4C.3

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Alicia Lin 2F
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Joined: Wed Sep 18, 2019 12:17 am

4C.3

Postby Alicia Lin 2F » Mon Jan 27, 2020 8:57 pm

This question has 2 parts but the answer key only gives one answer. What did you guys get for these parts?

Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am

Re: 4C.3

Postby Eileen Si 1G » Mon Jan 27, 2020 9:34 pm

For part a, I believe it's 343K for the final temperature and 935.1J for the change in enthalpy, and part b it's 374K for the final temperature and 947.6J for the change in enthalpy.

Sara Richmond 2K
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

Re: 4C.3

Postby Sara Richmond 2K » Tue Jan 28, 2020 12:27 pm

For part B how do we calculate the change in enthalpy? It is not included in the answer key

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

Re: 4C.3

Postby Morgan Carrington 2H » Thu Jan 30, 2020 1:46 am

Sara Richmond 2K wrote:For part B how do we calculate the change in enthalpy? It is not included in the answer key


The steps I followed were almost exactly the same as the ones shown n example 4C.1 on page 266 in the book. This explains how to find both the final temperature and the change in enthalpy.

Edmund Zhi 2B
Posts: 118
Joined: Sat Jul 20, 2019 12:16 am

Re: 4C.3

Postby Edmund Zhi 2B » Sat Feb 01, 2020 4:05 pm

Eileen Si 1G wrote:For part a, I believe it's 343K for the final temperature and 935.1J for the change in enthalpy, and part b it's 374K for the final temperature and 947.6J for the change in enthalpy.


Wouldn't the change in enthalpy just be 765J because in example 4C.1 on page 266, the solution states that "the enthalpy change at constant pressure is equal t the heat supplied," so deltaH = qp

Also, I'm getting 1275J for part (b), I'm doing deltaH = 765 + (0.82 * 8.3145 * 74.813) = 1275J


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