Hess¡¯ law says that when we add reactions together we can also add the enthalpies. Given the following set of reactions, use Hess¡¯ Law to calculate the enthalpy of the reaction given below.

Question

N2H4 + H2 ¡æ 2 NH3 ¥ÄH¡Æ = ?
N2 + 3 H2 ¡æ 2 NH3 ¥ÄH¡Æ = -92.38 kJ/mol
N2H4 + O2 ¡æ N2 + 2 H2O ¥ÄH¡Æ = -622.09 kJ/mol
2 H2 + O2 ¡æ 2 H2O ¥ÄH¡Æ = -483.6 kJ/mol

Answer 1

There are symbols which I don't understand in this problem that you typed, like these ¡æ and ¥ÄH¡Æ. But I understand the reactions. :3

We need to find the ΔH for this reaction:
N2H4 + H2 ---> 2 NH3

using the data for these reactions:
N2 + 3 H2 ---> 2 NH3 ; ΔH = -92.38 kJ/mol
N2H4 + O2 ---> N2 + 2 H2O ; ΔH = -622.09 kJ/mol
2 H2 + O2 ---> 2 H2O ; ΔH = -483.6 kJ/mol

What we'll do is reverse the third reaction. By doing this, we change the sign of its ΔH to its opposite. And then we add all the reactions, as well as the ΔH's.
N2 + 3 H2 ---> 2 NH3 ; ΔH = -92.38 kJ/mol
N2H4 + O2 ---> N2 + 2 H2O ; ΔH = -622.09 kJ/mol
2 H2O ---> 2 H2 + O2 ; ΔH = +483.6 kJ/mol
---------------------------------------------------------
N2H4 + H2 ---> 2 NH3 ; ΔH = 230.87 kJ/mol

hope this helps~ `u`

Answer 2

jae means ==>
i followed by upper case _ is '
That Ything is delta H = ?