Law of Mass Action: In chemistry textbooks, we often read about multiple reactions but have you ever wondered how a molecule converts to another and how the reaction is carried out? Molecules are always in motion. They are constantly undergoing collisions. And this is when a molecule with enough energy collides with another molecule and transforms into a different one.

What is Law of Mass Action?

The law of mass action was given by two Norwegian chemists, Cato Guldberg and Peter Waage, in 1864. The law gives the relationship between the rate of a reaction and the concentrations of the reactants. As per the law, the driving force of a chemical reaction is directly proportional to the active masses of the reacting substances. The law of mass action is stated below:

The rate of a chemical reaction at a temperature at any instant is proportional to the product of the active masses of the reactants at that instant.

Active mass is considered equivalent to molar concentration for gaseous reactions at low pressures or in solutions at low concentrations (dilute solutions). Thus, we can infer that the reaction rate is proportional to the product of the molar concentrations (active masses) of the reactants. The law applies to all reactions occurring in the gaseous phase or the liquid (or solution) phase.

Study Rate Law Expression: Law of Mass Action

Explanation of Law of Mass Action

The rate of the reaction is dependent on the number of possible collisions between reacting species.

In the above example, \(\rm{A}\) and \(\rm{B}\) undergo collisions to form products \(\rm{C}\) and \(\rm{D}\). The possibilities of collisions in I, II, III, and IV boxes between \(\rm{A}\) and \(\rm{B}\) molecules are \(1, 4, 6, 9\).

As per these possibilities between reacting molecules, \(\rm{A}\) and \(\rm{B}\) are equal to the number of molecules of each species per unit volume.

Rate of reaction \(\alpha [{\text{A}}][{\text{B}}]\)
\(=\text{K} [{\text{A}}][{\text{B}}]\)

Where \(\text{K}\) is called rate constant.

Application of Law of Mass Action to General Reversible Reaction

Let us consider a general reversible reaction
\({\text{aA}} + {\text{bB}} \leftrightarrow {\text{cC}} + {\text{dD}}\)

The law of mass action can be applied to both the forward reaction \({\text{aA}} + {\text{bB}} \to \left({{\text{products}}} \right)\) and the reverse reaction \({\text{cC}} + {\text{dD}} \to \left({{\text{reactants}}} \right)\), as shown below:

Forward reaction \( {\text{aA}} + {\text{bB}} \to \,……..\)
\({\vartheta _{\text{f}}}\alpha {[{\text{A}}]^{\text{a}}}{[{\text{B}}]^{\text{b}}}\)
\({\vartheta _{\text{f}}} = {{\text{K}}_{\text{f}}}{[{\text{A}}]^{\text{a}}}{[{\text{B}}]^{\text{b}}}\)

Reverse reaction \( {\text{cC}} + {\text{dD}} \to \,……..\)
\({\vartheta _{\text{b}}}\alpha {[{\text{C}}]^{\text{c}}}{[{\text{D}}]^{\text{d}}}\)
\({\vartheta _{\text{b}}} = {{\text{K}}_{\text{b}}}{[{\text{C}}]^{\text{c}}}{[{\text{D}}]^{\text{d}}}\)

Where \([{\text{A}}],[{\text{B}}],[{\text{C}}],[{\text{D}}]\) are the equilibrium concentrations or concentrations at equilibrium state.

Therefore, at equilibrium, the rate of forward reaction \(\left({{\vartheta _\rm{f}}} \right) = \) the rate of the reverse reaction \(\left({{\vartheta _\rm{b}}} \right) = \)

