# A reaction is exothermic and spontaneous. Is the change in entropy of the process positive, negative, or zero? Or can it not be determined from the information given?

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A reaction is exothermic and spontaneous. Is the change in entropy of the process positive, negative, or zero? Or can it not be determined from the information given?

A reaction is exothermic and spontaneous. Is the change in entropy of the process positive, negative, or zero? Or can it not be determined from the information given?

##### 1 Answer

The sign of

Recall the relationship between Gibbs' free energy (

#bb(DeltaG = DeltaH - TDeltaS)#

By definition:

- A
**spontaneous**reaction has#bb(DeltaG < 0)# for the particular#T# and#P# of the reaction (not necessarily#25^@ "C"# and#"1 atm"# ). - An
**exothermic**reaction has#bb(DeltaH < 0)# for the particular#T# and#P# of the reaction (not necessarily#25^@ "C"# and#"1 atm"# ).

When you treat each number as a sign, you have:

#(-)_1 = (-)_2 - T(?)#

**CASE I**

Suppose

#(-)_1 = (-)_2 - T(-)_3#

#= (-)_2 + T(+)_3#

When you evaluate the right side of the equation, there exists a dependence on

- If
#T# is large, then#-TDeltaS# is large and positive, and thus,#DeltaG > 0# . - If
#T# is small, then#-TDeltaS# is small and positive, and thus,#DeltaG < 0# .

*Therefore, when* *the reaction conditions are satisfied at low temperatures.*

**CASE II**

Suppose

#(-)_1 = (-)_2 - T(+)#

This means

*Therefore, when* *the reaction conditions are satisfied at all temperatures.*

**CASE III**

Suppose

#(-)_1 = (-)_2 - cancel(T(0))#

#(-)_1 = (-)_2#

Clearly, it means

*Therefore, when* *the reaction conditions are satisfied at all temperatures.*

We have just examined all the possible signs of

**Since all of the above can satisfy the reaction conditions with an unspecified temperature, the answer is that we cannot determine the sign of #bb(DeltaS)# from the information given.**