Heterogeneous Equilibria: Definition, Equations, examples

Heterogeneous equilibrium is an important part of chemical equilibrium. In this article, we will study some important aspects of heterogeneous equilibrium. We will also understand how to write equilibrium constant expressions for heterogeneous equilibrium. But before this, we will briefly discuss reversible reactions, chemical equilibrium, and equilibrium constant.

Reversible Reactions

Most of the chemical reactions do not proceed to completion. In most cases, reactants react to form products that react themselves to give back the reactants. Consider a reaction in which \({\rm{A}}\) and \({\rm{B}}\) react to form \({\rm{C}}\) and \({\rm{D}}.\) Now \({\rm{C}}\) and \({\rm{D}}\) may also react together to give back \({\rm{A}}\) and \({\rm{B}}:\)

\({\rm{A}} + {\rm{B}} \to {\rm{C}} + {\rm{D}}\quad {\rm{(Forward}}\,\,{\rm{reaction)}}\)

\({\rm{A}} + {\rm{B}} \leftarrow {\rm{C}} + {\rm{D}}\,\,\,\,{\rm{(Reverse}}\,\,{\rm{reaction)}}\)

Such reactions that go in the forward direction and in the backward direction simultaneously are known as Reversible reactions. To represent such reactions, a pair of the arrow is written between reactants and products.

\({\text{A}} + {\text{B}} \rightleftharpoons {\text{C}} + {\text{D}}\)

A few common examples of reversible reactions are given below:

\({{\text{H}}_{\text{2}}}{\text{(g) + }}{{\text{I}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2HI(g)}}\)

\({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH(l) + }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{(l) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}}\)

\({\text{2N}}{{\text{O}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{(g)}}\)

\({\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s)}} \rightleftharpoons {\text{CaO(s) + C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\)

Chemical Equilibrium: Definition

Consider the following general reaction:

\({\text{A}} + {\text{B}} \rightleftharpoons {\text{C}} + {\text{D}}\)

At the start of the reaction, there are only reactants (\({\rm{A}}\) and \({\rm{B}}\)). So only a forward reaction takes place. The backward reaction starts as soon as the products (\({\rm{C}}\) and \({\rm{D}}\)) are formed. The concentration of \({\rm{A}}\) and \({\rm{A}}\) declines as time passes, whereas \({\rm{C}}\) and \({\rm{D}}\) grow. As a result, the forward reaction rate falls while the backward reaction rate rises.

Ultimately a stage is attained where the forward reaction rate becomes equal to the rate of the backward reaction. Now the system is said to be in a state of equilibrium. As a result, chemical equilibrium can be described as the state of a reversible process in which the two opposing reactions occur at the same rate and the reactant and product concentrations do not change over time.

Equilibrium Constant

Chemical equilibrium is represented by the equilibrium constant. Consider the following general reaction:

\({\text{aA + bB}} \rightleftharpoons {\text{cC + dD}}\)

The equilibrium constant for this reaction may be represented as:

\({{\text{K}}_{\text{c}}} = \frac{{{{{\text{[C]}}}^{\text{c}}}{{{\text{[D]}}}^{\text{d}}}}}{{{{{\text{[A]}}}^{\text{a}}}{{{\text{[B]}}}^{\text{b}}}}}\)

Where the exponents \({\text{a,b,c,}}\) and \({\text{d}}\) have the same value as those in the balanced chemical equation, hence, the equilibrium constant can be defined as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants in the balanced chemical equation at a constant temperature, with each concentration term raised to a power equal to its stoichiometric coefficient.

The concentration of gases in gaseous reactions can also be described in terms of their partial pressures. For a gaseous reaction,

\({\text{lL(g) + mM(g)}} \rightleftharpoons {\text{yY(g) + zZ(g)}}\)

The equilibrium constant may be represented as:

\({{\text{K}}_{\text{p}}}{\text{ = }}\frac{{{{\left( {{{\text{p}}_{\text{Y}}}} \right)}^{\text{y}}}{{\left( {{{\text{p}}_{\text{Z}}}} \right)}^{\text{z}}}}}{{{{\left( {{{\text{p}}_{\text{L}}}} \right)}^{\text{l}}}{{\left( {{{\text{p}}_{\text{M}}}} \right)}^{\text{m}}}}}\)

Here \({{\text{K}}_{\text{p}}}\) is the equilibrium constant, the subscript \({\rm{p}}\) referring to partial pressure. Partial pressures are expressed in atmospheres.

Heterogeneous Equilibria

The equilibrium reactions in which the reactants and the products are not present in the same phase are called heterogeneous equilibrium reactions. Such equilibria in which the reactants and products are not all in the same phase are called heterogeneous equilibria. Heterogeneous equilibrium is demonstrated by the breakdown of calcium carbonate into calcium oxide and carbon dioxide when heated. The following equilibrium is obtained if the reaction is carried out in a closed vessel.

