Posts Tagged ‘mole fraction’

ae01

In my previous post JH van ‘t Hoff and the Gaseous Theory of Solutions, I related how van ‘t Hoff deduced a thermodynamically exact relation between osmotic pressure and the vapor pressures of pure solvent and solvent in solution, and then abandoned it in favor of an erroneous idea which seemed to possess greater aesthetic appeal, on account of a chance encounter with a colleague in an Amsterdam street.

Rendered in modern notation, the thermodynamically exact equation van ‘t Hoff deduced in his Studies in Chemical Dynamics (1884), was

ts06

Following his flawed moment of inspiration upon learning the results of osmotic experiments conducted by Wilhelm Pfeffer, he leaped to the conclusion that the law of dilute solutions was formally identical with the ideal gas law

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It would seem van ‘t Hoff was so enamored with the idea of solutions and gases obeying the same fundamental law, that he failed to notice that the latter equation is actually a special case of the former. Viewed from this perspective, the latter’s resemblance to the gas law is entirely coincidental; it arises solely from a sequence of approximations applied to the original equation.

As a footnote to history, CarnotCycle lays out the path by which the latter equation can be reached from the former, and shows how accuracy reduces commensurately with simplification.

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We begin with van ‘t Hoff’s thermodynamically exact equation from the Studies in Chemical Dynamics

ts06 (1)

where Π is the osmotic pressure, V1 is the partial molal volume of the solvent in the solution, p0 is the vapor pressure of the pure solvent and p is the vapor pressure of the solvent in the solution.

Assuming an ideal solution, in the sense that Raoult’s law is obeyed, then

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where x1 and x2 are the mole fractions of solvent and solute respectively. So for an ideal solution, equation 1 becomes

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If the ideal solution is also dilute, the mole fraction of the solute is small and hence

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so that

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For a dilute solution x2 approximates to n2/n1, where n2 and n1 are the moles of solute and solvent, respectively, in the solution. The above equation may therefore be written

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In dilute solution, the partial molal volume of the solvent V1 is generally identical with the ordinary molar volume of the solvent. The product V1n1 is then the total volume of solvent in the solution, and V1n1/n2 is the volume of solvent per mole of solute. Representing this quantity by V’, the above equation becomes

ae07 (2)

which is identical with the empirical equation proposed by HN Morse in 1905. For an extremely dilute solution the volume V’ may be replaced by the volume V of the solution containing 1 mole of solute; under these conditions we have

ts04 (3)

which is the van ‘t Hoff equation.

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It is instructive to compare the osmotic pressures calculated from the numbered equations shown above and those obtained by experiment. It is seen that Eq.1, which involves measured vapor pressures, is in good agreement with experiment at all concentrations. Eq.3 fails in all but the most dilute solutions, while Eq.2 represents only a modest improvement.

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These figures give a measure of van ‘t Hoff’s talent as a theoretician in deducing Eq.1, and the error into which he fell when abandoning it in favor of Eq.3.

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Mouse-over links to works referred to in this post

Jacobus Henricus van ‘t Hoff Studies in Chemical Dynamics

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P Mander June 2015

afr01

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

afr09

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

afr11

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

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P Mander March 2015

gibbs

It was the American mathematical physicist Josiah Willard Gibbs who introduced the concepts of phase and chemical potential in his milestone monograph On the Equilibrium of Heterogeneous Substances (1876-1878) with which he almost single-handedly laid the theoretical foundations of chemical thermodynamics.

In a paragraph under the heading “On Coexistent Phases of Matter” Gibbs mentions – in passing – that for a system of coexistent phases in equilibrium at constant temperature and pressure, the chemical potential μ of any component must have the same value in every phase.

This simple statement turns out to have considerable practical value as we shall see. But first, let’s go through the formal proof of Gibbs’ assertion.

An important result

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Consider a system of two phases, each containing the same components, in equilibrium at constant temperature and pressure. Suppose a small quantity dni moles of any component i is transferred from phase A in which its chemical potential is μ’i to phase B in which its chemical potential is μ”i. The Gibbs free energy of phase A changes by –μ’idni while that of phase B changes by +μ”idni. Since the system is in equilibrium at constant temperature and pressure, the net change in Gibbs free energy for this process is zero and we can write

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hence

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This result can be generalized for any number of phases: for a system in equilibrium at constant temperature and pressure, the chemical potential of any given component has the same value in every phase.

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Visualizing variance

The equality of pressure P, temperature T and component chemical potentials μn between coexistent phases in equilibrium provides a convenient way to visualize variance, or the number of degrees of freedom a system possesses. For example, the triple point of a single component system can be visualized as the array

pe05

where the solid, liquid and vapor phases are indicated by one, two and three primes respectively.

Each row represents a single variable, so the number of rows equates to the total number of variables. Each column lists the variables in a single phase. All but one of these may be independently varied; the last is determined by the Gibbs-Duhem relation

pr02

There are one of these for each phase, so the number of columns equates to the number of relations (=constraints) to which the system variables are subject. The variance, or number of degrees of freedom (f) of the system is defined

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For arrays of the kind presented above, this transposes into

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For the triple point of a single component system, there are three rows and three columns, so f =0. With zero degrees of freedom, the triple point is not subject to independent variation and is represented by a fixed point in the PT plane.

The above rule implies that a system of coexistent phases in equilibrium cannot have more phases than intensive system variables.

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Generating useful equations

For a component present in any pair of coexistent phases in equilibrium at constant temperature and pressure, the chemical potential of that component has the same value in both phases

pe03

From this general relation, equations may be deduced for computing various properties of thermodynamic systems such as ideal solutions, for example the elevation of boiling point, the depression of freezing point, and the variation of the solubility of a solute with temperature.

The key point to grasp 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)

where for a given phase, μi is the arbitrary chemical potential of i in the mixture, μ°i is the chemical potential of the pure substance, and xi is the mole fraction of the component.

As an example, let us take the relation

pe08 …(2)

where the chemical potential of the solid solvent is necessarily the standard potential because the mole fraction x is unity. The above relation will generate an equation for the depression of the solvent freezing point in a solution at a fixed pressure (p).

Substituting (1) for the liquid phase in (2) gives

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where by convention the subscript 1 refers to the solvent. Differentiating with respect to T at constant pressure

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using the quotient rule for ΔG/T gives

pe12 … (3)

Now since

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equation (3) simplifies to

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Integrating from the pure solvent state, where the mole fraction x1=1 and T0fus is the freezing point of the pure solvent, to the solution state where the mole fraction x1= x1 and Tfus is the freezing point of the solvent in the solution

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yields the equation for the depression of the solvent freezing point in a solution at a fixed pressure (p)

pe16

Since x1<1 in a solution, the logarithm is negative and therefore the freezing point of the solvent in the solution must be lower than the freezing point of the pure solvent.

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Ok, so maybe that wasn’t the simplest procedure for generating a useful thermodynamic equation. But the point to be made here is that the same procedure applies in the other cases, so you only need to understand the principle once.

For example, the equation for elevation of solvent boiling point in solution with a non-volatile solute at a fixed pressure (p) is

pe17

The similarity to the previous equation is evident.

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P Mander February 2015