"I assume the role of an attentive observer, carefully watching a chemical reaction unfold within a sealed container. This enclosed environment, completely closed ( no matter but energy can exchange) from external influences contains reactants A and B in gaseous State { which reacts as A(g) + B(g) -> C(g) , and one mole of each reactant is present there}. To ensure a stable environment, a piston is affixed to the container, regulating the pressure, while thermally conducting walls maintain a consistent temperature. The system is connected to a machine which is required 50 kilo joule energy.


As I am outside of the box , I really don't know what is going inside. So I will try to understand what's happening there based on what I have learnt so far.


• Reaction is carried on constant pressure 1 atm and at 300 k.

• There is no ionization energy , electron affinity , and any other kind of energies involved in reaction except Bond energies.

• Bond energy of molecule A = 100 kj / mol

• Bond energy of molecule B = 50 kj / mol

• Bond energy of molecule C = 200 kj / mol

• Entropy of formation of molecule A = Sa = 15 joule /mol

• Entropy of formation of molecule B = Sb = 5 joule /mol

• Entropy of formation of molecule C = Sc = 10 joule /mol


Mathmatical Calculation

Okay first I calculate the some mathematical terms to predict that energy released by system is enough or not to start my machine. Let's try to find out enthalpy of reaction ∆H.

∆H = -(energy released in formation of C - energy absorbed to break the bonds of A and B)

∆H = -[200 - ( 100 + 50 )] = -50 kilo joule

Feeling relieved, as the machine will work with this energy, but upon measurement ( by some measuring devices ), I discover that the machine only received 47 kilojoules, 3 kilojoules less than what the system released. There seems to be something we missed. I will use my brain to investigate the issue.


After careful thought, I realize that the amount of energy lost is equal in magnitude to the temperature multiplied by the change in entropy, T × ΔS = 300K × 10 joules/K = 3 kilojoules. Thus, TΔS amount of energy is lost

An Appeal to the Audience

Can someone help me figure out why 3 kilojoules of energy is missing, which is necessary to restart my machine?


"I would like to inquire why we cannot utilize the total enthalpy of reaction (∆H) to perform non-mechanical work (∆G). Specifically, I'm curious about why and where some of its energy (T∆S) is lost. Is it necessary to experience this energy loss (T∆S), or is there a way to prevent it?"

Why so much drama

The reason I asked this simple question in more complected way is that many questions are unanswered , some questions are closed and some previous accounts are also deleted , don't know what is reason behind it so I tried a new way to ask a question , hope I will not recieve downvotes in this question.

  • $\begingroup$ @Poutnik you are saying right there are so many gaps not covered by me yet because I am a online student so I have not any option to ask doubts and clarify concepts via connecting to teacher offline and also in online I have not purchased any app I just study from YouTube so I can't ask doubts in online also. I have already watched so many videos of high school on YouTube but couldn't find anything related to my doubts , I know there are some videos in which answer to my questions are explained but I can't understand anything after watching complete video because they use high mathematics. $\endgroup$ Commented Aug 2, 2023 at 14:46
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    $\begingroup$ @Poutnik sorry I have written isolated system instead of close let me correct it. $\endgroup$ Commented Aug 2, 2023 at 14:47
  • $\begingroup$ "just part of released enthalpy cannot be converted to work but remains as thermal energy" I am just asking about reason that why some part of enthalpy cannot be converted to heat $\endgroup$ Commented Aug 2, 2023 at 14:50
  • $\begingroup$ Hello Bharat Prajapat! I am glad you are an enthusiast. As a suggestion, get some money and buy a good book in thermodynamics. They are infinitely better than any video in YT, designed for entertainment and your attention. I will justify it. I can make a YT account and say nonsense about thermodynamics, and I am allowed. However, publishers that write about science, and write nonsense, they will have serious consequences. There is a reason those books are studied across the globe. I'll see if I can answer this later.. $\endgroup$ Commented Aug 2, 2023 at 14:52
  • $\begingroup$ @MetalStorm "I will acquire sufficient funds to purchase a good book , that is not a problem. My main concern lies in selecting the appropriate book for studying thermodynamics at the high school level or a slightly more advanced level. Many recommended books tend to cater to university-level content, making it challenging for me to grasp due to their heavy reliance on complex mathematics. Nevertheless, I am eager to explore books beyond my syllabus and learn new concepts, as long as they provide a clear understanding without overwhelming mathematical complexities." Could you suggest anything? $\endgroup$ Commented Aug 2, 2023 at 15:06

2 Answers 2


For this system, we have from the 1st law of thermodynamics, $$\Delta U=Q-W$$From the 2nd law of thermodynamics, we have $$Q=T\Delta S-T\sigma$$where $\sigma$ is the amount of entropy generated by irreversibility in the process. Combining these two equations, we obtain: $$\Delta U=T\Delta S-T\sigma-W$$or$$\Delta G=-(W-P\Delta V)-T\sigma$$or$$W=P\Delta V+(-\Delta G)-T\sigma$$So the maximum amount of work that this system can do over and above the PV work is $(\Delta G)=53\ kJ $. So, the maximum total amount of work that can be done with the system going between the initial and final state is $$W=P\Delta V-53\ kJ$$If the amount of work is less than this, it is due to process irreversibility. Part of the process irreversibility is due to the irreversibility of the heat transfer occurring with the average system temperature dropping lower than boundary temperature of 300 C (so there is a finite temperature driving force for heat transfer during the process). Additional irreversibility is due to the reaction proceeding at an uncontrolled (finite) rate. And, if the gas expansion and other work occurs very rapidly (due to the finite rate of the reaction), there will be additional irreversibility due to viscous dissipation.

