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An endotthermic reaction is non-spontane...

An endotthermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then:

A

`Delta H` is -ve, `Delta S` is + ve

B

`Delta H and Delta S` both are + ve

C

`Delta H and Delta S` both are - ve

D

`Delta H` is + ve, `Delta S` is - ve

Text Solution

Verified by Experts

The correct Answer is:
B
For a reaction to be spontaneous, `Delta G` must be negative. According to the equation -
`Delta G = Delta H- T. Delta S`
If `Delta H and DeltaS` both are positive, than term T. `Delta S` will be greater than `DeltaH` at high temperature and consequently `Delta G` will be negative at high temperature.
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Free enegry , G = H - TS , is state function that indicates whther a reaction is spontaneous or non-spontaneous. If you think of TS as the part of the system's enegry that is disordered already, then (H -TS) is the part of the system's energy that is still ordered and therefore free to cause spontaneous change by becoming disordered. Also, DeltaG = DeltaH - T DeltaS From the second law of thermodynamics, a reaction is spontaneous if Delta_("total")S is positive, non-spontaneous if Delta_("total")S is negative, and at equilibrium if Delta_('total")S is zero. Since, -T DeltaS = DeltaG and since DeltaG and DeltaS have opposite sings, we can restate the thermodynamic criterion for the spontaneity of a reaction carried out a constant temperature and pressure. IF DeltaG lt 0 , the reaction is spontaneous. If DeltaG gt 0 , the reaction is non-spontaneous. If DeltaG = 0 , the reaction is at equilibrium. Read the above paragraph carefully and answer the following questions based on the above comprehension. If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then

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