# How come when a reaction causes an increase of entropy, it is able to do more work?

Tags: entropy, increase, reaction, work
 P: 60 the total amount of energy able to do work is the gibbs free energy of a reaction. dG=dH-TdS according to this equation, the higher the entropy gain of a system(dS) in a reaction, the more work the system can achieve. I'm confused here. I thought entropy was the energy unable to do work?
 P: 24 Hold the phone, you're misinterpreting the equation. Now, by the way, I know NOTHING about thermodynamics, I'm waiting for a book to come in, so I can read up on it. But, according to your equation (and this is the first time I saw it) dG = dH - TdS, it seems that if dS is positive (i.e. an increase in entropy), then aren't you subtracting TdS from the differential of "Gibbs-free energy" dG? and therefore, it would seem that a large increase in entropy dS corresponds to a smaller increase (or decrease) in Gibbs-free energy dG? I feel bad even posting this because I know nothing about it, but unless T is a negative value in your equation (and I'm assuming it stands for some positive value like temperature), then I believe you misinterpreted it?
P: 60
 Quote by Mazerakham Hold the phone, you're misinterpreting the equation. Now, by the way, I know NOTHING about thermodynamics, I'm waiting for a book to come in, so I can read up on it. But, according to your equation (and this is the first time I saw it) dG = dH - TdS, it seems that if dS is positive (i.e. an increase in entropy), then aren't you subtracting TdS from the differential of "Gibbs-free energy" dG? and therefore, it would seem that a large increase in entropy dS corresponds to a smaller increase (or decrease) in Gibbs-free energy dG? I feel bad even posting this because I know nothing about it, but unless T is a negative value in your equation (and I'm assuming it stands for some positive value like temperature), then I believe you misinterpreted it?
Sorry I didnt mention that it is the magnitude of the reaction. The gibbs free energy of a reaction that is spontaneous (a reaction once started that does not need input of external factors) is always negative. It is the magnitude of dG I was talking about.

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P: 2,532
How come when a reaction causes an increase of entropy, it is able to do more work?

 Quote by gladius999 I thought entropy was the energy unable to do work?
It's not the best definition of entropy, and you're bumping up against the limits of it here. But one way to look at it is to compare two states, A and B, lower and higher entropy respectively, and consider the work that can be extracted by going from A to B.

Compare the two scenarios $S_B\approx S_A$ and $S_B\gg S_A$. We can't extract as much work in the first case, but are left in a configuration where we might extract work in the future starting with B, compared to the second case, where we can extract a lot of work, but we're less likely to get anything useful starting with B. Does this help answer your question?
 P: 5,462 Firstly this should be in the chemistry section, Gibbs Free Energy is a Chemistry thing. Secondly the equation you quote is the isothermal version. I agree that the Gibbs free energy is the energy available to do work and the equation is an accounting balance of (some) sources of this energy. However I do not agree that 'entropy is energy unavailable for work' Consider the expansion of a perfect gas into a vacuum, this can certtianly produce external work for instance in a turbine. $$\Delta {\rm H} = 0$$ The only available contribution to free energy is the increase in entropy. Another example would be the movement of solvent molecules, through a membrane against gravity.
P: 60
 Quote by Studiot Firstly this should be in the chemistry section, Gibbs Free Energy is a Chemistry thing. Secondly the equation you quote is the isothermal version. I agree that the Gibbs free energy is the energy available to do work and the equation is an accounting balance of (some) sources of this energy. However I do not agree that 'entropy is energy unavailable for work' Consider the expansion of a perfect gas into a vacuum, this can certtianly produce external work for instance in a turbine. $$\Delta {\rm H} = 0$$ The only available contribution to free energy is the increase in entropy. Another example would be the movement of solvent molecules, through a membrane against gravity.
 Quote by Mapes It's not the best definition of entropy, and you're bumping up against the limits of it here. But one way to look at it is to compare two states, A and B, lower and higher entropy respectively, and consider the work that can be extracted by going from A to B. Compare the two scenarios $S_B\approx S_A$ and $S_B\gg S_A$. We can't extract as much work in the first case, but are left in a configuration where we might extract work in the future starting with B, compared to the second case, where we can extract a lot of work, but we're less likely to get anything useful starting with B. Does this help answer your question?
sorry i just thought this was part of thermodynamics. I have heard some talk about how everytime a reaction releases energy that some of that energy is unavailable or irretrievable. Is this related to entropy? If it is, how is it related?? Can someone explain this further?

