Law of mass action and equilibrium constant

State the law of mass action in chemistry

The Law of mass action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864. This mass law state as the rate of a chemical reaction at equilibrium proportional to the product of the effective concentration of reacting species, each raised to a power equal to the corresponding stoichiometric number of the substance in the balanced chemical equation.

Effective concentration definition

The effective concentration is a thermodynamics quantity and it represents as,

• When the solution is dilute and the system behaves ideally the concentration represented as the molar concentration. The unit of molar concentration moles/lit.
• Partial pressure in the atmosphere unit for the gaseous system and when the pressure of the system is very low.
• But for pure solid and pure liquid, active mass assumed to be unity since their mass does not affect the rate of reaction.

The concentration of chemical reaction at equilibrium

Let us apply the law of mass action to a general chemical reaction.

A + B ⇆ C + D

As the reaction proceeds in the forward direction, the concentration of reactants decreases and the forward reaction rate also decreases. But when the products are getting accumulated in the system, the reverse reaction also starts.

Rf ∝ CA × CB
∴ Rf = Kf × CA × CB

Rb ∝ CC × CD
∴ Rb = Kb × CC × CD

Where CA, CB, and CC, CD are the concentration of forwarding and reverse reactions respectively in molar units. Kf and Kb are the rate constants at a given temperature.

Equilibrium point definition

As the reactions proceed in the forward reaction rate decreasing but reverse reaction rate increasing and attained a state when they are equal.

Thus the point where forward and reverse reaction rates equal is called the equilibrium point of the chemical reaction. There will be no further change in the concentration of the system.

Thus at equilibrium
Rf = Rb

or, Kf × CA × CB = Kb × CC × CD

$\therefore&space;\frac{K_{f}}{K_{b}}=\frac{C_{C}\times&space;C_{D}}{C_{A}\times&space;C_{B}}$

whare CA, CB, CC and CD are the equilibrium point concentration of reactants and products.

$\frac{K_{f}}{K_{b}}=K_{C}=\frac{C_{C}\times&space;C_{D}}{C_{A}\times&space;C_{B}}$

where Kc = concentration equilibrium constant of a chemical reaction.

Equilibrium constant KC formula

Let us consider a general chemical equation

Ɣ1A1 + Ɣ2A2 ⇆ Ɣ3A3 + Ɣ4A4
where Ɣ1, Ɣ2, Ɣ3, and Ɣ4 are stoichiometric coefficients.

$\therefore&space;K_{C}=\frac{C_{3}^{\gamma&space;_{3}}\times&space;C_{4}^{\gamma&space;_{4}}}{C_{1}^{\gamma&space;_{1}}\times&space;C_{2}^{\gamma&space;_{2}}}$

where KC = concentration equilibrium constant of the reaction and C1, C2, C3, and C4 are the equilibrium concentration of the reactants and product.

Thus the values of the KC of a chemical reaction depend on the mode of writing its stoichiometric balanced chemical equation.

Equilibrium concentration examples

The chemical reaction for the formation of ammonia from nitrogen and hydrogen, we can write the equation

N2 + 3 H2 ⇆ 2 NH3

Let the concentration equilibrium constant = KC for this chemical reaction.

$\therefore&space;K_{C}=\frac{C_{NH_{3}}^{2}}{C_{N_{2}}\times&space;C_{H_{2}}^{3}}$

If we write the chemical reaction

$\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightleftharpoons&space;NH_{3}$

Let the contraction equilibrium constant = K′C for the above chemical reaction.

$\therefore&space;{K}'_{C}=\frac{C_{NH_{3}}}{C_{N_{2}}^{\frac{1}{2}}\times&space;C_{H_{2}}^{\frac{3}{2}}}$

From the above studies, it is clear then KC, and K՛c are not equal in magnitude and concentration equilibrium constant depends on the modes of balancing chemical equation.

$\therefore&space;K_{C}=\sqrt{{K}'_{C}}=\left&space;({K}'_{C}&space;\right&space;)^{\frac{1}{2}}$

If a chemical equation multiplied by n then the general rule for writing equilibrium concentration n will be raised to the power.

