For a reaction M2O(s)→2M(s)+1/2O2(g) ΔH=30kJmol−1 and ΔS=0.07kJK−1mol−1 at 1 atm.  Calculate up to which temperature, the reaction would not be spontaneous?

Given for the changeΔH=30×103Jmol−1ΔS=70JK−1mol−1For a nonspontaneous reactionΔG=+veSince ΔG=ΔH−TΔS∴ ΔH−TΔS should be +ve  or ΔH>TΔS or T<ΔHΔS<30×10370T<428.57K.