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## Homework Statement

For a certain reaction, ΔG = 13580 + 16.1 T log

_{10}(T) - 72.59 T. Find ΔS and ΔH for the reaction at 298.15 K.

## Homework Equations

ΔG = ΔH - TΔS

[tex]\left[\frac{\partial (\Delta G)}{\partial T} \right]_P = - \Delta S[/tex]

## The Attempt at a Solution

For the sake of this thread's length I won't write too much details, I'm having a very specific question about the solutions of this problem.

First I calculated ΔG plugging T = 298.15 K into the given function and got ΔG = 3815.11 cal. Now, this result is telling me the reaction is non-spontaneous. Fine.

Then I calculated ΔS by differentiating the ΔG function and plugged T = 298.15 K in order to get ΔS = 25.76 cal K

^{-1}. This result is telling me the reaction is spontaneous. Both results are given by the textbook and are correct. I also plugged the numeric values of ΔS and ΔG into ΔH = ΔG + TΔS and got ΔH = 11495.45 cal, which is also a correct solution according to the textbook.

Aren't the signs of ΔG and ΔS supposed to be opposite, according to the spontaneity criteria? Thanks in advance for any input!