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.

You can't have an equation without arrows. No arrows means we don't know the reactants from the products.

95.4 kJ/mol

can you elaborate on how you got that browmf0x?

To find the enthalpy change for the formation of hydrazine (N2H4), we can use the given enthalpies of the reactions involving the formation of ammonia (NH3).

The first reaction given is:
N2H4(g) + H2(g) → 2 NH3(g) ΔH1 = –1876 kJ

The second reaction given is:
3 H2(g) + N2(g) → 2 NH3(g) ΔH2 = –922 kJ

We need to manipulate these two reactions to ultimately get the formation of hydrazine (N2H4).

To do this, we can rearrange the first reaction by flipping it and multiplying it by a factor of 3. This will allow us to cancel out the NH3 and H2, leaving us with 2 N2H4:
3 (2 NH3(g) → N2H4(g) + H2(g)) ΔH1 = 3(-1876 kJ)

Next, we can rearrange the second reaction and flip it as well. This will allow us to cancel out the NH3 and N2, leaving us with 2 H2:
2 (2 NH3(g) → 3 H2(g) + N2(g)) ΔH2 = 2(-922 kJ)

Now we have:
2 N2H4(g) = 3(-1876 kJ) + 2(-922 kJ)

Calculating the enthalpy change for the formation of hydrazine:
2 N2H4(g) = -5628 kJ + (-1844 kJ)
= -7472 kJ

Therefore, the enthalpy change for the formation of hydrazine (N2H4) is -7472 kJ/mol.