$K_{C}=\left&space;({K}'_{C}&space;\right&space;)^{n}$

Equilibrium partial pressure formula

When all the reactants and products are the gas-phase reacting system, the concentration expressed as partial pressure. Let the chemical equation for the above expression

Ɣ1A1 + Ɣ2A2 ⇆ Ɣ3A3 + Ɣ4A4
where Ɣ1, Ɣ2, Ɣ3, and Ɣ4 are stoichiometric coefficients.

$\therefore&space;K_{P}=\frac{P_{3}^{\gamma&space;_{3}}\times&space;P_{4}^{\gamma&space;_{4}}}{P_{1}^{\gamma&space;_{1}}\times&space;P_{2}^{\gamma&space;_{2}}}$

where KP = pressure equilibrium constant of the reaction.

P1, P2, P3, and P4 are the equilibrium partial pressure of reacting components.

Mole fraction equilibrium constant

Ɣ1A1 + Ɣ2A2 ⇆ Ɣ3A3 + Ɣ4A4
where Ɣ1, Ɣ2, Ɣ3, and Ɣ4 are stoichiometric coefficients.

$\therefore&space;K_{x}=\frac{x_{3}^{\gamma&space;_{3}}\times&space;x_{4}^{\gamma&space;_{4}}}{x_{1}^{\gamma&space;_{1}}\times&space;x_{2}^{\gamma&space;_{2}}}$

where Kx = mole fraction equilibrium constant and x1, x2, x3, and x4 are the equilibrium mole fraction of reacting components.

The relation between KP and KC

The relation between KP and KC  expressed for the chemical reaction in equilibrium

Ɣ1A1 + Ɣ2A2 ⇆ Ɣ3A3 + Ɣ4A4

$\therefore&space;K_{P}=\frac{P_{3}^{\gamma&space;_{3}}\times&space;P_{4}^{\gamma&space;_{4}}}{P_{1}^{\gamma&space;_{1}}\times&space;P_{2}^{\gamma&space;_{2}}}$

From the ideal gas law

$PV&space;=&space;nRT&space;=&space;\left&space;(&space;\frac{n}{V}&space;\right&space;)RT$

where C = concentration of gas.

$\therefore&space;K_{P}=\frac{\left&space;(&space;C_{3}RT&space;\right&space;)^{\gamma&space;_{3}}\times&space;\left&space;(&space;C_{4}RT&space;\right&space;)^{\gamma&space;_{4}}}{\left&space;(&space;C_{1}RT&space;\right&space;)^{\gamma&space;_{1}}\times&space;\left&space;(&space;C_{4}RT&space;\right&space;)^{\gamma&space;_{2}}}$

$=\frac{C_{3}^{\gamma&space;_{3}}\times&space;C_{4}^{\gamma&space;_{4}}}{C_{1}^{\gamma&space;_{1}}\times&space;C_{2}^{\gamma&space;_{2}}}\times&space;\left&space;(&space;RT&space;\right&space;)^{\left&space;(&space;\gamma&space;_{3}+\gamma&space;_{4}&space;\right&space;)-\left&space;(&space;\gamma&space;_{1}+\gamma&space;_{2}&space;\right&space;)}$

Thus KP = KC (RT)Δγ

Relation between KP and KC example

A chemical reaction in which the total number of reactant molecules and of resultant molecules are the same.

H2 + Cl2 ⇆ 2HCl
KP = KC (RT)2-(1+1) = KC

For the reactions in which the number of molecules of reactants differs from the resultant molecules of the chemical reaction.

PCl5 ⇆ PCl3 + Cl2
KP = KC (RT)(1+1) -1 = KC RT

This is the relation between KP and KC of the above chemical reaction.

Relationship between KP and KX

KP = Kx (P)Δγ

Thus the above study courses represent chemical equilibrium and the existence of an equilibrium constant. It represents the rate of some reversible chemical reactions by Mass Action Law before the thermodynamic approach developed.