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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

    \[A(g) \rightleftharpoons 2 B(g)\]

    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 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


    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 \(\Delta G_{rxn}^o\).


    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

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


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

    Contributors and Attributions

    • Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)