6.5 Heterogeneous Equilibria

Equilibrium in a system having more than one phase is called heterogeneous equilibrium. The equilibrium between water vapour and liquid water in a closed container is an example of heterogeneous equilibrium.
\(
\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})
\)
In this example, there is a gas phase and a liquid phase. In the same way, the equilibrium between a solid and its saturated solution,
\(
\mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s})+(\mathrm{aq}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})
\)
is a heterogeneous equilibrium.
Heterogeneous equilibria often involve pure solids or liquids. We can simplify equilibrium expressions for the heterogeneous equilibria involving a pure liquid or a pure solid, as the molar concentration of a pure solid or liquid is constant (i.e., independent of the amount present). In other words if a substance ‘ \(\mathrm{X}\) ‘ is involved, then \([\mathrm{X}(\mathrm{s})]\) and \([\mathrm{X}(\mathrm{l})]\) are constant, whatever the amount of ‘ \(\mathrm{X}\) ‘ is taken. Contrary to this, \([\mathrm{X}(\mathrm{g})]\) and \([\mathrm{X}(\mathrm{aq})]\) will vary as the amount of \(\mathrm{X}\) in a given volume varies. Let us take thermal dissociation of calcium carbonate which is an interesting and important example of heterogeneous chemical equilibrium.
\(
\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\rightleftharpoons} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \dots(7.16)
\)
On the basis of the stoichiometric equation, we can write,
\(
K_c=\frac{[\mathrm{CaO}(\mathrm{s})]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{CaCO}_3(\mathrm{~s})\right]}
\)
Since \(\left[\mathrm{CaCO}_3(\mathrm{~s})\right]\) and \([\mathrm{CaO}(\mathrm{s})]\) are both constant, therefore modified equilibrium constant for the thermal decomposition of calcium carbonate will be
\(
K_c^{\prime}=\left[\mathrm{CO}_2(\mathrm{~g})\right] \dots(7.17)
\)
\(
\text { or } K_p=p_{\mathrm{CO}_2} \dots(7.18)
\)
This shows that at a particular temperature, there is a constant concentration or pressure of \(\mathrm{CO}_2\) in equilibrium with \(\mathrm{CaO}\) (s) and \(\mathrm{CaCO}_3(\mathrm{~s})\). Experimentally it has been found that at \(1100 \mathrm{~K}\), the pressure of \(\mathrm{CO}_2\) in equilibrium with \(\mathrm{CaCO}_3\) (s) and \(\mathrm{CaO}(\mathrm{s})\), is \(2.0 \times 10^5 \mathrm{~Pa}\). Therefore, equilibrium constant at \(1100 \mathrm{~K}\) for the above reaction is:
\(
K_p=P_{\mathrm{CO}_2}=2 \times 10^5 \mathrm{~Pa} / 10^5 \mathrm{~Pa}=2.00
\)
Similarly, in the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel),
\(
\mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_4(\mathrm{~g})
\)
the equilibrium constant is written as
\(
K_c=\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}
\)
It must be remembered that for the existence of heterogeneous equilibrium pure solids or liquids must also be present (however small the amount may be) at equilibrium, but their concentrations or partial pressures do not appear in the expression of the equilibrium constant. In the reaction,
\(
\begin{aligned}
& \mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{HNO}_3(\mathrm{aq}) \rightleftharpoons 2 \mathrm{AgNO}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
& K_c=\frac{\left[\mathrm{AgNO}_3\right]^2}{\left[\mathrm{HNO}_3\right]^2}
\end{aligned}
\)

Units of Equilibrium Constant

The value of equilibrium constant \(K_{\mathrm{c}}\) can be calculated by substituting the concentration terms in \(\mathrm{mol} / \mathrm{L}\) and for \(K_p\) partial pressure is substituted in \(\mathrm{Pa}, \mathrm{kPa}\), bar or atm. This results in units of equilibrium constant based on molarity or pressure, unless the exponents of both the numerator and denominator are same. For the reactions, \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}, K_{\mathrm{c}}\) and \(K_p\) have no unit. \(\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}), K_{\mathrm{c}}\) has unit mol/L and \(K_0\) has unit bar.

Equilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. For a pure gas, the standard state is 1 bar. Therefore a pressure of 4 bar in standard state can be expressed as \(4 \mathrm{~bar} / 1\) bar \(=4\), which is a dimensionless number. Standard state \(\left(c_0\right)\) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numerical value of the equilibrium constant depends on the standard state chosen. Thus, in this system, both \(K_p\) and \(K_c\) are dimensionless quantities but have different numerical values due to different standard states.

Example 7.6: The value of \(K_p\) for the reaction,
\(
\mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) {\rightleftharpoons} 2 \mathrm{CO}(\mathrm{g})
\)
is 3.0 at \(1000 \mathrm{~K}\). If initially \(P_{\mathrm{CO}_2}=0.48 \mathrm{bar}\) and \(P_{\mathrm{CO}}=0\) bar and pure graphite is present, calculate the equilibrium partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_2\).

Answer: For the reaction,
let ‘ \(x\) ‘ be the decrease in pressure of \(\mathrm{CO}_2\), then
\(
\quad \quad \quad \quad \quad \quad  \mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})
\)
Initial pressure: \(0.48 \mathrm{~bar} \quad \quad \quad \quad \quad \quad 0\)
At equilibrium: \((0.48-\mathrm{x})\) bar \(\quad \quad \quad \quad 2 \mathrm{x}\) bar
\(
K_p=\frac{p_{\mathrm{CO}}^2}{p_{\mathrm{CO}_2}}
\)
\(
\begin{aligned}
& K_p=(2 \mathrm{x})^2 /(0.48-\mathrm{x})=3 \\
& 4 \mathrm{x}^2=3(0.48-\mathrm{x}) \\
& 4 \mathrm{x}^2=1.44-\mathrm{x} \\
& 4 \mathrm{x}^2+3 \mathrm{x}-1.44=0 \\
& \mathrm{a}=4, \mathrm{~b}=3, \mathrm{c}=-1.44
\end{aligned}
\)
\(
x=\frac{\left(-b \pm \sqrt{b^2-4 a c}\right)}{2 a}
\)
\(
\begin{aligned}
& =\left[-3 \pm \sqrt{ }(3)^2-4(4)(-1.44)\right] / 2 \times 4 \\
& =(-3 \pm 5.66) / 8
\end{aligned}
\)
\(
=(-3+5.66) / 8 \text { (as value of } x \text { cannot be }
\)
negative hence we neglect that value)
\(
\mathrm{x}=2.66 / 8=0.33
\)
The equilibrium partial pressures are,
\(
\begin{aligned}
& p_{\mathrm{CO}_2}=2 \mathrm{x}=2 \times 0.33=0.66 \mathrm{bar} \\
& p_{\mathrm{CO}_2}=0.48-\mathrm{x}=0.48-0.33=0.15 \mathrm{bar}
\end{aligned}
\)