How much heat will be released if a 98% (m/m) H2SO4 solution is diluted to 96% (m/m)

[nico1234]

I know i have to look at the reaction enthalpy. Normally for diluting sulphuric acid the following reactions occurs:

(1)     H2SO4 + H2O --> H+ HSO4- +H2O

(2)     HSO4- + H2O --> H+ SO42- +H2O

But in the above situation there is a shortage for the watermolecule

Does only the first reaction (1) take place? and therefore only heat is released by the first reaction?

Regards,

[mbdowell]

Hi Nico,

Before we go to "how hot does it get?", you asked whether the heat comes only from the first reaction. Yes, it does! Sulfuric acid is a strong acid, meaning that the first proton dissociates entirely and 1M H2SO4 has a [H+] of 1 M and pH of 0. All those protons in solution would keep HSO4- from dissociating to makesulfate,SO4-2. At high acid concentrations/ low pH, the second reaction doesn't happen.

Of course, protons would like to form hydronium ion if there is enough water: H+ + H2O = H3O+. That's the cation for acid concentrations ~1 M or less, but you have 96-98% H2SO4 which is more like 10M, and water is in short supply. You really do have more free protons running around than you have H2O to react with them so your reaction is really protonating the limited amount of water: 

H+ + H2O = H3O+ in excess H+, or you can consider it as hydrating anhydrous H2SO4:  H2SO4 + H2O = H2SO4:H2O. That really exists as a gas; it's harder to describe in solution.

So, back to "How Hot?" If you have to do this and not just calculate how hot, you've always been told " Put the acid into the water so the water doesn't boil and spatter you." You can't do that here. Wear goggles, use a shield and stir the acid like mad while you add a little water at a time. If you only have to calculate it, Google "heat of dissolution of concentrated sulfuric acid". The first entry, a PDF from a lab at DePauw University, shows that if you were to dilute concentrated H2SO4 "to infinite dilution" (concentration becomes very dilute but we also assume that all the heat raises temperature of the acid only), your temperature could go up ~50C. It's an easy calculation to follow, but you are only diluting from 98% H2SO4 to 96%. And you are only putting in a little water and the water might boil. So go to the PDF from Southern States Chemical's Chemical Safety Handbook, which unfortunately uses English units. If you dilute from 98% to 96% and the heat gets distributed uniformly throughout the acid, it doesn't get to boil water and spatter you. The acid temperature only goes up about 23C if my calculation is correct. In any case, stir like mad, wear goggles and gloves.

Mike Dowell

[gmbodner]

You might benefit from the following URL: ttps://www.thoughtco.com/add-sulfuric-acid-to-water-606099

Please note that this URL correctly points out that adding sulfuric acid to water can raise the temperature of the solution rom roughly 20 C (room temperature) to over 130 C.

I tell my students that everything they are likely to want to know about thermodynamic quantities can be found in J. Phys. Chem. Reference Data Volume 11, 1982. Also available as "The NBS tables of chemical thermodynamic properties: Selected values for inorganic and C1 and C2 organic substances in SI Units." by Wagman, et al. I keep a copy of this book on a shelve with several sets of reference data.

In theory, if you look up the KA's of H2SO4 and HSO4(-) you would predict 100% dissociation of H2SO4 and about10% dissociation of HSO$(-). But there are two problems with this. Contrary to popular belief, strong acids are not 100% dissociated in water. Here is the extent of dissociation of HCl, for which I happen to have the data.

0.001 M HCl 99% dissociated

0.01 M HCl 97% dissociated

0.1 M HCl 92% dissociated

1.0 M HCl 78% dissociated

Furthermore, you assume that the enthalpy change will result only from the dissociation of the sulfuric acid, which would be far from 100%. You've forgotten the possibility that there is an enthalpic contribution from the interaction between the acid and water.