2010 Final #2A RXN Enthalpy

RaquelAvalos1K
Posts: 35
Joined: Fri Sep 26, 2014 2:02 pm

2010 Final #2A RXN Enthalpy

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

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 $- \frac{3}{2}$ 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

Correct!

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