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From my understanding, the second law of thermodynamics states that entropy for a reversible reaction can not decrease and will remain constant between the system and the surroundings. What I didn't understand for the Midterm question was how that differs for an irreversible reaction? Do we disregard the 2nd law because it is irreversable or is it that any entropy gained or loss is constant and does not add up to 0 after combining it with the system and the surroundings? Thank you so much!
Because an irreversible reaction is spontaneous, we know that, according to the 2nd law of thermodynamics, the total entropy in the universe must increase. This tells us that the change in total entropy must be greater than 0.
In the midterm question the change in entropy of the system was not equal to the change in entropy of the surroundings in the irreversible reaction because the reaction occurred in a vacuum and if there are no molecules in the surroundings then you can't change their entropy. I'm not exactly sure how this applies to the second law of thermodynamics but that is why the total change in entropy was not equal to zero but rather positive.
Were no calculations required for the 2nd part? I understood the difference but I went about calculating the entropy different in each part because I knew one was reversible and the other was irreversible. How is it ok to not have to do another calculation when it is specified otherwise in different situations?
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