Change in Entropy for the Surroundings vs. System

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PeterTran1C
Posts: 30
Joined: Thu Jul 13, 2017 3:00 am

Change in Entropy for the Surroundings vs. System

Postby PeterTran1C » Wed Feb 14, 2018 11:54 pm

Is there a conceptual reason why is negative compared to which is a positive value?

Joshua Xian 1D
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am
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Re: Change in Entropy for the Surroundings vs. System

Postby Joshua Xian 1D » Thu Feb 15, 2018 12:10 am

∆S(system) = ∆H(system)/T since (assuming constant pressure) ∆H(system)=q(system), and if heat were added to the system, then entropy would increase. (+q(system) --> +∆S(system)). ∆S(surroundings) = -∆H(system)/T since if heat from the surroundings were added to the system, then the entropy of the surroundings would decrease (+q(system) --> -∆S(surroundings)).

mayapartha_1D
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: Change in Entropy for the Surroundings vs. System

Postby mayapartha_1D » Mon Feb 19, 2018 3:16 pm

The equations are opposite for the system vs the surroundings because heat is being transferred from one to the other, so the entropies must be opposite values to reflect this.


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