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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?

Amadeus
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1 Answers1

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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

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

Safdar Faisal
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    Things get more complicated, as delta H =f(T), because of the Hess law and different molar heat capacities of reagents and products. – Poutnik Jul 16 '20 at 07:52
  • I think OP assumed no phase change single component systems. – Safdar Faisal Jul 16 '20 at 07:54
  • My note does not consider phase changes. With a single component, there is no reaction. – Poutnik Jul 16 '20 at 07:57
  • @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. – Safdar Faisal Jul 16 '20 at 07:59
  • "Spontaneity of reaction" – Poutnik Jul 16 '20 at 07:59
  • Alright thanks. Should I add it to my answer or would you craft an answer for that.. It was my misunderstanding then. This was the simplest way to explain. – Safdar Faisal Jul 16 '20 at 08:01
  • Put it into your answer. :-) – Poutnik Jul 16 '20 at 08:07
  • I. You've left out the differential from your equations (1) and (2). Equation (1) (the incorrect one) should read: $dS= \frac{\text{đ}q}{T}$. And equation (2) (the correct one) should read: $dS= \frac{\text{đ}q_{rev}}{T}$ For more on this, see my answer here: https://chemistry.stackexchange.com/questions/136237/difference-between-Δq-and-dq/136254#136254 II. $\Delta S$ and $\Delta H$ for chemical reactions are not independent of T. – theorist Jul 17 '20 at 04:13
  • @theorist, i haven't said the $\Delta S$ is independant of T. The example given is for an gas expansion. The end of the answer states so/ – Safdar Faisal Jul 17 '20 at 04:19
  • You did say "$\Delta S$ is temperature independent". More broadly, his question was about chemical reactions ("spontaneity of a reaction"), so I'm not sure why you were using an ideal gas expansion here. $\Delta S$ for chemical reactions is temperature-dependent. – theorist Jul 17 '20 at 04:43