As we know, $$\Delta G=\Delta H - T\,\Delta S.$$

Both in my textbook and on the internet, it was given that temperature was a determining factor in the spontaneity of a reaction. For example, when $\Delta S$ is negative (unfavorable) even then the reaction can still be favorable at lower temperatures.

If $\Delta S = q/T,$ then wouldn't the entropy also decrease by the same factor on decreasing the temperature? In other words, why don't the temperature terms cancel leaving $\Delta G$ independent of temperature?

  • $\begingroup$ chemistry.stackexchange.com/questions/33953/… Does this help? $\endgroup$ Jul 16, 2020 at 4:04
  • 1
    $\begingroup$ Consider a melting point - the spontaneity of the reaction is clearly dependent on temperature. $\endgroup$
    – Jon Custer
    Jul 16, 2020 at 22:50
  • $\begingroup$ If you decrease T and keep q constant, won't the entropy change increase (according to your equation)? $\endgroup$
    – Buck Thorn
    Jul 18, 2020 at 20:52

1 Answer 1


The issue here is in your definition of entropy. According to your definition:

$$\mathrm dS = \frac{\text{đ}q}{T}.\tag{1}$$

However, there is a small difference.

As you may know, the entropy of a system is a state function and so it only depends on the initial and final states. Therefore, if the entropy of the system were to be calculated for a specific path for which the parameters are simple to find, then the value would not change irrespective of the path.

The actual formula of change in entropy is:

$$\mathrm dS = \frac{\text{đ}q_\mathrm{rev}}{T}.\tag{2}$$

This value is calculated for a reversible isothermal process.

Now, let us assume that we have taken such an isothermal reversible process and discovered the value of $\Delta S$ and now used an isobaric process to discover the value of $\Delta H.$ Now, we solve for $\Delta G$ in our isothermal process (this is done so that we can assume constant temperature in our formula).

Therefore, for such a process

$$q_\mathrm{rev} = nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{3}$$

Thus, we get change in entropy as

$$\Delta S = nR\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{4}$$

Now, enthalpy would be a constant value that we discerned from the isobaric process.

Therefore, since $\Delta G = \Delta H - T\Delta S,$ we get

$$\Delta G = \Delta H - nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{5}$$

As you can see, there is still a temperature dependence.

This is because $\Delta S$ is temperature independent and so you would still have a temperature dependence for $\Delta G.$

Another thing that you may have forgotten to take into consideration is the fact that $q$ is dependent on the temperature.

The above example is that for an ideal gas expansion.

As Poutnik stated in the comments in case of a reaction, the calculation of $\Delta H$ becomes more complicated as we have to take into consideration the different molar capacities of the reagents and products which would $\Delta H$ would also be a function of temperature.

For standard value of entropy, we use the formula

$$\Delta S^\circ_\mathrm{rxn} = \sum nS^\circ_\mathrm{products} - \sum nS^\circ_\mathrm{reactants}.\tag{6}$$

  • 1
    $\begingroup$ Things get more complicated, as delta H =f(T), because of the Hess law and different molar heat capacities of reagents and products. $\endgroup$
    – Poutnik
    Jul 16, 2020 at 7:52
  • $\begingroup$ I think OP assumed no phase change single component systems. $\endgroup$ Jul 16, 2020 at 7:54
  • $\begingroup$ My note does not consider phase changes. With a single component, there is no reaction. $\endgroup$
    – Poutnik
    Jul 16, 2020 at 7:57
  • $\begingroup$ @Poutnik, you misunderstand me. I meant that the OP was considering the expansion of ideal gases.. Adding this conception would be helpful to the question.. However I do not know this well enough and so cannot elaborate. $\endgroup$ Jul 16, 2020 at 7:59
  • $\begingroup$ "Spontaneity of reaction" $\endgroup$
    – Poutnik
    Jul 16, 2020 at 7:59

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