\({\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s)}} \rightleftharpoons {\text{CaO(s) + C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\)

Some other examples of heterogeneous equilibria are:

\({\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl(s)}} \rightleftharpoons {\text{N}}{{\text{H}}_{\text{3}}}{\text{(g) + HCl(g)}}\)

\({\text{3Fe(s) + 4}}{{\text{H}}_{\text{2}}}{\text{O(g)}} \rightleftharpoons {\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}{\text{(s) + 4}}{{\text{H}}_{\text{2}}}{\text{(g)}}\)

Quantitative Characteristics of Heterogeneous Equilibria

An important point to note is that in the case of heterogeneous equilibria, the expression for equilibrium constant does not include the concentrations of pure solids and pure liquids as their concentrations remain constant. In other words, the concentration (moles per unit volume) of a pure solid (or liquid) is fixed and cannot vary. Thus, the concentrations of pure solids or liquids are not included in the equilibrium constant expression. For example, consider the following heterogeneous reaction,

\({\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s)}} \rightleftharpoons {\text{CaO(s) + C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\)

The equilibrium constant expression for this reaction is:

\({{\text{K}}_{\text{c}}}{\text{ = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]{\text{[CaO]}}}}{{\left[ {{\text{CaC}}{{\text{O}}_{\text{3}}}} \right]}}\) \({\rm{C}}{{\rm{O}}_2}\)

But \({\text{CaC}}{{\text{O}}_{\text{3}}}\) and \({\text{CaO}}\) are pure solids. So, their concentrations may be assumed to be constant and can be ignored in the equilibrium constant expression. So the equilibrium constant expression for this reaction can be written as

\({{\text{K}}_{\text{c}}}{\text{ = }}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\)

Since the partial pressure of a gaseous component is proportional to its concentration,

\({{\text{K}}_{\text{p}}}{\text{ = }}{{\text{p}}_{{\text{c}}{{\text{o}}_{\text{2}}}}}\)

As a result, we can deduce that when \({\text{CaC}}{{\text{O}}_{\text{3}}}\) decomposes at a specific temperature, the equilibrium partial pressure of carbon dioxide gas equals its equilibrium constant in terms of partial pressure. In other words, there is constant pressure of carbon dioxide gas in equilibrium with calcium carbonate and calcium oxide at a particular temperature. So, it gives us a very easy method to find the value of \({{\text{K}}_{\text{p}}}{\text{.}}\) For this, we need to measure the pressure of carbon dioxide gas in equilibrium with calcium carbonate and calcium oxide at a particular temperature.

Similarly, consider the following reaction:

\({\text{2}}{{\text{H}}_{\text{2}}}{\text{O(l)}} \rightleftharpoons {\text{2}}{{\text{H}}_{\text{2}}}{\text{(g) + }}{{\text{O}}_{\text{2}}}{\text{(g)}}\)

After ignoring the concentration for liquid water, its equilibrium constant expression becomes:

\({{\text{K}}_{\text{c}}}{\text{ = }}{\left[ {{{\text{H}}_{\text{2}}}} \right]^{\text{2}}}\left[ {{{\text{O}}_{\text{2}}}} \right]\)

\({{\text{K}}_{\text{p}}}{\text{ = }}{\left( {{{\text{p}}_{{{\text{H}}_{\text{2}}}}}} \right)^{\text{2}}}\left( {{{\text{p}}_{{{\text{O}}_{\text{2}}}}}} \right)\)

Some Important Points Regarding Heterogeneous Equilibria

1. For heterogeneous chemical reactions, the concentration of pure solids and pure liquids are not included in the expression for the equilibrium constant. By convention
   [solid]\(=1\) and [liquid]\(=1\)
2. When the substance occurs in the liquid solutions, their concentrations are variable, and therefore their concentrations are given in the expression for the equilibrium constant.
3. It must be noted that pure substances must be present (however small it may be) for the equilibrium to exist, but they do not appear in the expression for the equilibrium constant.
4. The concentration of the solvent is considered constant because it is present in large quantities and will not change appreciably.

Heterogeneous Equilibria Summary

1. Chemical equilibrium is the state of a reversible reaction in which the two opposing reactions proceed at the same rate, and the reactant and product concentrations do not change over time. Chemical equilibrium is represented by the equilibrium constant. 
2. The equilibrium reactions in which the reactants and the products are not present in the same phase are called heterogeneous equilibrium reactions. 
3. Heterogeneous equilibrium is demonstrated by the breakdown of calcium carbonate into calcium oxide and carbon dioxide when heated.
4. Such equilibria in which the reactants and products are not all in the same phase are called heterogeneous equilibria.
5. The concentrations of pure solids and pure liquids are not included in the expression for the equilibrium constant in heterogeneous equilibria because their concentrations remain constant.

