### Law of Mass Action

**Law of mass action**was first formulated by two Norwegian chemists, Guldberg and Waage in 1864.

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 state the law 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 the atmosphere unit for the 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.

### Expression of Law of mass action

Let us Consider a reaction,

A + B ⇆ C + D

A + B ⇆ C + D

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

- According to the Law of mass action, the rate of the forward reaction,

Rf ∝ [A] × [B]

∴ Rf = Kf × [A] × [B]

∴ Rf = Kf × [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, the concentration of A and B decreases and Rf 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,

Rb ∝ [C] × [D]

∴ Rb = Kb × [C] × [D]

∴ Rb = Kb × [C] × [D]

- Here Kf and Kb 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 Rf is decreasing but Rb 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,

Rf = Rb

or, Kf × [A] × [B] = Kb × [C] × [D]

∴ Kf/Kb = [C] × [D]/[A] × [B]

Rf = Rb

or, Kf × [A] × [B] = Kb × [C] × [D]

∴ Kf/Kb = [C] × [D]/[A] × [B]

- [A], [B], [C] and [D] are the equilibrium concentration of A, B, C and D.

Again, Kf/Kb = Kc

where Kc is the concentration equilibrium constant of the reaction.

where Kc is the 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 = [C] × [D]/[A] × [B]

### Equilibrium constant

#### Concentration equilibrium constant

- If we write the equation as,

Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

Where Ɣ₁, Ɣ₂, Ɣ₃, and Ɣ₄ are stoichiometric coefficients.

∴ Kc = [A₃]

Where Ɣ₁, Ɣ₂, Ɣ₃, and Ɣ₄ are stoichiometric coefficients.

∴ Kc = [A₃]

^{Ɣ3}× [A₄]^{Ɣ4}/[A₁]^{Ɣ1}× [A₂]^{Ɣ2}- Where Kc is called the 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]

It is clear then Kc, and K՛c are not Equal in magnitude.

Thus, Kc = (K՛c)

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,

Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

Where Ɣ₁, Ɣ₂, Ɣ

Where Ɣ₁, Ɣ₂, Ɣ

_{3, }and Ɣ₄ are the stoichiometric coefficient∴ Kp = (P₃^{Ɣ₃} × P₄^{Ɣ₄})/(P₁^{Ɣ₁} × P₂^{Ɣ₂}) |

- Where Kp is called the pressure equilibrium constant of the reaction and P₁, P₂, P

_{3, }and P₄ are the equilibrium partial pressure of reacting components.

#### Mole fraction equilibrium constant

If we write the equation as,

Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

where Ɣ₁, Ɣ₂, Ɣ

Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

where Ɣ₁, Ɣ₂, Ɣ

_{3, }and Ɣ₄ are the stoichiometric coefficient∴ K_{x} = (x₃^{Ɣ₃} × x₄^{Ɣ₄})/(x₁^{Ɣ₁} × x₂^{Ɣ₂}) |

- Where K

_{x}is called a mole fraction equilibrium constant of the reaction and x₁, x₂, x₃, and x₄ are the equilibrium mole fraction of reacting components.

Mass action law |

#### Relation between Kp and Kc

The interrelations of these equilibrium constants are as follows,

Kp = (P₃

The 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.

= {(C₃RT)

Kp = (P₃

^{Ɣ₃}× P₄^{Ɣ₄})/(P₁^{Ɣ₁}× P₂^{Ɣ₂})The 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.

= {(C₃RT)

^{Ɣ₃}× (C₄RT)^{Ɣ₄}}/{(C₁RT)^{Ɣ₁}× (C₂RT)^{Ɣ₂}}∴ Kp = Kc(RT)^{ΔƔ}ΔƔ = (Ɣ₃ + Ɣ₄) - (Ɣ₁ + Ɣ₂) |

- 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

H₂ (g) + I₂ (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)]= Kc

Thus when (Ɣ₃ + Ɣ₄) = (Ɣ₁ + Ɣ₂), Kp = Kc

2SO₂(g) + O₂(g) ⇆ SO₃(g)

Here, Kp = Kc×(RT)

^{{1 - (2+1)}}= Kc×(RT)⁻²

Thus when, (Ɣ₃ + Ɣ₄) ≠ (Ɣ₁ + Ɣ₂), Kp ≠ Kc

Law mass action |

#### Relation between Kp and Kx

∴ Kp = K_{x}(P)^{ΔƔ}ΔƔ = (Ɣ₃ + Ɣ₄) - (Ɣ₁ + Ɣ₂) |

### Problems solutions

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

The reaction is, N₂O₄ ⇆ 2 NO₂

- 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 = (P

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

∴ Kp = (4x²P)/(a² - x²)

Kp = (P

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

∴ Kp = (4x²P)/(a² - x²)

- The fraction of the original N₂O₄ dissociated at equilibrium ɑ = x/a. Replacing (x/a) by ɑ, we have,

Kp = (4α²P)/(1 - α²) |

- Calculate the Kc value of the reaction N₂ + 3H₂ ⇆ 2NH₃ at 400⁰C, Given Kp at the same temperature 1.64 × 10⁻⁴.

- Kp = 0.5

- At 100⁰C the vapor density of N₂O₄ is 25 at 1 atm. Show that Kp = 9.6.

N₂O₄ (g) ⇆ 2NO₂ (g)

- Let 1 mole of N₂O₄ 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 have increased from 1 to (1+x). Let Volume increase from V₁ to V₂.

So, (1+x) = V₂/V₁

- As density and hence vapor density is inversely proportional to volume so vapor density will decrease from d₁ to d₂.

Hence, (1+x) = V₂/V₁= d₁/d₂.

The molecular weight of N₂O₄ 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

= (2×0.84)/1.84

= 0.913 atm

P

= 0.16/1.84= 0.087

∴ Kp = (PNO₂)2/PN₂O₄

= (0.913)2/0.087

≃ 9.6

The molecular weight of N₂O₄ 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

∴ Kp = (PNO₂)2/PN₂O₄

= (0.913)2/0.087

≃ 9.6