# Entropy plays a larger role in determining the Gibbs energy of reactions that take place at what?

Mar 17, 2018

Well, surely at higher temperature...

From the Gibbs' free energy Maxwell Relation,

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$

and the enthalpy Maxwell Relation,

$\mathrm{dH} = T \mathrm{dS} + V \mathrm{dP}$

We find then that

$\mathrm{dG} = - S \mathrm{dT} + \mathrm{dH} - T \mathrm{dS}$

At constant temperature, we would then get the Gibbs' isothermal relation:

$\mathrm{dG} = \mathrm{dH} - T \mathrm{dS}$

Integration from the initial to the final state gives:

${\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dG} = {\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dH} - T {\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dS}$

And this should be familiar:

$\implies \textcolor{b l u e}{\Delta G = \Delta H - T \Delta S}$

Since $\Delta S$ is multiplied by the temperature, higher temperature increases the contribution of $T \Delta S$ in determining $\Delta G$ for a given reaction.