Thanks
 P: 5,462 In order to offer an appropriate further explanation please let us know where you are coming from? Meanwhile a few thoughts, but first let me correct your equation and add another one. $$\Delta G\quad = \Delta H\quad - \quad T\Delta S$$ : This is at constant tmeperature $$\Delta G\quad = \Delta A\quad + \quad P\Delta V$$ : This is at constant pressure A is known as the work function I would not recommend the Wikipedia article it contains some glaring inconsistencies, including the statement that entropy is waste energy. A system is in equilibium when there is no further tendency to change its properties. Above we have a statements connecting temperature, energy ,volume, entropy.. Note energy here does not mean free energy. Consider an equilibrium system taken at constant volume and entropy: The energy is a minumum. This leads to the equilibrium conditions of ordinary everyday mechanics (force, moment equilibrium, work done, strain energy etc) where we do not take thermodynamic considerations into account. Now consider an equilibrium system taken at constant energy and volume : The entropy is a maximum. So we in a real system we have a tugging in opposite directions between energy and entropy. The concept of free energy and work functions were introduced to describe mathematically the resulting balance.
P: 60
 Quote by Studiot In order to offer an appropriate further explanation please let us know where you are coming from? Meanwhile a few thoughts, but first let me correct your equation and add another one. $$\Delta G\quad = \Delta H\quad - \quad T\Delta S$$ : This is at constant tmeperature $$\Delta G\quad = \Delta A\quad + \quad P\Delta V$$ : This is at constant pressure A is known as the work function I would not recommend the Wikipedia article it contains some glaring inconsistencies, including the statement that entropy is waste energy. A system is in equilibium when there is no further tendency to change its properties. Above we have a statements connecting temperature, energy ,volume, entropy.. Note energy here does not mean free energy. Consider an equilibrium system taken at constant volume and entropy: The energy is a minumum. This leads to the equilibrium conditions of ordinary everyday mechanics (force, moment equilibrium, work done, strain energy etc) where we do not take thermodynamic considerations into account. Now consider an equilibrium system taken at constant energy and volume : The entropy is a maximum. So we in a real system we have a tugging in opposite directions between energy and entropy. The concept of free energy and work functions were introduced to describe mathematically the resulting balance.
Hi, I am currently doing physics and chemistry first year in university. I apologize for any lack of competence in understanding any of previous explanations. What I've been taught so far is the gibbs free energy equation and that the magnitude of gibbs free energy is the energy in joules of work theoretcially obtainable from a reaction.

I have been given a simple explanation that entropy is a measure of disorder or the amound of ways energy can be distributed in a system. This explanation is not enough for me as I am very keen to fully understand it.

What I can make from your explanation is that entropy is just a quantity that helps satisfy the gibbs free equation? Entropy is given as a value say 100JK^-1. Does that value correspond to a measurement of work not being done or something? Does the value of entropy mean nothing else than to satisfy an equation?

