2017年1月10日 · $\ce O$ is a free oxygen atom and $\ce{O2}$ is two oxygen atoms chemically bound to form an oxygen molecule. There is no common analogy for $\ce C$, but $\ce{N2}$ is called nitrogen, $\ce{H2}$ is hydrogen and $\ce{Cl2}$ is chlorine, each having the same name as that of their constituent elements.

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2018年6月27日 · $\ce{O2} = \dfrac{20.291}{21}\times 8.47 = 8.18$ mg/L This is much better agreement with the Benson (1980) data. The OP's value of 1.26e-3 Mol/L/atm for the Henry's constant for O2 in water at 25C was evidently take from Table 8.23 of Warneck and Williams.

2016年2月28日 · It's called dioxygen, $\ce{O2}$, and its MO scheme is exactly the same as above except that there are two fewer electrons in the $\pi^*$ orbitals. Since there are only two electrons in the $\pi^*$ MOs as compared to four in the $\pi$ MOs, overall the $\pi$ and $\pi^*$ orbitals generate a net bonding effect.

$\begingroup$ The reaction $\ce{2H2 + O2 -> 2H2O}$ is a redox reaction (also if you go to atomic recombination), there is no Lewis acid or base, but electron transfer and this results in the contradiction you observed. $\endgroup$ –

2021年12月11日 · $\begingroup$ In the typical way these terms are used, the LUMO can't be the same as the HOMO. Since each of the pi spatial orbitals already have one electron, they are not unoccupied, so they can't be the LUMO.

2014年5月1日 · Is there a complete list of all the half equations for $\ce{H2O2}$ - both oxidation and reduction, in acidic and alkaline conditions?

Disclaimer: I am too lazy to check whether the reaction $\ce{CO2(g) <=> CO(g) + 1/2 O2(g)}$ is more favourable. There is no single tipping point where all carbon dioxide will suddenly be converted to carbon and oxygen. As you heat carbon dioxide, the percentage of oxygen present would gradually increase. You're interested in the reaction

2017年4月11日 · $$\ce{4 H2O -> O2 + 2 H2O + 4 e- + 4 H+}\tag{3}$$ Now, we notice that we again have spectators. Two water molecules turn up both on the reactant and on the product side (an additional two are unique to the reactant side). We can remove them like we did the protons before, giving us $(4)$. I will also perform some rearranging.

The free energy change of total reaction: $\ce{H2O -> 1/2 O2 + H2}$ is fixed at the experimentally found value of 2.46 eV per water molecule. This is done in order to avoid calculations of $\ce{O2}$, since this molecule has a complicated electronic structure, which is not described accurately with DFT.