A certain reaction has a ΔH = –11.7 kJ/mol and a ΔS = –105 J/mol•K. At what temperature (in K) would this rea?
A certain reaction has a ΔH = –11.7 kJ/mol and a ΔS = –105 J/mol•K. At what temperature (in K) would this reaction become spontaneous?
I think the answer us 111.42 K but just want to check. thanks!
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At equilibrium dG = 0 = dH - TdS so
dH = TdS
dH / dS = T
(-11.7kJ/mole) / (-0.105kJ/molK) = 111K.
At 0K dG = dH = -11.7kJ/mole which is spontaneous.
At 111K, dG = 0, which is equilibrium
Above 111K reaction is non-spontaneous.