Homogeneous Equilibria: Definition, Equation, Solved Examples

Homogeneous equilibrium is an important part of chemical equilibrium. In this article, we will study some important aspects of homogeneous equilibria. We will also understand how to write equilibrium constant expressions for homogeneous 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 \({\rm{B}}\).

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

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

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

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

A few common examples of reversible reactions are given below:

Chemical Equilibrium: Definition

Consider the following general reaction:

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

There are only reactants at the start of the reaction \({\rm{A}}\) and \({\rm{B}}\). As a result, only a forward reaction occurs. As soon as the products \({\rm{C}}\) and \({\rm{D}}\) are created, the reverse reaction begins. The concentration of \({\rm{A}}\) and \({\rm{B}}\) declines as time passes, whereas \({\rm{C}}\) and \({\rm{D}}\) increase. As a result, the forward reaction rate decreases 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.

Homogeneous Equilibria

Such equilibrium reactions in which all the reactants and products are in the same phase are called homogeneous equilibrium reactions. Such equilibria in which the reactants and products are all in the same phase are called homogeneous equilibria. Following are some of the examples of homogeneous chemical equilibrium:

In all these reactions, the reactants and the products are present in the same phase. Homogeneous equilibrium reactions can be further divided into two types:

1. Homogeneous equilibrium reactions with the same number of moles of reactants and products.
2. Homogeneous equilibrium reactions with different numbers of moles of reactants and products.

In the first type of homogeneous equilibrium reactions, the number of moles of reactants is the same as that of the products. For example,

In the second type of homogeneous equilibrium reaction, the number of moles of reactants is not equal to that of the products. For example,

Equilibrium Constant for Homogeneous Equilibrium Reactions

Chemical equilibrium is represented by the equilibrium constant. Consider the following general reaction in which all the reactants and products are present in the same phase:

The equilibrium constant for this reaction may be represented as: 

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

Where \({\rm{a,b,c}},\) and \({\rm{d}}\) have the same value as the exponents in the balanced chemical equation. Thus, 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 processes can also be described in terms of their partial pressures. For a gaseous reaction,

The equilibrium constant may be represented as:

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

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

Relation Between \({{\rm{K}}_{\rm{p}}}\) and \({{\rm{K}}_{\rm{c}}}\)

Consider the following general reaction in which all the reactants and products are gases.

We can write the value of \({{\rm{K}}_{\rm{p}}}\) for this reaction.

\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{\left( {{{\rm{p}}_{\rm{C}}}} \right)}^{\rm{I}}}{{\left( {{{\rm{p}}_{\rm{D}}}} \right)}^{\rm{m}}}}}{{{{\left( {{{\rm{p}}_{\rm{A}}}} \right)}^{\rm{j}}}{{\left( {{{\rm{p}}_{\rm{B}}}} \right)}^{\rm{k}}}}}\)

Now according to the ideal gas equation, the partial pressure of a gas can be written as:

\({\rm{p = }}\left( {\frac{{\rm{n}}}{{\rm{V}}}} \right){\rm{RT}}\)

Where \(\frac{{\rm{n}}}{{\rm{V}}}\) is the molar concentration.

So we can write:

The partial pressure of gas \({\rm{A = [A]RT}}\)

The partial pressure of gas \({\rm{B = [B]RT}}\)

The partial pressure of gas \({\rm{C = [C]RT}}\)

The partial pressure of gas \({\rm{D = [D]RT}}\)

Substituting the values of partial pressures in the expression of \({{\rm{K}}_{\rm{p}}}\), we get:

\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{l}}}{{{\rm{(RT)}}}^{\rm{l}}}{{{\rm{[D]}}}^{\rm{m}}}{{{\rm{(RT)}}}^{\rm{m}}}}}{{{{{\rm{[\;A]}}}^{\rm{j}}}{{{\rm{(RT)}}}^{\rm{j}}}{{{\rm{[\;B]}}}^{\rm{k}}}{{{\rm{(RT)}}}^{\rm{k}}}}}\)

\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{l}}}{{{\rm{[D]}}}^{\rm{m}}}}}{{{{{\rm{[\;A]}}}^{\rm{j}}}{{{\rm{[\;B]}}}^{\rm{k}}}}}{\rm{ \times }}\frac{{{{{\rm{(RT)}}}^{{\rm{l + m}}}}}}{{{{{\rm{(RT)}}}^{{\rm{j + k}}}}}}\)

