Ch4(g)+2o2(g)⇄co2(g)+2h2o(g) δh°rxn=−803kj/molrxn δs°rxn=−5j/(molrxn⋅k) the chemical equation above represents the exothermic reaction of ch4(g) with o2(g). which of the following best to explain why the reaction is thermodynamically favored (δg< 0) at 2000k and 1atm?

Solubility would be your answer.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.