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# Enthalpy Changes of Solution Question watch

1. Hello!

Today we were given a sheet and I was doing okay but now I'm a bit stuck. Well, maybe not stuck but need confirmation that I'm doing it right as I use these answers in the next part. We did an experiment where we dissolved different chlorides in water and measured the temperature increase.

We did this for LiCl, KCl, NaCl and CaCl2. The question asks that for each of these we work out the enthalpy change in solution in kJ for one mole of each. Because Moles = Mass / Mr here, Mr = mass.

We used 25cm^3 of water and in the previous question where we were working out enthalpy change m = mass of water + mass of salt in E=mcT

So for m in this can I just do Mr + 25 for each? and then plug that in with the corresponding temperature changes into the above equation?

Thanks in advance for any help.
2. Did you actually dissolve 1 mol of each solid?
3. (Original post by Pigster)
Did you actually dissolve 1 mol of each solid?
no
4. Therefore Mr doesn't = mass.

Anyhoo, you should really have measured the mass of the empty plastic cup and measured the mass of the solid and finally measured the mass of the cup, solid + water. But since you didn't, the m value in E = mcT is the mass of the water + mass of solid.

To work out DH, you'll do DH = - E / n where n = m (of solid) / Mr
5. (Original post by Pigster)
Therefore Mr doesn't = mass.

Anyhoo, you should really have measured the mass of the empty plastic cup and measured the mass of the solid and finally measured the mass of the cup, solid + water. But since you didn't, the m value in E = mcT is the mass of the water + mass of solid.

To work out DH, you'll do DH = - E / n where n = m (of solid) / Mr
What is DH sorry?
6. D = delta (I can't be bothered to type the greek letter)
H = enthalpy (I should have done it in italics)

DH = change in enthalpy
7. (Original post by Pigster)
D = delta (I can't be bothered to type the greek letter)
H = enthalpy (I should have done it in italics)

DH = change in enthalpy
Got it, awesome, thank you!
8. I am aware that this thread has been dead for a while now, but for anyone that remains confused;
Break the calculation for the enthalpy change of solution into 3 main parts:
1). Calculate the energy change of the solution, in KJ, using q=mcΔT.
In the question posed by the poster, this would be the mass of the water + the mass of the solid dissolved.
2). Calculate the moles of the solid dissolved - this is done by dividing the mass used by the mass number of the ionic compound.
3). Calculate the enthalpy change of solution by dividing the energy change, q, by the number of moles calculated in step 2.
(This will provide units of KJ/mol).
Hope this helps!
9. (Original post by TheModernDentist)
mass of the water + the mass of the solid dissolved.

(Do you teach?)

What would you use for the mass of solution is you reacted 1 g of CaCO3(s) with 25cm3 (an excess) of HCl?

How about if you used 1 g of Fe with 25 cm3 of (an excess) of CuSO4?
10. (Original post by Pigster)
[/b]

(Do you teach?)

What would you use for the mass of solution is you reacted 1 g of CaCO3(s) with 25cm3 (an excess) of HCl?

How about if you used 1 g of Fe with 25 cm3 of (an excess) of CuSO4?
Very good question.
Considering that in order for an enthalpy change of solution to occur, aqueous ions must form from gaseous ions.
In order for this to take place, the solvent molecules break the electrostatic attraction between the oppositely charged ions and then proceed to surround the separate ions. This can only happen as, with water as our solvent, the water is polar.
Now, as HCL(aq) is also polar, it will produce the same effect as water.
This means you follow the same method as you did when using water as the solvent; just simply add the two masses together!
e.g. for your example above for the reaction between Calcium carbonate and Hydrochloric acid...
Assuming the density of HCL is 1.00g/cm-3 = 25g + 1g = 26g.
Hope my explanation was of help!
(p.s. I'm afraid i do not teach ).
11. (Original post by TheModernDentist)
Assuming the density of HCL is 1.00g/cm-3 = 25g + 1g = 26g.

And here is my issue.

You can't add 1 g of CaCO3, since 0.44 g of that is lost as CO2.

i.e. the mass of solid added can't just be assumed to become part of the final solution.

Never mind the issue of the specific heat capacity of the solution having the same value as for water.

Never mind...
12. (Original post by Pigster)
[/b]

And here is my issue.

You can't add 1 g of CaCO3, since 0.44 g of that is lost as CO2.

i.e. the mass of solid added can't just be assumed to become part of the final solution.

Never mind the issue of the specific heat capacity of the solution having the same value as for water.

Never mind...
Again another very good question - I'm going to presume you are sitting A - levels and thus, will cater my response to your level of education;
At A - level, according to the current specifications, you will never be asked a question that includes something other than the formation of two aqueous ions from a solid ionic lattice - please direct me to the question by posting an image.
However, considering the level of knowledge needed for this topic at A - level, you would just minus the Carbon dioxide lost from the end solution;
1 - 0.44 = 0.56g + 25g (assuming 1g/cm3) = 25.56g for the total mass of the solution.
Hope this helps!
EDIT: Again, you will never be asked to calculate a specific heat capacity if the solvent used is not water.
EDIT 2: Ah, just realized you are 44 years of age...
EDIT 3: In actual fact, my reasoning should work, considering Oxygen and Calcium are left, both having ionic charges of 2....What do you think?

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