Solubility of a Solid in a Liquid Estimate the solubility of solid naphthalene in liquid n-hexane at 20°C. Data:^3 Naphthalene (C10H8, molecular weight = 128.19) Melting point: 80.2°C Heat of fusion: 18.804 kJ/mol Density of the solid: 1.0253 g/cc at 20°C Density of the liquid: 0.9625 g/cc at 100°C

Question

Solubility of a Solid in a Liquid

Estimate the solubility of solid naphthalene in liquid n-hexane at 20°C.

Data:[latex]^{3}[/latex]

Naphthalene ([latex]C _{10} H _{8}[/latex], molecular weight = 128.19)

Melting point: 80.2°C

Heat of fusion: 18.804 kJ/mol

Density of the solid: 1.0253 g/cc at 20°C

Density of the liquid: 0.9625 g/cc at 100°C

Sublimation pressure of the solid:

[latex]\log _{10} P^{ ext {sub }}( bar )=8.722-\frac{3783}{T}[/latex] (T in K)

The heat capacities of liquid and solid naphthalene may be assumed to be equal

[latex]^{3}[/latex]Reference: R. C. Weast, ed., Handbook of Chemistry and Physics, 68th ed., Chemical Rubber Publishing Co., Cleveland (1987), pp. C-357, D-214.

Accepted Answer

The solubility parameter and liquid molar volume for n-hexane are given in Table 9.6-1 as [latex]\delta_{2}=7.3 ext { and } \underline{V}_{2}^{ L }=132[/latex] cc/mol, respectively. Since the liquidmolar volume of naphthalene given in the data is for a temperature 80°C higher than the temperature of interest, and the volume change on melting of naphthalene is small, the molar volume of liquid naphthalene below its melting temperature will be taken to be that of the solid; that is,

[latex]\underline{V}_{1}^{ L }=\frac{128.19 g / mol }{1.0253 g / cc }=125 cc / mol[/latex]

The heat of sublimation of naphthalene is not given. However, we can compute this quantity from the vapor pressure curve of the solid and the Clausius-Clapeyron equation (Eq. 7.7-5a) by taking P to be equal to the sublimation pressure [latex]P^{ sub }, \Delta \underline{H}[/latex] o equal the heat of sublimation, and setting [latex]\Delta \underline{V}=\underline{V}^{\overline{ V }}-\underline{V}^{ S }=\Delta_{ sub } \underline{V} \cong R T / P^{ sub }[/latex]. Thus

[latex]\frac{d \ln P^{ vap }}{d T}=\frac{\Delta_{ vap } \underline{H}}{R T^{2}}[/latex] (7.7-5a)

[latex]\frac{\Delta_{ ext {sub } \underline{H}}}{R T^{2}}=\frac{d \ln P^{ sub }}{d T}=2.303 \frac{d \log _{10} P^{ sub }}{d T}=+2.303 \frac{(3783)}{T^{2}}[/latex]

and

[latex]\Delta_{ ext {sub }} \underline{H}=2.303(3783)(8.314 J / mol )=72434 J / mol[/latex]

Next,

[latex]\Delta_{ ext {vap }} \underline{U}=\Delta_{ ext {sub }} \underline{H}-\Delta_{ ext {fus }} \underline{H}-R T=72434-18804-8.314 imes 293.15=51193 J / mol[/latex]

so that

[latex]\delta_{1}=\left(\frac{51193 J / mol }{125 cc / mol imes 4.184 J / cal } ight)^{1 / 2}=9.9( cal / cc )^{1 / 2}[/latex]

Now using Eq. 12.1-7 with [latex]\Delta C_{ P }=0[/latex], and the regular solution expression for the activity coefficient, we obtain

[latex]\ln x_{1}=-\ln \gamma_{1}-\left\{\frac{\Delta_{ fus } \underline{H}\left(T_{t} ight)}{R T}\left[1-\frac{T}{T_{t}} ight]+\frac{\Delta C_{ P }}{R}\left[1-\frac{T_{t}}{T}+\ln \left(\frac{T_{t}}{T} ight) ight] ight\}[/latex] (12.1-7)

[latex]\ln x_{1}=-\frac{V_{1}^{ L }\left(\delta_{1}-\delta_{2} ight)^{2} \Phi_{2}^{2}}{R T}-\frac{\Delta_{ fus } H\left(T_{m} ight)}{R T}\left(1-\frac{T}{T_{m}} ight)[/latex]

As a first guess, assume that [latex]x_{1}[/latex] will be small, so that

[latex]\Phi_{2}=\frac{x_{2} \underline{V}_{2}^{ L }}{x_{1} \underline{V}_{1}^{ L }+x_{2} \underline{V}_{2}^{ L }} \approx 1[/latex]

In this case

[latex]\begin{aligned}\ln x_{1}=& \frac{-125 \frac{ cc }{ mol } imes(9.9-7.3)^{2} \frac{ cal }{ cc } imes 4.184 \frac{ J }{ cal }}{8.314 \frac{ J }{ mol K } imes 293.15 K } \&-\frac{18804 \frac{ J }{ mol }}{8.314 \frac{ J }{ mol K } imes 293.15 K }\left(1-\frac{293.15}{353.35} ight) \=&-1.451-1.314=-2.765\end{aligned}[/latex]

[latex]x_{1}=0.063[/latex]

With such a large value for [latex]x_{1}[/latex] we must go back and correct the value of [latex]\Phi_{2}[/latex] for the presence of the solute and repeat the computation. Thus

[latex]\Phi_{2}=\frac{0.937 imes 132}{0.937 imes 132+0.063 imes 125}=0.94[/latex]

and

[latex]\begin{aligned}\ln x_{1} &=1.282-1.314=2.596 \x_{1} &=0.0746\end{aligned}[/latex]

The results of the next two iterations are [latex]x_{1}=0.0768 ext { and } x_{1}=0.0772[/latex], respectively. This last prediction is in reasonable agreement with the experimental result of [latex]x_{1}=0.09 .^{4}[/latex]

Comment

Note that had we assumed ideal solution behavior, [latex]\gamma_{1}=1 ext { and } \ln \gamma_{1}=0[/latex], so that

[latex]\begin{aligned}\ln x_{1} &=-1.314 \x_{1} &=0.269\end{aligned}[/latex]

which is a factor of 3 too large.

[latex]^4[/latex]G. Scatchard, Chem. Rev., 8, 329 (1931).

Question 12.1.1: Solubility of a Solid in a Liquid Estimate the solubility of...

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