# Enthalpy using Hess's law (Chem)...

• October 26th 2010, 11:37 AM
butters
Enthalpy using Hess's law (Chem)...
This is the question:http://img831.imageshack.us/img831/4723/1004108.jpg

Now, I've gotten as far as this:http://img24.imageshack.us/img24/5396/1004109k.jpg

Don't know if it's right or wrong, but solve and post all the steps..
• November 4th 2010, 10:22 AM
Unknown008
Not all the equations show up... =/

But I suppose that's what are given and this is how I work them out:

http://p1cture.me/images/05962384827440712275.png

You are already given $\Delta H_1$ which is -91.8\ kJ/mol. (and this I think is the enthalpy of reaction and not formation and hence, you do not divide by 2)

Then, $\Delta H_2 = (3\times -241.8)\ kJ/mol$ (since you have 3 mol of water being formed from my Hess Cycle)

Then, $\Delta H_3 = (2\times -906.2)\ kJ/mol$ (I think they gave you the standard enthalpy of combustion of ammonia. So, for each mol of ammonia burnt, you get 906.2 kJ but since you burn 2 moles, it's twice)

Then, $\Delta H_4 = ? kJ/mol$ This is what you need to find I think.

Then, from Hess' Law, we find that by following the arrows:

$\Delta H_2 + \Delta H_4 = \Delta H_1 + \Delta H_3$

This will give you the answer.