carburo de dos hidrógenos octogonal

Oct[C_{8}]H_{2} + O_{2} + e^{+} ==> C_{8}O_{2} + H_{2}


C_{8}O_{2} + H_{2} + e^{-} ==> Oct[C_{8}]H_{2} + O_{2}


Entalpia de reacción:
[ Oct[C_{8}]H_{2} ] = 10
[ O_{2} ] = 4


[ C_{8}O_{2} ] = 12
[ H_{2} ] = 2


10+4 = 12+2 = 14