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 = H_f

thank you

Ask a New Question or answer this question.

Similar Questions

N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g)
1. 1 answer
N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g)
1. 3 answers
Please :)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
1. 0 answers
Please :)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
1. 1 answer