Is a reaction that is exothermic and becomes more positionally random spontaneous or non spontaneous? What about less positionally random? Is there enough information to tell?

1 Answer
Truong-Son N.
Jan 2, 2017

See this answer for the endothermic case.

For the exothermic case, the change in enthalpy #DeltaH < 0#, and if the reaction becomes more "positionally random", then the change in entropy #DeltaS > 0# since there is more motion.

Therefore, the Gibbs' free energy is:

#color(blue)(DeltaG) = DeltaH - TDeltaS#

#= (-) - (+)(+)#

#= (-) - (+)#

#= color(blue)((-))#

which if you recall, is negative for a spontaneous reaction.

So, an exothermic reaction with increased motion (more "positional randomness") is spontaneous at all temperatures.

Less "positionally random" implies less motion and thus #DeltaS < 0#. I think you are at the point where you can guess what the sign of #DeltaG# would conditionally be (hint: it's analogous to having #DeltaH > 0# and #DeltaS > 0#; how would the magnitude of #T# affect the sign of #DeltaG#?).

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