Question

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?

Answer 1

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.