Dissolving - entropy

username459260

16

how do I work out the ΔS total for dissolving?
How do I calculate the enthalpy change for reaction and ΔS system??

Original post by jsmith6131

how do I work out the ΔS total for dissolving?
How do I calculate the enthalpy change for reaction and ΔS system??

You can only work out ΔS if you know the absolute entropy of the solid, solvent and of the solution.

Enthalpy change can be found by experiment, or from Hess's law by considering the lattice enthalpy and hydration enthalpy of the ions.

Reply 2

username459260

OP

16

Here is the example of the question confusing me:

Which of SrSO4 and BaSO4 is more soluble? Justify your answer using the data below to calculate the difference in ΔS(total)

ΔH(solution) SrSO4 (s) = -9 kJ/mol
ΔH(solution) BaSO4 (s) = +19 kJ/mol
S(standard) Sr²⁺ = -33 kJ/mol
S(standard) Br²⁺ = +10 kJ/mol

Original post by jsmith6131

Here is the example of the question confusing me:

Which of SrSO4 and BaSO4 is more soluble? Justify your answer using the data below to calculate the difference in ΔS(total)

ΔH(solution) SrSO4 (s) = -9 kJ/mol
ΔH(solution) BaSO4 (s) = +19 kJ/mol
S(standard) Sr²⁺ = -33 kJ/mol
S(standard) Br²⁺ = +10 kJ/mol

The units given for S(standard) are wrong. Are you sure that you have copied the question correctly?

Reply 4

username459260

OP

16

sorry, units are J K^-1 mol ^-1

Reply 5

username459260

OP

16

even so I don;t know what to do?

Original post by jsmith6131

even so I don;t know what to do?

It is not possible to have a negative standard entropy, so I suggest that you have copied the question down incorrectly.

But I imagine they want you to proceed roughly as follows:

If you are given no other information then you have to make some assumptions.

1. Both of the solid crystal lattices have next to no entropy (or at least similar entropy)

So for a dissolution:

BaSO₄(s) + (aq) --> BaSO₄(s)

ΔS₁ = S(barium sulphate solution) - [S(barium sulphate solid) + S(aq)]

and

ΔS₂ = S(strontium sulphate solution) - [S(strontium sulphate solid) + S(aq)]

and as

[S(barium sulphate solid) + S(aq)] = [S(strontium sulphate solid) + S(aq)]

Then ΔS₁ - ΔS₂ = the difference in entropies of the dissolved ions (which you are given)

So you can estimate that the ΔS of disssolution is different by the difference in the entropy values of the two ions.

Reply 7

username459260

OP

16

here is the scan
My question is part c)

http://s359.photobucket.com/albums/oo40/jsmith613/?action=view&current=Question-1.jpg

Original post by jsmith6131

here is the scan
My question is part c)

http://s359.photobucket.com/albums/oo40/jsmith613/?action=view&current=Question-1.jpg

Effectively they have an error with the entropy value. :eek:

It is fundamentally impossible to have a negative absolute entropy. In the same way as it is impossible to have a negative absolute temperature.
Absolute scales measure from zero up.

That said, I would go about it as I outlined above.

As you are asked for difference in ΔS(total) you must also apply the change in entropy of the surroundings...

... which will be fun as they haven't given you the temperature either.

ΔS(total) = ΔS(system) + ΔS(surroundings)

and ΔS(surroundings) = ΔH/T

Reply 9

username459260

OP

16

I worked out the difference in Delta S(surrounding) from the difference between Delta H(solution)
difference = 28 kJ/mol
= 94 kJ/mol

But I cant work out how to do Delta S (system) as I don't have standard entropy values of SrSO4 (s) and BaSO4 (s)

Original post by jsmith6131

I worked out the difference in Delta S(surrounding) from the difference between Delta H(solution)
difference = 28 kJ/mol
= 94 kJ/mol

But I cant work out how to do Delta S (system) as I don't have standard entropy values of SrSO4 (s) and BaSO4 (s)

I think that you have to assume they are the same

Reply 11

username459260

OP

16

but thats stupid :s-smilie:

they are definitley not.
even so, the difference therefore in S(system) is 43
total difference = 51 J K^-1 mol ^-1
so SrSO4 is mole soluble?

Original post by jsmith6131

but thats stupid :s-smilie:

they are definitley not.
even so, the difference therefore in S(system) is 43
total difference = 51 J K^-1 mol ^-1
so SrSO4 is mole soluble?

They won't be very different as ionic crystalline lattices.

The solubility trend for sulphates in grop 2 is increased solubililty going up the group. Mg very soluble.... Ba very insoluble