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# 9.5: Degree of Dissociation

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Reactions such as the one in the previous example involve the dissociation of a molecule. Such reactions can be easily described in terms of the fraction of reactant molecules that actually dissociate to achieve equilibrium in a sample. This fraction is called the degree of dissociation. For the reaction in the previous example

$A(g) \rightleftharpoons 2 B(g) \nonumber$

the degree of dissociation can be used to fill out an ICE table. If the reaction is started with $$n$$ moles of $$A$$, and a is the fraction of $$A$$ molecules that dissociate, the ICE table will look as follows.

$$A$$ $$2 B$$
Initial $$n$$ $$0$$
Change $$-\alpha n$$ $$+2n\alpha$$
Equilibrium $$n(1 - \alpha)$$ $$2n\alpha$$

The mole fractions of $$A$$ and $$B$$ can then be expressed by

\begin{align*} \chi_A &= \dfrac{n(1-\alpha)}{n(1-\alpha)+2n\alpha} \\[4pt] &= \dfrac{1 -\alpha}{1+\alpha} \\[4pt] \chi_B &= \dfrac{2 \alpha}{1+\alpha} \end{align*}

Based on these mole fractions

\begin{align} K_x &= \dfrac{\left( \dfrac{2 \alpha}{1+\alpha}\right)^2}{\dfrac{1 -\alpha}{1+\alpha}} \\[4pt] &= \dfrac{4 \alpha^2}{1-\alpha^2} \end{align} \nonumber

And so $$K_p$$, which can be expressed as

$K_p = K_x(p_{tot})^{\sum \nu_i} \label{oddEq}$

is given by

$K_p = \dfrac{4 \alpha^2}{(1-\alpha^2)} (p_{tot}) \nonumber$

##### Example $$\PageIndex{1}$$

Based on the values given below, find the equilibrium constant at 25 oC and degree of dissociation for a system that is at a total pressure of 1.00 atm for the reaction

$N_2O_4(g) \rightleftharpoons 2 NO_2(g) \nonumber$

$$N_2O_4(g)$$ $$NO_2(g)$$
$$\Delta G_f^o$$ (kJ/mol) 99.8 51.3
###### Solution

First, the value of $$K_p$$ can be determined from $$\Delta G_{rxn}^o$$ via an application of Hess' Law.

\begin{align*} \Delta G_{rxn}^o &= 2 \left( 51.3 \, kJ/mol \right) - 99.8 \,kJ/mol &= 2.8\, kJ/mol \end{align*}

So, using the relationship between thermodynamics and equilibria

\begin{align*} \Delta G_f^o &= -RT \ln K_p \\[4pt] 2800\, kJ/mol &= -(8.314 J/(mol\,K) ( 298 \,K) \ln K_p \\[4pt] K_p &= 0.323 \,atm \end{align*}

The degree of dissociation can then be calculated from the ICE tables at the top of the page for the dissociation of $$N_2O_4(g)$$:

\begin{align*} K_p &= \dfrac{4 \alpha^2}{1-\alpha^2} (p_{tot}) \\[4pt] 0.323 \,atm & = \dfrac{4 \alpha^2}{1-\alpha^2} (1.00 \,atm) \end{align*}

Solving for $$\alpha$$,

$\alpha = 0.273 \nonumber$

Note: since a represents the fraction of N2O4 molecules dissociated, it must be a positive number between 0 and 1.

##### Example $$\PageIndex{2}$$

Consider the gas-phase reaction

$A + 2B \rightleftharpoons 2C \nonumber$

A reaction vessel is initially filled with 1.00 mol of A and 2.00 mol of B. At equilibrium, the vessel contains 0.60 mol C and a total pressure of 0.890 atm at 1350 K.

1. How many mol of A and B are present at equilibrium?
2. What is the mole fraction of A, B, and C at equilibrium?
3. Find values for $$K_x$$, $$K_p$$, and $$\Delta G_{rxn}^o$$.
###### Solution

Let’s build an ICE table!

A 2 B 2 C
Initial 1.00 mol 2.00 mol 0
Change -x -2x +2x
Equilibrium 1.00 mol - x 2.00 mol – 2x 2x = 0.60 mol

From the equilibrium measurement of the number of moles of C, x = 0.30 mol. So at equilibrium,

A 2 B 2 C
Equilibrium 0.70 mol 1.40 mol 0.60 mol

The total number of moles at equilibrium is 2.70 mol. From these data, the mole fractions can be determined.

\begin{align*} \chi_A &= \dfrac{0.70\,mol}{2.70\,mol} = 0.259 \\[4pt] \chi_B &= \dfrac{1.40\,mol}{2.70\,mol} = 0.519 \\[4pt] \chi_C &= \dfrac{0.60\,mol}{2.70\,mol} = 0.222 \end{align*}

So $$K_x$$ is given by

$K_x = \dfrac{(0.222)^2}{(0.259)(0.519)^2} = 0.7064 \nonumber$

And $$K_p$$ is given by Equation \ref{oddEq}, so

$K_p = 0.7604(0.890 \,atm)^{-1} = 0.792\,atm^{-1} \nonumber$

The thermodynamic equilibrium constant is unitless, of course, since the pressures are all divided by 1 atm. So the actual value of $$K_p$$ is 0.794. This value can be used to calculate $$\Delta G_{rxn}^o$$ using

$\Delta G_{rxn}^o = -RT \ln K_p \nonumber$

so

\begin{align*} \Delta G_{rxn}^o &= - (8.314 \, J/(mol\,K))( 1350\, K) \ln (0.792) \\[4pt] &= 2590 \, J/mol \end{align*}

This page titled 9.5: Degree of Dissociation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Patrick Fleming.

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