\({{\rm{K}}_{\rm{p}}}{\rm{ = }}{{\rm{K}}_{\rm{c}}}{\rm{ \times (RT}}{{\rm{)}}^{{\rm{(l + m) – (j + k)}}}}\)

\({{\rm{K}}_{\rm{p}}}{\rm{ = }}{{\rm{K}}_{\rm{c}}}{\rm{ \times (RT}}{{\rm{)}}^{{\rm{\Delta n}}}}\)

Where \({\rm{\Delta n}}\) is the difference between the number of moles of gaseous products and reactants, and \({\rm{R}}\) is the gas constant.

Solved Problems on Homogeneous Equilibria

Q.1. Write the equilibrium constant expression \(\left( {{{\rm{K}}_{\rm{c}}}} \right)\) for the following homogeneous equilibrium reaction.

Ans: The equilibrium constant expression for the given reaction is: 
\({{\rm{K}}_{\rm{c}}}{\rm{ = }}\frac{{{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]}^{\rm{2}}}}}{{\left[ {{{\rm{N}}_{\rm{2}}}} \right]{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]}^{\rm{3}}}}}\)

Q.2. Write the equilibrium constant expression \(\left( {{{\rm{K}}_{\rm{c}}}} \right)\) for the following homogeneous equilibrium reaction.

Ans:  First of all, we need to balance the given chemical equation. The balanced chemical equation is:

The equilibrium constant expression can be written as:
\({{\rm{K}}_{\rm{c}}}{\rm{ = }}\frac{{{{\left[ {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right]}^{\rm{4}}}\left[ {{{\rm{O}}_{\rm{2}}}} \right]}}{{{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}} \right]}^{\rm{2}}}}}\)

Q.3. Write the equilibrium constant expression (K_p) for the following homogeneous equilibrium reaction.

Ans: The equilibrium constant expression for the given reaction is: 
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{\left( {{{\rm{p}}_{{\rm{HI}}}}} \right)}^{\rm{2}}}}}{{\left( {{{\rm{p}}_{{{\rm{H}}_{\rm{2}}}}}} \right)\left( {{{\rm{p}}_{{{\rm{I}}_{\rm{2}}}}}} \right)}}\)

Some Important Points Regarding Homogeneous Equilibria

1. For homogeneous chemical reactions, all the reactants and products are in the same phase. Such equilibria in which the reactants and products are all in the same phase are called homogeneous equilibria.

2. Homogeneous equilibrium reactions can be further divided into two types. In the first type of homogeneous equilibrium reactions, the number of moles of reactants is the same as that of the products.

3. In the second type of homogeneous equilibrium reaction, the number of moles of reactants is not equal to that of the products.

4. Homogeneous equilibrium reactions can be represented by the equilibrium constant in terms of molar concentration and the equilibrium constant in terms of partial pressure.

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. 
2. Chemical equilibrium is represented by the equilibrium constant. 
3. The equilibrium reactions in which the reactants and the products are present in the same phase are called homogeneous equilibrium reactions. 
4. Such equilibria in which the reactants and products are all in the same phase are called homogeneous equilibria. 
5. Homogeneous equilibrium reactions can be further divided into two types. In the first type of homogeneous equilibrium reactions, the number of moles of reactants is the same as that of the products.
6. In the second type of homogeneous equilibrium reaction, the number of moles of reactants is not equal to that of the products.

Frequently Asked Questions (FAQs) on Homogeneous 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 \({\rm{B}}\).
\(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}\) (Forward reaction)
\(\mathrm{A}+\mathrm{B} \longleftarrow \mathrm{C}+\mathrm{D}\) (Reverse 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:

Q.2. What is “Chemical Equilibrium”?
Ans: Consider the following general reaction:
\(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\)
With the passage of time, the concentration of \({\rm{A}}\) and \({\rm{B}}\) decreases and \({\rm{C}}\) and \({\rm{D}}\) increase. 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. What are homogeneous equilibria? Give some examples.
Ans: Such equilibrium reactions in which all the reactants and products are in the same phase are called homogeneous equilibrium reactions. Such equilibria in which the reactants and products are all in the same phase are called homogeneous equilibria. Following are some of the examples of homogeneous chemical equilibrium:

In all these reactions, the reactants and the products are present in the same phase. Homogeneous equilibrium reactions can be further divided into two types:
1. Homogeneous equilibrium reactions with the same number of moles of reactants and products.
2. Homogeneous equilibrium reactions with different numbers of moles of reactants and products.

