Question 9.19


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Question 9.19

Postby 404975170 » Sat Mar 16, 2019 7:14 pm

In the question you are calculating the standard entropy of vaporization of water at 85 degrees C, given that it’s standard entropy of vaporization is 100 degreees C.
The three steps you sum together are:
1. the entropy change heat the reactants to 100 C
2. the entropy of vaporization of h2o at 100 C
3. cooling the products to 85 C

Basically, I need clarification of what is happening to the water so I can better visualize what is going on and understand why these are the steps because I don’t get if the water is starting at 85 then returning to 85 or starting at a different temperature then going to 100 then to 85?

Ashish Verma 2I
Posts: 59
Joined: Fri Sep 28, 2018 12:28 am

Re: Question 9.19

Postby Ashish Verma 2I » Sat Mar 16, 2019 7:42 pm

I believe you are heating the liquid water to its boiling point, vaporizing the liquid water at its boiling point into a gas, then cooling the gaseous to the desired temperature, while adding together the entropy for each of these separate stages. Though I can't be sure as I can't seem to find the 9.19 question you are referring to.

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