Enthalpy H2S2O7+H2O=>2H2SO4

[kenhung123]

Is this reaction exo or endo?

[98.40_for_sure]

As far as i know, this reaction is a simple dilution and no energy is produced or absorbed. And i can't find an enthalpy value for it anywhere, so it's neither exo nor endo

[kenhung123]

Yea but SO3+H2O=H2SO4 is also dilution and is exo though

[happyhappyland]

its exothermic.

Its not a dilution.

SO3 + H2O = H2SO4 releases too much energy thus the sulfuric acid forms into a mist which is too hard to collect therefore we split this equation in to 2 and effectively you split the energy released. Lets call energy released here amount "C"

So we put it into two equations the first being
SO3 + H2SO4= H2S2O7 ENERGY IS RELEASED but less than "C" so lets call it "A"

Then
H2S2O7 + H2O = 2H2SO4 energy again is released but ALSO less than "C" so no mist of h2so4 is formed Call energy released "B"

A + B must = C. but by splitting up the equation you allow less heat generated in such short amount of time.

[kenhung123]

Wow that was very intelligent of you to figure that out. Thanks.
Is the chamber which H2S2O7+H2O not called the diluter?

[superflya]

dilution tower?? the oleum is simply hydrated to produce sulfuric acid, dont know where u got dilution tower from :/

[kenhung123]

Ok thanks

[akira88]

Yeah it is called the diluter.