6.3 Law of Chemical Equilibrium and Equilibrium Constant

During a chemical process, chemical equilibrium refers to the state in which the concentrations of both reactants and products have no tendency to fluctuate over time. When the forward and reverse reaction rates are equal, a chemical reaction is said to be in chemical equilibrium. The state is known as a dynamic equilibrium, and the rate constant is known as the equilibrium constant because the rates are equal and there is no net change in the concentrations of the reactants and products. To answer these questions, let us consider a general reversible reaction:
\(
\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}
\)
where A and B are the reactants, C and D are the products in the balanced chemical equation. On the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato Maximillian Guldberg and Peter Waage proposed in 1864 that the concentrations in an equilibrium mixture are related by the following equilibrium equation,
\(
K_c=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \dots(7.1)
\)
where \(K_c\) is the equilibrium constant and the expression on the right side is called the equilibrium constant expression. 

Law of mass action

The equilibrium equation is also known as the law of mass action because in the early days of chemistry, concentration was called “active mass“. In order to appreciate their work better, let us consider the reaction between gaseous \(\mathrm{H}_2\) and \(\mathrm{I}_2\) carried out in a sealed vessel at \(731 \mathrm{~K}\).

Six sets of experiments with varying initial conditions were performed, starting with only gaseous \(\mathrm{H}_2\) and \(\mathrm{I}_2\) in a sealed reaction vessel in first four experiments (1, 2, 3 and 4) and only \(\mathrm{HI}\) in other two experiments (5 and 6). Experiments 1, 2, 3 and 4 were performed taking different concentrations of \(\mathrm{H}_2\) and / or \(\mathrm{I}_2\), and with time it was observed that intensity of the purple colour remained constant and equilibrium was attained. Similarly, for experiments 5 and 6, the equilibrium was attained from the opposite direction.

Data obtained from all six sets of experiments are given in Table 7.2. Data obtained from all six sets of experiments are given in Table 7.2. It is evident from the experiments \(1,2,3\) and 4 that number of moles of dihydrogen reacted \(=\) number of moles of iodine reacted \(=\) \(1 / 2\) (number of moles of HI formed). Also, experiments 5 and 6 indicate that,
\(
\left[\mathrm{H}_2(\mathrm{~g})\right]_{\mathrm{eq}}=\left[\mathrm{I}_2(\mathrm{~g})\right]_{\mathrm{eq}}
\)
Knowing the above facts, in order to establish a relationship between concentrations of the reactants and products, several combinations can be tried. Let us consider the simple expression,
\(
[\mathrm{HI}(\mathrm{g})]_{\mathrm{eq}} /\left[\mathrm{H}_2(\mathrm{~g})\right]_{\mathrm{eq}}\left[\mathrm{I}_2(\mathrm{~g})\right]_{\mathrm{eq}}
\)

It can be seen from Table 7.3 that if we put the equilibrium concentrations of the reactants and products, the above expression is far from constant. However, if we consider the expression,
\(
[\mathrm{HI}(\mathrm{g})]_{\mathrm{eq}}^2 /\left[\mathrm{H}_2(\mathrm{~g})\right]_{\mathrm{eq}}\left[\mathrm{I}_2(\mathrm{~g})\right]_{\mathrm{eq}}
\)
we find that this expression gives constant value (as shown in Table 7.3) in all the six cases.

It can be seen that in this expression the power of the concentration for reactants and products are actually the stoichiometric coefficients in the equation for the chemical reaction. Thus, for the reaction
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}),
\)
following equation 7.1, the equilibrium constant \(K_c\) is written as,
\(
K_c=[\mathrm{HI}(\mathrm{g})]_{\mathrm{eq}}^2 /\left[\mathrm{H}_2(\mathrm{~g})\right]_{\mathrm{eq}}\left[\mathrm{I}_2(\mathrm{~g})\right]_{\mathrm{eq}} \dots(7.2)
\)
Generally, the subscript ‘eq’ (used for equilibrium) is omitted from the concentration terms. It is taken for granted that the concentrations in the expression for \(K_c\) are equilibrium values. We, therefore, write,
\(
K_c=[\mathrm{HI}(\mathrm{g})]^2 /\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right] \dots(7.3)
\)
The subscript ‘ \(c\) ‘ indicates that \(K_c\) is expressed in concentrations of \(\mathrm{mol} \mathrm{~L}^{-1}\).

