Heat of Mixing 
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Enthalpy Changes Upon Mixing

If seems to you that there are a lot of disconnected pieces of information in this section of material, you should reinforce in your own mind the link between all of these pages. That link is simply that in order to do energy balances we must be able to calculate changes in enthalpy associated with the process.

We always have the enthalpy of the outlet streams minus the enthalpy of the inlet streams appearing in our equations. At this point we know several ways to get the enthalpy change between inlet and outlet streams for specific situations. For example, we have seen how to use steam tables to get this change when steam is the working fluid. We have used a PH diagram to get enthalpies of a refrigerant. We have found that we can get changes in enthalpy associated with sensible heat changes (changes in temperature) by integrating CpdT.  And, on the last page, we saw how the psychrometric chart can be used to get enthalpies for air + water systems.

But, what about mixtures? When two liquids are mixed, the final enthalpy is not necessarily the sum of the pure component enthalpies. This is because the unlike interactions between molecules is most likely different than the like interactions. Thus, if the A-B interactions are stronger than the A-A and B-B interactions, then the mixing process will be exothermic (heat will be released because the more tightly bound A-B interactions are at a lower energy). Again, what we want is the change in enthalpy that occurs when the mixing occurs.

Let us examine the situation of the change in enthalpy upon mixing two liquid streams isothermally as shown in the diagram at the right. n1 moles of liquid 1 are mixed with n2, and the product liquid mixture leaves the mixer at the same temperature as the two inlet streams. We will have to supply or remove a certain amount of heat, Q, in order to maintain the isothermal process. A calorimeter is used to measure this quantity of heat that is given off or absorbed by the mixing process. The energy balance for this process can be written as,

Q =  (n1 + n2) hmn1h1 - n2h2
If we write this in terms of the heat absorbed or released per mole of product solution, we obtain
Q/m hm - x1h1 - x2h2 = Dhmix
This heat release or gain per mole of product solution for isothermal mixing is called the heat of mixing, Dhmix. It may be either negative (exothermic - heat given off because the mixture has a lower enthalpy than the pure components) or positive (endothermic - heat absorbed because the mixture has a higher enthalpy than the pure components).  We can, of course, turn the equation around and write that for an isothermal mixing process, the enthalpy of the mixture is the sum of the pure component enthalpies (weighted by their mole fractions) plus the heat of mixing:
 hmx1h1 + x2h2 + Dhmix    (isothermal)

Ideal Mixtures

An ideal mixture is one in which the interactions in the mixture are the same as for the two pure components. Thus, for an ideal mixture,
 hmx1h1 + x2h2         (ideal mixture)
Dhmix = 0                     (ideal mixture)
There are many mixtures that mix ideally or nearly ideally. All gases mix virtually ideally, because the interactions are small between gas molecules. Even in liquids, the interactions may be very similar between the components such that the heat of mixing is small. For example, the heat of mixing of hexane and heptane would be nearly zero. When ideal mixing can be assumed, one simply adds the contributions from the pure components to obtain the enthalpy of the mixture.

Nonideal mixtures

As you can infer from what was said above, often you can tell from the types of molecular interactions whether or not heats of mixing will contribute significantly to the energy balance. Liquids that can associate, form hydrogen bonds, or even break bonds as solvation occurs will often have fairly large heats of mixing. For example the mixing of H2SO4 and H2O produces so much heat that you must be careful to add the acid to the water very slowly so that it doesn't get hot enough to vaporize some of the water.

In this class, we will ignore the heat of mixing term unless we are dealing with fairly large quantities such as the solvation of acids and bases. Table B.11 on page 653 of your text lists heats of mixing for HCl, NaOH, and H2SO4 with water. That same table is also reproduced and available through the course tool links. Below is an example of how one can use these tabular data in solving energy balance problems.

Open Table B.11: Heat of Mixing Data

Problem Statement:
What is the heat duty for a mixer that mixes 9.2 moles of H2O with 1 mole of 0.2 mole fraction H2SO4 if the inlet and outlet streams are all to be at 25oC?

The heats of mixing given in Table B.11 are recorded in terms of a variable r which is the moles of water per mole of solute. We must first find this value of r (shown, however, as N in the figure below) in order to obtain the value from the table.

The energy balance for this system is

Q = n3h3 - n1h1 - n2h2
We need to find the enthalpy of each stream relative to a reference enthalpy. We can do this because we are always interested in a difference in enthalpies and so the reference will always cancel out. But, we must be sure that we choose the same reference for each stream so that the cancellation really does occur. In this case, we choose the pure components at 25oC as the reference enthalpy relative to which we will find the values of all the streams.

Stream 1:
Stream 1 is a mixture of sulfuric acid and water at 25oC. It's enthalpy relative to the pure references is simply the heat of mixing for this mixture since the pure enthalpies are identically zero by definition of the reference: hmx1h1 + x2h2 + Dhmix = 0 + 0 + Dhmix = Dhmix. Therefore, we find r for this mixture in the following way, where A represents acid and W water:

xA = nA/(nW + nA) = 0.2 = nA/1 mol =====>       nA = 0.2 mol         and     nW = 0.8 mol

so r = nW /nA = (0.8 mol)/(0.2 mol) = 4.0

From Table B.11, we therefore obtain h1 = -54.06 kJ/mol A

Stream 2:
This stream is pure water which is the reference state. Therefore, relative to the reference state,

h2 = 0
Stream 3:
We now do a mass balance to find r for the outlet stream. The acid balance tells us that 0.2 moles of A end up in the mixture and the water balance says that 9.2 + 0.8, or 10 mol of water ends up in the final mixture. Thus,
r = nW /nA = (10 mol)/(0.2 mol) = 50
From Table B.11, we obtain h3 = -73.34 kJ/mol A

Now we complete the energy balance to obtain:

Q = (0.2 mol A)(-73.34 kJ/mol A + 54.06 kJ/mol A) = -3.9 kJ

Continue to next section:  Mixing Process