How many watt-hours are theoretically necessary to produce 2kg of methane from sea water and air? Given the composition of sea water in mol:
\begin{array}{lr}
  \ce{H2O}    &amp;:&amp; \pu{53.6 mol}\\
  \ce{Cl-}    &amp;:&amp; \pu{0.54 mol}\\
  \ce{Na+}    &amp;:&amp; \pu{0.469 mol}\\
  \ce{Mg^2+}  &amp;:&amp; \pu{0.053 mol}\\
  \ce{SO4^2-} &amp;:&amp; \pu{0.028 mol}\\
  \ce{Ca^2+}  &amp;:&amp; \pu{0.010 mol}\\
  \ce{K+}     &amp;:&amp; \pu{0.010 mol}\\
  \ce{Ct}~(\ce{CO2 + HCO3- + CO3^2-})
              &amp;:&amp; \pu{0.002 mol}\\
\end{array}

And the composition of air: 
\begin{array}{lr}
  \ce{N2}    &amp;:&amp; \pu{78.08 mol}\\
  \ce{O2}    &amp;:&amp; \pu{20.95 mol}\\
  \ce{Ar}    &amp;:&amp; \pu{0.934 mol}\\
  \ce{CO2}   &amp;:&amp; \pu{0.04 mol}\\
\end{array}

My general problem: considering that these concentration do not change during the reaction (excess of reactive) and that the product ($\ce{CH4}$) is taken away immediately, how much energy is theoretically needed to produce $\pu{2 kg}$ of $\ce{CH4}$ from the optimal mix of these two phases?

My specific questions: 


How can I be sure of which ions/molecule to definitely "get out" of the equation ?
How can I be sure which reactions are the best ones for my case ?
How do I do a dynamic equilibrium with more than 1 phase ?
How do I do a dynamic equilibrium with added energy ?
Too much for making it of any practical interest

This question cannot be answered easily, there are too many factors. Bytheway it would not be practically convenient making such a process. With $\ce{CO2}$ as your only source of carbon, you would need an extremely energy dispendious air separation apparatus to get some $\ce{CO2}$. You can get then hydrogen from water electrolysis and use  Sabatier reaction (you need a Nickel catalyst) to get finally methane.