Applications of Equilibrium Constant

Applications of Equilibrium Constant: The numerical value of the equilibrium constant is critical because it tells us whether we should expect a reaction mixture at equilibrium to include high or low product concentrations. Such information is vital in the industrial process because it allows us to predict if a planned industrial process will yield sufficient output. The equilibrium constant has a variety of applications. Following are important applications of the equilibrium constant:

1. It allows us to estimate the extent of a reaction based on its magnitude.
2. It predicts the direction of the reaction.
3. It helps to calculate the equilibrium concentration.

In this article, all these applications of the equilibrium constant have been discussed in detail.

Prediction of Extent of Reaction

The value of \({\rm{K}},\) the equilibrium constant, denotes how far a reaction can progress. To put it another way, it’s a criterion for evaluating whether a reversible reaction is complete.

However, it should be observed that an equilibrium provides no information on the rate at which it is reached. The magnitude of \({{\rm{K}}_{\rm{c}}}\) or \({{\rm{K}}_{\rm{p}}}\) is directly proportional to the concentration of the products and inversely proportional to the reactant concentration, as we all know. 

As a result, a high \({\rm{K}}\) value indicates a higher concentration of products (or a low concentration of reactants) and vice versa. Thus, the higher the value of \({\rm{K}},\) the higher the equilibrium concentration of the products compared to the reactants, implying that the reaction proceeds fast.

For example, the following reaction takes place at \({\rm{300}}\,{\rm{K}}\)

The value of \({\rm{K}}\) is very large for this reaction, indicating that the molar concentration of \({\rm{HBr}}\) in the equilibrium mixture is very high. In contrast, the molar concentrations of \({{\rm{H}}_{\rm{2}}}\) and \({{\rm{Br}}_{\rm{2}}}\) are very low. Thus, the reaction is nearly complete, and it can be argued that under these conditions, hydrogen bromide is more stable than hydrogen and bromine.

Let us consider another reaction at \({\rm{298}}\,{\rm{K}}\)

The value of \({\rm{K}}\) is very small for this reaction, indicating that the molar concentration of \({\rm{NO}}\) in the equilibrium mixture is very low. In contrast, the molar concentrations of \({{\rm{N}}_{\rm{2}}}\) and \({{\rm{O}}_{\rm{2}}}\) are very high. In other words, the forward reaction has proceeded to a small extent only. Thus, we can say under these conditions, \({{\rm{N}}_{\rm{2}}}\) and \({{\rm{O}}_{\rm{2}}}\) are more stable than \({\rm{NO}}.\)

The composition of the equilibrium mixture and the equilibrium constant \({\rm{K}}\) can be generalized as follows:

1. At equilibrium, products predominate over reactants if \({\rm{K > 1}}{{\rm{0}}^{\rm{3}}}{\rm{.}}\) The reaction goes almost to completion if \({\rm{K}}\) is sufficiently large.
2. At equilibrium, reactants predominate over products if \({\rm{K}}\) is less than \({\rm{1}}{{\rm{0}}^{\rm{-3}}}{\rm{.}}\) The reaction proceeds to a very small extent if \({\rm{K}}\) is very small.
3. Appreciable concentrations of both reactants and products are present if \({\rm{K}}\) is in the range of \({\rm{1}}{{\rm{0}}^{\rm{-3}}}{\rm{.}}\) to \({\rm{1}}{{\rm{0}}^{\rm{3}}}{\rm{.}}\)

Predicting the Direction of the Reaction

The equilibrium constant can also determine which direction an arbitrary reaction mixture of reactants and products will take. We use the reaction quotient \({\rm{Q}}\) to do this. The reaction quotient is the ratio of product concentrations (or partial pressures) to the reactant concentrations (or partial pressures) at any point in the reaction. In other words, the reaction quotient is calculated in the same way as the equilibrium constant is calculated with molar concentrations \(\left( {{{\rm{Q}}_{\rm{c}}}} \right)\) or partial pressure \(\left( {{{\rm{Q}}_{\rm{p}}}} \right)\)

At equilibrium, \({{\rm{Q}}_{\rm{c}}}{\rm{ = }}{{\rm{K}}_{\rm{c}}}\)

Thus,

1. If \(\left( {{{\rm{Q}}_{\rm{c}}}} \right)\) is more than \(\left( {{{\rm{K}}_{\rm{c}}}} \right),\) the reaction will continue from right to left in the direction of the reactants (reverse reaction)
2. If \(\left( {{{\rm{Q}}_{\rm{c}}}} \right)\) is less than \(\left( {{{\rm{K}}_{\rm{c}}}} \right),\) the reaction will proceed from left to right in the direction of the products (forward reaction)
3. If When \(\left( {{{\rm{Q}}_{\rm{c}}}} \right)\) is equal to \(\left( {{{\rm{K}}_{\rm{c}}}} \right),\) the reaction mixture is already at equilibrium.