\({\vartheta _{\text{f}}} = {{\text{K}}_{\text{f}}}{[{\text{A}}]^{\text{a}}}{[{\text{B}}]^{\text{b}}} = {\vartheta _{\text{b}}} = {{\text{K}}_{\text{b}}}{[{\text{C}}]^{\text{c}}}{[{\text{D}}]^{\text{d}}}\)
i.e., \({{\text{K}}_{\text{f}}}{[{\text{A}}]^{\text{a}}}{[{\text{B}}]^{\text{b}}} = {{\text{K}}_{\text{b}}}{[{\text{C}}]^{\text{c}}}{[{\text{D}}]^{\text{d}}}\)
\(\frac{{{{\text{K}}_{\text{f}}}}}{{{{\text{K}}_{\text{b}}}}} = {[{\text{C}}]^c}{[{\text{D}}]^{\text{d}}}/{[{\text{A}}]^{\text{a}}}{[{\text{B}}]^{\text{b}}}\)
\(\frac{{{{\text{K}}_{\text{f}}}}}{{{{\text{K}}_{\text{b}}}}} = {{\text{K}}_{\text{c}}}\)

Here \(\rm{K}_\rm{c}\) is the equilibrium constant.

Since molar concentrations are employed in writing the equilibrium constant, this equilibrium constant is generally written as \(\rm{K}_\rm{c}\).

Thus, the equilibrium constant, \({{\text{K}}_{\text{c}}} = {[{\text{C}}]^{\text{c}}}{[{\text{D}}]^{\text{d}}}/{[{\text{A}}]^{\text{a}}}{[{\text{B}}]^{\text{b}}}\)

Hence, the concentration terms must be raised to the powers equivalent to the coefficients of the corresponding reactants and products in the stoichiometric equation.

Equilibrium Constant for Gases

Like concentration equilibrium constant, when the reactants and products in a reversible reaction are in the gaseous state, the partial pressure can be used instead of concentrations at a definite temperature, as the partial pressure of a substance is proportional to the concentration in the gas phase. It is called a partial pressure equilibrium constant.

It represented as \(\rm{K}_\rm{p}\).

\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{\rm{product\;of\;equilibrium\;partial\;pressure\;of\;the\;products}}}}{{{\rm{product\;of\;equilibrium\;partial\;pressure\;of\;the\;reactants}}}}\)

Derivation of \(\rm{K}_\rm{p}\)

In reversible gaseous reactions, the equilibrium constant \(\rm{K}_\rm{p}\) is written by using the partial pressure of the reactants and products in place of the molar concentrations.

Example \( {\text{xA + yB}} \leftrightarrow {\text{mC + nD}}\)

The equilibrium constant \({{\text{K}}_{\text{p}}} = \frac{{{\text{P}}_{\text{C}}^{\text{m}} \times {\text{P}}_{\text{D}}^{\text{n}}}}{{{\text{P}}_{\text{A}}^{\text{x}} \times {\text{P}}_{\text{B}}^{\text{y}}}}\)

Some examples for \(\rm{K}_\rm{p}\)
1. \({{\text{N}}_2} + 3{{\text{H}}_2} \leftrightarrow 2{\text{N}}{{\text{H}}_3}\)
Therefore \({{\text{K}}_{\text{p}}}{\text{=}}\frac{{{\text{P}}_{{\text{N}}{{\text{H}}_{\text{3}}}}^{\text{2}}}}{{{{\text{P}}_{{{\text{N}}_{\text{2}}}}}{\times \text{P}}_{{{\text{H}}_{\text{2}}}}^{\text{3}}}}\)

2. \({\text{CaC}}{{\text{O}}_3}({\text{s}}) \leftrightarrow {\text{CaO}}({\text{s}}) + {\text{C}}{{\text{O}}_2}({\text{g}})\)
Therefore, \({{\text{K}}_{\text{p}}} = \frac{{{{\text{P}}_{{\text{CaO}}}} \times {{\text{P}}_{{\text{C}}{{\text{O}}_2}}}}}{{{{\text{P}}_{{\text{CaC}}{{\text{O}}_3}}}}}\)

