Gibbs free energy and entropy inconsistency

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Homework Statement


For a certain reaction, ΔG = 13580 + 16.1 T log10(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 lenght 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!
 

Answers and Replies

  • #2
Bystander
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signs of ΔG and ΔS supposed to be opposite
A reaction can be "entropically driven," or, it may be "enthalpically driven." No big deal. Entropically driven examples, vaporization, solution, mmmm ---- what else? Enthalpically driven, oxidation, solidification, anything that gets hot. Throw sulfuric acid into water and watch enthalpy and entropy at work. Dilute a concentrated sodium chloride solution and watch entropy at work and enthalpy opposing the process (it cools, at room T).
 
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  • #3
Ygggdrasil
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A reaction is spontaneous if it results in an increase in the entropy of the universe (i.e. the system and its surroundings). The ΔS in the equation reflects only the change in entropy of the system. For a reaction at constant temp and pressure, the enthalpy will be proportional to the change in entropy of the surroundings (with more negative ΔH leading to a larger increase in entropy). Thus, if a reaction increases the entropy of the system at the expense of the entropy of the surroundings (by converting thermal energy into chemical potential energy), the reaction can still be non-spontaneous overall.
 
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The ΔS in the equation reflects only the change in entropy of the system.
I was ignoring this little detail, now I can be at peace again. Thank you both!
 
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In my question the temperature given is 27°C.. should I just go ahead with the same temperature or should I convert it into Kelvin?
 
  • #6
mjc123
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If you are given an empirical equation like this:
ΔG = 13580 + 16.1 T log10(T) - 72.59 T.
the source should tell you whether T is in °C or K. Sometimes it's one, sometimes the other. If the question asks "Find ΔS and ΔH for the reaction at 298.15 K" then presumably T is in K for this equation.
For fundamental thermodynamic equations like ΔG = ΔH - TΔS, T is always in K.
 

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