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.

Turn equation 1 around and change the sign of H1.
Add to equation 2, cancel like molecules appearing on both sides, and add the two delta Hs. H1(with sign changed) + H2 = Hf

thank you

To find the enthalpy change for the formation of hydrazine (N2H4), you need to manipulate the given equations and their enthalpy values. Here's a step-by-step guide:

1. Turn equation 1 around:
N2H4(g) + H2(g) -> 2 NH3(g)

2. Change the sign of H1:
Since H1 = -1876 kJ, after changing the sign, it becomes H1 = +1876 kJ.

3. Add the modified equation 1 to equation 2:
2 NH3(g) + 3 H2(g) -> N2(g) + 3 H2(g) + 2 NH3(g)
This equation cancels out the NH3(g) on both sides.

Now, you can sum up the enthalpy changes:
H1 + H2 = Hf (enthalpy change for the formation of hydrazine)

4. Substitute the values:
Hf = +1876 kJ + (-922 kJ)
Hf = +954 kJ

Therefore, the enthalpy change (Hf) for the formation of hydrazine (N2H4) is +954 kJ/mol. Note that the positive sign indicates that the reaction is endothermic.