In reversible reactions, if there are solids and liquids taking part in the reaction, their concentrations are taken as unity. So, for these type of reactions \(\rm{K}_\rm{p}\) is written as \({{\text{K}}_{\text{p}}} = {{\text{P}}_{{\text{C}}{{\text{O}}_2}}}\left({{\text{CaO}},{\text{CaC}}{{\text{O}}_3}} \right.\) are solids, \(\left. {{{\text{P}}_{{\text{CaO}}}} = {{\text{P}}_{{\text{CaC}}{{\text{O}}_3}}} = 1.} \right)\)

Characteristics of Equilibrium Constant

The critical characteristics of the equilibrium constant are listed below:

1. The equilibrium constant has a definite value at a particular temperature for every reaction. The value of the equilibrium constant increases with an increase in temperature for endothermic reactions. In the case of an exothermic reaction  \(\rm{K}_\rm{c}\), decreases with the rise of temperature.
2. The equilibrium constant value depends on the original concentration or pressure of the reactants at a constant temperature.
3. The equilibrium constant has a greater dependence on the nature of reactants. However, it is independent of catalysts, as the catalyst affects the rate of both forward and backward reactions to the same extent.
4. The value of the equilibrium constant is a measure of the extent to which a reaction proceeds in the forward or backward direction. A large value of constant denotes a greater extent of the reaction, i.e., in the forward direction. Conversely, if the \(\rm{K}\) value is small, the backward reaction proceeds to a larger extent.
5. The equilibrium constant is dependent on stoichiometric coefficients. If the equilibrium is reversed, the equilibrium constant is inversed. Thus, for a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the reverse reaction.
6. The numerical value of the equilibrium constant is an important measure to predict information regarding the reversible reactions in the laboratory and industries. Some applications of the equilibrium constant are discussed below:

Application 1

The magnitude of the equilibrium constant helps to predict the extent of the reaction. The larger the value of \(\rm{K}\), the greater will be the equilibrium concentrations of the products relative to those of the reactants. The \(\rm{K}_\rm{c}\) for the reaction,

\(2{\text{CO}} + {{\text{O}}_2} \leftrightarrow 2{\text{C}}{{\text{O}}_2}\)

at \(1000\, {\text{k}}\) is \(2.3 \times {10^{22}}{\rm{L}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\).
This high value of \(\rm{K}\) denotes that the molar concentration of \(\rm{CO}_2\) in the equilibrium mixture is quite large \(1000\,\rm{K}\).

The \(\rm{K}_\rm{c}\) for the reaction, \(2{\text{N}}{{\text{O}}_2} \leftrightarrow {{\text{N}}_2} + 2{{\text{O}}_2}\)

at \(298\,\rm{K}\) is \(6.7 \times {10^{ – 16}}{\rm{mol}}\,{{\rm{L}}^{ – 1}}.\)
The low value of \(\rm{K}\) denotes that the molar concentration of \(\rm{NO}_2\) in the equilibrium mixture is large, suggesting that \(\rm{NO}_2\) is quite stable at room temperature.

If \(\rm{K} >10^3\), products are predominant over reactants at equilibrium. The reaction proceeds nearly to completion. If \(\rm{K} < 10^{- 3}\) reactants predominate over products at equilibrium. The reaction hardly proceeds. If \(\rm{K}\) is in the range \(10^{- 3}\) and \(10^3,\) But both reactants and products are present at equilibrium.

Application 2

The magnitude of the equilibrium constant helps in predicting the direction of the reaction. The value of \(\rm{K}\) helps to find the direction in which an arbitrary reaction mixture of reactants and products will proceed.

Reaction Quotient or Concentration Quotient

The ratio of the concentration of products to that of reactants is called the reaction quotient \(\rm{(Q)}\). At equilibrium, the reaction quotient \(\rm{(Q)}\) and the equilibrium constant \(\rm{(K)}\) are equal.

1. If \(\rm{Q} = \rm{K}\), the reaction is at equilibrium, and no net reaction occurs.
2. If \(\rm{Q} > \rm{K}\), the reverse reaction will proceed predominantly before attaining equilibrium.
3. If \(\rm{Q} < \rm{K}\), the forward reaction process predominately before attaining equilibrium.

The relationship between \(\rm{Q}\) and \(\rm{K}\) is illustrated below:

A chemical reaction tends to form products if \( {\text{Q < K}}\) and form reactants if \( {\text{Q > K}}\).

Application 3

The value of the equilibrium constant helps to calculate the equilibrium concentrations and partial pressures. If the concentration or partial pressure of substances are known at equilibrium, the equilibrium constant of a reaction can be calculated. On the other hand, if the equilibrium constant is known, either concentrations or partial pressures of equilibrium mixtures can be calculated.

Application 4

The spontaneous nature of the reaction can be obtained using the value of the equilibrium constant.

The standard energy change \((∆\rm{G}^°)\) is related to the equilibrium constant \((\rm{K}_\rm{c})\) as \(\Delta {\rm{G}}^\circ = \, – 2.303\,{\rm{RT}}{\mkern 1mu} \,\log \,{{\rm{K}}_{\rm{c}}}.\) This equation helps to predict spontaneity in terms of the value of \({\text{∆ }}{{\text{G}}^ \circ }\).

When \(\rm{K} > 1\), the standard energy change \( {\text{∆ }} { {\text{G}}^ \circ } < 0\).

It implies a spontaneous reaction- the reaction proceeds in the forward direction to such an extent that the products are presently predominantly.

When \(\rm{K} <1\), the standard energy change \( {\text{∆ }} { {\text{G}}^ \circ } > 0\).

It implies a non-spontaneous reaction. The reaction proceeds in a backward direction to such an extent that only a small quantity of products is present. One can generally expect that exothermic solid reactions are spontaneous redox reactions with a large equilibrium constant value. Such reactions go to near completion on their own.

Law of Mass Action in Solids and Liquids

The law of mass action can be applied to a reversible reaction to derive a mathematical expression for the equilibrium constant, known as the law of chemical equilibrium. The Law of mass action cannot be applied for pure solids and pure liquids, and even solid-state or liquid state substances.

Summary

We understood the concept of the law of mass action and the importance of reversible and irreversible reactions. The derivation of equilibrium constant and its uses and application in various reactions.

FAQs on Law of Mass Action

Let’s look at some of the commonly asked questions about law of mass action:

Q.1. Who proposed the law of mass action?
Ans: The law was stated by two Norwegian chemists, Cato Guldberg and Peter Waage, in 1864.

Q.2. What is meant by the law of mass action?
Ans: At constant temperature, the reaction rate is directly proportional to the product of the active masses (concentration) of the reactants taking part in it.

Q.3. Why is the law of mass action important?
Ans: The law of mass action is important to understand the rate of reaction. The reaction rate is proportional to the product of the molar concentrations (active masses) of the reactants. It explains and predicts the behaviour of solutions in dynamic equilibrium.

Q.4. What is the equilibrium constant for the law of mass action?
Ans: Equilibrium constant is the ratio of rate constants of forward and backward reactions.
The equilibrium constant of a reversible reaction is given as \({\text{K=}}\frac{{{{\text{k}}_{\text{f}}}}}{{{{\text{k}}_{\text{b}}}}}\)
Here \(\rm{k}_\rm{f}\) is the forward rate constant and \(\rm{k}_\rm{b}\) is the backward rate constant of a reversible reaction.

Q.5. What is the law of mass action examples?
Ans: This law states that in any chemical reaction, the frequency is proportional to the sum of the masses of the reacting materials; here, each mass is being elevated to a power equal to the coefficient in the chemical equation.

Q.6. What is the mass action equation?
Ans: The law of mass action equation is represented as
\({\text{A + B}}\overset{{{{\text{V}}_{\text{1}}}}}{\mathop  \Leftrightarrow } {\text{C + D}}\)

Learn Equilibrium Constant Expression Here

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