Enthalpy of Sublimation

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Sumaiyah I 1E
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

Enthalpy of Sublimation

Postby Sumaiyah I 1E » Fri Jan 08, 2016 5:47 pm

Is the enthalpy of sublimation a combination of fusion and vaporization? I'm super confused on which formula to use, the Hvapor-Hsolid or the Hfus+Hvap?

IsabellaTous_1B
Posts: 10
Joined: Fri Sep 25, 2015 3:00 am

Re: Enthalpy of Sublimation

Postby IsabellaTous_1B » Fri Jan 08, 2016 6:18 pm

The enthalpy of sublimation can be expressed as a combination of the enthalpy of fusion and the enthalpy of vaporization given that sublimation is the process of a solid vaporizing and bypassing the liquid phase and enthalpy is a state property. They are essentially both the same formula if you expand the Hsub=Hvap+Hfus into Hsub=(Hvapor-Hliquid)+(Hliquid-Hsolid) by substituting in the equations for Hvap and Hfus respectively. In this case the Hliquid terms cancel out and give you Hsub= Hvapor-Hsolid.

Angela Kim 2H
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

Re: Enthalpy of Sublimation

Postby Angela Kim 2H » Fri Jan 08, 2016 6:21 pm

You can use either equation; they both work. The enthalpy of sublimation is basically a two-step process: solid -->liquid and then liquid --> vapor. The equation for deltaH(vapor)= H(vapor) - H(liquid) and the equation for deltaH(fusion)= H(liquid) - H(solid). When adding those two equations together you will be left with deltaH(sub)= H(vapor) - H(solid) because the two H(liquid)s cancel. In summary, deltaH(sub) = Hvapor - Hsolid and deltaHsub = deltaHfusion + deltaH vapor essentially mean the same thing and yield the same results.


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