How do you begin from the Gibbs-Helmholtz equation to determine the new temperature at which a reaction ceases to become spontaneous? What fundamental assumptions are involved?

The Gibbs-Helmholtz equation at a chosen standard pressure (#@#) (of #"1 atm"# in this case) is:

#((del(G^@//T))/(delT))_P = -H^@/T^2#

To check spontaneity, one would need #DeltaG^@#, keeping in mind that ALL of this is NOT at #"298.15 K"#, i.e. NOT at standard temperature.

The alternative version is:

#((del(DeltaG^@//T))/(delT))_P = -(DeltaH^@)/T^2#

Keeping in mind that this process is at constant pressure:

#d(DeltaG^@//T) = -(DeltaH^@)/T^2dT#

Integrating both sides, we get:

#int_((1))^((2)) d(DeltaG^@//T) = -int_(T_1)^(T_2) (DeltaH^@)/T^2dT#

Assuming that #DeltaH^@# is constant in the temperature range (and is defined at standard pressure):

#(DeltaG^@(T_2))/T_2 - (DeltaG^@(T_1))/T_1 = DeltaH^@(P^@)[1/T_2 - 1/T_1]#

A reaction is "no longer spontaneous" when it reaches equilibrium. For that scenario, #DeltaG^@(T_2) = 0#. So, we solve for the #T_2# where that is the case.

#-(DeltaG^@(T_1))/(DeltaH^@)1/T_1 = 1/T_2 - 1/T_1#

#-(DeltaG^@(T_1))/(DeltaH^@)1/T_1 + 1/T_1 = 1/T_2#

This becomes:

#color(blue)(T_2) = [-(DeltaG^@(T_1))/(DeltaH^@)1/T_1 + 1/T_1]^(-1)#

#= [(1 - (DeltaG^@(T_1))/(DeltaH^@))1/T_1]^(-1)#

#= T_1/[1 - (DeltaG^@(T_1))/(DeltaH^@)]#

#= color(blue)((DeltaH^@T_1)/[DeltaH^@ - DeltaG^@(T_1)])#

Or, if we use the relation that #DeltaG^@(T_1) = DeltaH^@ - T_1DeltaS^@# at the first temperature, also assuming the change in entropy is constant in the temperature range,

#color(blue)(T_2) = (DeltaH^@T_1)/[cancel(DeltaH^@ - DeltaH^@) + T_1DeltaS^@]#

#= color(blue)(DeltaH^@//DeltaS^@)#

which should be a familiar result.