Law of Chemical Equilibrium and Equilibrium Constants with Examples

Law of Chemical Equilibrium and Equilibrium Constants: Chemical equilibrium is of great importance in chemistry. The state of a reversible reaction in which the two opposing reactions proceed at the same pace and the concentrations of reactants and products do not change over time is known as a chemical equilibrium. Furthermore, a reaction’s actual equilibrium can be achieved from both sides. The concentrations of the reactants and products are constant at equilibrium. In this article, we will study the law of chemical equilibrium and equilibrium constant.

Chemical Equilibrium is Dynamic Equilibrium

Let us consider the reaction:

\({\rm{A}} + {\rm{B}}⇌{\rm{C}} + {\rm{D}}\)

When \({\rm{A}}\) and \({\rm{B}}\) are combined in a closed vessel, the forward reaction produces \({\rm{C}}\) and \({\rm{D}}.\) The concentrations of \({\rm{A}}\) and \({\rm{B}}\) are steadily decreasing, while those of \({\rm{C}}\) and \({\rm{D}}\) are steadily increasing. As a result, the forward reaction rate drops while the reverse reaction rate increases. The pace of the two opposing reactions eventually equalises, and the system achieves equilibrium.

We’ve seen that once the reaction \({\rm{A}} + {\rm{B}}⇌{\rm{C}} + {\rm{D}}\) reaches equilibrium, the concentrations of \({\rm{A}}\) and \({\rm{B}},\) as well as \({\rm{C}}\) and \({\rm{D}},\) remain constant over time. It looks that the equilibrium has been lost. However, this is not the case. The equilibrium is dynamic. In reality, the forward and reverse processes are in equilibrium, but the concentrations are unaffected.

On the basis of the kinetic molecular model, the dynamic nature of chemical equilibrium may be easily understood. \({\rm{C}}\) and \({\rm{D}}\) are formed when the molecules of \({\rm{A}}\) and \({\rm{B}}\) collide in the equilibrium mixture.

Similarly, \({\rm{C}}\) and \({\rm{D}}\) collide to return \({\rm{A}}\) and \({\rm{B}}.\) Molecule collisions in a closed system are a constant occurrence. As a result, even at equilibrium, collisions between \({\rm{A}}\) and \({\rm{B}}\) produce \({\rm{C}}\) and \({\rm{D}}\) (forward reaction) while collisions between \({\rm{C}}\) and \({\rm{D}}\) produce \({\rm{A}}\) and \({\rm{B}}\) (reverse reaction), although concentrations remain unaltered.

Law of Chemical Equilibrium: Equilibrium Constant

Let us consider a general reaction:

\({\rm{A}} + {\rm{B}}⇌{\rm{C}} + {\rm{D}}\)

And let \(\left[ {\rm{A}} \right]\left[ {\rm{B}} \right]\left[ {\rm{C}} \right]\) and \(\left[ {\rm{D}} \right]\) represent the molar concentrations of \({\rm{A,}}\,{\rm{B,}}\,{\rm{C}}\) and \({\rm{D}}\) at the equilibrium point. According to the Law of Mass action.

Suppose

\(\left[ {\rm{A}} \right] = \) Equilibrium molar concentrations of \({\rm{A}}\)

\(\left[ {\rm{B}} \right] = \) Equilibrium molar concentrations of \({\rm{B}}\)

\(\left[ {\rm{C}} \right] = \) Equilibrium molar concentrations of \({\rm{C}}\)

\(\left[ {\rm{D}} \right] = \) Equilibrium molar concentrations of \({\rm{D}}\)

As per law of mass action,

Forward reaction rate \({\rm{\alpha }}\,\left[ {\rm{A}} \right]\left[ {\rm{B}} \right] = {{\rm{k}}_1}\left[ {\rm{A}} \right]\left[ {\rm{B}} \right]\)

Backward reaction rate \({\rm{\alpha }}\,\left[ {\rm{C}} \right]\left[ {\rm{D}} \right] = {{\rm{k}}_2}\left[ {\rm{C}} \right]\left[ {\rm{D}} \right]\)

Where, \({{\rm{k}}_1} = \) Forward reaction rate constant, \({{\rm{k}}_2} = \) Backward reaction rate constant

At equilibrium, rate of forward reaction \(=\) rate of backward reaction. Therefore,

