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Since entropy increases when temperature increases, the entropy of surroundings increases during an exothermic reaction. When I apply the same logic, it seems like the entropy of a chemical system increases in an endothermic reaction. Is this true?

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Not necessarily.

Exo- and endothermic refer to the direction in which heat is exchanged between system and surroundings, not to changes in the entropy of the system, or of the temperature of either system or surroundings. Your logic though is partly correct: exothermic means the entropy of the surroundings increases, as it would do with an increase in T - the catch is that we usually assume that the surroundings has constant T, but that's because we say it has infinite heat capacity.

It is a recurrent subject of confusion to conflate changes in the entropy of the surroundings, of the system, and of the sum of both. Always keep the second law in mind. What determines whether a reaction will be spontaneous is the sum of the changes in entropy of the surroundings and the system.

When a reaction occurs reversibly, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}=0$.

When spontaneous, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}>0$.

When a reaction is exothermic, $\Delta S_{\mathrm{surr}} >0$.

When a reaction is endothermic, $\Delta S_{\mathrm{surr}} <0$.

Nothing about the thermicity (whether the reaction is endo or exothermic) tells you what happens to $\Delta S_{\mathrm{sys}}$, unless you already know whether the reaction is spontaneous or not (in which case the sum of the entropy changes is positive).

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All spontaneous chemical reactions are described by the following formula:

$$\Delta_\mathrm{r}G < 0$$

If the reaction is endothermic, it means that

$$\Delta_\mathrm{r}H > 0$$

As a consequence, $\pu{T\Delta_\mathrm{r}S}$ is the sum of two positive terms:

$$T\Delta_\mathrm{r}S = - \Delta_\mathrm{r}G + \Delta_\mathrm{r}H > 0$$

and as the temperature is always positive, it means that for all spontaneous endothermic reactions, the entropy of the system increases:

$$\Delta_\mathrm{r}S > 0$$

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One way to carry out a reversible ideal gas process where you start with stoichiometric quantities of pure reactants at T and P and end up with corresponding stoichiometric quantities of pure products at T and P, would be to use a three step sequence. The middle step would involve using an van't Hoff equilibrium box in which a reaction mixture is already at chemical equilibrium within the box at T and P, and very slowly injecting stoichiometric quantities of pure reactants at the same temperature and partial pressures as in the reaction mixture while, at the same time, very slowly removing corresponding stoichiometric quantities of pure products at the same temperature and partial pressures as in the reaction mixture. In this open system step, the heat you add is equal to the enthalpy change of the reaction, and the entropy change is equal to the enthalpy change divided by T: $$\Delta S_{middle\ step}=\frac{\Delta H}{T}$$

The first step of the overall process is very different from this. In the first step, you take each reactant individually and expand it isothermally at T from its initial pressure to its partial pressure in the equilibrium box. This step is not at constant pressure P, and the enthalpy change of the gas is not equal to the heat added. Since it is at constant temperature, the enthalpy of the gas is constant, but its entropy increases (because of the pressure change). So in this step of the process, the enthalpy changes of the gases are zero, but not their entropies increase.

Similarly, for the final third step of the process, the pure product gases are compressed isothermally from their partial pressures in the equilibrium mixture to the total pressure P. Here again, the enthalpy change is zero, but, in this case, the entropy changes of the gases are negative.

Overall, the entropy of the gases changes in all three steps of the process, but the enthalpy of the gases only changes in the middle step. So, overall, the entropy change for the entire process is not equal to the enthalpy change divided by the temperature. We say that the overall process, the pressure is constant (i.e., the same at the end as in the beginning) even though in steps 1 and 3, the pressure does change.

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