Question
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
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
Answers
drwls
You left out reaction arrows
N2H4(g) + H2(g) -> 2 NH3(g)
H1 = –1876 kJ
3 H2(g) + N2(g) -> 2 NH3(g)
H2 = –922 kJ
Subtract reaction 2 from 1, and move "-" reactants to the product side:
N2H4 -> 2H2 + N2
H3 = -2798 kJ
For the reverse (formation) reaction, the heat of formation is +2798 kJ
N2H4(g) + H2(g) -> 2 NH3(g)
H1 = –1876 kJ
3 H2(g) + N2(g) -> 2 NH3(g)
H2 = –922 kJ
Subtract reaction 2 from 1, and move "-" reactants to the product side:
N2H4 -> 2H2 + N2
H3 = -2798 kJ
For the reverse (formation) reaction, the heat of formation is +2798 kJ