12.3: Binary Mixture in Equilibrium with a Pure Phase

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Page ID: 20641

Howard DeVoe
University of Maryland

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This section considers a binary liquid mixture of components A and B in equilibrium with either pure solid A or pure gaseous A. The aim is to find general relations among changes of temperature, pressure, and mixture composition in the two-phase equilibrium system that can be applied to specific situations in later sections.

In this section, $\mu\A$ is the chemical potential of component A in the mixture and $\mu\A^*$ is for the pure solid or gaseous phase. We begin by writing the total differential of $\mu\A/T$ with $T$, $p$, and $x\A$ as the independent variables. These quantities refer to the binary liquid mixture, and we have not yet imposed a condition of equilibrium with another phase. The general expression for the total differential is \begin{equation} \dif(\mu\A/T) = \bPd{(\mu\A/T)}{T}{p,x\A}\!\dif T + \bPd{(\mu\A/T)}{p}{T,x\A}\!\difp + \bPd{(\mu\A/T)}{x\A}{T,p}\!\dx\A \tag{12.3.1} \end{equation} With substitutions from Eqs. 9.2.49 and 12.1.3, this becomes \begin{equation} \dif(\mu\A/T) = -\frac{H\A}{T^2}\dif T + \frac{V\A}{T}\difp + \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A \tag{12.3.2} \end{equation}

Next we write the total differential of $\mu\A^*/T$ for pure solid or gaseous A. The independent variables are $T$ and $p$; the expression is like Eq. 12.3.2 with the last term missing: \begin{equation} \dif(\mu\A^*/T) = -\frac{H\A^*}{T^2}\dif T + \frac{V\A^*}{T}\difp \tag{12.3.3} \end{equation}

When the two phases are in transfer equilibrium, $\mu\A$ and $\mu\A^*$ are equal. If changes occur in $T$, $p$, or $x\A$ while the phases remain in equilibrium, the condition $\dif(\mu\A/T) = \dif(\mu\A^*/T)$ must be satisfied. Equating the expressions on the right sides of Eqs. 12.3.2 and 12.3.3 and combining terms, we obtain the equation \begin{equation} \frac{H\A-H\A^*}{T^2}\dif T - \frac{V\A-V\A^*}{T}\difp = \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A \tag{12.3.4} \end{equation} which we can rewrite as \begin{gather} \s{ \frac{\Delsub{sol,A}H}{T^2}\dif T - \frac{\Delsub{sol,A}V}{T}\difp = \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A } \tag{12.3.5} \cond{(phases in} \nextcond{equilibrium)} \end{gather} Here $\Delsub{sol,A}H$ is the molar differential enthalpy of solution of solid or gaseous A in the liquid mixture, and $\Delsub{sol,A}V$ is the molar differential volume of solution. Equation 12.3.5 is a relation between changes in the variables $T$, $p$, and $x\A$, only two of which are independent in the equilibrium system.

Suppose we set $\difp$ equal to zero in Eq. 12.3.5 and solve for $\dif T/\dx\A$. This gives us the rate at which $T$ changes with $x\A$ at constant $p$: \begin{gather} \s{ \Pd{T}{x\A}{\!p} = \frac{T^2}{\Delsub{sol,A}H} \bPd{(\mu\A/T)}{x\A}{T,p} } \tag{12.3.6} \cond{(phases in} \nextcond{equilibrium)} \end{gather} We can also set $\dif T$ equal to zero in Eq. 12.3.5 and find the rate at which $p$ changes with $x\A$ at constant $T$: \begin{gather} \s{ \Pd{p}{x\A}{T} = -\frac{T}{\Delsub{sol,A}V}\bPd{(\mu\A/T)}{x\A}{T,p} } \tag{12.3.7} \cond{(phases in} \nextcond{equilibrium)} \end{gather}

Equations 12.3.6 and 12.3.7 will be needed in Secs. 12.4 and 12.5.

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