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If there is a system which while undergoing a process produced 30J energy and wasted 40J energy therefore the useful energy was -10J and this explains the process was non spontaneous because ∆G is coming out to be positive where ∆G is negative of useful work done. According to me the system which already had 30J of energy which was not sufficient for the process therefore to complete the process it gained 10J energy more and got its process completed and from this I understood that the process was non spontaneous because to complete it extra 10J was given to the system so the whole process took 40J to complete in which 30J was used by the system itself but 10J was given to the system. So according to me 40J energy is used. But my teacher said that if a system produces less energy and wasted more then the useful work done is negative and he gave this example that a system produced 30J and wasted 40J so what I wanted to ask was I didn't got that how 40J was wasted if the system produced only 30J? According to me 30J was contributed by the system to the process and extra 10J was given to the system for the process....so according to me 40J total was used and not wasted!

Sukriti Sharma
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    It is always good to summarize and/or complement the verbal description by describing the problem/task with symbolic algebraic expressions. And, keeping it that way until all is ready to just plug-in literal numbers with units. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. //

    You may find useful formatting mathematical/chemical expressions/formulas.

     – Poutnik Nov 05 '23 at 12:13
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    What is the meaning of this post ? If a system produces $30$ J and wastes $40$ J, it can be understood that the system produces $70$ J, but that only $30$ J is a useful work, the rest ($40$ J) is lost as heat given to the environment. Is it right ? – Maurice Nov 05 '23 at 17:01
  • @maurice Can you give me an example where the useful work done is negative. So that ∆G comes out to be positive. – Sukriti Sharma Nov 05 '23 at 17:18
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    The first sentence about 55 words, the second one about 80 words. Such sentences are very hard to comprehend. One of guidelines for writing technical texts is that average sentence length should be 20-25 words, with recommendation to avoid sentences longer than 35 words. – Poutnik Nov 05 '23 at 17:24
  • Any reaction like $\ce{Zn + 2 H+ -> Zn^{2+} + H2}$ can produce useful work. Why should Delta G be positive ? – Maurice Nov 05 '23 at 17:41
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    Even James Joyce used paragraphs in Finnegans Wake. Riverrunonsentences. – DrMoishe Pippik Nov 05 '23 at 17:47
  • @Maurice if in any reaction ∆G comes out to be positive then in that case the reaction is non spontaneous. If the reaction is non spontaneous then the useful work done should be negative such that ∆G comes out to be positive because ∆G is negative of useful workdone – Sukriti Sharma Nov 05 '23 at 18:58

2 Answers2

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The 40 Joules of energy wasted in a non-spontaneous process by the system is due to the fact that the system must overcome a certain energy barrier to initiate the process. In a non-spontaneous process, the system doesn't naturally proceed from its initial state to the final state, so energy input is required to drive the process against the natural tendencies.

This energy input could be in the form of work or heat supplied to the system to make it change its state. Essentially, it's the energy needed to make the process happen even though it is not favored by the system's inherent thermodynamics. In contrast, spontaneous processes occur without the need for external energy input because they are energetically favorable based on the system's conditions.

So, the 40 Joules of energy represents the additional energy required to force the system through a non-spontaneous change or to push it over an energy barrier preventing the process from occurring on its own.

Abhinav
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  • can you pls explain me a process where a system itself produces less energy but wastes more energy than produced which will lead to negative useful work done and hence positive ∆G – Sukriti Sharma Nov 05 '23 at 17:25
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Consider a system consisting of two heat reservoirs of temperatures $T_2 = \frac 43 T_1$.

The first process would be the external thermal engine, taking heat $Q_2=\pu{40 J}$ of heat from the warmer reservoir $T_2$, producing work $W=x \lt \pu{10 J}$ and passing the heat $Q_2 = 40 - x\ \pu{J}$ to the colder reservoir $T_1$. In this case is $\Delta G \lt 0$.

The second process would be the external thermal pump, taking heat $Q_1=40 - x \ \pu{J}$ of heat from the colder reservoir $T_1$, work being externally provided $W=x \gt \pu{10 J}$ and passing the heat $Q_2 = \pu{40 J}$ to the warmer reservoir $T_2$. In this case is $\Delta G \gt 0$. The work provided by the system is negative, what means it is work provided to the system.

Poutnik
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