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 *C*pd*T*. 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,

If we write this in terms of the heat absorbed or released per mole of product solution, we obtainQ= (n_{1}+n_{2})_{ }h_{m}-n_{1}h_{1}-n_{2}h_{2}

This heat release or gain per mole of product solution for isothermal mixing is called theQ/m=_{ }h_{m}-x_{1}h_{1}-x_{2}h_{2}= Dh_{mix}

_{ }h_{m}=x_{1}h_{1}+x_{2}h_{2}+ Dh_{mix }(isothermal)

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._{ }h_{m}=x_{1}h_{1}+x_{2}h_{2}(ideal mixture)Dh_{mix }= 0 (ideal mixture)

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 H_{2}SO_{4} 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 H_{2}O
with 1 mole of 0.2 mole fraction H_{2}SO_{4} if the inlet
and outlet streams are all to be at 25^{o}C?

__Solution:__

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

We need to find the enthalpy of each streamQ=n_{3}h_{3}-n_{1}h_{1}-n_{2}h_{2}

__Stream 1:__

Stream 1 is a mixture of sulfuric acid and water at 25^{o}C.
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: *h*_{m} = *x*_{1}*h*_{1}
+ x_{2}*h*_{2} + D*h*_{mix}
= 0 + 0 + D*h*_{mix}
= D*h*_{mix}.
Therefore, we find *r* for this mixture in the following way, where
A represents acid and W water:

From Table B.11, we therefore obtainx_{A}=n_{A}/(n_{W}+n_{A}) = 0.2 =n_{A}/1 mol =====>n_{A}= 0.2 mol andn_{W }= 0.8 molso

r=n_{W }/n_{A}= (0.8 mol)/(0.2 mol) = 4.0

__Stream 2:__

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

h_{2}= 0

We now do a mass balance to find

From Table B.11, we obtainr=n_{W }/n_{A}= (10 mol)/(0.2 mol) = 50

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