### What is the 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 sodium chloride intake water and calcium carbonate 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,

- 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 gas 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.

#### Chemical equilibrium of a reaction

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.

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]

### Chemical equilibrium of a reaction

#### Concentration equilibrium of a chemical reaction

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.

∴ Kc = [A₃]

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 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 ammonia from nitrogen and hydrogen, we can write the equation as,

N₂ + 3 H₂ ⇆ 2 NH₃

The equilibrium constant

Kc = [NH₃]²/[N₂] [H₂]².

If the equilibrium point 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 general rule is that if the equation is multiplied by n then the relation of equilibrium concentration n will be raised to the power. The equilibrium constant

Kc = [NH₃]²/[N₂] [H₂]².

If the equilibrium point 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}Kc = (K՛c)^{n} |

#### Pressure equilibrium of a chemical reaction

When all the reactants and products are gases (that is, gas-phase reacting system), the expression of the equilibrium constant for the equation,
Æ”₁A₁ + Æ”₂A₂ ⇆ Æ”₃A₃ + Æ”₄A₄

Where Æ”₁, Æ”₂, Æ”

Where Æ”₁, Æ”₂, Æ”

_{3, }and Æ”₄ are the stoichiometric coefficient∴ Kp = (P₃^{Æ”₃} × P₄^{Æ”₄})/(P₁^{Æ”₁} × P₂^{Æ”₂}) |

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

#### Mole fraction equilibrium of a chemical reaction

If we write the equation as,

Æ”₁A₁ + Æ”₂A₂ ⇆ Æ”₃A₃ + Æ”₄A₄

where Æ”₁, Æ”₂, Æ”

Æ”₁A₁ + Æ”₂A₂ ⇆ Æ”₃A₃ + Æ”₄A₄

where Æ”₁, Æ”₂, Æ”

_{3, }and Æ”₄ are the stoichiometric coefficient∴ K_{x} = (x₃^{Æ”₃} × x₄^{Æ”₄})/(x₁^{Æ”₁} × x₂^{Æ”₂}) |

_{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 |

#### Pressure and concentration equilibrium

The interrelations of these equilibria are as follows,

Kp = (P₃

The ideal gas law,

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

^{Æ”₃}× P₄^{Æ”₄})/(P₁^{Æ”₁}× P₂^{Æ”₂})The ideal gas law,

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)

^{Æ”₃}× (C₄RT)^{Æ”₄}}/{(C₁RT)^{Æ”₁}× (C₂RT)^{Æ”₂}}∴ Kp = Kc(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

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 (Æ”₃ + Æ”₄) = (Æ”₁ + Æ”₂), Kp = Kc

2SO₂(g) + O₂(g) ⇆ SO₃(g)

Here, Kp = Kc×(RT)

^{{1 - (2+1)}}
= Kc×(RT)⁻²

Thus when, (Æ”₃ + Æ”₄) ≠ (Æ”₁ + Æ”₂), Kp ≠ Kc

Law mass action |

#### Pressure and mole fraction equilibrium

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

### Problems solutions

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

Solution

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 É‘ = x/a. Replacing (x/a) by É‘, we have,

Kp = (4Î±²P)/(1 - Î±²) |

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

Solution

Kp = 0.5

Problem

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

Solution

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