\({{\rm{k}}_1}\,\left[ {\rm{A}} \right]\left[ {\rm{B}} \right] = {{\rm{k}}_2}\left[ {\rm{C}} \right]\left[ {\rm{D}} \right]\)

or \(\frac{{{{\rm{k}}_1}}}{{{{\rm{k}}_2}}} = \frac{{\left[ {\rm{C}} \right]\left[ {\rm{D}} \right]}}{{\left[ {\rm{A}} \right]\left[ {\rm{B}} \right]}}\) …(i)

Because both \({{{\rm{k}}_1}}\) and \({{{\rm{k}}_2}}\) are constants, \({{{\rm{k}}_1}/{{\rm{k}}_2}}\) is constant at any given temperature. The ratio of rate constants for forward reaction and rate constant for backward reaction is called equilibrium constant. Equilibrium constant is represented by the sign \({{\rm{K}}_{\rm{c}}},\) or simply \({\rm{K}}.\) The subscript ‘\({\rm{c}}\)’ denotes that the value is in terms of reactant and product concentrations. Equation (i) may be expressed as

The Equilibrium constant expression or Equilibrium law is the term given to this equation.
Consider the reaction

The forward reaction is reliant on the collisions of each of the two \({\rm{A}}\) molecules in this case. As a result, each molecule is treated as a separate entity when writing the equilibrium expression, i.e.,
\({\rm{A + A}}⇌{\rm{C + D}}\)

Then the equilibrium constant expression is:

In general, if a chemical equation contains two or more molecules of the same substance, the concentration is raised to the power equal to the numerical coefficient of the substance in the equation.

Equilibrium Constant Expression for a Reaction in General Terms

The general reaction may be written as:

\({\rm{aA + bB}}⇌{\rm{cC + dD}}\)

Where \({\rm{a,}}\,{\rm{b,}}\,{\rm{c}}\) and \({\rm{d}}\) are numerical quotients of the substance, namely \({\rm{A,}}\,{\rm{B,}}\,{\rm{C}}\) and \({\rm{D}}.\) The expression for the equilibrium constant is

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

\({{\rm{K}}_{\rm{c}}}\) stands for the Equilibrium Constant. The equilibrium constant is defined as the product of the equilibrium concentrations of the products divided by the product of the equilibrium concentrations of the reactants, with each concentration term raised to a power equal to the substance’s coefficient in the balanced equation.

Steps to Write Equilibrium Constant Expression

• 1. Write the equilibrium reaction’s balanced chemical equation. The chemicals present on the left of the equation are referred to as reactants, while those on the right are referred to as products.
• 2. Write the product of the concentrations of the ‘products’ and elevate the concentrations of each substance to the power of its numerical quotient.
• 3. Write the product of the concentrations of the ‘reactants,’ and elevate the concentration of each substance to the power of its numerical quotient.
• 4. Put the product concentrations in the numerator and the reactant concentrations in the denominator to write the equilibrium constant expression. That is to say,

\({{\rm{K}}_{\rm{c}}} = \frac{{{\rm{Product}}\,{\rm{of}}\,{\rm{concentrations}}\,{\rm{of}}\,{\rm{‘products’}}\,{\rm{form}}\,{\rm{Step}}\,\left( {\rm{2}} \right)}}{{{\rm{Product}}\,{\rm{of}}\,{\rm{concentrations}}\,{\rm{of}}\,{\rm{‘reactants’}}\,{\rm{form}}\,{\rm{Step}}\,\left( {\rm{3}} \right)}}\)

Equilibrium Constant Expression in Terms of Partial Pressures

When all of the reactants and products are gases, the equilibrium constant expression can alternatively be expressed in terms of partial pressure. For this, we make use of the ideal gas equation, which shows the relationship between pressure and concentration.

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

The quantity \({\frac{{\rm{n}}}{{\rm{V}}}}\) is simply the molar concentration. Thus,

\({\rm{p}} = \left( {{\rm{Molar}}\,{\rm{concentration}}} \right) \times {\rm{RT}}\)

i.e., at a given temperature, the partial pressure of a gas in an equilibrium mixture is directly proportional to its molar concentration. As a result, rather than molar concentrations, we can formulate the equilibrium constant equation in terms of partial pressure. Consider the following general reaction:

\({\rm{lL}}\left( {\rm{g}} \right) + {\rm{mM}}\left( {\rm{g}} \right)⇌{\rm{yY}}\left( {\rm{g}} \right) + {\rm{zZ}}\left( {\rm{g}} \right)\)

The equilibrium law or the equilibrium constant may be written as

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

The equilibrium constant \({{\rm{K}}_{\rm{p}}}\) is used here, with the subscript \({\rm{p}}\) denoting partial pressure.

