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Thread: Heat of formation question

07022010, 12:22 AM #1
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 Mar 2010
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Heat of formation question
I learned from the chad video to calculate the heat of formation in easy way. For example,
Q. given that the delta H of the following reaction is 92.2 kj/mol, what is the heat of formation of ammoni a gas?
2NH3 > N2 (g) + 3H2 (g)  92kj/mol
N2 (g) + 3H2 > 2NH3 92kj/mole, so heat of formation of ammonia gas is 92/2 = 46kj/mole
Today, I have faced with similar problem and I solve it with same manner as above example, but I got it wrong. The question was
Q. The combustion of ammonia is represented by this equation.
4NH3 (g) + 5O2 (g) > 4NO (g) + 6H2O (g) Delta H = 904.8KJ
(Enthalpy of formation data, NO(g) = 90 KJ/mol, H2O (g) = 241.8kj/mol)
What is the enthalpy of formation of NH3(g)
I solved it with exactly same nanner.
4NO + 6H2O  4NH3 + 5O2, Delta H = (+) 904.8
divided by 4 = 227, but the correct answer is 46 KJ per mole
I figured out how to reach the correct answer, but I don't know why my calculation doesn't get the correct answer?

07022010, 10:33 AM #2
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 Jun 2007
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 1,871
The first question is an application of Hess' Law.
You first need to figure out what you're being asked. The heat of formation of ammonia corresponds to forming 1 mole of NH3 from it's individual elements in their standard states leaving us with: 1/2N2(g) + 3/2H2(g) > NH3(g)
We can easily see that this is similar the rxn provided and all we need to do is reverse the one provided and cut it in half to yield the above formation rxn.
But the 2nd question is a little different based upon the data they supply you with. Again you first need to figure out what you're being asked. The heat of formation of ammonia still corresponds to forming 1 mole of NH3 from it's individual elements in their standard states leaving us with: 1/2N2(g) + 3/2H2(g) > NH3(g)
Simply dividing the provided rxn by 4 doesn't get us this formation rxn for NH3. So we'll have to take a slightly different approach. Here we'll use the fact that the deltaH rxn = sum (delta H,products)  sum (delta H, reactants) and so:
904.8 = [4(x)+5(0)][4(90)+6(241.8)]
Solving for x gives 46.5kJ.
Hope this helps.

08062010, 04:13 PM #3PreDental
 Join Date
 Aug 2010
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 3
Chad, I hope you don't mind if I follow up on your answer to this poster's question because I'm confused about it and am also doing heats of formation in my gen chem II course.
I don't understand how the math works out on the last calculation.
If I substitute in your value of 46.5 for every instance of x:
904.8 = [4(46.5)+5(0)][360+(1450.8)]
904.8 = [186][1090.8]
904.8 = [186]+[1090.8]
904.8 ≠ 904.8
Should the enthalpy of formation of NH3 be 498.9 instead of 46.5?
I [think] my algebra is correct so I wonder if I'm misunderstanding the problem setup.
Thanks.

08062010, 06:24 PM #4
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 Jun 2007
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 1,871
Hey Jake,
Don't mind at all...especially when I've done something wrong (I lost sight of the original question). To set the record straight, the question is:
Q. The combustion of ammonia is represented by this equation.
4NH3 (g) + 5O2 (g) > 4NO (g) + 6H2O (g) Delta H = 904.8KJ
(Enthalpy of formation data, NO(g) = 90 KJ/mol, H2O (g) = 241.8kj/mol)
What is the enthalpy of formation of NH3(g)
904.8 = [4(90)+6(241.8)][4(x)+5(0)]
Solving for x does indeed yield 46.5kJ
In the original thread I'd set up the calculation as reactants minus products whereas it should have been products minus reactants.
Thanks for catching my error and hopefully this clears this one up.Last edited by chad; 08062010 at 08:07 PM.

08062010, 07:00 PM #5PreDental
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 Aug 2010
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 3
Thanks Chad! Just to clarify though, x should be 46.5 kJ right, because it's exothermic? You had that in the initial answer.
Thanks for the VESPR video also, it is great. You're much better at explaining it than my prof is.
Later,
Jake

08062010, 08:09 PM #6
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 Jun 2007
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 1,871
Yes, you caught me again. I corrected it in the post above.