## Homework Help on 8.67

John Huang 1G
Posts: 46
Joined: Thu Jul 13, 2017 3:00 am
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### Homework Help on 8.67

8.67) Use the information in Tables 8.3, 8.6, 8.7 to estimate the enthalpy of formation of each of the following compounds in the liquid state. The standard enthalpy of sublimation of carbon is +717 kj/mol.
(a)H20
(b) Methanol, CH3OH
(c) benzene, C6H6 (without resonance)
(d) benzene, C6H6 (with resonance)

The solution manual says to find the enthalpy of formation in the liquid state of a compound:

(Enthalpy of Formation of Liquid Substance) = (Enthalpy of Formation of Gas Substance) - (Enthalpy of Vaporization of Substance)

Can someone explain to me how this equation works? Why don't we add instead?

Jared Smith 1E
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### Re: Homework Help on 8.67

I think the reason that the enthalpy of vaporization is subtracted is due to the fact that changing from a gas to a liquid is an exothermic process, meaning that it would be the negative version of the enthalpy of vaporization. The fact that it is an exothermic process determines that it is subtracted not added.

Pooja Nair 1C
Posts: 55
Joined: Thu Jul 13, 2017 3:00 am

### Re: Homework Help on 8.67

Another way to think of it is thinking of deriving the enthalpy of formation of gas. To do so, we would add (enthalpy of formation of liquid) + (enthalpy of vaporization) = (enthalpy of formation of gas). Therefore, to find the enthalpy of formation of liquid, you would have to subtract: (enthalpy of formation of gas) - (enthalpy of vaporization)