For spontaneous reaction or the reaction at equilibrium which of the following is wrong?

Question

For spontaneous reaction or the reaction at equilibrium which of the following is wrong? (A)  Δ H < T Δ S reaction is exothermic and TΔ S is + ve (B)  Δ H=-ve, TΔ S=0  (C)  Δ H= TΔ S  (D)  Δ H=-ve, TΔ S=-ve

Tardigrade Admin · Accepted Answer

For a reaction to be spontaneous on the reaction at equilibrium: Δ H=-v e Δ S=+v e Δ G=Δ H-T Δ S=-v e So, Δ H < T Δ S, reaction is exothermic and T Δ S is + ve statement is wrong

Q. For spontaneous reaction or the reaction at equilibrium which of the following is wrong?

$Δ H < T Δ S$ reaction is exothermic and $T Δ S$ is + ve

$Δ H = - v e, T Δ S = 0$

$Δ H = T Δ S$

$Δ H = - v e, T Δ S = - v e$

For a reaction to be spontaneous on the reaction at equilibrium:
$Δ H = - v e Δ S = + v e$
$Δ G = Δ H - T Δ S = - v e$
So, $Δ H < T Δ S$ ,
reaction is exothermic and $T Δ S$ is $+ v e$ statement is wrong

Q. For spontaneous reaction or the reaction at equilibrium which of the following is wrong?

ΔH<TΔS reaction is exothermic and TΔS is + ve

ΔH=−ve,TΔS=0

ΔH=TΔS

ΔH=−ve,TΔS=−ve

For a reaction to be spontaneous on the reaction at equilibrium: ΔH=−veΔS=+ve ΔG=ΔH−TΔS=−ve So, ΔH<TΔS, reaction is exothermic and TΔS is +ve statement is wrong

$Δ H < T Δ S$ reaction is exothermic and $T Δ S$ is + ve

$Δ H = - v e, T Δ S = 0$

$Δ H = T Δ S$

$Δ H = - v e, T Δ S = - v e$

For a reaction to be spontaneous on the reaction at equilibrium:
$Δ H = - v e Δ S = + v e$
$Δ G = Δ H - T Δ S = - v e$
So, $Δ H < T Δ S$ ,
reaction is exothermic and $T Δ S$ is $+ v e$ statement is wrong