FAQs on Heterogeneous Equilibria

Q.1. What are reversible reactions? Give some examples
Ans: Most of the chemical reactions do not proceed to completion. In most cases, reactants react to form products that react themselves to give back the reactants. Consider a reaction in which \({\rm{A}}\) and \({\rm{B}}\) react to form \({\rm{C}}\) and \({\rm{D}}\). Now, \({\rm{C}}\) and \({\rm{D}}\) may also react together to give back \({\rm{A}}\) and \({\rm{B}}\).
\({\rm{A}} + {\rm{B}} \to {\rm{C}} + {\rm{D}}\quad {\rm{(Forward}}\,\,{\rm{reaction)}}\)
\({\rm{A}} + {\rm{B}} \leftarrow {\rm{C}} + {\rm{D}}\,\,\,\,{\rm{(Reverse}}\,\,{\rm{reaction)}}\)
Such reactions which go in the forward direction as well as in the backward direction simultaneously are known as Reversible reactions. A few common examples of reversible reactions are given below:
\({{\text{H}}_{\text{2}}}{\text{(g) + }}{{\text{I}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2HI(g)}}\)
\({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH(l) + }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{(l) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}}\)
\({\text{2N}}{{\text{O}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{(g)}}\)
\({\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s)}} \rightleftharpoons {\text{CaO(s) + C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\)

Q.2. How will you explain the term “Chemical Equilibrium”?
Ans: Consider the following general reaction:
\({\text{A}} + {\text{B}} \rightleftharpoons {\text{C}} + {\text{D}}\)
With the passage of time, the concentration of \({\text{A}}\) and \({\text{B}}\) decreases, and that of \({\text{C}}\) and \({\text{D}}\) increases. As a result, the rate of forward reaction decreases, and that of backward reaction increases. Ultimately a stage is attained where the rate of forward reaction becomes equal to the rate of the backward reaction. Now the system is said to be in a state of chemical equilibrium.

Q.3. How can we write the expression for the equilibrium constant?
Ans: A chemical equilibrium is represented by the equilibrium constant. Consider the following general reaction:
\({\text{aA + bB}} \rightleftharpoons {\text{cC + dD}}\)
The equilibrium constant for this reaction may be represented as
\({{\text{K}}_{\text{c}}}{\text{ = }}\frac{{{{{\text{[C]}}}^{\text{c}}}{{{\text{[D]}}}^{\text{d}}}}}{{{{{\text{[A]}}}^{\text{a}}}{{{\text{[B]}}}^{\text{b}}}}}\)
where the exponents \({\text{a,b,c,}}\) and \({\text{d}}\) have the same value as those in the balanced chemical equation. Thus, the equilibrium constant may be defined as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants in the balanced chemical equation at a constant temperature, with each concentration term raised to a power equal to its stoichiometric coefficient.

Q.4. What are heterogeneous equilibria? Give some examples.
Ans: The equilibrium reactions in which the reactants and the products are not present in the same phase are called heterogeneous equilibrium reactions. Such equilibria in which the reactants and products are not all in the same phase are called heterogeneous equilibria. Heterogeneous equilibrium is demonstrated by the breakdown of calcium carbonate into calcium oxide and carbon dioxide when heated. The following equilibrium is obtained if the reaction is carried out in a closed vessel.
\({\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s)}} \rightleftharpoons {\text{CaO(s) + C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\)
Some other examples of heterogeneous equilibria are:
\({\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl(s)}} \rightleftharpoons {\text{N}}{{\text{H}}_{\text{3}}}{\text{(g) + HCl(g)}}\)
\({\text{3Fe(s) + 4}}{{\text{H}}_{\text{2}}}{\text{O(g)}} \rightleftharpoons {\text{F}}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}{\text{(s) + 4}}{{\text{H}}_{\text{2}}}{\text{(g)}}\)

Q.5. Why are the concentration terms for pure solids and pure liquids ignored in the expression for the equilibrium constant for heterogeneous equilibria?
Ans: In the case of heterogeneous equilibria, the expression for equilibrium constant does not include the concentrations of pure solids and pure liquids as their concentrations remain constant. Thus, the concentrations of pure solids or liquids are not included in the equilibrium constant expression. For example,
\({\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s)}} \rightleftharpoons {\text{CaO(s) + C}}{{\text{O}}_{\text{2}}}{\text{(g)}}\)
The equilibrium constant expression for this reaction is:
\({{\text{K}}_{\text{c}}}{\text{ = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]{\text{[CaO]}}}}{{\left[ {{\text{CaC}}{{\text{O}}_{\text{3}}}} \right]}}\)
But \({\text{CaC}}{{\text{O}}_{\text{3}}}\) and \({\text{CaO}}\) are pure solids. So, their concentrations may be assumed to be constant and can be ignored in the equilibrium constant expression. So the equilibrium constant expression for this reaction can be written as:
\({{\text{K}}_{\text{c}}}{\text{ = }}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\)
and
\({{\text{K}}_{\text{p}}}{\text{ = }}{{\text{p}}_{{\text{c}}{{\text{o}}_{\text{2}}}}}\)

Study Mass Percent Formula Here

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