You are Here: Home

# Dissolving - entropy watch

1. how do I work out the ΔS total for dissolving?
How do I calculate the enthalpy change for reaction and ΔS system??
2. (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.
3. 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
4. (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?
5. sorry, units are J K-1 mol -1
6. even so I don;t know what to do?
7. (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.
8. here is the scan
My question is part c)

http://s359.photobucket.com/albums/o...Question-1.jpg
9. (Original post by jsmith6131)
here is the scan
My question is part c)

http://s359.photobucket.com/albums/o...Question-1.jpg
Effectively they have an error with the entropy value.

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
10. 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)
11. (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
12. but thats stupid 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?
13. (Original post by jsmith6131)
but thats stupid 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
14. also, just to check is it correct to say that if we know Delta H(hydration) that
ΔS(surr) = -ΔH(hydartion)*1000/T
15. (Original post by jsmith6131)
also, just to check is it correct to say that if we know Delta H(hydration) that
delta S(surr) = -Delta H(hydartion)*1000/T
yes
16. and also, is ΔS(system) = ΔS(cation [aq]) + ΔS(anion [aq])- ΔS(ionic crystal [s])
17. I presume this because
ΔS system = ΔS products - ΔS reactants
18. (Original post by jsmith6131)
I presume this because
ΔS system = ΔS products - ΔS reactants
ΔS system = S products - S reactants
19. oh right, yes

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: September 25, 2011
Today on TSR

### University open days

Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University