1 Gases
2 Microscopic Energies
3 First Law
4 2nd & 3rd Law
5 Phase transitions
6 Mixtures
7 Phase Diagrams
8 Equilibrium
9 Molecular Interactions
Partial molar Quantities
partial molar quantities can be succinctly described as the effect that a change in a single component has on the particular thermodynamic property.
Partial Molar Volume
the molar volume of water Vm is about 18 cm3/mol. Thus, if we add a mole of water to a larger body of water, it's volume will increase by 18 cm3. However, if we add that same amount of water to a large volume of ethanol the total volume will only increase by 14 cm3. This would be due to the solvation effects between ethanol solvent and water solute in which the entropy of the ethanol is reduced somewhat as solvation shells are formed around the water molecules. The increase in volume of the solution due to the change in the amount of water is called the partial molar volume of water in the ethanol/water solution. This number is always taken as the limit of a relatively small change in solute, compared to a large amount of solvent.
Fig. 5.1 (from Atkins) The partial molar
volumes of water and ethanol at 25°C. Note the
different scales (water on the left, ethanol on
the right). |
We can generalize the form of the partial molar volume Vj of component j as:
In our simplified binary mixtures, we can thus write the total volume dependence as
if we can assume that the partial molar volumes do not change (for example, if we keep the compositions of the mixture constant) the total volume of the system can be calculated easily via
In general, partial molar volumes do not remain constant if the composition of the mixture changes. However, since V is a state function, the final equation should be valid no matter how the solution was prepared.
While volumes are always positive, partial molar volumes need not be. for example, the partial molar volume of MgSO4 in a large volume of water is –1.4 cm3/mol. This means that as we add this salt to water, the volume actually goes down. This makes sense if we recall that the ordering effect of solvation tends to reduce the volume of the solvent and in this case, the solid solute contributes very little. So the net effect is a small reduction of the total volume.
Partial Molar Gibbs Energy
We have already seen this quantity if the form of a chemical potential. The additional factor is that there are more than one components in the state in question, thus, for component j of the mixture
 Fig. 5.4 (from Atkins) The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this case, both chemical potentials are positive. |
Following similar logic as we did for partial molar volumes, we can determine the total Gibbs energy as
Since Gibbs energy depends on pressure and temperature as well as on composition, our overall equation for Gibbs energy is now
Which, for constant temperature and pressure is simplified to be
Under constant temperature and pressure, The change is Gibbs energy can be replaced by the maximum additional (non-expansion) work available from the system wadd,max. So,
This latter equation is useful for determining the work we can extract from an electrochemical process as the composition changes, for example.
We recall the Gibbs defining equation can be written
as G = U + pV – TS.
Thus, we can write
U = –pV + TS + G.
The equation for an infinitesimal change in
U is
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dU |
= –pdV – Vdp + SdT + T dS + dG = –pdV – Vdp + SdT + T dS + (Vdp – SdT + mAdnA + mB dnB + ...) = –pdV + T dS + mAdnA + mB dnB + ... and, for constant volume and entropy, = mAdnA + mB dnB + ... |
It follows thus that
similarly, we can recast the chemical potential in terms of enthalpy (when entropy and pressure are constant) or in terms of Helmholtz energy when volume and temperature are constant.
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so, depending on the conditions, the chemical potential tells us about all the extensive thermodynamic properties of state. Hence the importance of chemical potential in chemistry.
In a multi-component system, the sum of the chemical potentials must remain constant. In other words, if the chemical potential of one component of a binary system goes up, the chemical potential of the other component must go down by the same amount.
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This is the Gibbs-Duhem equation.
For the simpler binary case, we can relate the two chemical potential infinitesimals as
We can use this equation for all the partial molar quantities. In general, the change in the partial molar quantity for component A can be used to determine the change for component B. If the ratio of A to B is large then a small change in A will give a large change in B.
Thermodynamics of Mixing
Consider two ideal gases, A and B with amounts nA and nB , respectively both at temperature T and pressure p. The chemical potential for each pure gas can be calculated via equation 5.4 where, by comparing with equation 4.47 and substituting m directly for Gm. we get
In this case, mº is the chemical potential of the pure substance at standard pressure pº (= 1 bar). Thus, the Gibbs energy for the system is given using equation 6.5 as
After mixing, the individual partial pressures change because of the change in volumes so we get
After mixing, the change in Gibbs energy can be easily determined as
Now, we replace pi/p with xi and ni with xin
Here, we see that since the mole fractions must be less than one that the entropy of mixing will always be negative for two gases mixing.