 P: 5,462 Hello Gladius, Now we know where we are going, let's take a step back to the second law or before. It's good to see someone obsessed with academic work, but you need to realise that mechanical work is only one form of energy so don't get too hung up on it. When the concepts of mechanical work and energy were being developed it took a long time and a lot of effort to prove that mechanical work was the same as heat energy. This is actually pretty fundamental, but often rushed these days. Entropy was introduced before statistical thermodynamics came along and again a deal of work was needed to prove that the two definitions were equivalent ie they referred to the same thing. It is a good idea to study a few actual situations and do a few actual calculations to help get the hang of most subjects. Also in high school you are shielded from akward examples and situations but now you are in university you can expect to meet some. Whichever is is most important that you take note of the conditions under which a law or equation is stated or defined. Many problems arise simply because some law is applied outside its conditions of validity. In thermodynamic terms this means adiabatic, isothermal, reversible, irreversible, and with due regard to properly defining the 'system' under consideration. Now entropy is not, repeat not, any form of energy. Entropy depends upon the quantity of matter present in the system. Physicists tend to regard the quantity of matter as fixed and therefore take it as the unit so measure entropy in joules per degree. Chemists tend to consider the quantity of matter and divide by this so measure in joules per degree per mole. So 18 grammes or 1 mole of ice is melted 6025 joules are absorbed. This is a reversible isothermal process therefore $$\Delta S = \quad \frac{{\Delta H}}{T}\quad = \quad \frac{{6025}}{{273}}\quad = \quad 22.1joules{K^{ - 1}}mol{e^{ - 1}}$$ Note that that there is no change in free energy so $$\Delta G = 0$$. Since this is a reversible process this heat energy is, in principle, fully recoverable upon freezing the water. However if we tried to turn this heat into mechanical work we would run into trouble. This is where the idea of Gibbs free energy and 'useful work' comes in. Since $$\Delta G = 0$$ there is no available free energy ie no work is available. Please confirm you have followed the story so far or ask questions if needed. I will do part 2 in the next post and develop free energy and the equation you first quoted further. Incidentally, I misread your original, you did get it correct.
P: 60
 Quote by Studiot Hello Gladius, Now we know where we are going, let's take a step back to the second law or before. It's good to see someone obsessed with academic work, but you need to realise that mechanical work is only one form of energy so don't get too hung up on it. When the concepts of mechanical work and energy were being developed it took a long time and a lot of effort to prove that mechanical work was the same as heat energy. This is actually pretty fundamental, but often rushed these days. Entropy was introduced before statistical thermodynamics came along and again a deal of work was needed to prove that the two definitions were equivalent ie they referred to the same thing. It is a good idea to study a few actual situations and do a few actual calculations to help get the hang of most subjects. Also in high school you are shielded from akward examples and situations but now you are in university you can expect to meet some. Whichever is is most important that you take note of the conditions under which a law or equation is stated or defined. Many problems arise simply because some law is applied outside its conditions of validity. In thermodynamic terms this means adiabatic, isothermal, reversible, irreversible, and with due regard to properly defining the 'system' under consideration. Now entropy is not, repeat not, any form of energy. Entropy depends upon the quantity of matter present in the system. Physicists tend to regard the quantity of matter as fixed and therefore take it as the unit so measure entropy in joules per degree. Chemists tend to consider the quantity of matter and divide by this so measure in joules per degree per mole. So 18 grammes or 1 mole of ice is melted 6025 joules are absorbed. This is a reversible isothermal process therefore $$\Delta S = \quad \frac{{\Delta Q}}{T}\quad = \quad \frac{{6025}}{{273}}\quad = \quad 22.1joules{K^{ - 1}}mol{e^{ - 1}}$$ Note that I have used the heat energy absorbed, Q, and that there is no change in enthalpy so $$\Delta H = 0$$. Since this is a reversible process this heat energy is, in principle, fully recoverable upon freezing the water. However if we tried to turn this heat into mechanical work we would run into trouble. This is where the idea of Gibbs free energy and 'useful work' comes in. Please confirm you have followed the story so far or ask questions if needed. I will do part 2 in the next post and develop free energy and the equation you first quoted further. Incidentally, I misread your original, you did get it correct.
Hi Studiot,

Thank you very much for your very detailed reply. I cannot find any of this material online or in books. Either they are too complicated, contradicted each other or were too vague.

I managed to follow most of what you wrote. Only part I don't understand is when you said "there is no change in enthalpy". Wasn't there a change in enthalpy of 6025J? Also when you were talking about the change in entropy, was it the change in entropy of the surroundings or system? Does it matter which one? Also how come we cannot transfer the heat into work? Isnt it just a conversion of energy??

I'm very keen for Part 2!
P: 5,462
 Wasn't there a change in enthalpy of 6025J?
You are absolutely right, what a silly thing for me to say, perhaps I should go back to growing roses. I have corrected my post, lest some other reader is mislead.

