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Sarah Rutzick 1L
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Postby Sarah Rutzick 1L » Sat Feb 10, 2018 8:06 pm

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?

Ammar Amjad 1L
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Re: 9.15

Postby Ammar Amjad 1L » Sat Feb 10, 2018 8:36 pm

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

Postby Rachel Formaker 1E » Sat Feb 10, 2018 9:26 pm

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

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Re: 9.15

Postby Manvir2K » Sun Feb 11, 2018 2:36 pm

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

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