9.5: Degree of Dissociation
- Page ID
- 84348
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
\[A(g) \rightleftharpoons 2 B(g)\]
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
\[ \begin{align} K_x &= \dfrac{\left( \dfrac{2 \alpha}{1+\alpha}\right)^2}{\dfrac{1 -\alpha}{1+\alpha}} \\[4pt] &= \dfrac{4 \alpha^2}{1-\alpha^2} \end{align}\]
And so \(K_p\), which can be expressed as
\[K_p = K_x(p_{tot})^{\sum \nu_i} \label{oddEq}\]
is given by
\[ K_p = \dfrac{4 \alpha^2}{(1-\alpha^2)} (p_{tot})\]
Example \(\PageIndex{1}\)
Based on the values given below, find the equilibrium constant at 25 ^{o}C 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 N_{2}O_{4} 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.
- How many mol of A and B are present at equilibrium?
- What is the mole fraction of A, B, and C at equilibrium?
- 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
\[ \begin{align*} \Delta G_{rxn}^o &= - (8.314 \, J/(mol\,K))( 1350\, K) \ln (0.792) \\[4pt] &= 2590 \, J/mol \end{align*}\]
Contributors and Attributions
Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)