TBR chem example 8.11

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altered states

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Hey guys idk if im missing something here but i thought that when calculating standard enthalpy change and using heat of formation, when a reactant has a negative heat of formation in order to properly sum all of the heat of formations you are supposed to reverse it so that it is positive...idk if that just made sense lol but here is the example in the book:

What is the standard enthalpy change for the water shift reaction?
CO2(g) + H2(g)-------> CO(g) + H2O(g)

Here are the pertinent Hf:
CO(g): -111.2
H2O(g): -242
CO2(g): -393.5

So basically what I was thinking is that since CO2 is on the reactant side, it's Hf should be taken as 393.5 instead of -393.5?

But in the solution they keep it negative.

Any help clearing this up would be appreciated!
 
Hey guys idk if im missing something here but i thought that when calculating standard enthalpy change and using heat of formation, when a reactant has a negative heat of formation in order to properly sum all of the heat of formations you are supposed to reverse it so that it is positive...idk if that just made sense lol but here is the example in the book:

What is the standard enthalpy change for the water shift reaction?
CO2(g) + H2(g)-------> CO(g) + H2O(g)

Here are the pertinent Hf:
CO(g): -111.2
H2O(g): -242
CO2(g): -393.5

So basically what I was thinking is that since CO2 is on the reactant side, it's Hf should be taken as 393.5 instead of -393.5?

But in the solution they keep it negative.

Any help clearing this up would be appreciated!
Why would you do that since it's already given to you...Because CO2 is in the reactant, that does does not mean its Hf has to be positive.
http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation
Scroll down in that link, you will see that the Hf for CO2 is exatly -393.5
 
oh okay thank you for your response, for some reason that just seemed too straightforward for me lol but I think I got it all cleared up!
 
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