# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Topic started by: spring_in_bloom on April 02, 2004, 05:50:58 PM

Title: enthalpy change
Post by: spring_in_bloom on April 02, 2004, 05:50:58 PM
We started a new unit and i don't really understand the homework. If you could help lead me in the right direction that would be great. Thanks!

1. Calculate the standard stae enthalpy change (?H ? ) when ammonia is oxidized.
Use the following thermochemical equations.
i) i) ½ N2 (g)      + 3/2 H2 (g)    ?  NH3 (g)                       ?H ? = -45.9 KJ
ii) ½ N2 (g)    + O2 (g)  ?  NO2  (g)         ?H ? = 33.2 KJ
iii) H2  (g)  + ½ O2 (g)  ?  H2O (g)                 ?H ? = -241.8 KJ
So far I have the equation when ammonia is oxidized being:
NH 3 (g)    + 7/4 O2 (g)       ?  NO 2  (g)   +  3/2   H2  (g)

Thanks so much is you are able to help me.
Title: Re:enthalpy change
Post by: jdurg on April 02, 2004, 06:36:37 PM
This one's pretty easy, and once you get into the discussion of enthalpy a bit more you'll find that it's not too bad.   ;D  Enthalpy is the "heat" associated with a reaction, and although we can't measure it directly we can measure the changes.  An important thing to remember is that the enthalpy of formation for a pure element in its standard state is zero.  This will be very important later on.

To calculate the enthalpy of a reaction, you simply take the sum of the enthalpy of formation of the products and subtract the sum of the enthalpy of formation of the reactants.

?Hrxn = ?Hf(products) - ?Hf(reactants).

The next thing you'll need is the proper reaction of what you're investigating.  Since in your example they want you to determine the oxidation of ammonia based upon the equations they gave you, you simply have to rearrange the three equations into one equation.  The reaction you have is almost there, but it's missing the water that's given in equation number 3.  I think what you'll need for this question is:

NH3 (g) + 3/2 O2 (g) ? 1/2 H2 (g) + NO2 (g) + H2O (g)

Now, you just take the ?Hfs of the products and subtract the ?Hfs of the reactant; the ammonia.  I won't do the math for you, but it should be pretty easy to figure out.  Let me know what you come up with.   ;D