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For the melting of Benzene, since it is melting, it feels intuitive to think that the change in entropy will be positive (more disorder) and enthalpy will be positive (exothermic). However, when I look in my textbook, it says that the sign for delta S will be negative. Why is this? Wouldn't any process of melting from solid to liquid have a positive change in entropy? Here is the original question: "The normal melting point of benzene, C6 H6 , is 5.5°C. For the process of melting, what is the sign of each of the following? (a) ∆H°, (b) ∆S°"

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It would seem that either your textbook is mistaken or that you are interpreting your textbook incorrectly, for benzene has a positive entropy of fusion according to the NIST Chemistry WebBook.

Most substances will indeed have a positive entropy of fusion, as is suggested by your reasoning above. One notable exception, however, is helium at near-zero Kelvin temperatures, where it has a negative entropy of fusion. Presumably this is a quantum-mechanical effect, though I have no further explanation as to why this would be the case.

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  • $\begingroup$ I'm going to assume that I'm not smarter than my textbook..lol. Here is the original question. "The normal melting point of benzene, C6 H6 , is 5.5°C. For the process of melting, what is the sign of each of the following? (a) ∆H°, (b) ∆S°" The answer says a) positive b)negative. Am I reading something wrong here? $\endgroup$ – Joe Nov 20 '17 at 6:07
  • $\begingroup$ @Joe, it seems to me that your textbook is incorrect, though I could be missing something important---for example, they ask for ∆S° where I would have expected ∆S. I recommend that you update your original post with the question statement and see if someone else responds. $\endgroup$ – a-cyclohexane-molecule Nov 20 '17 at 6:58

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