## ∆Hsub=∆Hfus+∆Hvap

Beatrice Petelo 1F
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

### ∆Hsub=∆Hfus+∆Hvap

How does ∆Hsub=∆Hfus+∆Hvap work?
Thanks!

Sarah Kiamanesh 1D
Posts: 30
Joined: Fri Sep 28, 2018 12:22 am

### Re: ∆Hsub=∆Hfus+∆Hvap

Sublimation is the phase change from solid to vapor. Fusion is the transition from solid to liquid, and vaporization is the transition from liquid to gas. Because these transitions are state properties, their enthalpies can be added together. The enthalpy of fusion added to the enthalpy of vaporization equals the enthalpy of sublimation, because these two processes are intermediary steps from solid to vapor.

Angela Grant 1D
Posts: 67
Joined: Fri Sep 28, 2018 12:25 am

### Re: ∆Hsub=∆Hfus+∆Hvap

∆Hfus = Hliquid- Hsolid
∆Hvap = Hvapor - Hliquid
∆Hsub = Hvapor - Hsolid
Since sublimation goes straight from a solid to a vapor state, it'll have the same total change as the individual changes in fusion and vaporization put together. Therefore, you can combine them all to get ∆Hsub=∆Hfus+∆Hvap
Hope this helped!

Casandra
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am

### Re: ∆Hsub=∆Hfus+∆Hvap

Beatrice Petelo 1F wrote:How does ∆Hsub=∆Hfus+∆Hvap work?
Thanks!

Because Enthalpy is a state function, its properties can be added or subtracted. We know that ∆Hvap= Hvapor-Hliguid and that ∆Hfus=Hliquid-Hsolid. For sublimation we are going from a solid to a liquid and a liquid to a vapor, which is two steps, therefore we add the H of fusion(solid to a liquid) and the H of vaporization ( liquid to a vapor) to give us ∆Hsub=∆Hfus+∆Hvap.