# H2SO4, dissolution

by moleman1985
Tags: dissolution, h2so4
 P: 20 Hello, I have the following problem and am unsure how to solve it, although I believe the heat of dissolution to be -71.76kJ/mol, but I can only see how this would help me if my acid was infinetly dissoluted which is not the case so I need to know how the calculate a temperature rise when this is not the case. My acid starts off at 98% wt. total wieght 115.985041 kg, (1158.9 moles H2SO4, and 128.8 moles H2O). The acid ends up at 75% wt. total weight 151.55284 kg, (1158.9 moles H2SO4, and 2103.1 moles H20). Both the acid and water initial temps can be taken as 30oC, So basically 115.98kg of 98% wt. H2SO4 at 30oC, with 35.5kg water at 30oC
 P: 20 the water is also in vapour phase at the beggining, so will I also have to consider the heat of condensation of the water going from the vapour phase to the liquid phase? thankyou.
 P: 116 I vote no to the second question, but I'm not sure. It's not like it's steam? If the pressure is 1 atm of course... They are little particles of water at 30 °C. And, the solution of H2SO4 seems 18 M so, I think you may use the formular of: http://people.depauw.edu/harvey/Chem...ongAcidKey.pdf Which you posted yourself. So if all H2SO4 dissociates it will give: 83170 kJ of energy to the solution. For the warmth capacity I'd say approximately: H2SO4: 3,5 J/gC . 113000 g = 400 kJ/C H2O: 4,18 J/gC . 35,3 kg = 150 kJ/C So dQ = (cp1m1 + cp2m2) dT dT = 152 ° C Oh boy, there must be something wrong with the calculation??? Hey I'm not a professional... yet. Your reactor is going to explode... do you have some kind of cooling medium? The boiling point of H2SO4 was something at 350°C so maybe, just maybe your solution won't evaporate at 180 °C. Otherwise, you're in a big trouble. I don't know how your installation looks like, so that info would be interesting too. I'm curious myself how you're going to handle this calculation! Please post everything you got.
P: 20

## H2SO4, dissolution

Hi can I just say first thanks loads for your reply, I will be using a cooling loop in the system, sorry but I gave you all the information I got, but it going to 150oC sounds correct. So do you recon that I would be able to half or maybe even third the heating due to it not beening infinet dilution woukld you be able to hazard a guess with a vague explanation, if not it doesn't matter too much all I need to do is put in my report that I presume that it is infinet dissolution, its just that I dont want to be using more cooling than I sholud as I have to do a detailed costing on the system.

One other thing I also found the heat of dissolution to be 800kJ/mol, but that was on wiapedia.com, or however it's spelt, so it's probaly not reliable.