Discussion:
Formation Enthalpy of Urea
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n***@gmail.com
2005-12-13 00:48:10 UTC
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I couldn't find much on google, so if any of you guys would like to
dirty your hands I would greatly appreciate it. :)

My textbook gives the following info:

2*NH3(g) + CO2(g) --> CO(NH2)2(g) + H20(l)
The reaction is exothermic == 133.5kJ

So, I am supposed to find the enthalpy of formation for urea, I have
the rest of the enthalpys on a chart. So I end up getting:

Delta H(r) = Delta H(f) Products - Delta H(f) reactants

X kJ(CO(NH2)2) + (-285.5kJ)(H2O) - (2*(-46kJ))(2*NH3) - (-393.5kJ)(CO2)
= (-133.5kJ)

X + 200kJ = (-133.5kJ)

X = (-333.5kJ/mol)

Unfortunatly the textbook gives me (-533.5 Kj/mol) our answers differ
by 200kJ meaning my calculation is wrong by 2x... I don't get it.
Anyone have any insight where I am messing up?
Ron Jones
2005-12-13 01:15:03 UTC
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Post by n***@gmail.com
I couldn't find much on google, so if any of you guys would like to
dirty your hands I would greatly appreciate it. :)
2*NH3(g) + CO2(g) --> CO(NH2)2(g) + H20(l)
The reaction is exothermic == 133.5kJ
So, I am supposed to find the enthalpy of formation for urea, I have
Delta H(r) = Delta H(f) Products - Delta H(f) reactants
X kJ(CO(NH2)2) + (-285.5kJ)(H2O) - (2*(-46kJ))(2*NH3) -
(-393.5kJ)(CO2) = (-133.5kJ)
X + 200kJ = (-133.5kJ)
X = (-333.5kJ/mol)
Unfortunatly the textbook gives me (-533.5 Kj/mol) our answers differ
by 200kJ meaning my calculation is wrong by 2x... I don't get it.
Anyone have any insight where I am messing up?
Possibly nowhere. Misprint in book? Wouldn't be the first...
http://www.chemistry.oregonstate.edu/courses/ch121-3s/ch121/2004/Nafshun%20Worksheet%201.doc
= "The heat of formation of urea, CO(NH2)2 (s) is -333.19 kJ/mol"
--
--
Ron Jones

Don't repeat history, see unreported near misses in chemical lab/plant
at http://www.crhf.org.uk
Only two things are certain: The universe and human stupidity; and I'm
not certain about the universe. ~ Albert Einstein
n***@gmail.com
2005-12-13 01:23:09 UTC
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Thanks for the reply all.

With the amount those bloody books cost you'd think they would
proofread them...
t***@gmail.com
2005-12-13 01:36:01 UTC
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Post by n***@gmail.com
With the amount those bloody books cost you'd think
they would proofread them...
They do proofread them, but in a bloody book that size,
errors are inevitable. How well would you do proofreading
a sizable table of formation enthalpies and related thermodynamic
functions?
G. R. L. Cowan
2005-12-13 01:34:12 UTC
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Post by n***@gmail.com
Thanks for the reply all.
With the amount those bloody books cost you'd think they would
proofread them...
Up to 3 percent of proofreaders don't know the delta 'H' sub f
of urea ...


--- Graham Cowan, former Ontario, Canada hydrogen fan
http://www.eagle.ca/~gcowan/Paper_for_11th_CHC.html
boron as energy carrier: real-car range, nuclear cachet
t***@gmail.com
2005-12-13 01:15:23 UTC
Permalink
Post by n***@gmail.com
X = (-333.5kJ/mol)
Unfortunatly the textbook gives me (-533.5 Kj/mol) our
answers differ by 200kJ meaning my calculation is wrong
by 2x... I don't get it.
Anyone have any insight where I am messing up?
If your number is exactly correct except for the first digit,
think "typo".

According to Atkins, the enthalpy of formation of urea is
indeed -333.51 kJ/mol.
r***@gmail.com
2005-12-13 11:33:02 UTC
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Post by n***@gmail.com
2*NH3(g) + CO2(g) --> CO(NH2)2(g) + H20(l)
The reaction is exothermic == 133.5kJ
Redo your calculation. I get -333.35 kJ/mol. Your error is doubling
up on the 2*(-46) for ammonia, you're doing 4*(-46) hence the ~200
kJ/mol difference.

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