is increases from V1 to V2.
So, (1+x) =  V2/V1. 
As density and hence vapour density is inversely proportional to volume so vapour density will decrease from d1 to d2.
Hence (1+x) =  V2/V1d1/d2.
Molecular weight of a is 92 and vapour density = 92/2 = 46.
Due to dissociation it is = 25.
∴ 1+x = 46/25
or, x = 0.84
Now partial pressure are, PNO2 
= {2x/(1+x)}P = (2×0.84)/1.84 = 0.913 atm 
and PN2O4  = {(1-x)/(1+x)}P = 0.16/1.84 = 0.087
∴ KP = (PNO2)2/PN2O4 = (0.913)2/0.087 ≃ 9.6

Relation Between Kand Kx :

Kp = Kx × (P)Δn 
Where, ΔƔ = (Ɣ3 + Ɣ4) - (Ɣ1 + Ɣ2)