Confusion about activity coefficients

[Matthew99]

Dear community!

I've just found a few contradictions in a topic I've never put much thought in. In order to calculate the thermodynamic activity, different formulas are being used. According to Wikipedia and common sense, these formulas should return the same value:

According to IUPAC, the standard values for concentration and molality are 1 mole/l and 1 mole/kg respectively. Thus, this relation is only possible if the activity coefficients for x,b,c and so on are different. Consider a simple example, 0.5 mole of a substance J dissolved in 1 l of water (~55 mole):

As you can see, the values of the activity coefficients depend on which formula I use. This leads to a few confusions:

• Often, the activity coefficient is interpreted in a quantitative way, for example, what a activity coefficient greater than 1 means. However, when I look at my calculations, there is no point in interpreting y as its value is quite random and depends on the equation I am using.
• When calculating the activity coefficient, for example with the Debye-Hückel equation, just looking at the equation, how do I know which y (y_x,y_c,y_b) it returns?

Edited September 29, 2020 by Matthew99
Latex issue

[studiot]

I don't follow you calculation. Where does the 55 come from ?

do you mean a molecular mass of 55 ?

So I dissolve 1/2 mole or 27.5g of J in 1 litre of water.

So the molality would be

[math]molality = \frac{{0.5}}{{1 + 0.0275}}mole/kg\;ofsolution = 0.486618mole/kg[/math]

 

To obtain the molarity, we don't know the density of the solution, so the amounts will be slightly different.

But since it is very dilute let us say that there is no change of volume on solution so we have 1 litre of solution.

So the molarity is

[math]molarity = 0.5/1 = 0.5mole/l[/math]

 

 

Edited September 29, 2020 by studiot

[Matthew99]

Dear studiot,

my calculation is meant to display the following exemplary scenario. I dissolve 0.5 mole of J in one liter of water. Therefore, I've got:

• 0.5 moles of J
• 1000g/18 g/mole = 55 moles of H2O

Further, as you pointed out, I assume that the volume does not change. Thus, I can calculate the mole fraction x_J and the molarity of J. This in turn shows that the activity coefficients y_x and y_c are different, what leads to the confusions I pointed out above.

[studiot]

1 hour ago, Matthew99 said:

Dear studiot,

my calculation is meant to display the following exemplary scenario. I dissolve 0.5 mole of J in one liter of water. Therefore, I've got:

• 0.5 moles of J
• 1000g/18 g/mole = 55 moles of H2O

Further, as you pointed out, I assume that the volume does not change. Thus, I can calculate the mole fraction x_J and the molarity of J. This in turn shows that the activity coefficients y_x and y_c are different, what leads to the confusions I pointed out above.

OK so I see where you are coming from now. I can't do any more for a couple of days as I am preparing for a funeral.

So my best suggestion is to get hold of a copy of

Chemical Equilibrium

Denbigh

Cambridge University Press

and read chapter 9 which is devoted to this.

I can't post the whole chapter but here are the first 4 pages.

 

Note the discussion of two different conventions in use.

 

Edited September 29, 2020 by studiot

[Matthew99]

Dear studiot,

thank you for your answer. I'll see what I can find. I've had look in my atkins as well, but they never mention that there are different conventions at all.

I've found a copy of the book and this chapter tells very clearly what I was searching for. If anyone else who reads this is interested:

Activity coefficients for molality are indeed no good tool to measure deviations from ideal behaviour. This is what Denbigh says: