Chemistry Asked on October 6, 2021
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?
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
$$q_mathrm{rev} = nRTlnfrac{V_ce{B}}{V_ce{A}}.tag{3}$$
Thus, we get change in entropy as
$$Delta S = nRlnfrac{V_ce{B}}{V_ce{A}}.tag{4}$$
Now, enthalpy would be a constant value that we discerned from the isobaric process.
Therefore, since $Delta G = Delta H - TDelta S,$ we get
$$Delta G = Delta H - nRTlnfrac{V_ce{B}}{V_ce{A}}.tag{5}$$
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
$$Delta S^circ_mathrm{rxn} = sum nS^circ_mathrm{products} - sum nS^circ_mathrm{reactants}.tag{6}$$
Correct answer by Safdar Faisal on October 6, 2021
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