0
$\begingroup$

I have some confusion in my mind about the difference between an excess property and a property change of mixing. I know that the Excess property is defined as the difference between the actual value and those of an ideal mixture: $$\ M^E(T,P,x_i)=M^{actual}-M^{id}(T,P,x_i)$$ On the other hand the property changes of a mixture is: $$\ M^{mix}=M{actual}-\sum n_i*M_i$$ So the latter seems to me that the whole change of a mixing process is with respect to the pure, ideal gas, species.

Excess and mixing are the same for enthalpy and volume but are different for entropy and Gibbs free energy.

Now come the problem: property changes of mixing and excess properties are related each other by the residuals properties (deviations from ideal gas)?

Otherwise I do not understand why they are different.

Thanks a lot

$\endgroup$
1
$\begingroup$

The difference is subtle, however they are different as you suspect.

You can look at the property of mixing as the change in property resulting from mixing the pure components (with everything at the same temperature and pressure) together to form the mixture.

The excess property is deviation between the actual property and the property in an ideal mixture at the same temperature and pressure.

They are the same for volume and enthalpy because the volume of mixing and the enthalpy of mixing in an ideal solution are both 0. This is not the case for G or S.

In other words the difference arises because $M^{id}=\Sigma x_i M_i$ is not true for all properties. It is only true for enthalpy and volume. So the second term in your expression for the excess property above, cannot in general be written as $M^{id}=\Sigma x_i M_i$ (if it could, the excess property and mixture property would be the same). It should instead be $M^{id}=\Sigma x_i M_i + M_{mix}^{id}$.

The extra term for mixing the components would be be given by $$ \frac{G^{id}_{mix}}{RT} = \Sigma x_iln(x_i) $$ and $$ \frac{S^{id}_{mix}}{R} = -\Sigma x_iln(x_i) $$ for Gibbs free energy and entropy, respectively.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.