# Law of mass action

### Define the Law of mass action in chemistry

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

The study of the law of mass action was formulated by 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 + Na2CO3 → CaCO3 + 2NaCl

They recognized that chemical equilibrium is a dynamic and not static equilibrium. The fact that the forward rate and reverse rate of reactions become the same.

The rate of a chemical reaction proportional to the product of effective concentrations of reacting species, each raised to a power equal to the corresponding stoichiometric number of the substance in the balanced chemical equation.

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

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

#### Forward and reverse reaction rates 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.

When the products are getting accumulated in the system, the reverse reaction also starts. Thus according to the law of mass action the forward and reverse reaction rates

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 of the forward and reverse reaction and they do not depend on concentration at a given temperature.

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

The state where forward and reverse reaction rates equal is called the chemical equilibrium. There will be no further change in the composition of the system.

At equilibrium of a chemical reaction,
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 [A], [B], [C] and [D] are the equilibrium concentration of reactants and products.

$\frac{K_{f}}{K_{b}}=K_{C}$

where Kc = concentration equilibrium constant of a chemical reaction.

At a given temperature Kc is constant does not depend on the concentration of the reacting components.

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

#### How does concentration affect the rate of a chemical reaction?

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.

The values of the equilibrium constant of a chemical reaction depend on the mode of writing its stoichiometric balanced chemical equation.

#### The equilibrium constant for ammonia formation

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

N2 + 3 H2 ⇆ 2 NH3

The concentration equilibrium constant = Kc.

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

If we write the chemical reaction

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

The contraction equilibrium constant = K′C.

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

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 writing a balanced 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.

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

#### How to calculate KP in chemical equilibrium

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.

Problem
For the dissociation of N2O4 ⇆ 2NO2, how to calculate the fraction of original N2O4 dissociated at equilibrium in terms of Kp and total pressure?

Solution

N2O4 ⇆ 2 NO2

Let one mole of N2O4 taken initially and x moles of N2O4 dissociated at equilibrium then mole number of N2O4 and NO2 at equilibrium (a – x) and 2x.

Total moles number at equilibrium = (a -x + 2x) = (a + x).

Equilibrium partial pressure of the components

$\frac{\left&space;(&space;a-x&space;\right&space;)}{\left&space;(a+x&space;\right&space;)}P\,&space;and\,&space;\frac{2x}{\left&space;(&space;a+x&space;\right&space;)}P$

where P = total pressure of the reacting system.

$\therefore&space;K_{P}=\frac{\left&space;(&space;P_{NO_{2}}&space;\right&space;)^{2}}{P_{N_{2}O_{4}}}$

$=\frac{\left&space;(&space;\frac{2x}{a+x}P&space;\right&space;)^{2}}{\frac{\left&space;(&space;a-x&space;\right&space;)}{\left&space;(&space;a+x&space;\right&space;)}P}$

$=\frac{4x^{2}P}{\left&space;(&space;a^{2}-x^{2}&space;\right&space;)}$

The fraction of the original N2O4 dissociated at equilibrium ɑ = x/a.

$\therefore&space;K_{P}=\frac{4\alpha&space;^{2}P}{\left&space;(&space;a^{2}-\alpha&space;^{2}&space;\right&space;)}$

#### Mole fraction in chemical equilibrium

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

#### The relation between Kp and Kc

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

Ɣ1A1 + Ɣ2A2 ⇆ Ɣ3A3 + Ɣ4A4

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

The ideal gas law for n moles ideal gas
PV = nRT
or, P = (n/V)RT = CRT
where C = concentration of gas moles per unit volume.

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

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

$Thus,K_{P}=K_{C}\left&space;(&space;RT&space;\right&space;)^{\Delta&space;\gamma&space;}$

#### How to calculate Kp from Kc?

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

$H_{2}+Cl_{2}\leftrightharpoons&space;2HCl$

$K_{P}=K_{C}\left&space;(&space;RT&space;\right&space;)^{2-\left&space;(&space;1+1&space;\right&space;)}=K_{C}$

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

$PCl_{5}\leftrightharpoons&space;PCl_{3}+Cl_{2}$

$K_{P}=K_{C}\left&space;(&space;RT&space;\right&space;)^{\left&space;(&space;1+1&space;\right&space;)-1}=K_{C}RT$

Problem
How to calculate Kp form Kc for the chemical reaction, N2 + 3H2 ⇆ 2NH3 at 400⁰C?
KC at 400°C = 1.64 × 10-4.

Solution

$N_{2}+3H_{2}\leftrightharpoons&space;2NH_{3}$

$K_{P}=K_{C}\left&space;(&space;RT&space;\right&space;)^{2-\left&space;(&space;1+3&space;\right&space;)}=\frac{K_{C}}{\left&space;(RT&space;\right&space;)^{2}}$

#### Relationship between Kp and Kx

$K_{P}=K_{x}\left&space;(&space;P&space;\right&space;)^{\Delta&space;\gamma&space;}$

The above courses represent chemical equilibrium and the existence of an equilibrium constant.

The rate of some reversible chemical reactions before the thermodynamic approach developed by Law of mass action.

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