## 9.15

$\Delta S = \frac{q_{rev}}{T}$

Sarah Rutzick 1L
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Joined: Tue Oct 10, 2017 7:13 am

### 9.15

For 9.15 part a, the answer book gives the heat of fusion as -6.01 kJ/mol. How do we know that the heat of fusion should be negative?

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### Re: 9.15

The book gives the heat of fusion of water as 6.01 kj.mol^-1, meaning thats how much energy is required to go from solid to liquid water. However, the question is asking you to calculate the entropy for the freezing of water; therefore, you would use negative heat of fusion, because that's how much energy the system will release.

Rachel Formaker 1E
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### Re: 9.15

Heat is always required to melt a substance, so ∆H is positive.

Heat is released when a substance freezes, so ∆H is negative.

∆Hfreezing = ∆Hfusion, and since you are given ∆Hfusion from Table 8.3, you just reverse the sign of ∆Hfusion to get ∆Hfreezing = -6.01 kJ/mol

Manvir2K
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: 9.15

How do you get -6.01 kJ/mol or 4.35 kJ/mol from appendix 2A?

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