6.4 Homogeneous Equilibria

Types of Chemical Equilibrium

• Homogeneous Equilibrium
• Heterogeneous Equilibrium

Homogeneous Equilibria

In a homogeneous system, all the reactants and products are in the same phase. For example, in the gaseous reaction,
\(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\), reactants and products are in the homogeneous phase.
Similarly, for the reactions,
\(
\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq})
\)
\(
\text { and, } \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}(\mathrm{SCN})^{2+}(\mathrm{aq})
\)
all the reactants and products are in a homogeneous solution phase.

Equilibrium Constant in Gaseous Systems

So far we have expressed the equilibrium constant of the reactions in terms of molar concentration of the reactants and products, and used the symbol, \(K_c\) for it. For reactions involving gases, however, it is usually more convenient to express the equilibrium constant in terms of partial pressure. In a gaseous system, the equilibrium constant is denoted by \(K_P\) and it is related to \(K_C\) by the expression:
\(
K_p=K_c(\mathrm{R} T)^{\Delta n}
\)

Derivation:

The ideal gas equation is written as,
\(
\begin{aligned}
& p V=n \mathrm{R} T \\
& \Rightarrow p=\frac{n}{V} \mathrm{R} T
\end{aligned}
\)
Here, \(p\) is the pressure in \(\mathrm{Pa}, n\) is the number of moles of the gas, \(V\) is the volume in \(m^3\) and \(T\) is the temperature in Kelvin.
Therefore, \(n / V\) is concentration expressed in \(\mathrm{mol} / \mathrm{m}^3\)

If concentration \(\mathrm{c}\), is in \(\mathrm{mol} / \mathrm{L}\) or \(\mathrm{mol} / \mathrm{dm}^3\), and \(p\) is in bar then
\(
p=c \mathrm{R} T
\)
We can also write \(p=[\text { gas }] R T\)
At constant temperature, the pressure of the gas is proportional to its concentration i.e., \(p \propto\) [gas]
For the reaction in equilibrium
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})
\)
We can write either
\(
K_{\mathrm{c}}=\frac{[\mathrm{HI}(\mathrm{g})]^2}{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right]}
\)
\(
\text { or } K_c=\frac{\left(p_{H I}\right)^2}{\left(p_{H_2}\right)\left(p_{I_2}\right)} \dots(7.12)
\)
Further, since
\(
\begin{aligned}
& p_{\mathrm{HI}}=[\mathrm{HI}(\mathrm{g})] \mathrm{R} T \\
& p_{\mathrm{I}_2}=\left[\mathrm{I}_2(\mathrm{~g})\right] \mathrm{R} T \\
& p_{\mathrm{H}_2}=\left[\mathrm{H}_2(\mathrm{~g})\right] \mathrm{R} T
\end{aligned}
\)

Therefore,
\(
\begin{aligned}
K_p= & \frac{\left(p_{\mathrm{HI}}\right)^2}{\left(p_{\mathrm{H}_2}\right)\left(p_{\mathrm{I}_2}\right)}=\frac{[\mathrm{HI}(\mathrm{g})]^2[\mathrm{R} T]^2}{\left[\mathrm{H}_2(\mathrm{~g})\right] \mathrm{R} T \cdot\left[\mathrm{I}_2(\mathrm{~g})\right] \mathrm{R} T} \\
= & \frac{[\mathrm{HI}(\mathrm{g})]^2}{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right]}=K_c \dots(7.13)
\end{aligned}
\)
In this example, \(K_p=K_c\) i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)
\(
\begin{aligned}
K_p & =\frac{\left(p_{N H_3}\right)^2}{\left(p_{N_2}\right)\left(p_{H_2}\right)^3} \\
& =\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2[\mathrm{R} T]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right] \mathrm{R} T \cdot\left[H_2(\mathrm{~g})\right]^3(\mathrm{R} T)^3}
\end{aligned}
\)
\(
=\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2[\mathrm{R} T]^{-2}}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[H_2(\mathrm{~g})\right]^3}=K_c(\mathrm{R} T)^{-2}
\)
or \(K_p=K_c(\mathrm{R} T)^{-2} \dots(7.14)\)
Similarly, for a general reaction
\(
\mathrm{aA}+\mathrm{bB} \rightleftharpoons \mathrm{cC}+\mathrm{dD}
\)
\(
K_p=\frac{\left(p_C^c\right)\left(p_D^d\right)}{\left(p_A^a\right)\left(p_B^b\right)}=\frac{[\mathrm{C}]^c[\mathrm{D}]^d(\mathrm{R} T)^{(c+d)}}{[\mathrm{A}]^a[\mathrm{~B}]^b(\mathrm{R} T)^{(a+b)}}
\)
\(
=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{~A}]^a[\mathrm{~B}]^b}(\mathrm{R} T)^{(c+d)-(a+b)}
\)
\(
K_p=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{~A}]^a[\mathrm{~B}]^b}(\mathrm{R} T)^{\Delta n}=K_c(\mathrm{R} T)^{\Delta n} \dots(7.15)
\)
where \(\Delta n=\) (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. It is necessary that while calculating the value of \(K_p\), pressure should be expressed in the bar because the standard state for pressure is \(1 \mathrm{bar}\). We know from Unit 1 that:
1pascal, \(\mathrm{Pa}=1 \mathrm{~Nm}^{-2}\), and \(1 \mathrm{~bar}=10^5 \mathrm{~Pa}\)
\(K_p\) values for a few selected reactions at different temperatures are given in Table 7.5

