If an endothermic reaction is non-spontaneous at freezing point of water becomes feasible at its boiling point, then

Question

If an endothermic reaction is non-spontaneous at freezing point of water becomes feasible at its boiling point, then (A) Δ H is -ve, Δ S is +ve (B) Δ H and Δ S both are +ve (C) Δ H and Δ S both are -ve (D) Δ H is +ve, Δ S is -ve

Tardigrade Admin · Accepted Answer

Δ G=Δ H-TΔ S, since Δ G=+ve at T = 273 K and Δ H=+ve and at high temperature 373 K, it becomes spontaneous (.Δ G=-ve.) even if Δ H=+ve . This is possible only, if Δ S=+ve because at high temperature. TΔ S>Δ H to give Δ G=-ve .

Q. If an endothermic reaction is non-spontaneous at freezing point of water becomes feasible at its boiling point, then

$Δ H$ is -ve, $Δ S$ is +ve

$Δ H$ and $Δ S$ both are +ve

$Δ H$ and $Δ S$ both are -ve

$Δ H$ is +ve, $Δ S$ is -ve

$Δ G = Δ H - T Δ S,$ since $Δ G = + v e$ at T = 273 K and $Δ H = + v e$ and at high temperature 373 K, it becomes spontaneous $(Δ G = - v e)$ even if $Δ H = + v e$ . This is possible only, if $Δ S = + v e$ because at high temperature. $T Δ S > Δ H$ to give $Δ G = - v e$ .

Q. If an endothermic reaction is non-spontaneous at freezing point of water becomes feasible at its boiling point, then

ΔH is -ve, ΔS is +ve

ΔH and ΔS both are +ve

ΔH and ΔS both are -ve

ΔH is +ve, ΔS is -ve

ΔG=ΔH−TΔS, since ΔG=+ve at T = 273 K and ΔH=+ve and at high temperature 373 K, it becomes spontaneous (ΔG=−ve) even if ΔH=+ve . This is possible only, if ΔS=+ve because at high temperature. TΔS>ΔH to give ΔG=−ve .

$Δ H$ is -ve, $Δ S$ is +ve

$Δ H$ and $Δ S$ both are +ve

$Δ H$ and $Δ S$ both are -ve

$Δ H$ is +ve, $Δ S$ is -ve

$Δ G = Δ H - T Δ S,$ since $Δ G = + v e$ at T = 273 K and $Δ H = + v e$ and at high temperature 373 K, it becomes spontaneous $(Δ G = - v e)$ even if $Δ H = + v e$ . This is possible only, if $Δ S = + v e$ because at high temperature. $T Δ S > Δ H$ to give $Δ G = - v e$ .