Thus, a reaction tends to form products if \({\rm{Q < K}}\) and reactants if \({\rm{Q > K}}.\)

Calculation of Equilibrium Concentrations

The equilibrium constant can be calculated if the concentrations of reactants and products are known at equilibrium in a reaction. On the other hand, the concentration in the equilibrium mixture may be determined if the equilibrium constant is known. This method can be understood by looking at the following solved problems:

Solved Problems on Equilibrium Concentrations

Q.1. In a one-litre closed glass container, one mole of hydrogen and one mole of iodine were heated at \({\rm{490}}\,^\circ {\rm{C}}\) until equilibrium was established. Find the final concentrations of \({{\rm{H}}_{\rm{2}}}{\rm{,}}{{\rm{l}}_{\rm{2}}}{\rm{,}}\) and \({\rm{Hl,}}\) assuming the equilibrium constant is \(45.9.\)
Ans: The equation for the given reaction is
\({{\text{H}}_2} + {{\text{I}}_2} \rightleftharpoons 2{\text{HI}}\)
Suppose the number of moles of \({{\rm{H}}_{\rm{2}}}\) reacted at equilibrium\({\text{ = x}}\)
Moles of all components at equilibrium are
\(\mathop {{{\text{H}}_2}}\limits_{\left( {1 – {\text{x}}} \right)} + \mathop {{{\text{I}}_2}}\limits_{\left( {1 – {\text{x}}} \right)} \rightleftharpoons \mathop {2{\text{HI}}}\limits_{{\text{2x}}} \)
Since the volume of the vessel is one litre, the concentrations at equilibrium are
\(\left[ {{{\text{H}}_{\text{2}}}} \right]{\text{ = 1 – x}}\)
\(\left[ {{{\text{l}}_{\text{2}}}} \right]{\text{ = 1 – x}}\)
\({\text{[HI] = 2x}}\)
The equilibrium constant expression becomes
\({\text{K = }}\frac{{{{{\text{[HI]}}}^{\text{2}}}}}{{\left[ {{{\text{H}}_{\text{2}}}} \right]\left[ {{{\text{I}}_{\text{2}}}} \right]}}{\text{ = }}\frac{{{{{\text{(2x)}}}^{\text{2}}}}}{{{\text{(1 – x)(1 – x)}}}}{\text{ = 45}}{\text{.9}}\)
Or
\(\frac{{4{x^2}}}{{1 – 2x + {x^2}}} = 45.9\)Or
\({\text{41}}{\text{.9}}{{\text{x}}^{\text{2}}}{\text{ – 91}}{\text{.8x + 45}}{\text{.9 = 0}}\)
On solving we get \({\text{x}} – 0.79,\) So equilibrium concentrations are
\(\left[ {{{\text{H}}_2}} \right] = 1 – 0.79 = 0.21\,{\text{mole}}\,{\text{per}}\,{\text{litre}}\)
\(\left[ {{{\text{I}}_2}} \right] = 1 – 0.79 = 0.21\,{\text{mole}}\,{\text{per}}\,{\text{litre}}\)
\([{\text{HI}}] = 2x = 2(0.79) = {\mathbf{1}}.{\mathbf{58}}\,{\mathbf{mole}}\,{\mathbf{per}}\,{\mathbf{litre}}\)

Q.2. Value of \({\text{K}}\) for the reaction \(3{{\text{C}}_2}{{\text{H}}_2}(\;{\text{g}}) \rightleftharpoons {{\text{C}}_6}{{\text{H}}_6}(\;{\text{g}})\) is four at a given temperature. What is the concentration of \({{\text{C}}_6}{{\text{H}}_6}\) if the equilibrium concentration of \({{\text{C}}_2}{{\text{H}}_2}\) is \(0.5\,{\text{mole/litre}}\)?
Ans: The given reaction is
\(3{{\text{C}}_2}{{\text{H}}_2}(\;{\text{g}}) \rightleftharpoons {{\text{C}}_6}{{\text{H}}_6}(\;{\text{g}})\)
The equilibrium constant expression for the given reaction is
\({\text{K}} = \frac{{\left[ {{{\text{C}}_6}{{\text{H}}_6}} \right]}}{{{{\left[ {{{\text{C}}_2}{{\text{H}}_2}} \right]}^3}}}\)
Following information has been given in the question.
\({\text{K = 4}}\)
\(\left[ {{{\text{C}}_2}{{\text{H}}_2}} \right] = 0.5\,{\text{mole/litre}}\)
Putting these values in the expression for the equilibrium constant, we get
\(4 = \frac{{\left[ {{{\text{C}}_6}{{\text{H}}_6}} \right]}}{{{{[0.5]}^3}}}\)
\(\left[ {{{\text{C}}_6}{{\text{H}}_6}} \right] = 4 \times {(0.5)^3} = 0.5\,{\text{moles/litre}}\)

