Question #d46e8

For spontaneous processes, the change on the free energy #color(green)(DeltaG)# has to be negative.

The relationship between free energy #color(green)(DeltaG)# and the enthalpy #color(blue)(DeltaH)# is the following:

#color(green)(DeltaG)=color(blue)(DeltaH)-Tcolor(red)(DeltaS)#

For endothermic processes, #color(blue)(DeltaH)# is positive. Therefore, for the process to be spontaneous:
#color(green)(DeltaG)=color(blue)(DeltaH)-Tcolor(red)(DeltaS)<0#

#=>color(blue)(DeltaH)< Tcolor(red)(DeltaS)#

Since #color(blue)(DeltaH)>0#, for the term #Tcolor(red)(DeltaS)# to be greater than #color(blue)(DeltaH)#, the change on the entropy has to be positive as well (#color(red)(DeltaS)>0#).

#=>T > (color(blue)(DeltaH))/(color(red)(DeltaS))#

Therefore, we can say that endothermic processes can be spontaneous at high temperature.

Note that, endothermic reactions are never spontaneous if the change on entropy is negative.

#"If " color(blue)(DeltaH)>0 and color(red)(DeltaS)<0 =>color(green)(DeltaG)>0 " Always"#

You can follow this figure to predict the spontaneity of the reaction based on the signs of #color(blue)(DeltaH) and color(red)(DeltaS)#

Here is a video that further explains this topic:
Thermodynamics | Spontaneous Process & Entropy.