Is Delta G naught Dependent on Temperature?

Question

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1. What the student is really asking

The student tried to obtain the Gibbs‑free‑energy expression from a kinetic picture and ended up with

[ \Delta G^{\circ}= -E , ]

i.e. the standard free energy change is identified with a fixed molecular‑energy difference (E).
Because (E) was taken as a constant, the result predicts that (\Delta G^{\circ}) does not depend on temperature, which contradicts the well‑known relation

[ \Delta G^{\circ}= \Delta H^{\circ} -T\Delta S^{\circ}= -RT\ln K . ]

The question therefore is:

Why does the standard Gibbs free energy change depend on temperature, and where did the derivation go wrong?

We will give a complete, step‑by‑step thermodynamic derivation, point out the conceptual errors in the kinetic approach, and end with the correct temperature‑dependence of (\Delta G^{\circ}).

2. Full derivation (thermodynamic route)

2.1 Chemical potential and the definition of (\Delta G)

For any component (i) in a homogeneous phase the chemical potential is

[ \mu_i = \mu_i^{\circ} + RT\ln a_i , \tag{1} ]

where

(\mu_i^{\circ}) – standard chemical potential (defined for the standard state, e.g. 1 bar gas, 1 M solution)
(a_i) – activity (≈ concentration or partial pressure for ideal solutions/gases)

The Gibbs free energy of a reacting system containing (N_i) moles of each species is

[ G = \sum_i N_i\mu_i . \tag{2} ]

If a reaction is written in the conventional form

[ \underbrace{aA}{\text{reactants}}\;\longrightarrow\;\underbrace{bB}{\text{products}}, ]

the stoichiometric coefficients are taken negative for reactants and positive for products. The change in Gibbs energy that accompanies an infinitesimal advancement (\mathrm d\xi) of the reaction is

[ \mathrm dG = \sum_i \nu_i \mu_i \,\mathrm d\xi . \tag{3} ]

The term in parentheses is defined as the reaction Gibbs energy

[ \boxed{\Delta_r G \equiv \sum_i \nu_i \mu_i } . \tag{4} ]

2.2 Introducing activities → the reaction quotient

Insert Eq. (1) into Eq. (4):

[ \Delta_r G = \sum_i \nu_i\bigl(\mu_i^{\circ}+RT\ln a_i\bigr) = \underbrace{\sum_i \nu_i\mu_i^{\circ}}_{\displaystyle\Delta_r G^{\circ}} + RT\sum_i \nu_i\ln a_i . \tag{5} ]

The first sum is the standard reaction Gibbs energy (\Delta_r G^{\circ}) (a constant for a given temperature and pressure).

The second sum can be rewritten using logarithm rules as

[ RT\sum_i \nu_i\ln a_i = RT\ln!\Bigl(\prod_i a_i^{\,\nu_i}\Bigr) \equiv RT\ln Q , \tag{6} ]

where

[ Q = \frac{a_B^{\,b}}{a_A^{\,a}} ]

is the reaction quotient (the instantaneous “ratio of activities”).

Thus the general expression is

[ \boxed{\Delta_r G = \Delta_r G^{\circ} + RT\ln Q } . \tag{7} ]

At equilibrium, (\Delta_r G=0) and (Q) becomes the equilibrium constant (K); therefore

[ \Delta_r G^{\circ}= -RT\ln K . \tag{8} ]

Equation (7) is the textbook formula that the student was trying to obtain, but now we have a clear thermodynamic foundation.

2.3 Connecting (\Delta_r G^{\circ}) to enthalpy and entropy

From the definition of Gibbs free energy

[ G = H - TS \quad\Longrightarrow\quad \Delta_r G = \Delta_r H - T\Delta_r S . \tag{9} ]

If the temperature range of interest is moderate, the standard enthalpy (\Delta_r H^{\circ}) and entropy (\Delta_r S^{\circ}) are often taken as temperature‑independent (they are, strictly speaking, functions of (T) but their variation is small). Substituting the standard quantities gives

[ \boxed{\Delta_r G^{\circ}(T)= \Delta_r H^{\circ} - T\Delta_r S^{\circ}} . \tag{10} ]

Equation (10) shows explicit temperature dependence: the slope of a (\Delta_r G^{\circ}) vs. (T) plot is (-\Delta_r S^{\circ}).

Differentiating Eq. (10) with respect to (T) (holding pressure constant) yields the fundamental thermodynamic relation

[ \left(\frac{\partial \Delta_r G^{\circ}}{\partial T}\right)_p = -\Delta_r S^{\circ}. \tag{11} ]

Thus, unless the reaction entropy change is zero, (\Delta_r G^{\circ}) must vary with temperature.