Example: two containers of equal volume contain 3.0 mol H2(g) and 1.0 mol N2(g), respectively. The divider separating the two gases is removed and they are allowed to mix. What is the Gibbs energy of mixing?
We need to start with the individual Gibbs energies since we have different starting pressures. Let the pressure of the N2 be p. The pressure of the H2 will therefore be 3p. After mixing, the partial pressures will each drop by a factor of 2 because each individual gas now sees double the volume. Thus, the equation 6.17 needs to be modified a bit since we started with different initial pressures.
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Entropy and Enthalpy of mixing:
Recall equation 4.40 () from the Maxwell equations. It allows us to use our thermodynamically derived Gibbs energy of mixing to get to the Entropy of mixing from pure thermodynamics considerations.
Exactly the same as the entropy of mixing equation we derived earlier (equation 4.4a) from pure statistical considerations.
Using equation 6.18 and equation 6.19 it is quite simple to see, in conjunction with ΔG = ΔH – TΔS, that ΔmixH = 0. This makes sense since the gas molecules do not interact (ideal gas).
The Chemical Potential of Liquids
We need to understand the way Gibbs energy changes with composition. These concepts will be important in understanding chemical equilibrium
Ideal Solutions
We will designate thermodynamic properties of pure substances with an asterisk *. Thus, the chemical potential of pure A at non standard conditions is
If there is a second substance present then the equation reverts back to equation 6.14 (without the *). We can combine these two equations to eliminate the need for the standard conditions.
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Now, if our system is that of a gas in equilibrium with a liquid, the pressures are vapour pressures, pA*, being the vapour pressure of the pure liquid A and p being the vapour pressure of A in solution.
Raoult's Law states that the partial pressure of a vapour component in equilibrium with it's solution is proportional to the mole fraction in solution, pA = c pA*. Thus, we can rewrite equation 6.21 to be
This equation defines an ideal solution. Solutions that follow this relationship tend to be composed of individual components that are very similar structurally, such as benzene/methylbenzene or heptane/octane. At the molecular level, the intermolecular forces of the individual components are quite similar to each other and to the solution at all concentrations. (Ideally, they are identical)
Ideal dilute solutions
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While most solutions do not obey this law exactly, the quality of the fit between experiment and this law improves as the mixture tends to purity of the given component.
On the other hand, solutions that are very dilute in the given component can also be approximated by a straight line; just not the same slope (constant of proportionality). Such solutions follow Henry's Law. Thus, for a component, B, in a dilute solution of B Henry's Law states pB = cB KB. Obviously, Henry's Law and Raoult's Law merge as the solution tends to ideal behaviour. If the solution is not ideal then if K > p* we have a positive deviation from Raoult's Law and if K < p* we have a negative deviation.
Some solutions deviate quite far from Raoult's Law to the extent that they have a maximum or minimum in the vapour pressure at some intermediate concentration.
Consider the solution of carbon disulfide with acetone. The intermolecular forces of the solution are significantly reduced compared to the individual pure liquids and hence, the vapour pressures are increased for both components well above what Raoult's Law predicts. This solution is quite far from ideal behaviour.
Example:
A series of solutions of propanone (acetone) and trichloromethane (chloroform) were measured at 35ºC. The following vapour pressures were measured.
| xc | 0 | 0.20 | 0.40 | 0.60 | 0.80 | 1 |
| pC/kPa | 0 | 4.7 | 11 | 18 | 26.7 | 36.4 |
| pA/kPa | 46.3 | 33.3 | 22 | 12.3 | 4.9 | 0 |
What is Henry's Law constants for acetone and chloroform in this solution and show that the solution behaves like Raoult's law for dilute solutions.
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By plotting the data and tracing straight lines along Raoult's Law positions (red), we see that the solution follows Raoult's Law best when the particular component is nearly pure.
When the given component is very dilute, the line does not follow Raoult's Law but can still be approximated by straight lines (blue). The two slopes (K constants) are approximately 23.3 for acetone and 22.0 for chloroform.
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