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$$\ce{C6H12O6(s) + 6 O2(g) -> 6 CO2(g) + 6 H2O(g)}$$ When looking at glucose combustion I found that the change in Gibbs free energy is greater than the change in enthalpy. If we assume the process is reversible we can set the change in Gibbs free energy to be the work available and if we assume there is no work we can set the enthalpy to be the heat. Why then would the work be less than the heat? I suspect that $\Delta G = \Delta H-\Delta (TS)$ is at work.

Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0? Also would you please explain how an increase in entropy would imply the discrepancy between heat and work (not mathematically).

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  • $\begingroup$ Yes, and since you are going from a large molecule and some oxygen gas to lots of carbon dioxide and water, you can guess that the entropy increases. $\endgroup$ – Karsten Theis Apr 30 at 15:46
  • $\begingroup$ Cool, so can you substitute work, $W$ and heat, $Q$ into the equation and write $W = Q - T \Delta S$ to directly show that $\Delta S >0$? Also would you please explain how an increase in entropy would imply the discrepancy between heat and work (not mathematically). $\endgroup$ – UnhookedSchnook Apr 30 at 15:56
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For the process to go forward, the entropy of the universe has to increase. For this specific case, the reaction is exothermic and the entropy of reaction (entropy change of the system) is positive. If all of the enthalpy is released as heat, the entropy of the universe will increase (the entropy of the surrounding increases because of the heat transfer, and the entropy of the system increases because molecules are more dispersed in the gas phase).

If there is a mix of work and heat exchange, you just have to make sure the entropy of the universe still increases. In this case, the work possible exceeds the enthalpy change. When you do maximal work, you have heat transfer into the system. The entropy decrease of the surroundings is offset by the entropy increase in the system.

So in this case $$|\text{work}| > |\Delta H|$$

For a picture, see https://chemistry.stackexchange.com/a/112958 scenario 2.

Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0?

No, where there is work, enthalpy change is no longer equal to heat.

Also, you have to specify which ΔS you are talking about. If it is the entropy change in the system (due to the reaction) $\Delta S_\text{sys}$, the equation below is a variation of the second law (the left-hand side is equal to the entropy increase of the universe times temperature):

$$T \Delta S_\text{sys} - q_\text{rev} > 0$$

And if you want to relate work to enthalpy, you use the first law:

$$\Delta H = q + w$$

and insert it into equation above to get:

$$T \Delta S_\text{sys} + w - \Delta H > 0$$

or for maximal work with a bit of rearrangement,

$$ w_\text{max} = \Delta H - T\Delta S_\text{sys}$$

So you can say that if the work you get out of a reaction is larger than the enthalpy change, the entropy of the reaction has to be positive.

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