Chemistry » Fundamental Equilibrium Concepts » Equilibrium Constants

# Homogeneous Equilibria

## Homogeneous Equilibria

A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this tutorial, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here.

$${\text{C}}_{2}{\text{H}}_{2}\left(aq\right)+2{\text{Br}}_{2}\left(aq\right)⇌{\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{C}}_{2}{\text{H}}_{2}{\text{Br}}_{4}\right]}{\left[{\text{C}}_{2}{\text{H}}_{2}\right]\phantom{\rule{0.2em}{0ex}}{\left[{\text{Br}}_{2}\right]}^{2}}$$

$${\text{I}}_{2}\left(aq\right)+{\text{I}}^{\text{−}}\left(aq\right)⇌{\text{I}}_{3}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{I}}_{3}{}^{\text{−}}\right]}{\left[{\text{I}}_{2}\right]\left[{\text{I}}^{\text{−}}\right]}$$

$${\text{Hg}}_{2}{}^{2+}\left(aq\right)+{\text{NO}}_{3}{}^{\text{−}}\left(aq\right)+3{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)⇌2{\text{Hg}}^{2+}\left(aq\right)+{\text{HNO}}_{2}\left(aq\right)+4{\text{H}}_{2}\text{O}\left(l\right)$$

$${K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{Hg}}^{\text{2+}}\right]}^{2}\left[{\text{HNO}}_{2}\right]}{\left[{\text{Hg}}_{2}{}^{2+}\right]\left[{\text{NO}}_{3}{}^{\text{−}}\right]{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}^{3}}$$

$$\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{\text{−}}\right]}{\left[\text{HF}\right]}$$

$${\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{NH}}_{4}{}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{NH}}_{3}\right]}$$

In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H2O(l) is the solvent for these solutions, its concentration does not appear as a term in the Kc expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.

Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.

$${\text{C}}_{2}{\text{H}}_{6}\left(g\right)⇌{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{C}}_{2}{\text{H}}_{4}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}{\left[{\text{C}}_{2}{\text{H}}_{6}\right]}$$

$$3{\text{O}}_{2}\left(g\right)⇌2{\text{O}}_{3}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{O}}_{3}\right]}^{2}}{{\left[{\text{O}}_{2}\right]}^{3}}$$

$${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{2}\right]}^{3}}$$

$${\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{CO}}_{2}\right]}^{3}{\left[{\text{H}}_{2}\text{O}\right]}^{4}}{\left[{\text{C}}_{3}{\text{H}}_{8}\right]{\left[{\text{O}}_{2}\right]}^{5}}$$

Note that the concentration of H2O(g) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.

Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, $$\cfrac{n}{V}.$$

$$PV=nRT$$

$$P=\left(\cfrac{n}{V}\right)RT$$

$$=MRT$$

Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.

Using the partial pressures of the gases, we can write the reaction quotient for the system $${\text{C}}_{2}{\text{H}}_{6}\left(g\right)⇌{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)$$ by following the same guidelines for deriving concentration-based expressions:

$${Q}_{P}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{{\text{C}}_{2}{\text{H}}_{4}}{P}_{{\text{H}}_{2}}}{{P}_{{\text{C}}_{2}{\text{H}}_{6}}}$$

In this equation we use QP to indicate a reaction quotient written with partial pressures: $${P}_{{\text{C}}_{2}{\text{H}}_{6}}$$ is the partial pressure of C2H6; $${P}_{{\text{H}}_{2}},$$ the partial pressure of H2; and $${P}_{{\text{C}}_{2}{\text{H}}_{6}},$$ the partial pressure of C2H4. At equilibrium:

$${K}_{P}={Q}_{P}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{{\text{C}}_{2}{\text{H}}_{4}}{P}_{{\text{H}}_{2}}}{{P}_{{\text{C}}_{2}{\text{H}}_{6}}}$$

The subscript P in the symbol KP designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, KP, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.

Conversion between a value for Kc, an equilibrium constant expressed in terms of concentrations, and a value for KP, an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).

The equation relating Kc and KP is derived as follows. For the gas-phase reaction $$m\text{A}+n\text{B}⇌x\text{C}+y\text{D:}$$

$${K}_{P}=\phantom{\rule{0.2em}{0ex}}\cfrac{{\left({P}_{C}\right)}^{x}{\left({P}_{D}\right)}^{y}}{{\left({P}_{A}\right)}^{m}{\left({P}_{B}\right)}^{n}}$$

$$=\cfrac{([\text{C}]\;\times\;RT)^x([\text{D}]\;\times\;RT)^y}{([\text{A}]\;\times\;RT)^m([\text{B}]\;\times\;RT)^n}$$

$$=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[\text{C}\right]^{x}{\left[\text{D}\right]}^{y}}{\left[\text{A}\right]^{m}{\left[\text{B}\right]}^{n}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{{\left(RT\right)}^{x\text{+}y}}{{\left(RT\right)}^{m\text{+}n}}$$

$$={K}_{c}{\left(RT\right)}^{\left(x\text{+}y\right)-\left(m\text{+}n\right)}$$

$$={K}_{c}{\left(RT\right)}^{\text{Δ}n}$$

The relationship between Kc and KP is

$${K}_{P}={K}_{c}{\left(RT\right)}^{\text{Δ}n}$$

In this equation, Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction $$m\text{A}+n\text{B}⇌x\text{C}+y\text{D,}$$ we have

$$\text{Δ}n=\left(x\text{+}y\right)-\left(m\text{+}n\right)$$

## Example

Calculation of KPWrite the equations for the conversion of Kc to KP for each of the following reactions:

(a) $${\text{C}}_{2}{\text{H}}_{6}\left(g\right)⇌{\text{C}}_{2}{\text{H}}_{4}\left(g\right)+{\text{H}}_{2}\left(g\right)$$

(b) $$\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)$$

(c) $${\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)⇌2{\text{NH}}_{3}\left(g\right)$$

(d) Kc is equal to 0.28 for the following reaction at 900 °C:

$${\text{CS}}_{2}\left(g\right)+4{\text{H}}_{2}\left(g\right)⇌{\text{CH}}_{4}\left(g\right)+2{\text{H}}_{2}\text{S}\left(g\right)$$

What is KP at this temperature?

### Solution

(a) Δn = (2) − (1) = 1

KP = Kc (RT)Δn = Kc (RT)1 = Kc (RT)

(b) Δn = (2) − (2) = 0

KP = Kc (RT)Δn = Kc (RT)0 = Kc

(c) Δn = (2) − (1 + 3) = −2

KP = Kc (RT)Δn = Kc (RT)−2 = $$\cfrac{{K}_{c}}{{\left(RT\right)}^{2}}$$

(d) KP = Kc (RT)Δn = (0.28)[(0.0821)(1173)]−2 = 3.0 $$×$$ 10−5