Q.4. How can we write the expression for the equilibrium constant for homogeneous equilibrium reactions?
Ans: Chemical equilibrium is represented by the equilibrium constant. Consider the following general reaction in which all the reactants and products are present in the same phase:

The equilibrium constant for this reaction may be represented as: 
\({{\rm{K}}_{\rm{c}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{c}}}{{{\rm{[D]}}}^{\rm{d}}}}}{{{{{\rm{[\;A]}}}^{\rm{a}}}{{{\rm{[\;B]}}}^{\rm{b}}}}}\)
For a gaseous reaction:

The equilibrium constant may be represented as:
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{\left( {{{\rm{p}}_{\rm{Y}}}} \right)}^{\rm{y}}}{{\left( {{{\rm{p}}_{\rm{Z}}}} \right)}^{\rm{z}}}}}{{{{\left( {{{\rm{p}}_{\rm{L}}}} \right)}^{\rm{l}}}{{\left( {{{\rm{p}}_{\rm{M}}}} \right)}^{\rm{m}}}}}\)
Here \({{\rm{K}}_{\rm{p}}}\) is the equilibrium constant, the subscript \({\rm{P}}\) referring to partial pressure. Partial pressures are expressed in atmospheres.

Q.5. How will you derive the relation between K_p and K_c?
Ans: Consider the following general reaction in which all the reactants and products are gases.

We can write the value of \({{\rm{K}}_{\rm{p}}}\) for this reaction.
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{\left( {{{\rm{p}}_{\rm{C}}}} \right)}^{\rm{I}}}{{\left( {{{\rm{p}}_{\rm{D}}}} \right)}^{\rm{m}}}}}{{{{\left( {{{\rm{p}}_{\rm{A}}}} \right)}^{\rm{j}}}{{\left( {{{\rm{p}}_{\rm{B}}}} \right)}^{\rm{k}}}}}\)
Now according to the ideal gas equation, the partial pressure of a gas can be written as:
\({\rm{p = }}\left( {\frac{{\rm{n}}}{{\rm{V}}}} \right){\rm{RT}}\)
Where \(\frac{{\rm{n}}}{{\rm{V}}}\) is the molar concentration.
So we can write:
The partial pressure of gas \({\rm{[A]RT}}\)
The partial pressure of gas \({\rm{[B]RT}}\)
The partial pressure of gas \({\rm{[C]RT}}\)
The partial pressure of gas \({\rm{[D]RT}}\)
Substituting the values of partial pressures in the expression of \({{\rm{K}}_{\rm{p}}}\), we get:
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{l}}}{{{\rm{(RT)}}}^{\rm{l}}}{{{\rm{[D]}}}^{\rm{m}}}{{{\rm{(RT)}}}^{\rm{m}}}}}{{{{{\rm{[\;A]}}}^{\rm{j}}}{{{\rm{(RT)}}}^{\rm{j}}}{{{\rm{[\;B]}}}^{\rm{k}}}{{{\rm{(RT)}}}^{\rm{k}}}}}\)
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{l}}}{{{\rm{[D]}}}^{\rm{m}}}}}{{{{{\rm{[\;A]}}}^{\rm{j}}}{{{\rm{[\;B]}}}^{\rm{k}}}}}{\rm{ \times }}\frac{{{{{\rm{(RT)}}}^{{\rm{l + m}}}}}}{{{{{\rm{(RT)}}}^{{\rm{j + k}}}}}}\)
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}{{\rm{K}}_{\rm{c}}}{\rm{ \times (RT}}{{\rm{)}}^{{\rm{(l + m) – (j + k)}}}}\)
\({{\rm{K}}_{\rm{p}}}{\rm{ = }}{{\rm{K}}_{\rm{c}}}{\rm{ \times (RT}}{{\rm{)}}^{{\rm{\Delta n}}}}\)

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