How Does Mixing Affect Entropy Calculations in Chemical Reactions?

[tomothy]

I know this is more of a chemistry question but it's come to this I'm afraid.
When calculating entropy changes in chemical reactions sometimes we students of chemistry use standard molar entropies of the reactants and products. But this surely doesn't account for the mixing effect. e.g.
N2 + 3 H2 --> 2 NH3
calculating the entropy change reaction for this using standard molar entropies would surely yield an entropy change too large because the reactants would in fact start off mixed, thus at a higher entropy than just the N2 and 3 H2 together.

 

[Chestermiller]

I know this is more of a chemistry question but it's come to this I'm afraid.
When calculating entropy changes in chemical reactions sometimes we students of chemistry use standard molar entropies of the reactants and products. But this surely doesn't account for the mixing effect. e.g.
N2 + 3 H2 --> 2 NH3
calculating the entropy change reaction for this using standard molar entropies would surely yield an entropy change too large because the reactants would in fact start off mixed, thus at a higher entropy than just the N2 and 3 H2 together.

This is not true. The standard entropy change and the standard free energy change are based on the use of a van't Hopf equilibrium box, which inherently includes entropy of mixing.