N2H4(g) + H2(g) ---> 2 NH3(g) H1 = –1876 kJ
3 H2(g) + N2(g) ---> 2 NH3(g) H2 = –922 kJ
The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be _________ kJ/ mol.
Use Hess' Law.
To find the value of ΔHf for the formation of hydrazine (N2H4), we need to consider the given equations for the reactions of hydrazine formation and determine the appropriate combination of those equations to yield the formation reaction.
Given reactions:
1) N2H4(g) + H2(g) → 2 NH3(g) ΔH1 = -1876 kJ
2) 3 H2(g) + N2(g) → 2 NH3(g) ΔH2 = -922 kJ
We can see that in reaction 1, N2H4 is being consumed, and in reaction 2, N2H4 is being produced. Therefore, we can combine these two reactions to obtain the formation reaction of hydrazine:
N2H4(g) + 3 H2(g) + H2(g) + N2(g) → 2 NH3(g) + 2 NH3(g)
By canceling out the common compounds on the left and right side, we get:
N2H4(g) + 4 H2(g) + N2(g) → 4 NH3(g)
Now that we have the formation reaction, we can calculate the ΔHf for hydrazine by summing up the enthalpy changes of the individual reactions:
ΔHf = (4 * ΔH1) + (ΔH2)
Substituting the given values:
ΔHf = (4 * -1876 kJ) + (-922 kJ)
ΔHf = -7504 kJ + (-922 kJ)
ΔHf = -8426 kJ
Therefore, the ΔHf for the formation of hydrazine (N2H4) is -8426 kJ/mol.