HW 8.67 b)  [ENDORSED]

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AshleyAguilar_1F
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HW 8.67 b)

Postby AshleyAguilar_1F » Fri Jan 20, 2017 2:19 pm

I thought that the formation of methanol was CO + 2H2. While it makes sense to me why it is actually set up C (gr) + 2H2 (g) + 1/2O2, I don't understand how we are supposed to know that this is the way it's done for this particular problem. CO is bonded by a triple bond, which is a mean bond enthalpy not given in the book so is that the reason why? Is it because it's always best to use the most stable form of the element?

Katelyn Li 2J
Posts: 27
Joined: Wed Sep 21, 2016 3:00 pm

Re: HW 8.67 b)  [ENDORSED]

Postby Katelyn Li 2J » Fri Jan 20, 2017 5:52 pm

When doing problems regarding the standard enthalpy of a substance, the reactants are in their pure elemental forms. So although the reaction you used may be a possibility, given the context of the question, you would not want to use CO because it is a compound.


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