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 equation as is (using arrows to separate reactants from products is a MUST if you are to do this properly) and the H as is.

Reverse equation 2 and change the sign of H2.
Now add equation 1 to the new equation 2 and add H1 to the reversed equation 2 for delta Hf for the product.