3) At low temperatures the entropic term (-TΔS) is greater than zero and less in absolute value at the enthalpic term (ΔH) The entropy is always minor than enthalpy. Then ΔG< 0 so the reaction is spontaneous. At high temperatures the entropic term is positive larger in absolute value than the enthalpy term, which is negative. Then the entropic term is over enthalpic term. As ΔG>0, the reaction is not spontaneous .
4) We need to calculate ΔGº and apply spontaneity and equilibrium criteria. ΔGºrxn= ΔGºfC6H6(g) -ΔGºfC2H2(g) ΔGºrxn= 1mol(129.72 kJ/mol)-3mol(147.92kJ/mol) ΔGºrxn= -314.04 kJ/mol
As ΔGº is negative, this reaction is spontaneous, so the method is recommended. This do not means this reaction take little time, in Thermodynamics you do not care a lot the reaction time, but if is possible or not.