KOH dilution exothermic heat

[jamesframes]

I am trying to calculate the temperature rise that will ocuur when I take 45% KOH solution and dilute it (with water) to a 20% (weight) solution.
I have found figures of 1020J/g Heat of Solution, but I can't figure out how to apply this to the partial dilution occuring during 45% to 20%.
I found a good chart for NaOH in the Oxy lierature, but not for KOH.

 

[siretb]

I have at 18°C
enthalpy of dilution ( source Landolt Bornstein)
3 mol KOH 5 mol water -1496 kcal/mol KOH
5 mol KOH 7 mol water -2095
3 mol KOH 100 water -2748
---this is not the enthalpy of dissolution---

average between 45% and 20% : about 2500 kcal/mol

for 1000 kilos of 45%wt 8 kmols KOH --> 20000 kcal
final mass (approx) 2300 kilos
temperature rise 20000/2300/1 8.7 deg C
(quick and dirty, assuming a Cp of 1)

a process simulator yields T rise 6.5°C

so temperature elevation is very limited

 

[jamesframes]

Thanks for the reply Siretb,
The figures you show for heat of dilution look way to large. Did you make an error there?
I have a figure of 13.8kcal/mol for total solution heat.
If you take your figure of 2500 kcal/mol and multiply by 8,000 mols you get 20,000,000 kcal (not 20,0000 you show).
That being said, the 6.5C sounds about right.

 

[siretb]

Yes, I messed up 3 zeros when typing the reference
the total enthalpy of dissolution I have is 13.3 kcal/mol almost for infinite dilution, at 18°C

the enthalpies of dilution are
-1.496, -2.095, -2.748 kcal/mol KOH

the quick calculation by hand (that yielded 8.7 was OK for a quick estimate. Apologies for skipping the decimal point.
I agree with your 13.8 figure