## Homework: Chapter 8, #67

megan_thi_2j
Posts: 10
Joined: Wed Sep 21, 2016 2:55 pm

### Homework: Chapter 8, #67

We discussed this question in my discussion section, and I noticed that we used the enthalpy of vaporization and enthalpy of condensation of water to solve the problem. The enthalpy of vaporization and enthalpy of condensation were the same number, just different signs, i.e. vaporization = 44.0 kJ/mol & condensation = -44.0 kJ/mol. Is it like this for all compounds besides water?

E_Villavicencio 2N
Posts: 26
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Homework: Chapter 8, #67

Yes, this happens with all the compounds. The enthalpy of reverse processes only differs in the sign. In this case, the process of vaporization (from liquid to gas) requires heat, that's why the sign is positive, whereas condensation, which is the reverse process (from gas to liquid), releases heat, hence the negative sign. Because enthalpy is a state function (doesn't depend on the path, only on the initial and final states), the value will be the same, just with a change in sign. The same will happen with fusion and freezing. Hope it helps!

Jonathan Sarquiz 3F
Posts: 32
Joined: Fri Jul 15, 2016 3:00 am

### Re: Homework: Chapter 8, #67

You are correct! They are equal and opposite values. The amount of heat it takes to go from a solid to a liquid is the same amount of heat released when going from a liquid to a solid. So if for a specific element $s\rightarrow l$ = +10 kJ/mol (it is positive because it is absorbing heat), then for that same element to go from $l\rightarrow s$ = -10kJ/mol (it is negative because it is releasing heat).

To summarize:
$\Delta H$ $s\rightarrow l$ = - $\Delta H$ $l\rightarrow s$
$\Delta H$ $l\rightarrow g$ = - $\Delta H$ $g\rightarrow l$

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

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