Archive for November, 2015

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If the man who almost single-handedly invented chemical thermodynamics – the American mathematical physicist Josiah Willard Gibbs – had owned an automobile, he would have had no trouble figuring out the action of antifreeze.

“The problem reduces to consideration of a binary solution in equilibrium with solid solvent,” I can hear old Josiah saying. “Such a thermodynamic system has two degrees of freedom, so at constant pressure there must be a relation between temperature and composition.”

And indeed there is. The relation corresponds to the observed depression of the freezing point of a solvent by a solute. What’s more, its exact form confirms how antifreeze really works.

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Computing chemical potential

We have Josiah Willard Gibbs to thank for introducing the concept of chemical potential (μ) as a sort of generalized force driving the flow of chemical components between coexistent phases.

When the phases are in equilibrium at constant temperature and pressure, the chemical potential of any component has the same value in each phase

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The key point to note here is that μi is the chemical potential of component i in an arbitrary state, i.e. in a mixture of components. In order to compute this potential we need to know two things: the chemical potential of the pure substance μi0 at a pressure p (such as that of the atmosphere), and the mole fraction (xi) of the component in the mixture. Assuming an ideal solution, use can then be made of the textbook formula

dce12 …(1)

With pressure and temperature fixed, this equation has a single variable (xi), from which we can draw the conclusion that the variation in chemical potential of a component in an ideal solution is determined solely by its own mole fraction.

The significance of this fact can be appreciated by considering the following diagrams

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Here is water in equilibrium with ice at 273K. The chemical potentials of the solid and liquid phases are equal; there is no net driving force in either direction. Now consider the effect of adding an antifreeze agent to the liquid phase

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Assuming the temperature held constant at 273K, the addition of antifreeze reduces the mole fraction of water, lowering its chemical potential in accordance with equation 1. The coexistent solid phase now has a higher potential, providing the driving force to transform ice into water. Since the temperature is held constant, this equates to the lowering of the freezing point of water in the mixture.

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Deducing a formula for freezing-point depression

To obtain a formula for the freezing point of water in a solution containing antifreeze, we start with the equilibrium relation

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where the zero superscript indicates a standard potential, i.e. that the solid phase consists of pure ice whose mole fraction x is unity. Substituting the left hand side with

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we obtain

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which after differentiation with respect to temperature at constant pressure and subsequent integration yields the formula for the freezing point of water in a solution containing antifreeze at 1 atmosphere pressure:

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The terms on the right are the molar enthalpy of fusion of water (ΔHf0), the freezing point of pure water (Tf0), the gas constant R and the mole fraction of water (xH2O) in the solution containing antifreeze.

The latter is the only variable, confirming that the freezing point of water in a solution containing antifreeze is determined solely by the mole fraction of water in the mixture – in other words the extent to which the water is diluted by the antifreeze agent.

This is how antifreeze works. There is nothing active about its action. It exerts its effect passively by being miscible and thereby reducing the mole fraction of water in the liquid mixture. There’s really nothing more to it than that.

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Using the formula

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Values for constants

Enthalpy of fusion of water ΔHf0 = 6.02 kJmol-1
Freezing point of pure water Tf0 = 273.15 K
Gas constant R = 0.008314 kJmol-1K-1

Example

651 grams of the antifreeze agent ethylene glycol (molecular weight 62.07) are added to 1.5 kg of water (molecular weight 18.02). What is the freezing point of water in this solution?

Strategy

1. Calculate the mole fraction of water in the solution

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Number of moles of water = 1500/18.02 = 83.2
Number of moles of ethylene glycol = 651/62.07 = 10.5
Mole fraction of water = 83.2/(83.2 + 10.5) = 0.89

2. Calculate the freezing point of water in the solution

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The solution will give antifreeze protection down to 261.65K or –11.5°C

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P Mander March 2015

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