- #1
pzona
- 234
- 0
This is probably something I should know, but I don't want to wait and ask my professor. In the combustion of H2, assuming O2 is plentiful enough not to be a limiting reactant, H2O forms as a result of the O-H bond being more stable than the bonds between two H or two O. Is this due to the electronegativity difference?
On a related note, although the answer might be the same, in a hydroxide's dissociation in a solution (let's say the dissociation of NaOH into Na[tex]^{+}[/tex] and OH[tex]^{-}[/tex] in water), the OH[tex]^{-}[/tex] ion remains held together because of this bond, which is due to the difference in electronegativities, correct (aside from the fact that there isn't enough energy to break it)? And if so, what equation would I use to calculate the temperature or kinetic energy of the water needed to dissociate it further into H[tex]^{+}[/tex] and O[tex]^{2-}[/tex]?
On a related note, although the answer might be the same, in a hydroxide's dissociation in a solution (let's say the dissociation of NaOH into Na[tex]^{+}[/tex] and OH[tex]^{-}[/tex] in water), the OH[tex]^{-}[/tex] ion remains held together because of this bond, which is due to the difference in electronegativities, correct (aside from the fact that there isn't enough energy to break it)? And if so, what equation would I use to calculate the temperature or kinetic energy of the water needed to dissociate it further into H[tex]^{+}[/tex] and O[tex]^{2-}[/tex]?