[the_zom]

So I have a question in this text book I'm working from and I'm not sure I understand how to approach it.

"The fertilizer urea is produced along with liquid water by the reaction of ammonia and carbon dioxide. Calculate the standard molar enthalpy of reaction for ammonia."

So this is what I'm thinking...

1) Write a balanced equation (2NH3 + CO2 --> CH4N2O + H2O)

2) From a enthalpy of formation table determine that Urea's enthalpy of formation is -333.5 kJ/mol and water's is -285.8 kJ/mol (a total of -619.3 kJ/mol)

3) From the same table determine the enthalpy of formation of carbon dioxide, subtract it from the -619.3 kJ/mol and divide the remainder in two to determine the value for ammonia?

[rabolisk]

The question is asking for the enthalpy of reaction for one mole of ammonia. You have to use the standard enthalpy of formation of all 4 species to get to that.

[the_zom]

I'm not sure I follow.  The textbook is a real piece of crap and I'm doing this chem course by mail.

Could you maybe expand on that and give me general idea of how to tackle this?

[the_zom]

 So these are the values I'm working with.

 :delta: H_f Urea = -333.5 kJ/mol
 :delta: H_f Water = -285.8 kJ/mol
 :delta: H_f Ammonia = -45.9 kJ/mol
 :delta: H_f Carbon Dioxide = -393.5 kJ/mol

I have 2NH₃ + CO₂ --> CH₄N₂O + H₂O

 :delta: H_r will be equal to [(-333.5 kJ/mol) + (285.8 kJ/mol)] - [(2 x -45.9 kJ/mol) + (-393.5 kJ/mol)] = -134 kJ


I just simply do not understand how to obtain  :delta: H_r for ammonia given the information I have.