Point to remember: At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

Equilibrium constant for a general reaction

The equilibrium constant for a general reaction,
\(
\mathrm{aA}+\mathrm{bB} \rightleftharpoons \mathrm{cC}+\mathrm{dD}
\)
is expressed as,
\(
K_c=[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}} /[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}} \dots(7.4)
\)
where [A],[B],[C] and [D] are the equilibrium concentrations of the reactants and products.
Equilibrium constant for the reaction, \(4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\) is written as
\(
K_c=[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}\right]^6 /\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5
\)
Molar concentration of different species is indicated by enclosing these in the square bracket and, as mentioned above, it is implied that these are equilibrium concentrations.
Let us write equilibrium constant for the reaction,
\(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \dots(7.5)\)
as, \(K_c=[\mathrm{HI}]^2 /\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]=\mathrm{x} \dots(7.6)\)
The equilibrium constant for the reverse reaction, \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})\), at the same temperature is,
\(
K_c^{\prime}=\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right] /[\mathrm{HI}]^2=1 / \mathrm{x}=1 / K_c \dots(7.7)
\)
\(
\text { Thus, } K_c^{\prime}=1 / K_c \dots(7.8)
\)

Point to remember: Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

If we change the stoichiometric coefficients in a chemical equation by multiplying throughout by a factor then we must make sure that the expression for equilibrium constant also reflects that change. For example, if the reaction (7.5) is written as,
\(
1 / 2 \mathrm{H}_2(\mathrm{~g})+1 / 2 \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrm{g}) \dots(7.9)
\)
the equilibrium constant for the above reaction is given by
\(
K_c^{\prime \prime}=[\mathrm{HI}] /\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}=\left\{[\mathrm{HI}]^2 /\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]\right\}^{1 / 2}=\mathrm{x}^{1 / 2}=K_c^{1 / 2} \dots(7.10)
\)
On multiplying the equation (7.5) by \(n\), we get
\(
\mathrm{nH}_2(\mathrm{~g})+\mathrm{nI}_2(\mathrm{~g}) \mathrm{D} \rightleftharpoons 2 \mathrm{nHI}(\mathrm{g}) \dots(7.11)
\)
Therefore, equilibrium constant for the reaction is equal to \(K_c{ }^{\mathrm{n}}\). These findings are summarised in Table 7.4. It should be noted that because the equilibrium constants \(K_c\) and \(K_c^{\prime}\) have different numerical values, it is important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant.

Example 7.1: The following concentrations were obtained for the formation of \(\mathrm{NH}_3\) from \(\mathrm{N}_2\) and \(\mathrm{H}_2\) at equilibrium at \(500 \mathrm{~K}\). \(\left[\mathrm{N}_2\right]=1.5 \times 10^{-2} \mathrm{M} .\left[\mathrm{H}_2\right]=3.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{NH}_3\right]=1.2 \times 10^{-2} \mathrm{M}\). Calculate the equilibrium constant.

Answer: The equilibrium constant for the reaction, \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\) can be written as,
\(
\begin{aligned}
K_c & =\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3} \\
& =\frac{\left(1.2 \times 10^{-2}\right)^2}{\left(1.5 \times 10^{-2}\right)\left(3.0 \times 10^{-2}\right)^3} \\
& =0.106 \times 10^4=1.06 \times 10^3
\end{aligned}
\)

Example 7.2: At equilibrium, the concentrations of \(\mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M}\) and \(\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}\) in a sealed vessel at \(800 \mathrm{~K}\). What will be \(K_c\) for the reaction
\(
\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})
\)

Answer: For the reaction equilibrium constant, \(K_c\) can be written as,
\(
\begin{aligned}
K_c & =\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \\
& =\frac{\left(2.8 \times 10^{-3} \mathrm{M}\right)^2}{\left(3.0 \times 10^{-3} \mathrm{M}\right)\left(4.2 \times 10^{-3} \mathrm{M}\right)} \\
= & 0.622
\end{aligned}
\)