2.4 Van’t Hoff equation – an alternative route

From Eq. (8) we can write

[ \ln K = -\frac{\Delta_r G^{\circ}}{RT}. \tag{12} ]

Insert Eq. (10) for (\Delta_r G^{\circ}) and differentiate with respect to (T):

[ \frac{\mathrm d\ln K}{\mathrm dT}= \frac{\Delta_r H^{\circ}}{RT^{2}} . \tag{13} ]

Equation (13) is the van’t Hoff equation. It tells us that a non‑zero standard enthalpy change also forces the equilibrium constant – and therefore (\Delta_r G^{\circ}) – to change with temperature.

3. Where the original derivation went wrong

Step in the student’s attempt	Why it is incorrect (or incomplete)
1. Treating the internal energy difference (E) as the reaction free energy	(E) (the difference in “chemical energy” of isolated molecules) is an internal energy term. The Gibbs free energy also contains the (PV) work and the (TS) (entropy) contribution. Ignoring the (TS) term forces the result to be temperature‑independent.
2. Writing the backward rate as (k[B]^b e^{-(E-W)/(k_B T)})	The exponential factor in a rate constant is the activation energy (or the free‑energy barrier), not the overall reaction free energy. The forward and reverse rate constants are related by the microscopic reversibility condition (\displaystyle \frac{k_{\text f}}{k_{\text r}} = K = e^{-\Delta_r G^{\circ}/RT}). Using (E) in the exponent mixes kinetic and thermodynamic quantities incorrectly.
3. Multiplying a single‑molecule energy by (N_A) to obtain a per‑mole quantity	This step is algebraically fine, but it does not convert a microscopic internal‑energy difference into a macroscopic standard free energy. The standard free energy also contains contributions from configurational and thermal degrees of freedom that scale with temperature.
4. Setting the work term (W = 0) and concluding (\Delta G^{\circ} = -E)	Work of expansion/compression ((PV) work) and especially entropy change are always present in a chemical transformation. Even if no external electrical work is extracted, the system does (P\Delta V) work on its surroundings and exchanges heat, giving the (-T\Delta S) term.
5. Equating (\Delta G) to (RT\ln(R_B/R_F))	(R_B) and (R_F) are reaction rates, not activities. The correct equilibrium condition involves the ratio of activities (the reaction quotient (Q)), not the ratio of rates. Only at equilibrium does the ratio of rates equal the equilibrium constant, and then the relation reduces to Eq. (8).
6. Assuming (\Delta G^{\circ}) is a “purely intrinsic molecular potential”	The standard Gibbs energy is a state function that includes both energetic (enthalpy) and entropic contributions. It is not a single‑molecule potential energy; it is a thermodynamic property of the bulk system at a given temperature and pressure.

Because of these conceptual slips, the derivation lost the temperature‑dependent entropy term, yielding the erroneous conclusion that (\Delta G^{\circ}) is constant.

4. Final answer

Yes, (\displaystyle \Delta G^{\circ}) depends on temperature.
The correct expression is

[ \boxed{\displaystyle \Delta_r G^{\circ}(T)=\Delta_r H^{\circ} - T\Delta_r S^{\circ} = -RT\ln K(T)} . ]

The temperature dependence arises from the (-T\Delta S^{\circ}) term (or, equivalently, from the temperature‑dependence of the equilibrium constant via the van’t Hoff equation).
The student’s derivation omitted the entropy contribution and mis‑identified the microscopic energy difference (E) with the macroscopic standard Gibbs free energy, leading to the (incorrect) prediction of temperature independence.

5. Common mistakes for this type of problem

Mistake	How to avoid it
Confusing activation energy with reaction free energy – using the same exponential factor for both forward and reverse rate constants.	Remember: (k_{\text f}=A_{\text f}\,e^{-E_{\text f}^{\ddagger}/RT}), (k_{\text r}=A_{\text r}\,e^{-E_{\text r}^{\ddagger}/RT}). Their ratio gives (e^{-\Delta_r G^{\circ}/RT}), not the individual barriers.
Treating the internal (potential) energy difference as the Gibbs energy.	Use the definition (G = H - TS). Any change in Gibbs energy must contain an enthalpic term and an entropic term.
Setting the work term to zero and forgetting the (PV) and (TS) work.	Even in a closed, isothermal vessel the system does expansion work and exchanges heat. Include the full (P\Delta V) and (T\Delta S) contributions.
Using concentrations directly in kinetic expressions for thermodynamic quantities.	Replace concentrations (or partial pressures) with activities in the thermodynamic derivation; they reduce to concentrations only for ideal solutions/gases.
Assuming (\Delta H^{\circ}) and (\Delta S^{\circ}) are zero because the reaction looks “simple”.	Always check literature values or calculate them

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