**was first formulated by two Norwegian chemists, Guldberg and Waage in 1864.**

*Law of Mass Action*The basis of their formulation is the observation of a huge deposit of Sodium Carbonate on Egyptian take shore. A large amount of NaCl intake water and CaCO₃ on the take shore made the reverse reaction possible.

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

- The forward 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,

- Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally.
- Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low.
- For pure solid and pure liquid, active mass is assumed to be unity since their mass does not affect the rate of reaction.

###
*Mathematical expression of Law of mass action*

*Mathematical expression of Law of mass action*

- Let us Consider a reaction, A + B ⇆ C + D

- Let the reacting system contains reactants only and [A] and [B] are their Concentrations in molar units.

- According to the

**, the rate of the forward reaction,**

*Law of mass action*
R

∴ R

_{f}∝ [A] × [B]∴ R

_{f}= K_{f}× [A] × [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 stoichiometric coefficients are raised in the of the concentration term(this is true for the elementary or one-step reaction).

- As the reaction proceeds in the forward direction, concentration of A and B decreases and R

_{f}also decreases.

- When the products are getting accumulated in the system, the backward reaction also starts and the rate of the backward reaction according to the

**,**

*law of mass action*
R

∴ R

_{b}∝ [C] × [D]∴ R

_{b}= K_{b}× [C] × [D]- Here K

_{f}and K

_{b}are the rate constants of the forward and backward reaction and they do not depend on concentration at a given temperature.

- As the reactions proceed in the forward reactions R

_{f}is decreasing but R

_{b}is increasing. A state is then attained 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, R

_{f}= R

_{b}

or, K

∴ K

_{f}× [A] × [B] = K_{b}× [C] × [D]∴ K

_{f}/K_{b}= [C] × [D]/[A] × [B]- [A], [B], [C] and [D] are the equilibrium concentration of A, B, C and D.

- Again, K

_{f}/K

_{b}= K

_{c}Where K

_{c}is the concentration equilibrium constant of the reaction.

- At a given temperature for a reaction, K

_{c}is constant does not depend on the concentration of the reacting components.

K_{c} = [C] × [D]/[A] × [B] |

###
*Equilibrium constant*

*Equilibrium constant*

*Concentration equilibrium constant*

- If we write the equation as,

Æ”₁A₁ + Æ”₂A₂ ⇆ Æ”₃A₃ + Æ”₄A₄

Where Æ”₁, Æ”₂, Æ”₃, and Æ”₄ are stoichiometric coefficients.

Where Æ”₁, Æ”₂, Æ”₃, and Æ”₄ are stoichiometric coefficients.

∴ Kc = [A₃]^{Æ”3} × [A₄]^{Æ”4}/[A₁]^{Æ”1} × [A₂]^{Æ”2} |

- Where Kc is called concentration equilibrium constant of the reaction and [A₁], [A₂], [A₃], and [A₄] is the equilibrium concentration of A, B, C, and D.

- However, the values of the equilibrium constant of a chemical reaction depend 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 = [NH3]²/[N2] [H2]²

- But if the equation is written as, ½ N2 + 3/2 H2 ⇆ NH3

- Then, K՛c = [NH3]/[N2]

^{3/2}[H2]

^{½}

- It is clear then Kc and K՛c are not Equal in magnitude. Thus, Kc = (K՛c)

^{½}

- 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.

Kc = (K՛c)^{n} |

*Pressure equilibrium constant*

- When all the reactants and products are gases (that is, gas-phase reacting system), the expression of the 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

_{p}is called Pressure equilibrium constant of the reaction and P

_{1}, P

_{2}, P

_{3, }and P

_{4}are the equilibrium partial pressure of reacting components.

*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 the stoichiometric coefficient

∴ K_{x} = (x_{3}^{Æ”3} × x_{4}^{Æ”4})/(x_{1}^{Æ”1} × x_{2}^{Æ”2}) |

- Where K

_{x}is called Mole fraction Equilibrium Constant of the reaction and x

_{1}, x

_{2}, x

_{3}and x

_{4}are the equilibrium mole fraction of reacting components.

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

K

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

PV = nRT, may be written as,

P = (n/V)RT = CRT

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

__ideal gas Equation__,PV = nRT, may be written as,

P = (n/V)RT = CRT

Where C is the concentration of gas expressed as an 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_{c}×(RT)^{{1 - (2+1)}}
= K

_{c}×(RT)^{-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} |

####
*Relation between Kp and Kx*

*Relation between Kp and Kx*

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

###
*Problems solutions*

*Problem*

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

*Solution*

__Vant Hoff reaction__isotherm is Î”G = - RT lnK

_{a}+ RT lnQ

_{a}But when the reaction attain equilibrium,Qa = Ka

Thus, Î”G = 0.

*Problem*

- For the dissociation N

_{2}O

_{4}⇆ 2 NO

_{2}, obtain an expression for the fraction of original N

_{2}O

_{4}dissociated at equilibrium in terms of K

_{p}and total pressure.

*Solution*

- The reaction is, N

_{2}O

_{4}⇆ 2 NO

_{2}

- Let 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. 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,

K

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

∴ K

K

_{p}= (P_{NO2})^{2}/P_{N2O4}= {2x/(a + x)}P/{(a - x)/(a + x)}P

∴ K

_{p}= (4x^{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} = (4Î±^{2}P)/(1 - Î±^{2}) |

*Problem*

- Calculate the K

_{c}value of the reaction N

_{2}+ 3H

_{2}⇆ 2NH

_{3}at 400

^{0}C, Given Kp at the same temperature 1.64 × 10

^{-4}.

*Solution*

- K

_{p}= 0.5

*Problem*

- At 100

^{0}C the vapor density of N

_{2}O

_{4}is 25 at 1 atm. Show that K

_{p}= 9.6.

*Solution*

- N

_{2}O

_{4}(g) ⇆ 2NO

_{2}(g)

- Let 1 mole of N

_{2}O

_{4}is taken initially (t = 0) and x mole of N

_{2}O

_{4}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 increases from V

_{1}to V

_{2}.

So, (1+x) = V

_{2}/V_{1}- As density and hence vapor density is inversely proportional to volume so vapor density will decrease from d

_{1}to d

_{2}.

Hence, (1+x) = V

The molecular weight of N

= 92/2

= 46

Due to dissociation, it is = 25.

∴ 1+x = 46/25

or, x = 0.84

Now partial pressure are,

P

= (2×0.84)/1.84

= 0.913 atm

P

= 0.16/1.84= 0.087

∴ K

= (0.913)2/0.087

≃ 9.6

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

_{2}O_{4}is 92 and vapor 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

_{p}= (PNO2)2/PN2O4= (0.913)2/0.087

≃ 9.6