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 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,
  1. Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally.
  2. Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low.
  3. 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

    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 Law of mass action, the rate of the forward reaction,
Rf ∝ [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, 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]
    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]
    [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.
    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]Ɣ3 × [A₄]Ɣ4/[A₁]Ɣ1 × [A₂]Ɣ2
    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,
    Ɣ1A1 + Ɣ2A2 Ɣ3A3 + Ɣ4A4
    Where Ɣ1, Ɣ2, 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.
  • Mole fraction equilibrium constant
    If we write the equation as,
    Ɣ1A1 + Ɣ2A2 Ɣ3A3 + Ɣ4A4
    Where Ɣ1, Ɣ2, Ɣ3, and Ɣ4 are the stoichiometric coefficient
∴ Kx = (x3Ɣ3 × x4Ɣ4)/(x1Ɣ1 × x2Ɣ2)
    Where Kx is called Mole fraction Equilibrium Constant of the reaction and x1, x2, x3 and x4 are the equilibrium mole fraction of reacting components.
Mathematical expression of law of mass action
Equilibrium constant

Relation between Kp and Kc

The interrelations of these Equilibrium constants are as follows,
Kp = (P3Ɣ3 × P4Ɣ4)/(P1Ɣ1 × P2Ɣ2)
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)
  1. For the reaction in which total number of reactant molecules and of resultant molecules are same
  2. H2 (g) + I2 (g) 2HI
    KP = (PHI)2/{(PH2)(PI2)}
    = (CHIRT)2/(CH2RT) (CI2RT)
    = [(CHI)2/{(CH2) (CI2)}] × [(RT)2/(RT)(RT)]
    = Kc
    Thus when (Ɣ3 + Ɣ4) = (Ɣ1 + Ɣ2), Kp = Kc
  3. For the reactions in which the number of molecules of reactants differ from that of the resultant
  4. 2SO2(g) + O2(g) SO3(g)
    Here, Kp = Kc×(RT){1 - (2+1)}
    = Kc×(RT)-2
    Thus when, (Ɣ3 + Ɣ4) ≠ (Ɣ1 + Ɣ2), Kp ≠ Kc
(Gaseous System)
Relation between
Kp and Kc
H2 + Cl2  2HCl Kp = Kc
CO H2O   2HCO2 Kp = Kc
CO + NO2  NO CO2 Kp = Kc
PCl5  PCl3 Cl2 Kp = Kc RT
2H2 + O2  2H2O Kp = Kc RT -1
2CO + O2  2CO2 Kp = Kc RT -1
N2 + 3H2  2NH3 Kp = Kc RT -2

Relation between Kp and Kx

Kp = Kx(P)ΔƔ
ΔƔ = (Ɣ3 + Ɣ4) - (Ɣ1 + Ɣ2)

Problems solutions

  • Problem
    Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
  • Solution
Vant Hoff reaction isotherm is ΔG = - RT lnKa + RT lnQaBut when the reaction attain equilibrium,Qa = Ka
Thus, ΔG = 0.
  • 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.
  • 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 = (PNO2)2/PN2O4
= {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)
  • Problem
    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 has increased from 1 to (1+x). Let Volume increases 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,
PNO2= {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
PN2O4 = {(1-x)/(1+x)}P
= 0.16/1.84= 0.087
Kp = (PNO2)2/PN2O4
= (0.913)2/0.087
≃ 9.6

Mathematical expression of law of mass action for the determination of equilibrium constant of a reaction and relation between pressure, concentration and mole fraction equilibrium constant.

Inorganic Chemistry

[Inorganic chemistry][column1]

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