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"#?

Question

Avery Adams · Accepted Answer

Well, if you notice, you are at two different temperatures. You are given #DeltaH_"rxn"^@# and #DeltaS_"rxn"^@# at #25^@ "C"#, and yet, you are asked to calculate #DeltaG_"rxn"^@# at #37^@ "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 #bb(25^@ "C")# to #bb(37^@ "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 #DeltaG = 0# (not #DeltaG^@#!). Therefore, from #DeltaG = DeltaG^@ + RTlnQ#, we have that #Q = K# and: #color(blue)(DeltaG_"rxn"^@ = -RTlnK_(eq))# After you determine #DeltaG_"rxn"^@#, 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.

Peyton Adams · Answer

undefined You can use the equation ΔG_rxn^@ = ΔH_rxn^@ - TΔS_rxn^@ and the Van 't Hoff equation ln(K_eq2/K_eq1) = ΔH_rxn^@/R * (1/T1 - 1/T2), where T1 = 25°C + 273.15 and T2 = 37°C + 273.15. Calculate ΔG_rxn^@ using the given values at 25°C, then