Hello, for part b and c of this question, it asks (b) the melting of
1.00 mol C2H5OH(s) at its melting point; (c) the freezing of
1.00 mol C2H5OH(l) at its freezing point.
These two are calculated in the exact same way but I was wondering if there are any conceptual differences I should be aware of? Is there a significance in these two being the same value?
Textbook 4I.7
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 124
- Joined: Wed Sep 30, 2020 9:35 pm
- Been upvoted: 6 times
Re: Textbook 4I.7
These two are calculated in the same way but you have to remember to change the sign of the delta H when you freeze something. When you freeze ethanol at its freezing point, you are allowing it to lose heat meaning delta H will be negative. Besides that, conceptually just be careful of the sign of delta H fusion when you calculate the melting or freezing of a substance by recognizing whether the system will give off heat or absorb heat for the phase change.
-
- Posts: 92
- Joined: Wed Sep 30, 2020 9:59 pm
Re: Textbook 4I.7
The phase changes are tricky in this problem. If there is more heat required to achieve a specific phase or state, then the enthalpy will increase, such as solid to liquid. But if the phase needs to cool down, it needs to release heat and thus will have a negative enthalpy change, -q. For example, a gas condensing to a liquid.
Return to “Concepts & Calculations Using Second Law of Thermodynamics”
Who is online
Users browsing this forum: No registered users and 2 guests