  • $\begingroup$ Thank you so much for answer but I have some doubts 1. How did you calculated Gibbs free energy change ∆G = -53 kj. 2. "heat transfer occurring with the average system temperature dropping lower" why temperature is dropping lower ? And note that the process is a ideal and reversible so viscous and other additional irreversibility makes no sense in this case so try to explain assuming it a ideal and reversible process. $\endgroup$ Commented Aug 3, 2023 at 3:32
  • $\begingroup$ please answer what I have asked above $\endgroup$ Commented Aug 3, 2023 at 8:09
  • $\begingroup$ $\Delta G=\Delta H-T\Delta S=50-(-10)(300)=53\ kJ$. I will correct the sign error in my answer. Temperature in the bulk of the reaction mixture during the change drops lower than the boundary temperature because the change is occurring at finite rate and, in addition, $\Delta H$ is positive (meaning that heat has to be added irreversibly to the system during the process to try to bring its temperature back uo to the boundary temperature). You can't just make the process reversible by saying that it is. What are you doing to guarantee that the reaction proceeds at infinitesimal rate? $\endgroup$ Commented Aug 3, 2023 at 10:21
  • $\begingroup$ Hm, did not you get the Delta H reversed? Bond energies 100 + 50 -> 200 would mean reaction is exothermic. Delta_r H = -50 kJ/mol ( or rather Delta U, otherwise we have to involve delta (pV), being it a reaction in gaseous phase)). $\endgroup$
    – Poutnik
    Commented Aug 3, 2023 at 10:43
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    $\begingroup$ It is (-200) - (-100 + -50) = -50, as bond energies, expressed as enthalpies of bond formation, are negative. By other words, breaking 50 kJ/mol bonds and 100 kJ/mol bonds, forming 200 kJ/mol bonds, releases 50 kJ/mol of energy. Unles I have missed something. $\endgroup$
    – Poutnik
    Commented Aug 3, 2023 at 11:10

Note that at constant $T,p$, the known formula:

$$\Delta G_\text{sys} = \Delta H_\text{sys} - T \Delta S_\text{sys}$$

divided by $-T$ gives:

$$- \frac{\Delta G_\text{sys}}{T} = - \frac{\Delta H_\text{sys}}{T} + \Delta S_\text{sys} = \Delta S_\text{surr} + \Delta S_\text{sys}=\Delta S_\text{universe}$$

Therefore, $\Delta G_\text{sys}$ related to the maximal amount of possibly done work without violating the 2nd law of thermodynamics which says in one of its formulations that the total entropy cannot decrease.

Response to feedback:

"just part of released enthalpy cannot be converted to work but remains as thermal energy" I am just asking about reason that why some part of enthalpy cannot be converted to heat.

It IS converted to heat. At least that rest of enthalpy that could be converted to work due the 2nd law of thermodynamics.

Further studying:

  • $\begingroup$ thank you for answer. But you really didn't not understand what I was asking. I was just asking why there is heat transfer in this process in terms of T∆S , what is need of heat generation here ? Like expansion work is done by or to the system to maintain pressure similarly heat exchange to maintain temperature. But in this case temperature is constant so no need to transfer heat or if temperature is changing in between the reaction then what is the reason behind of this change in temperature? $\endgroup$ Commented Aug 3, 2023 at 3:39
  • $\begingroup$ This is a rather unusual question: if the term 'entropy' had not been coined by scientists, implying a lack of understanding about the concept, how would we then calculate the heat transfer in this reaction?" I want to no the reason for heat transfer without using entropy, is it possible ? $\endgroup$ Commented Aug 3, 2023 at 3:43
  • $\begingroup$ at p,T const, H change means energy released and exchanged as heat unless part of energy is used for nonvolume work. Theoretially, all that energy can be used for work but the T delta S part. That does not mean it is lost, just it remains as thermal energy and cannot be used. $\endgroup$
    – Poutnik
    Commented Aug 3, 2023 at 3:59
  • $\begingroup$ " Temperature is constant....." Since the process is reversible let's consider an infinitesimal amount dA and dB of reactant A and B use for production of product dC. In this process temperature increased by dT and to maintain this temperature constant dQ amount of heat flows out of system. Until the process is done this dQ heat will continuously flow and at the end net heat flown will be equal to ΣdQ = T∆S. Is it what you want to say ? But why there is a change in temperature dT in infinitesimal change (dA + dB -> dC) ? Could you explain it without using the concept of entropy ? $\endgroup$ Commented Aug 3, 2023 at 4:57
  • $\begingroup$ Sum of dq = Delta H. You need not the concept of entropy for that. $\endgroup$
    – Poutnik
    Commented Aug 3, 2023 at 5:25

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