Enthalpy...when to multiply by coefficients?

[bchang57]

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Thought I had this but then a question came up from "Practice Source X" and it went like this...

Calculate the enthalpy change for the following:
2Fe(s) + 3H2O(g) --> Fe2O3(s) + 3H2(g)

1] Fe(s) + H2O(g) --> FeO(s) + H2(g) H= -24.7 kJ
2] 2FeO + 1/2O2(g) --> Fe2O3 H= -289.2 kJ
3] H2(g) + 1/2O2(g) --> H2O(g) H= -241.8 kJ

Answer:
Eq. 1] keep sign, multiply by 2. -49.4 kJ
Eq. 2] keep sign, value does not change. -289.2 kJ
Eq. 3] flip sign, value DOES NOT change. 241.8 kJ

This gives you the answer of -96.8 kJ (if you do the math)


Why did the text not multiply Eq.3 by 3 (which is what I did, and ended up getting none of the answer choices) even though it multiplied Eq.1 by 2??

 

[Dotoday]

Because you already have 2 H2O and 2 H2 from the first equation. If you do not multiply the last one that would make 3 H2O and 3 H2, just what the overall reaction needs. If you would have multiplies the last equation by 3, then this would bring thelecture values of water and H2 up to 5. That would have been a wrong answer. Hope this helps.

Sent from my MB855 using Tapatalk

 

[ushaseos]

Exactly. The third equation isn't multiplied by 3 it is flipped so its H2O -> H2 + 1/2 O2 that way you get the 3 H2O (the other 2 come from 2x the first equation).

This is why the third isn't multiplied by 3, but instead multiplied by -1 (or sign changed!)

 

[bchang57]

oh man...i feel like... /facepalm