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Topic: heat of dilution condensation and reaction  (Read 7378 times)

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Offline moleman1985

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heat of dilution condensation and reaction
« on: March 26, 2007, 06:47:40 PM »
hello I am slight confused to what actually happen when sulphuric acid is used to absorbe water vapour in a dryer.
I know that the temperature increases due to the 'heat of dissolution', and also because of the 'heat of condensation' of water, but!

Is the 'heat of dissolution' the 'heat of reaction', or does the 'heat of reaction' also need to be considered further increasing the temperature of the system, or does the 'heat of reaction' not actually exist as I primarily thought, until coming accross this web page:-

http://people.depauw.edu/harvey/Chem%20260/pdf%20files/Worksheets/EnthalpyStrongAcidKey.pdf

which clearly say that there is a heat of reaction, so do I need to consider all three "heats of ."

thankyou.

Offline xiankai

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Re: heat of dilution condensation and reaction
« Reply #1 on: March 27, 2007, 05:05:40 AM »
the heat of dissolution in this case is the heat of reaction for the dissolution of concentrated sulphuric acid so yes, you are right.

however, the heat of condensation is quite inapplicable because no water is condensing, rather water is being evaporated (involving the heat of vaporisation which is endothermic).
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Offline moleman1985

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Re: heat of dilution condensation and reaction
« Reply #2 on: March 27, 2007, 10:45:15 AM »
thankyou very much for clearing that up, although I did get it from a very reliable source that there was also heat of condensation not evapouration of water I though that this would be correct as the water vapour in the gas stream will condense into the liquid form in the acid. But I am not 100% sure and you will probally know better than me. So if any one can clear this up it would be great.

Offline eugenedakin

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Re: heat of dilution condensation and reaction
« Reply #3 on: March 27, 2007, 11:35:30 AM »
Hello moleman1985,

This is one-of-those questions which can be severely over-analyzed and are good for discussions :)

If the temperature of the final mixture of Sulphuric Acid and water are below the boiling point of water (assuming that the temperature of the Sulphuric Acid-Water mixture does not increase above the boiling point of water at 1 atm - which is another topic we could discuss  ;) ), then condensation of air-borne water into the Sulphuric Acid-Water mixture takes place. 

When water is condensing, it is now diluting the acid (by dissolution), which creates heat.  In this case, condensation of water into a Sulphuric Acid-Water mixture will cause the mixture temperature to increase by two factors: 1) The heat from condensed water is transferred to the mixture (phase change), and 2) water (as condensation) being added to the mixture will impart heat due to the heat of dissolution.

When the temperature of the Sulphuric Acid-Water mixture is above the boiling point of water (again, with the above assumptions), then water is evaporating from the mixture.  With this situation, heat is lost due to: 1) water with a phase change (liquid to gas), and 2) a negative heat of dissolution (further concentrating the acid).

I hope this explanation helps.  Cheers!

Eugene

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