Let's try this again...
(posting pictures below)

Question

Let's try this again...
(posting pictures below)

danjs · Answer

k

anonymous · Answer

Here's the question...

anonymous · Answer

and answers...

danjs · Answer

ok

danjs · Answer

http://assets.openstudy.com/updates/attachments/549e1af2e4b0b8a54faf6f81-catseyeglint911-1419647868074-mc0235.jpg

danjs · Answer

See the reactants at the bottom, goes to the product with a net enthalpy change of 40.2 kJ per Mole

danjs · Answer

+ H is endothermic and -H for heat releasing Exothermic Reactions.

anonymous · Answer

Quick question: How do you calculate that it is 40.2 kJ per Mole? 
Sorry, I haven't done this type of problem in a while.

danjs · Answer

me either.. lol 

The net enthalpy change, would be +393 - 110.4 - 242.4

The diagram is set up so + is up and - is down

danjs · Answer

The overall net effect is after moving up 393 and down 110.4 and 242.4 you end up at a level of enthalpy of +40.2

anonymous · Answer

Okay I see.
How do you know which equation goes where on the chart. (like the ones on the 2nd and 3rd are flipped)

danjs · Answer

right, 

Look at the reactants, CO2 and H2, the entire process has a net Enthalpy change of +40.2 kJ / mol. 
The arrow should point from the Reactants (CO2 + H2) to the Products, (CO and H2O)

danjs · Answer

The Products enthalpy overall is at the+ 40.2 level on that diagram

danjs · Answer

think of the base line as zero

anonymous · Answer

I understand now. Thank you!