Summary 

1. The equilibrium constant gives important information about chemical equilibrium reactions. It has many applications.
2. At equilibrium, products predominate over reactants if \({\text{K > 1}}{{\text{0}}^{\text{3}}}.\) The reaction goes almost to completion if \({\text{K}}\) is sufficiently large.
3. At equilibrium, reactants predominate over products if \({\text{K}}\) is less than \({\text{1}}{{\text{0}}^{{\text{ – 3}}}}.\) The reaction proceeds to a very small extent if \({\text{K}}\) is very small.
4. Appreciable concentrations of both reactants and products are present if \({\text{K}}\) is in the range of \({\text{1}}{{\text{0}}^{{\text{ – 3}}}}.\) to \({\text{1}}{{\text{0}}^{{\text{ 3}}}}.\)
5. If \({{\text{Q}}_{\text{c}}}\) is greater than \({{\text{K}}_{\text{c}}},\) the reaction will proceed in the direction of the reactants, from right to left (reverse reaction)
6. If \({{\text{Q}}_{\text{c}}}\) is less than \({{\text{K}}_{\text{c}}},\) the reaction will proceed in the direction of the products, from left to right (forward reaction)
7. If When \({{\text{Q}}_{\text{c}}}\) is equal to \({{\text{K}}_{\text{c}}},\) the reaction mixture is already at equilibrium.
8. The equilibrium constant can be calculated if the concentrations of reactants and products are known at equilibrium in a reaction. On the other hand, the concentration in the equilibrium mixture may be determined if the equilibrium constant is known.

FAQs on Applications of Equailibrium Constant

Q.1. What is the importance of an equilibrium constant in a chemical equilibrium system?
Ans: Equilibrium constant is extremely important because it tells us whether we should expect a reaction mixture at equilibrium to include high or low product concentrations. This data is critical in the industrial process because it allows us to forecast whether or not a planned industrial process will produce enough output. The equilibrium constant has several interesting uses. Following are important applications of equilibrium constant.
1. It allows us to estimate the extent of a reaction based on its magnitude.
2. It predicts the direction of the reaction.
3. It helps to calculate the equilibrium concentration.

Q.2. How can we predict the extent of reaction with the help of the equilibrium constant?
Ans: We can predict the extent of reaction with the help of an equilibrium constant. It is used for evaluating whether a reversible reaction is complete. The magnitude of \({{\text{K}}_{\text{c}}}\) or \({{\text{K}}_{\text{p}}}\) is directly proportional to the concentration of the products and inversely proportional to reactants concentration, as we all know. As a result, a high \({\rm{K}}\) value indicates a higher concentration of products (or a low concentration of reactants) and vice versa. As a result, the higher the value of \({\rm{K}},\) the higher the equilibrium concentration of the products compared to the reactants, implying that the reaction proceeds fast.

Q.3. How can we predict the direction of the reaction with the help of the equilibrium constant?
Ans: We can predict the direction of the reaction as explained below:
1. If \({{\text{Q}}_{\text{c}}}\) is greater than \({{\text{K}}_{\text{c}}},\) the reaction will proceed in the direction of the reactants, from right to left (reverse reaction)
2. If \({{\text{Q}}_{\text{c}}}\) is less than \({{\text{K}}_{\text{c}}},\) the reaction will proceed in the direction of the products, from left to right (forward reaction)
3. If When \({{\text{Q}}_{\text{c}}}\) is equal to \({{\text{K}}_{\text{c}}},\) the reaction mixture is already at equilibrium.
Thus, a reaction tends to form products if \({\rm{Q}}\) is less than \({\rm{K}}\) and reactants if \({\rm{Q}}\) is more than \({\rm{K}}.\)

Q.4. What is the reaction quotient?
Ans: The equilibrium constant can also be used to determine which direction an arbitrary reaction mixture of reactants and products will take. We use the reaction quotient Q to do this. The reaction quotient is the ratio of the product of the product concentrations (or partial pressures) to the reactant concentrations (or partial pressures) at any point in the reaction. In other words, the reaction quotient is calculated in the same way as the equilibrium constant is calculated with molar concentrations \(\left( {{{\text{Q}}_{\text{c}}}} \right)\) or partial pressure \(\left( {{{\text{Q}}_{\text{p}}}} \right).\)
\({\text{aA}} + {\text{bB}} \rightleftharpoons {\text{cC}} + {\text{dD}},\)
\({{\text{Q}}_{\text{c}}}{\text{ = }}\frac{{{{{\text{[C]}}}^{\text{c}}}{{{\text{[D]}}}^{\text{d}}}}}{{{{{\text{[A]}}}^{\text{a}}}{{{\text{[B]}}}^{\text{b}}}}}\)
\({{\text{Q}}_{\text{p}}}{\text{ = }}\frac{{{\text{p}}_{\text{c}}^{\text{c}}{\text{p}}_{\text{d}}^{\text{d}}}}{{{\text{p}}_{\text{a}}^{\text{a}}{\text{p}}_{\text{b}}^{\text{b}}}}\)
At equilibrium, \({{\text{Q}}_{\text{c}}}{\text{ = }}{{\text{K}}_{\text{c}}}{\text{.}}\)

Q.5. What are the relationship between equilibrium constant and equilibrium concentration?
Ans: The equilibrium constant can be calculated if the concentrations of reactants and products are known at equilibrium in a reaction. On the other hand, the concentration in the equilibrium mixture may be determined if the equilibrium constant is known.

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