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KOH dilution exothermic heat

KOH dilution exothermic heat

KOH dilution exothermic heat

(OP)
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.

RE: KOH dilution exothermic heat

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         

RE: KOH dilution exothermic heat

(OP)
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.

RE: KOH dilution exothermic heat

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

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