Wednesday, 23 September 2015

thermodynamics - How can -G/T and -A/T be thought of as 'disguised' entropies?


I understand that entropy is defined as dS=dq/T for a reversible change but I fail to see why G/T and A/T can be thought of as 'disguised' entropies. This is a question on my problem sheet at university and my tutor has hinted: "You need to think about the total entropy change for a process".



Answer



The fundamental equation of (chemical) thermodynamics \ce{\Delta G} = \Delta H – {T\Delta S} may be divided by
(-T):


\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}


In this equation \frac{\Delta H_{sys}}{T} is the entropy change of the system by “heat” exchange with the surroundings, hence \frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}. On this basis, the above equation may be identified as:


\Delta S_{total} ={\Delta S_{surroundings} + \Delta S}_{sys}


\Delta S_{sys} and \Delta S_{surroundings} strive mutually to a maximum of \Delta S_{total}.


\ce{\Delta G} is a disguised entropy change, because \ce{\Delta H} is intrinsically an entropy change too, as explained above.



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