The correct thermodynamic conditions for the spontaneous reaction at all temperatures is

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ΔH<0 and ΔS>0

ΔH<0 and ΔS<0

ΔH<0 and ΔS=0

ΔH>0 and ΔS<0

answer is A.

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Detailed Solution

ΔG=ΔH-TΔS If ΔH<0 and ΔS>0ΔG=(-ve)-T(+ve) then at all temperatures, ΔG=-ve, spontaneous reaction. If ΔH<0 and ΔS=0ΔG=(-ve)-T(0)=-ve at all temperatures. Option 3 can also be condition for spontaneity.

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