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 CaCO_{3}on the take shore made the reverse reaction possible.CaCl_{2}(aq) + Na_{2}CO_{3}(aq) |

↓ |

CaCO_{3}(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,

(moles/lit) When the solution is dilute, that is when the system behaves ideally.__Molar Concentration__in atmosphere unit for gaseous system and when the pressure of the system is very low.__Partial pressure__- For pure solid and pure liquid, active mass is assumed to be
since their mass does not effect the rate of reaction.__unity__

###
__Application of Low of Mass Action to the Chemical 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

**C**and_{A}**C**are their Concentrations in molar units._{B}
According to the mass action low, the rate of the foreword reaction,

R_{f} ∝ C_{A} × C_{B}∴ R_{f} = K_{f} × C_{A} × C_{B} |

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).

As the reaction proceeds in the foreword direction, conc. of

**A**and**B**decreases and**R**also decreases. When the products are getting accumulated in the system, backwards reaction also starts and the rate of the backward reaction,_{f}R_{b} ∝ C_{C} × C_{D}∴ R_{b} = K_{b} × C_{C} × C_{D} |

Here

**K**and_{f}**K**are the rate constants of the foreword and backward reaction and they do not depends on Conc. at a given temperature._{b}
As the reactions proceeds in the foreword reactions

**R**is decreasing but_{f}**R**is increasing. A state is then attain when they are equal. This state is called the_{b}**. There will be no further change in the composition of the system.**__chemical equilibrium__
Thus at equilibrium,

**R**

_{f}= R_{b}
or, K

_{f}× C_{A}× C_{B}= Kb × C_{C}× C_{D}∴ K_{f}/K_{b} = (C_{C} × C_{D})/(C_{A} × C_{B}) |

C

_{A}, C_{B}, C_{C}and C_{D}are the equilibrium concentration of A, B, C and D.
Again,

**K**, called_{f}/K_{b}= K_{c}**of the reaction.**__concentration equilibrium constant__
At a given temperature for a reaction,

**K**is constant does not depend on the concentration of the reacting components._{c}K_{c} = (C_{C} × C_{D})/(C_{A} × C_{B}) |

###
__Equilibrium Constant:__

__Equilibrium Constant:__

__Concentration Equilibrium Constant:__
If we write the equation as,

Æ”

_{1}A_{1}+ Æ”_{2}A_{2}⇆ Æ”_{3}A_{3}+ Æ”_{4}A_{4}
Where Æ”

_{1}, Æ”_{2}, Æ”_{3}and Æ”_{4}are stoichiometric coefficient.∴ K_{c} = (C_{3}^{Æ”3} × C_{4}^{Æ”4})/(C_{1}^{Æ”1} × C_{2}^{Æ”2}) |

Where

**K**is called concentration equilibrium constant of the reaction and_{c}**C**,_{1}**C**,_{2}**C**and_{3}**C**are the equilibrium concentration of A, B, C and D._{4}
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 **NH**from_{3}**N**and_{2}**H**, we can write the equation as,_{2}**N**

_{2}+ 3 H_{2}⇆ 2 NH_{3}
The equilibrium constant can be written as,

K = (C_{c}_{NH3})^{2}/(C_{N2}) (C_{H2})^{2} |

But if the equation written as,

^{1}/

_{2}N

_{2}+

^{3}/

_{2}H

_{2}⇆ NH

_{3}

Then, K՛= (C_{c} _{NH3})/(C_{N2})^{3/2}(C_{H2})^{1/2} |

It is clear then

**K**and_{c}**K՛**are not Equal in magnitude._{c}Thus, K_{c} = (K՛_{c})^{1/2} |

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, K_{c} = (K՛_{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,

Æ”

_{1}A_{1}+ Æ”_{2}A_{2}⇆ Æ”_{3}A_{3}+ Æ”_{4}A_{4}
Where Æ”

_{1}, Æ”_{2}, Æ”_{3}and Æ”_{4}are stoichiometric coefficient.∴ K_{p} = (P_{3}^{Æ”3} × P_{4}^{Æ”4})/(P_{1}^{Æ”1} × P_{2}^{Æ”2}) |

Where

**K**is called Pressure equilibrium constant of the reaction and_{p}**P**,_{1}**P**,_{2}**P**and_{3}**P**are the equilibrium partial pressure of reacting components._{4}
For the dissociation

**N**, obtain an expression for the fraction of original_{2}O_{4}⇆ 2 NO_{2}**N**dissociated at equilibrium in terms of_{2}O_{4}**K**and total pressure._{p}
The reaction is, N

_{2}O_{4}⇆ 2 NO_{2}
Let

Total moles number at equilibrium = (a -x + 2x) = **a**mole of N_{2}O_{4}is taken initially and**x**moles of N_{2}O_{4}is dissociated at equilibrium then mole number of N_{2}O_{4}and NO_{2}at equilibrium is**(a - x)**and**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,

**K**= (P

_{p}_{NO2})

^{2}/P

_{N2O4}

= {2x/(a + x)}P/{(a - x)/(a + x)}P

∴

**K**= (4x_{p}^{2}P)/(a^{2}- x^{2})
The fraction of the original N

_{2}O_{4}dissociated at equilibrium**É‘ = x/a**.
Replacing (

**x/a**) by**É‘**, we have,K_{p} = (4É‘^{2}P)/(1 - É‘^{2}) |

__Mole Fraction Equilibrium Constant:__
If we write the equation as,

Æ”

