How do you solve this Given the enthalpies of reaction

S(s)+O2(g)→SO2(g) ΔH = -297 kJ

2S(s)+3O2(g)→2SO3(g) ΔH=-791kJ

Calculate the enthalpy change (ΔH) for the following reaction using the data above and Hess's Law:

2SO2+O2⟶2SO3

i think is conclusions.i think so

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.