Thermochemistry long problem help asap!?

Question

calculat ethe enthalpy for the following reaction:

2N2(g) + 5O2(g) ---> 2N2O5(g)

using this info:

H2(g) + (1/2)O2(g) ---> H2O(l) H = -285.8 kJ

N2O5(g) + H2O(l) ----> 2HNO3(l) H = -76.6 kJ

(1/2)N2(g) + (3/2) O2(g) + (1/2) H2(g) ---> HNO3(l) H= -174.1 kJ

answer is 28.5kJ

please explain the steps that need to be taken

Answer 1

This is what you do.  Try to inverse and add integers (2,3,4) to each of the 3 equations at the bottom to make them cancel out and look exactly at the top.  

Therefore, you want N2 and O2 on the left side, and N2O5 on the right.

So, reverse the 2nd EQ to say 2HNO3=N2O5 + H20 .  By doing this you make the number +76.6 instead of minus. Then make it 4HNO3=2N2O5 +H20 since u need 2N2O5's in the top reaction.  This makes it +153.2 (double 76.6 cuz u double all the things in the EQ)

Last equation should be times by 4 to get 2N2 + 6O2 + 2 H2 = 4HNO3.  And make it  -696.4 KJ.

Now the first equation should be times by 10 to get 10H2 + 5 O2= 10H20, and so make it   -2858 KJ

See how i changed the formulas to get exactly 2N2O5 on the right side of EQ 2.  And I got 2N2 on the left side for equation 3.  And I got 5O2 on the left side on equation one.

Now you got:

10H2 + 5 O2= 10H20                         -2858 KJ

4HNO3 = 2N2O5 +H20                         +153.2 KJ

2N2 + 6O2 + 2 H2 = 4HNO3               -696.4 KJ.

Now cancel the equations: look at 2 and 3.  There is 4HNO3 as a reactant in 2nd EQ, there is 4HNO3 as product in 3rd.  So you cancel them out.  Now you can change the formulas by changing the numbers in front, until you can cancel everything out except for the original formula 2N2(g) + 5O2(g) ---> 2N2O5(g) but make sure you change the enthalpy numbers too.

It's a long process but its easier with practice.