2010 Final #2A RXN Enthalpy

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RaquelAvalos1K
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Joined: Fri Sep 26, 2014 2:02 pm

2010 Final #2A RXN Enthalpy

Postby RaquelAvalos1K » Fri Mar 13, 2015 6:41 pm

Hey Everyone!

Why was the standard enthalpy for 3BaO(s)-->3Ba(s)+3/2O2 deltaH=1660.5 kJ
when all we changed originally was flipping the equation, so only the sign should change?


Thank you!!

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: 2010 Final #2A RXN Enthalpy

Postby Neil DSilva 1L » Fri Mar 13, 2015 6:56 pm

I don't think you labeled the problem correctly. I found this in my course reader in the 2010 Midterm, Q2B (you may want to edit your title).

To answer your question: you're right that you have to flip the original equation that you have, but notice the coefficients are different. The original equation has a coefficient of 2 on the Ba(s) and the reaction we need has a coefficient of 3 on the Ba(s). You have to multiply the standard reaction enthalpy by to get the correct reaction enthalpy for the adjusted reaction.

Enthalpy is an extensive property, so we need to account for the change in coefficients.

Chem_Mod
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Re: 2010 Final #2A RXN Enthalpy

Postby Chem_Mod » Fri Mar 13, 2015 8:56 pm

Correct!


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