Relation Between K_p and K_c

Let us consider a general reaction:

\({\rm{jA}} + {\rm{kB}}⇌{\rm{lC + mD}}\)

All reactants and products are gases in this case. In terms of partial pressures, the equilibrium constant expression can be written as:

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

From the Ideal gas equation:

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

Where \({\frac{{\rm{n}}}{{\rm{V}}}}\) is the molar concentration, as a result, partial pressures of individual gases \({\rm{A,}}\,{\rm{B,}}\,{\rm{C}}\) and \({\rm{D}}\) are calculated as:

\({{\rm{P}}_{\rm{A}}} = \left[ {\rm{A}} \right]{\rm{RT;}}\,{{\rm{P}}_{\rm{B}}} = \left[ {\rm{B}} \right]{\rm{RT;}}\,{{\rm{P}}_{\rm{C}}} = \left[ {\rm{C}} \right]{\rm{RT;}}\,{{\rm{P}}_{\rm{D}}} = \left[ {\rm{D}} \right]{\rm{RT}}\)

Substituting these values in expression for \({{\rm{K}}_{\rm{p}}},\) we have,

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

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

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

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

Where \(\Delta {\rm{n}} = \left( {{\rm{l}} + {\rm{m}}} \right) – \left( {{\rm{j}} + {\rm{k}}} \right),\) the difference in the sums of the coefficients for the gaseous products and reactants.

Units of Equilibrium Constant

The partial pressure is given in atmospheres \(\left( {{\rm{atm}}} \right),\) and the concentrations are given in moles/litre or \({\rm{mol/L}}\) in the equilibrium expression for a specific reaction. The units of \({{\rm{K}}_{\rm{c}}}\) and \({{\rm{K}}_{\rm{p}}}\) are determined as shown below:

• (1) When the total number of moles of reactants are equal to the total number of moles of products.
• The concentration terms or pressure terms present in the numerator and denominator of the equilibrium expression for these reactions exactly cancel out. As a result, \({{\rm{K}}_{\rm{c}}}\) or \({{\rm{K}}_{\rm{p}}}\) for such a reaction has no units. Using the reaction as an example
• \({{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)⇌2{\rm{HI}}\left( {\rm{g}} \right)\)
• \({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{HI}}} \right]}^2}}}{{\left[ {{{\rm{H}}_2}} \right]\left[ {{{\rm{I}}_2}} \right]}} = \frac{{{{\left( {{\rm{mol}}/{\rm{L}}} \right)}^2}}}{{\left( {{\rm{mol}}/{\rm{L}}} \right)\left( {{\rm{mol}}/{\rm{L}}} \right)}}\) (No units)
• \({{\rm{K}}_{\rm{p}}} = \frac{{{{\left( {{{\rm{P}}_{{\rm{HI}}}}} \right)}^2}}}{{\left( {{{\rm{P}}_{{{\rm{H}}_{\rm{2}}}}}} \right)\left( {{{\rm{P}}_{{{\rm{I}}_{\rm{2}}}}}} \right)}} = \frac{{{{\left( {{\rm{atm}}} \right)}^2}}}{{\left( {{\rm{atm}}} \right)\left( {{\rm{atm}}} \right)}}\) (No units)