Example 7.3: \(\mathrm{PCl}_5, \mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) are at equilibrium at \(500 \mathrm{~K}\) and having concentration \(1.59 \mathrm{M}\) \(\mathrm{PCl}_3, 1.59 \mathrm{M} \mathrm{~Cl}_2\) and \(1.41 \mathrm{M} \mathrm{~PCl}_5\). Calculate \(K_c\) for the reaction,
\(
\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2
\)

Answer: The equilibrium constant \(K_c\) for the above reaction can be written as,
\(
K_{\mathrm{c}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{(1.59)^2}{(1.41)}=1.79
\)

Example 7.4: The value of \(K_c=4.24\) at \(800 \mathrm{~K}\) for the reaction, \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\)
Calculate equilibrium concentrations of \(\mathrm{CO}_2, \mathrm{H}_2, \mathrm{CO}\) and \(\mathrm{H}_2 \mathrm{O}\) at \(800 \mathrm{~K}\), if only \(\mathrm{CO}\) and \(\mathrm{H}_2 \mathrm{O}\) are present initially at concentrations of \(0.10 \mathrm{M}\) each.

Answer:For the reaction,
\(
\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})
\)
Initial concentration:
\(
\begin{array}{llll}
0.1 \mathrm{M} & \quad 0.1 \mathrm{M} & & & 0 & & & & 0
\end{array}
\)
Let \({x}\) mole per litre of each of the product be formed.
At equilibrium:
\(
(0.1-x) M \quad(0.1-x) M \quad x M \quad x M
\)
where \(\mathrm{x}\) is the amount of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\) at equilibrium.
Hence, the equilibrium constant can be written as,
\(
\begin{aligned}
& K_c=\mathrm{x}^2 /(0.1-\mathrm{x})^2=4.24 \\
& \mathrm{x}^2=4.24\left(0.01+\mathrm{x}^2-0.2 \mathrm{x}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{x}^2=0.0424+4.24 \mathrm{x}^2-0.848 \mathrm{x} \\
& 3.24 \mathrm{x}^2-0.848 \mathrm{x}+0.0424=0
\end{aligned}
\)
\(
\mathrm{a}=3.24, \mathrm{~b}=-0.848, \mathrm{c}=0.0424
\)
(for quadratic equation \(a x^2+b x+c=0\),
\(
x=\frac{\left(-b \pm \sqrt{b^2-4 a c}\right)}{2 a}
\)
\(
\begin{aligned}
& x=0.848 \pm \sqrt{ }(0.848)^2-4(3.24)(0.0424) / (3.24 \times 2) \\
&
\end{aligned}
\)
\(
\begin{aligned}
& x=(0.848 \pm 0.4118) / 6.48 \\
& x_1=(0.848-0.4118) / 6.48=0.067 \\
& x_2=(0.848+0.4118) / 6.48=0.194
\end{aligned}
\)
the value 0.194 should be neglected because it will give a concentration of the reactant which is more than the initial concentration.  Hence the equilibrium concentrations are,
\(
\begin{aligned}
& {\left[\mathrm{CO}_2\right]=\left[\mathrm{H}_2\right]=x=0.067 \mathrm{M}} \\
& {[\mathrm{CO}]=\left[\mathrm{H}_2 \mathrm{O}\right]=0.1-0.067=0.033 \mathrm{M}}
\end{aligned}
\)

Example 7.5: For the equilibrium,
\(
2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g})
\)
the value of the equilibrium constant, \(K\) is \(3.75 \times 10^{-6}\) at \(1069 \mathrm{~K}\). Calculate the \(K_p\) for the reaction at this temperature.

Answer:

We know that,
\(
K_p=K_c(\mathrm{R} T)^{\Delta n}
\)

For the above reaction,
\(
\begin{aligned}
& \Delta n=(2+1)-2=1 \\
& K_p=3.75 \times 10^{-6}(0.0831 \times 1069) \\
& K_p=0.033
\end{aligned}
\)