9.15a  [ENDORSED]


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Grace Ramey 2K
Posts: 57
Joined: Thu Jul 27, 2017 3:01 am

9.15a

Postby Grace Ramey 2K » Mon Jan 29, 2018 10:04 am

9.15
Use data in Table 8.3 or appendix 2A to calculate the entropy change for a) the freezing of 1.00 mol H20 (l) at 0 degrees C.

I know in class we talked about delta S = (q(rev))/T) which in this case would equal ((delta H fus)/ T(melting)). So I understand where the solutions manual got (6.01 kJ/mol)(1 mol)/273 K from, but I am not sure why the actual calculation is (-6.01 kJ/mol)(1mol)/273 K. Can someone please explain where the negative sign comes from in "-6.01 kJ/mol"?

Thank you for your help!

Janet Nguyen 2H
Posts: 51
Joined: Fri Sep 29, 2017 7:05 am

Re: 9.15a

Postby Janet Nguyen 2H » Mon Jan 29, 2018 10:49 am

delta H fus is 6.01 for melting so for freezing you would have to reverse the sign and make it -6.01 kJ/mol!

RyanS2J
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

Re: 9.15a  [ENDORSED]

Postby RyanS2J » Mon Jan 29, 2018 10:51 am

Since freezing is an exothermic process (heat is released from the system), the deltaH in the equation would take a negative sign. Also, as in Table 8.3, the standard enthalpy of melting is +6.01 kJ/mol, so since you are concerned with the reverse reaction (freezing), you would flip the sign and make it negative.


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