_{1}A_{1}+ Æ”_{2}A_{2}⇆ Æ”_{3}A_{3}+ Æ”_{4}A_{4}
Where Æ”

_{1}, Æ”_{2}, Æ”_{3}and Æ”_{4}are stoichiometric coefficient.∴ K_{x} = (x_{3}^{Æ”3} × x_{4}^{Æ”4})/(x_{1}^{Æ”1} × x_{2}^{Æ”2}) |

Where

**K**is called Mole fraction Equilibrium Constant of the reaction and_{x}**x**,_{1}**x**,_{2}**x**and_{3}**x**are the equilibrium mole fraction of reacting components._{4}Equilibrium Constant. |

###
__Relation Between __K_{p} __and__ K_{c}__:__

__Relation Between__K

_{p}

__and__K

_{c}

__:__

The interrelations of these Equilibrium constants are as follows,

K

_{p}= (P_{3}^{Æ”3}× P_{4}^{Æ”4})/(P_{1}^{Æ”1}× P_{2}^{Æ”2})
The

__ideal gas Equation__, PV = nRT, may be written as,
P = (n/V)RT = CRT

Where C is the concentration of gas expressed as amount per unit volume.

= {(C

_{3}RT)^{Æ”3}× (C_{4}RT)^{Æ”4}}/{(C_{1}RT)^{Æ”1}× (C_{2}RT)^{Æ”2}}∴ K_{p} = K_{c}(RT)^{Î”Æ”}Î”Æ” = (Æ” _{3} + Æ”_{4}) - (Æ”_{1} + Æ”_{2}) |

__For the reaction in which total number of reactant molecules and of resultant molecules are same____:____For the reactions in which the number of molecules of reactants differ from that of the resultant:__

**H**

_{2}(g) + I_{2}(g) ⇆ 2HI**KP**= (P

_{HI})

^{2}/{(P

_{H2})(P

_{I2})}

= (C

_{HI}RT)^{2}/(C_{H2}RT) (C_{I2}RT)
= [(C

_{HI})^{2}/{(C_{H2}) (C_{I2})}] × [(RT)^{2}/(RT)(RT)]
=

**K**_{c}
Thus when (Æ”

_{3}+ Æ”_{4}) = (Æ”_{1}+ Æ”_{2}),**K**=_{p}**K**_{c}**2SO**

_{2}(g) + O_{2}(g) ⇆ SO_{3}(g)
Here,

**K**=_{p}**K**×(RT)_{c}^{{1 - (2+1)}}
=

**K**×(RT)_{c}^{-2}
Thus when, (Æ”

_{3}+ Æ”_{4})**≠**(Æ”_{1}+ Æ”_{2}),**K**_{p}**≠****K**_{c}Reaction(Gaseous System) |
Relation between K _{p} and K_{c} |

H_{2} + Cl_{2} ⇆ 2HCl |
K = _{p}K_{c} |

CO + H_{2}O ⇆ 2H_{2 }+ CO_{2} |
K = _{p}K_{c} |

CO + NO_{2} ⇆ NO + CO_{2} |
K = _{p}K_{c} |

PCl_{5} ⇆ PCl_{3} + Cl_{2} |
K = _{p}K _{c}RT |

2H_{2} + O_{2} ⇆ 2H_{2}O |
K = _{p}K _{c}RT^{ -1} |

2CO + O_{2} ⇆ 2CO_{2} |
K = _{p}K _{c}RT^{ -1} |

N_{2} + 3H_{2} ⇆ 2NH_{3} |
K = _{p}K _{c}RT^{ -2} |

Calculate the

**K**value of the reaction_{c}**N**at_{2}+ 3H_{2}⇆ 2NH_{3}**400**, Given^{0}C**Kp**at the same temperature**1.64 × 10**.^{-4}
At

**100**the vapour density of^{0}C**N**is_{2}O_{4}**25**at 1 atm. Show that**K**._{p}= 9.6
N

_{2}O_{4}(g) ⇆ 2NO_{2}(g)
Let

**1**moles of**N**is taken initially (_{2}O_{4}**t = 0**) and**x**mole of**N**has reacted at equilibrium._{2}O_{4}
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

**V**to_{1}**V**._{2}
So,

**(1+x) = V**_{2}/V_{1}
As density and hence vapour density is inversely proportional to volume so vapour density will decrease from

**d**to_{1}**d**._{2}
Hence, (1+x) = V

_{2}/V_{1}= d_{1}/d_{2}.
Molecular weight of N

_{2}O_{4}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,

P

_{NO2}= {2x/(1+x)}P
= (2×0.84)/1.84

**= 0.913 atm**

P

_{N2O4}= {(1-x)/(1+x)}P
= 0.16/1.84

**= 0.087**

∴

**K**= (PNO2)2/PN2O4_{p}
= (0.913)2/0.087

**≃ 9.6**

###
__Relation Between __K_{p} __and__ K_{x}__:__

__Relation Between__K

_{p}

__and__K

_{x}

__:__

∴ K_{p} = K_{x}(P)^{Î”Æ”}Î”Æ” = (Æ” _{3} + Æ”_{4}) - (Æ”_{1} + Æ”_{2}) |

Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.

__Vant Hoff reaction__isotherm is

**Î”G = - RT ln**

**K**

_{a}+ RT lnQ_{a}
But when the reaction attain equilibrium, Q

_{a}= K_{a}**Thus, Î”G = 0.**