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?

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Answer 1 · 2016-11-03T03:10:52

The sign of $DeltaS$ cannot be determined because it could be all of the above...

Recall the relationship between Gibbs' free energy ( $DeltaG$ ), entropy ( $DeltaS$ ), and enthalpy ( $DeltaH$ ):

$bb(DeltaG = DeltaH - TDeltaS)$

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 $DeltaS < 0$ . Then we have the following:

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

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

When you evaluate the right side of the equation, there exists a dependence on $T$ that would change whether $DeltaH - TDeltaS$ is negative or positive.

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 $bb(DeltaS < 0)$ , the reaction conditions are satisfied at low temperatures.

CASE II

Suppose $DeltaS > 0$ . Then we have the following:

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

This means $DeltaH - TDeltaS < 0$ , and thus, $DeltaG < 0$ .

Therefore, when $bb(DeltaS > 0)$ , the reaction conditions are satisfied at all temperatures.

CASE III

Suppose $DeltaS = 0$ . Then we have the following:

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

$(-)_1 = (-)_2$

Clearly, it means $DeltaG = DeltaH$ , and $DeltaG < 0$ and $DeltaH < 0$ .

Therefore, when $bb(DeltaS = 0)$ , the reaction conditions are satisfied at all temperatures.

We have just examined all the possible signs of $DeltaS$ , and all are valid under specific temperature ranges, or under all temperatures.

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.

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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?