TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Given
∆H‌‌=+30.0kJmol−1
∆S‌‌=0.06kJK−1mol−1
p‌‌=1atm
∆G‌‌=0
We know that,
if, we put temperature 227∘C or
227+273=500K
In equation then the value of free energy is obtained will be zero.
Hence, ∆G=30−500×0.06
∆G=30−30=0
and nature of reaction will be non-spontaneous.