Mol mass of LiBr= 7+80= 87

Mass of pure LiBr present FINALLY in 100[87+13]kg of water

=48000 C48/100= 23040gm

No of moles of pure LiBr=23040/87=264.827

Enthaly of hydration[H(hyd)] of LiBr= - 350.0 kJmol^-1[From literature survey]

According to the definition of enthalpy of hydration( Wikipedia):

Enthalpy of hydration, (Hhyd), of an ionic compound is the amount of energy released when a mole of the substance dissolves in a large amount of water forming an infinite dilute solution in the process

When you dissolve 264.8275 moles of LiBr in100(83+17) kg, you have dissolved it in =100000gm or 100000/18=5555.55 moles of water; meaning therby this solution of LiBr has been infinitely diluted.

Now:

The energy released for one mole of LiBr=350kJ

And total energy released for 264.8275 moles will be= 350C 264.8275

= 92689.63 kJ

Mol mass of LiBr= 7+80= 87

Mass of pure LiBr present FINALLY in 100[87+13]kg of water

=48000 C48/100= 23040gm

No of moles of pure LiBr=23040/87=264.827

Enthaly of hydration[H(hyd)] of LiBr= - 350.0 kJmol^-1[From literature survey]

According to the definition of enthalpy of hydration( Wikipedia):

Enthalpy of hydration, (Hhyd), of an ionic compound is the amount of energy released when a mole of the substance dissolves in a large amount of water forming an infinite dilute solution in the process

When you dissolve 264.8275 moles of LiBr in100(83+17) kg, you have dissolved it in =100000gm or 100000/18=5555.55 moles of water; meaning therby this solution of LiBr has been infinitely diluted.

Now:

The energy released for one mole of LiBr=350kJ

And total energy released for 264.8275 moles will be= 350C 264.8275

= 92689.63 kJ

Thank you for your response, Dr. Sehgal.

I had a couple of questions:

1)  When we speak about the enthalpy of hydration at infinite dilution, doesn't it mean that the concentration of LiBr in the final solution has to be negligible?

2) The calculations indicate that the amount of energy released is only a function of the mass of LiBr and its Hydration enthalpy. Does that mean that the energy released will be the same no matter how much water I add?

Dr.Sehgal,

How do you define in 'large' in the definition of Hydration Enthalpy? There must be some relation on the basis of the no.of moles of the solute present.

For Nilanjan Chaudhuri

I wait for you comments and then will answer to my young friendSaumitra Vinay Joshi. 

My friend, the world Large can be one litre or 10 litres. What I mean is that you go adding water and side wise go on noting the heat change[Change in temperature]. At a certain addition of volume of water, there will be no more heat change{ Temperature becomes constant} and this is what we mean by the term LARGE VOLUME .

Hello all!

I did some reading up on solution chemistry. The enthalpy of solution can be measured in two ways:

As is evident, the differential heat of concentration between c1 and c2 can be calculated by subtracting the Integral heat of solution at c2 from the Integral heat of solution at c1. i.e.

DeltaH = IH(c2) - IH(c1)

The closest data that I could find for the Integral heat of solution for H2O+LiBr has been attached in this comment.

In case a better data source is available, please share. Thank you!

Dr. Sehgal,

There seems to be a small error in the calculation of the number of moles of LiBr. The mass of LiBr present in the solution never changes (i.e. is fixed at 48 kg). Only the mass of water changes from 83 kg to 100 kg.

Thus 48 kg of LiBr = 48*1000/87 = 551.742 moles of LiBr.

Now, (Moles of LiBr) / (Moles of Water) = 551.742/5555.56 = 9.93%

Infinite dilution does not hold in this case.

 Now the answer to the second question from Saumitra Vinay Joshi. As I have explained to Dr Nilanjan Chaudhuri, when you have added so much of water{ Imply Large] then a stage will come when their will be no further evolution of heat. So what I mean that the energy released will be the same no matter how much water you may add.

Sir, I perfectly understand that. But what I'm trying to say is, that the amount of water added is not "very large" in this case, because even after adding 17 kg of water, the molar concentration of LiBr in the solution is 9.93% (as shown in my previous post). Thus we haven't yet reached the stage where the evolution of heat stops, and thus we cannot use the molar heat of infinite dilution in this case.

You need to use the tabulated thermodynamic property called the relative apparent molar enthalpy, which you can find in, e.g., Supplement 2 of J. Phys. Chem. Ref. Data 11 (1982), by D.D. Wagman, et al., "The NBS tables of chemical thermodynamic properties:  Selected values for inorganic and C1 and C2 organic substances in SI units.  (Look at page 2-293.)  If you really mean that your initial solution was 48 kg of LiBr and 83 kg of water, it was actually 36.6%; and your final solution was 32.4%.  To use the tables for LiBr (aq), the initial solution was 8.35 mol water per mol LiBr, and the final about 10:1.  The table includes values for 8 and 10, so I'll use those.  (You can fit more of the data to a smooth polynomial function to get better results.)  For 8:1 the quantity (shown as DH formation for the soln) is -395.568 kJ/mol; for 10:1 it is -396.426 kJ/mol.  Thus your heat of dilution is -0.858 kJ/mol.  Multiply by the total mols of LiBr to get your total heat.

That is very helpful, Dr. Tellinghuisen. Thank you!