bi-hidróxido nítrico

H_{3}N + 2·H_{2}O + e^{+} ==> HN(OH)_{2} + 2·H_{2}


HN(OH)_{2} + 2·H_{2} + e^{-} ==> H_{3}N + 2·H_{2}O


Entalpia de reacción:
[ H_{3}N ] = 5
[ 2·H_{2}O ] = 6


[ HN(OH)_{2} ] = 7
[ 2·H_{2} ] = 4


5+6 = 7+4 = 11