Textbook 4I.7

[Anna Yakura 2F]

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?

[sophie esherick 3H]

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.

[edward_brodell_2I]

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.