• (2) When the total number of moles of the reactants and products are unequal.
• \({{\rm{K}}_{\rm{c}}}\) will have units \({\left( {{\rm{mol}}/{\rm{litre}}} \right)^{\rm{n}}},\) and \({{\rm{K}}_{\rm{p}}}\) will have units \({\left( {{\rm{atm}}} \right)^{\rm{n}}}\) in such reactions, where \({\rm{n}}\) equals the total number of moles of products minus the total number of moles of reactants. As a result, for the reaction:
• \({{\rm{N}}_2}{{\rm{O}}_4}\left( {\rm{g}} \right)⇌2{\rm{N}}{{\rm{O}}_2}\left( {\rm{g}} \right)\)
• \({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]}^2}}}{{\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right]}} = \frac{{{{\left( {{\rm{mol}}/{\rm{L}}} \right)}^2}}}{{\left( {{\rm{mol}}/{\rm{L}}} \right)}} = {\rm{mol}}/{\rm{L}}\)
• \({{\rm{K}}_{\rm{p}}} = \frac{{{{\left[ {{{\rm{p}}_{{\rm{NO}}}}_{_2}} \right]}^2}}}{{\left[ {{{\rm{P}}_{\rm{N}}}{{_{_2}}_{\rm{O}}}_{_4}} \right]}} = \frac{{{{\left( {{\rm{atm}}} \right)}^2}}}{{\left( {{\rm{atm}}} \right)}} = {\rm{atm}}\)
• Thus, units for \({{\rm{K}}_{\rm{c}}}\) are \({\rm{mol/L}}\) and for \({{\rm{K}}_{\rm{p}}},\) units are \({\rm{atm}}.\)
• For the reaction:
• \({{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)⇌2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)\)
• the units for \({{\rm{K}}_{\rm{c}}}\) and \({{\rm{K}}_{\rm{p}}}\) may be found as follows:
• \({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{N}}{{\rm{H}}_3}} \right]}^2}}}{{\left[ {{{\rm{N}}_2}} \right]{{\left[ {{{\rm{H}}_2}} \right]}^3}}} = \frac{{{{\left( {{\rm{mol}}/{\rm{L}}} \right)}^2}}}{{\left( {{\rm{mol}}/{\rm{L}}} \right){{\left( {{\rm{mol}}/{\rm{L}}} \right)}^3}}} = {\rm{mo}}{{\rm{l}}^{ – 2}}{{\rm{L}}^2}\)
• \({{\rm{K}}_{\rm{p}}} = \frac{{{{\left( {{{\rm{p}}_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}} \right)}^2}}}{{\left( {{{\rm{p}}_{{{\rm{N}}_{\rm{2}}}}}} \right)\left( {{{\rm{p}}_{{{\rm{H}}_{\rm{2}}}}}} \right)}} = \frac{{{{\left( {{\rm{atm}}} \right)}^2}}}{{\left( {{\rm{atm}}} \right){{\left( {{\rm{atm}}} \right)}^3}}} = {\rm{at}}{{\rm{m}}^{ – 2}}\)
• Thus, units of \({{\rm{K}}_{\rm{c}}}\) are \({\rm{mo}}{{\rm{l}}^{ – 2}}{{\rm{L}}^2}\) and units of \({{\rm{K}}_{\rm{p}}}\) are \({\rm{at}}{{\rm{m}}^{ – 2}}.\)

Summary

1. The state of a reversible reaction in which the two opposing reactions proceed at the same pace and the concentrations of reactants and products do not change over time is known as a chemical equilibrium.
2. The equilibrium constant is calculated by dividing the product of the equilibrium concentrations of the products by the product of the equilibrium concentrations of the reactants, with each concentration term raised to a power equal to the coefficient of the substance in the balanced equation.
3. When all of the reactants and products are gases, the equilibrium constant expression can alternatively be expressed in terms of partial pressure.

FAQs on Law of Chemical Equilibrium and Equilibrium Constants

Q.1. Describe the dynamic nature of chemical equilibrium.
Ans: The dynamic equilibrium of chemical equilibrium can be understood easily. Products are formed when the molecules of reactants collide in the equilibrium mixture. Similarly, molecules of products collide to return reactants. Molecule collisions in a closed system are a constant occurrence. As a result, even at equilibrium, collisions between molecules of reactants produce products (forward reaction) while collisions between molecules of products produce reactants (reverse reaction), although concentrations remain unaltered. This proves that equilibrium is dynamic in nature.

Q.2. What is the equilibrium constant?
Ans: The product of the equilibrium concentrations of the products divided by the product of the equilibrium concentrations of the reactants, with each concentration term raised to a power equal to the coefficient of the substance in the balanced equation, is known as equilibrium constant.

Q.3. What are the steps to write the equilibrium constant expression?
Ans: (1) Write the equilibrium reaction’s balanced chemical equation. The chemicals on the left of the equation are referred to as reactants, while those on the right are referred to as products.
(2) Write the product of the concentrations of the ‘products’ and elevate the concentrations of each substance to the power of its numerical quotient.
(3) Write the product of the concentrations of the ‘reactants,’ and elevate the concentration of each substance to the power of its numerical quotient.
(4) Put the product concentrations in the numerator and the reactant concentrations in the denominator to write the equilibrium constant expression.

Q.4. What are the units of equilibrium constant when the total number of moles of reactants are equal to the total number of moles of products?
Ans: The concentration terms or pressure terms present in the numerator and denominator of the equilibrium expression for these reactions exactly cancel out. As a result, \({{\rm{K}}_{\rm{c}}}\) or \({{\rm{K}}_{\rm{p}}}\) for such a reaction has no units.

Q.5. What is chemical equilibrium?
Ans: The state of a reversible reaction in which the two opposing reactions proceed at the same pace and the concentrations of reactants and products do not change over time is known as a chemical equilibrium.

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