# T∆S implies that heat flows which way?

From Schroeder's Introduction to Thermal Physics.

Parts $$(a)$$ and $$(b)$$ are quite simple, and we find that $$\Delta H =-2808\ \mathrm{kJ}$$ and $$\Delta G =-2879\ \mathrm{kJ}$$, so by $$\Delta G = \Delta H -T\,\Delta S$$, we find that $$T\,\Delta S = 71\ \mathrm{kJ}$$ at room temperature.

What I absolutely cannot understand is how $$(c)$$ and $$(d)$$ work. It intuitively makes sense that entropy is increasing. Assuming $$1$$ mole of glucose, we start the reaction with $$7$$ moles of reactants and end with $$12$$ moles of products. Entropy obviously increases, as we can arrange these molecules in many more ways. However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment. But this isn't true in real life – metabolism reactions obviously release heat. What's going on here?

The fact that the book is using $$\Delta G$$ as the relevant state variable for free energy implies this is being carried out at constant T and p. So, proceeding:

"However, this implies that we are getting usable work from increases in entropy...."

Yes, that's correct. The positive $$\Delta S$$ lowers $$\Delta G$$

"However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment."

No, that's incorrect. The heat flow here is purely due to $$\Delta H$$, and it is flowing out of the system.

What you're probably thinking is that, at constant T (which we have here), $$\Delta S = \frac{q}{T},$$ and thus, since $$\Delta S >0$$, q must be $$> 0$$ as well, i.e., that the positive $$\Delta S$$ means heat is flowing into the system.

This is wrong. It is not the case that $$\Delta S = \frac{q}{T}$$. Rather:

$$\Delta S = \frac{q_{rev}}{T},$$

and this reaction is not happening reversibly. Instead, $$\Delta S$$ is determined from the difference between $$S_{products}$$ and $$S_{reactants}$$, and each of these individual values are calculated from:

$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$

[For more on this, see my answer here: Infinite Increase in Entropy when Energy added to Absolute Zero ]

In summary, in a chemical reaction at constant pressure (and with no non-pV work),

$$\text{đ}q_{sys}=dH_{sys}$$

Thus it is the sign of $$\Delta H$$, not $$\Delta S$$, that determines the direction of heat flow.

Finally, the answer to (d) is that, as thermal energy is generated by the reaction, it heats up the system, and the 2nd law of thermodynamics says that thermal energy naturally flows from hotter to cooler regions. So there is heat flow into the surroundings until the reaction mixture returns to the temperature of the surroundings.

Let's start from the famous expression $$\ce{\Delta G° = - zEF}$$. In the case of glucose, it means that the combustion of glucose in a galvanic cell produces 2879 kJ/mol. It is more than the heat that could be released if the reaction does not produce electricity (2808 kJ/mol). In other words, of all the energy available from this reaction (2879 kJ), a part of it (71 kJ) is used to increase the entropy of the system due to the chemical reaction, and this part will not be converted into heat. The heat released (2808 kJ/mol) will be used for heating the system, then for heating the environment and increasing its entropy, if the heat can get out of the system. This amount of heat is slightly smaller than the total available energy $$\ce{\Delta G°}$$.