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  • Help me answer: If an endothermic reaction is non­spontaneous at freezing point of water and becomes feasible at its boiling point, then

If an endothermic reaction is non­spontaneous at freezing point of water and becomes feasible at its boiling point, then

  • Option 1)

    \Delta H\; is -ve,\Delta S\; is\; +ve

  • Option 2)

    \Delta H\; and \, \Delta S\; both \: \: are\; +ve

  • Option 3)

    \Delta H\; and \, \Delta S\; both \: \: are\; -ve

  • Option 4)

    \Delta H\; is +ve,\Delta S\; is\; -ve

 

Answers (1)

best_answer

As we learnt in

Spontanous process -

\Delta G= \Delta H-\Delta T

 

- wherein

For spontanouse process \Delta G must be negative.

 

 \Delta G=-ve\\ \Delta G=\Delta H-T.\Delta s\\ \Delta H-\:for \:thermodynamics\:process\:is\:positive

At lower temperature \Delta s+ve\:hence \:\Delta G=+ve

But at high temperature T\Delta s\:will\:be\:greater\: than

\Delta H, hence \Delta G=-ve\:spontaneous


Option 1)

\Delta H\; is -ve,\Delta S\; is\; +ve

This solution is incorrect 

Option 2)

\Delta H\; and \, \Delta S\; both \: \: are\; +ve

This solution is correct 

Option 3)

\Delta H\; and \, \Delta S\; both \: \: are\; -ve

This solution is incorrect 

Option 4)

\Delta H\; is +ve,\Delta S\; is\; -ve

This solution is incorrect 

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Aadil

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