# How can I do this problem, even though I am given DeltaH and DeltaS at the wrong temperature? For a certain reaction, DeltaH_(rxn)^@ = . . .  and DeltaS_(rxn)^@ = . . .  at 25^@ "C". Calculate DeltaG_(rxn)^@ and K_(eq) at 37^@ "C"?

Apr 9, 2017

Well, if you notice, you are at two different temperatures. You are given $\Delta {H}_{\text{rxn}}^{\circ}$ and $\Delta {S}_{\text{rxn}}^{\circ}$ at ${25}^{\circ} \text{C}$, and yet, you are asked to calculate $\Delta {G}_{\text{rxn}}^{\circ}$ at ${37}^{\circ} \text{C}$.

Since $H$, $S$, and $G$ are functions of at least temperature (among other natural variables), it follows that at different temperatures, $H$, $S$ and $G$ will be different numbers.

Therefore, you assume that all three thermodynamic functions do not vary significantly going from $\boldsymbol{{25}^{\circ} \text{C}}$ to $\boldsymbol{{37}^{\circ} \text{C}}$ (in reality that is a decent assumption, actually). Hence, you can use those literature values as they are.

As a head start, at equilibrium we have that $\Delta G = 0$ (not $\Delta {G}^{\circ}$!). Therefore, from

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$,

we have that $Q = K$ and:

$\textcolor{b l u e}{\Delta {G}_{\text{rxn}}^{\circ} = - R T \ln {K}_{e q}}$

After you determine $\Delta {G}_{\text{rxn}}^{\circ}$, see if it's negative. That would mean the reaction is spontaneous at THAT temperature and pressure. If that is the case, it logically means that the REVERSE reaction is nonspontaneous at THAT temperature and pressure.