jh311

10+ Year Member
Jan 20, 2009
37
0
utah
Status
Pre-Dental
Using Hess's Law, what is the ▲H rxn for:

2C(s) + O2(g) >>>>> 2CO(g)

C(s) + O2 >>>>> CO2(g) ▲H= -393.7KJ/mol
CO(g) + ½O2 >>> CO2(g) ▲H= -283.3KJ/mol

When I did this problem I got an answer that was way off apparently. By using ▲H = Σ▲H(products) - Σ▲H(rxtnts) Shouldn't you write(?):
-((2(-283)) - 2(-393)? This would give me +566 -(-786) = +1352KJ. I was really confused to see that the answer was -220KJ and the way that it was set up seemed opposite to me. The answer said 2(-393KJ)-(-2(283KJ)) = -220KJ. To me it looks like the ▲H = reactants - products. Anyways, for this particular problem there's something i'm not seeing... please help!! Thanks
 

doc1986

10+ Year Member
Mar 2, 2009
134
0
Status
Pre-Dental
First of all, Hess's Law states that enthalpies of reactions are additive. Therefore, after you manipulate the reactions to look like the reaction that you want, you add them, not subtract.

So, the first reaction should be multiplied by 2 and the second should be flipped (multiply by -1) and multiplied by 2. When you do this to both equations you will see that the 2CO2's will cancel out and 2O2 and O2 will cancel to become 1O2 on reactant side. You will also still have 2C on the reactant side to give you a net reaction of:
2C + O2 --> 2CO which is what you want.

So, now adding what you manipulated you get:
(2(-393.7)) + (2(283.3)) = -220.8 kJ

Hope this helps.
 
OP
J

jh311

10+ Year Member
Jan 20, 2009
37
0
utah
Status
Pre-Dental
Thanks!! That helps a lot. I was using the wrong equation for the problem and didn't quite understand that Hess's Law is a summation of the ▲H. I understand it now, thank you!!
 

gatorPredent

10+ Year Member
Nov 9, 2008
111
0
Florida
Status
Pre-Dental
so which problems do you do (delta H of the products - delta H of the reactants) ?
Thanks