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When mixing two fluids of different composition at constant pressure when no heat can escape, is the enthalpy additive?
If not, how to compute the enthalpy of the mixture resulting from mixing two fluids with given composition after equilibrium is re-established?
The enthalpy is not additive under isobar, adiabatic conditions for non-ideal mixtures.
The change in the enthalpy is called the excess enthalpy (enthalpy of mixing). It can be computed with models such as the Margules, van Laar, Wilson, NRTL, and UNIQUAC.
These model give equations for the excess Gibbs energy. The excess enthalpy can be derived from these as described in 6.2 Fundamental Relations of Excess Functions in Molecular Thermodynamics of Fluid-Phase Equilibria (3rd Edition). (F'x already referred to this excellent book.)
One has to be careful with the conflicting sign convetions in the literature.
As Ali said, unless the mixture is ideal the enthalpy of mixing (or excess enthalpy) is not zero. This is the basis for a whole field of thermodynamics (and in particular, statistical thermodynamics): solution theories. To quote only a few of the most famous ones:
In this literature, you'll frequently encounter the name of John M. Prausnitz, whose book are considered classics (though I have not read them myself), including:
The above are for mixture of molecular fluids… polymer solutions are a bit of a special case of this, and specific models have been developed for them, the best-known being Flory–Huggins.
Also, you did not specify what kind of computation of the mixing enthalpy you were considering, so I have to add that several molecular simulation techniques have been developped to addres this question. For a basic introduction on this topic, I would recommend Singh and Gubbins’ review in Molecular-Based Study of Fluids.
This is intended as an addendum to the other answers, not a full answer in itself.$\newcommand{\b}[2]{\ce{#1\bond{...}#2}}$
Let's say we have a solution of molecule A, and a solution of molecule B. I'm denoting the strength of intramolecular forces by $\b AA,\b BB$ etc.
Whenever we create a real (i.e. non-ideal) solution, we classify the solution on the basis of deviation from Raoult's law:
$\Delta V_\text{mix}>0$
$\Delta H_\text{mix}>0$
$VP>VP_\text{expected}$
$\b AB<\b AA, \b BB$. (cohesive forces are stronger than adhesive forces)
Example: water and benzene.
Here, the adhesive forces are weaker. This leads to a net "expansion" of the mixture, and, it is endothermic due to the fact that we are partially replacing strong cohesive "bonds" with weaker, adhesive ones.
$\Delta V_\text{mix}<0$
$\Delta H_\text{mix}<0$
$VP<VP_\text{expected}$
$\b AB>\b AA, \b BB$. (cohesive forces are weaker than adhesive forces)
Example: ethanol and chloroform.
Here, the adhesive forces are stronger. This leads to a net "contraction" of the mixture, and, it is exothermic due to the fact that we are partially replacing weak cohesive "bonds" with stronger, adhesive ones.
So yes, the enthalpy is not additive for non-ideal solutions.
Manish's explanation is very nice and clear. Going back a step, there's an initial confusion that sometimes arises (it did for me). You can mix two fluids and the first law tells us that the mixture total enthalpy has to equal the sum of the two fluids quantities x their respective specific enthalpies.
To avoid the potential confusion, we should use the term 'isothermal' heat of mixing. When the mixing of two fluids at the same temperature is non-ideal, there is a change in temperature. If mixing is endothermic, the temperature falls and there is a positive heat required to restore the mixture to the initial temperature and vice versa. This heat is the Isothermal Heat of Mixing.
