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Water freezes at 0$^{\circ}$ C, ethylene glycol freezes at -12$^{\circ}$ C, but my car starts at -20$^{\circ}$ C. Mixing them together prevents my car engine's coolant from freezing down to -30$^{\circ}$ C, (at least that's what it says on the label of the can).

This question Freezing Point Depression : Enthalpy and Entropy is related, but unfortunately, it is not answered.

Salt and water freezing point is also related but not a duplicate, as far as I can tell.

The concepts behind enthalpy and entropy are both familiar to me, but their role in freezing point depression is not.

I do know that the antifreeze process is based on the prevention of ice crystals in the water, and from Wikipedia Ethylene Glycol I read that:

Ethylene glycol disrupts hydrogen bonding when dissolved in water. Pure ethylene glycol freezes at about −12 $^{\circ}$ C but when mixed with water, the mixture does not readily crystallize, and therefore the freezing point of the mixture is depressed. Specifically, a mixture of 60% ethylene glycol and 40% water freezes at −45 $^{\circ}$ C.

My question is: how would this process be explained (briefly, of course) in thermodynamical terms.

If this is too broad a question to be asked here, my apologies and I will delete the question.

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What I am going to write here is taken from Wedler: Lehrbuch der Physikalischen Chemie. I am unfortunately unaware of an English translation of this book - similar explanations in less formal-mathematical style might perhaps be contained in Atkins/de Paula: Physical Chemistry.

For the chemical potential of component $i$ in an ideal mixture we can write $$\mu_i (p,T) = \mu_i^*(p,T)+RT \ln(x_i) $$ Here $\mu_i^*(p,T)$ stands for the chemical potential of the pure component (i.e. the same substance unmixed with anything else) and $x_i$ for the mole fraction of component $i$ in the mixture. To account for deviations from ideal behaviour, the concept of Thermodynamic activity $a_i$ is used: $$\mu_i (p,T) = \mu_i^*(p,T)+RT \ln(a_i) $$ The thermodynamic activity $a_i$ of component $i$ is defined via this equation and usually given as product of an activity coefficient $f_i$ and the mole fraction $x_i$.

Let us now consider a mixture of a liquid (component 1) with one component (component 2) that does (idealised) not evaporate. The liquid phase $\beta$ is in equilibrium with a solid phase $\alpha$ of component 1. Thermodynamic equilibrium conditions are: $$\frac{\mu_1^{\alpha}}{T} = \frac{\mu_1^{\beta}}{T} $$ For the solid phase, we just have $\mu_1^{* \alpha}=\mu_1^{\alpha}$ as it is a pure phase. For the liquid phase, we can write as above: $$\mu_1^{\beta} = \mu_1^{* \beta} + RT \ln a_1 $$ Combining this we obtain: $$\textrm{d}\left(\frac{\mu_1^{* \alpha}}{T}\right) = \textrm{d}\left(\frac{\mu_1^{* \beta}}{T}\right) + \textrm{d}\left(R \ln a_1 \right) $$ Using $dp=0$ and the properties of the chemical potential, this is equivalent to: $$\frac{-h_1^{* \alpha}}{T^2} dT = \frac{-h_1^{* \beta}}{T^2} dT + \textrm{d}\left(R \ln a_1 \right) $$ or: $$\left( \frac{\partial T}{\partial \ln a_1} \right)_{p} = \frac{RT^2}{\Delta_m H} $$ Here, $\Delta_m H$ is the molar melting enthalpy. Stopping here for a second, we see that there is not a simple mixing rule for the melting temperatures but rather that the magnitude of melting point lowering is proportional to the amount by which the chemical activity of the main species is lowered by mixing in the additional species.
For ideal diluted mixtures we can set $\ln a_i = \ln (1-x_2) \approx -x_2$ and under the assumption of temperature indepent melting enthalpy, we can integrate above equation to obtain: $$T_m(x_2)-T_m(x_2=0) = - \frac{RT^2_m(x_2=0)}{\Delta_m H} x_2$$ This stresses the point that, for ideal or very dilute mixtures, the amount by which the melting temperature is lowered is dependant on the amount of second species that is put in but not on the melting point of that species. In reality, the properties of the second species do of course matter via the activity coefficient.

I hope I understood your question correctly and what I wrote gives you an idea. I am very thankful for constructive criticism and questions about things I have remained unclear about.

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    $\begingroup$ Thank you very much for your time and effort Sanya, I had not realised the wide range of subjects covered by statistical mechanics and the necessity to study it until recently. I want to keep learning QFT but I know I need to study posts exactly like yours. $\endgroup$ – user108787 Sep 18 '16 at 18:04
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    $\begingroup$ @CountTo10 The problem with physics is that there are too many interesting fields :) And you are very welcome, I had been wanting to look at this in detail for some time, so it was a good opportunity :) $\endgroup$ – Sanya Sep 18 '16 at 20:09

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