So I will go back to the first law, which is just an energy account, and reintroduce enthalpy properly.

Initially the first law was introduced to link the heat added to (dq) to the work done on (dw) a system.

dE = dq + dw wh

This was used to define a property called the internal energy E of a system.

Later this definition was extended to include eletromagnetic, kinetic gravitational energies and finally energies involved in nuclear reactions. Nuclear energy changes are not usually included in thermodynamics, however.

It is convenient to distinguish at this stage between

A closed system. - This has no mass crossing its boundaries.

An isolated system - This has no energy crossing its boundaries.

Now it is all very well to include all the additional forms of energy we can think of, but it makes the accounts cumbersome and often some of these do not contribute in any case.

So a new property called enthalpy was defined.

Lots of processes we consider are open to the ordinary atmosphere. That is they occur at constant pressure.
Consider adding heat to a system at constant pressure.

Going back to dE=dq+dw
dq is heat added so is positive
p is constant so any work done by the system volume change and is negative.

It is important to get the signs correct here. We are putting heat, but no work, into the system. Yet work may be done within the system.

dE=dq-pdV

or dq=dE+pdV

This is a differential form we call this heat the enthalpy change and define enthalpy

H = E + PV

If (when) you have done enough maths you will recognise that dE and dH are total differentials, dq and dw are not.

So enthalpy is the total heat input, some of which goes into heat or bond (breaking) energy some of which converts to mechanical work 'on' the system even though we do not directly do any mechanical work.

I will come back to introduce the free energy and work function in a similar fashion in the next installment.
P: 60
 Quote by Studiot You are absolutely right, what a silly thing for me to say, perhaps I should go back to growing roses. I have corrected my post, lest some other reader is mislead. So I will go back to the first law, which is just an energy account, and reintroduce enthalpy properly. Initially the first law was introduced to link the heat added to (dq) to the work done on (dw) a system. dE = dq + dw wh This was used to define a property called the internal energy E of a system. Later this definition was extended to include eletromagnetic, kinetic gravitational energies and finally energies involved in nuclear reactions. Nuclear energy changes are not usually included in thermodynamics, however. It is convenient to distinguish at this stage between A closed system. - This has no mass crossing its boundaries. An isolated system - This has no energy crossing its boundaries. Now it is all very well to include all the additional forms of energy we can think of, but it makes the accounts cumbersome and often some of these do not contribute in any case. So a new property called enthalpy was defined. Lots of processes we consider are open to the ordinary atmosphere. That is they occur at constant pressure. Consider adding heat to a system at constant pressure. Going back to dE=dq+dw dq is heat added so is positive p is constant so any work done by the system volume change and is negative. It is important to get the signs correct here. We are putting heat, but no work, into the system. Yet work may be done within the system. dE=dq-pdV or dq=dE+pdV This is a differential form we call this heat the enthalpy change and define enthalpy H = E + PV If (when) you have done enough maths you will recognise that dE and dH are total differentials, dq and dw are not. So enthalpy is the total heat input, some of which goes into heat or bond (breaking) energy some of which converts to mechanical work 'on' the system even though we do not directly do any mechanical work. I will come back to introduce the free energy and work function in a similar fashion in the next installment.
Great I understood that! 1 question though. If the pressure is kept the same, is all the energy exchanged turned into heat?
P: 5,462
 If the pressure is kept the same, is all the energy exchanged turned into heat?
Not necessarily. Work can still be done by change of volume.

It is worth getting these ideas clear before proceeding further. So let us look at a few real world examples.

Suppose we have a calorimeter and we measure the 'heat of a chemical reaction'. What does this mean i.e. what have we measured? Well as I keep stressing it depends upon the conditions.

Consider the neutralisation of hydrchloric acid by sodium hydroxide. When we take 1 mole of each in solution and allow them to react we measure the same heat whether we do this in an open beaker or in a vessel full to the brim and sealed.

$${\rm{HCl(aq) + NaOH(aq) = NaCl(aq) + }}{{\rm{H}}_2}{\rm{O(l)}}$$

Whichever way we measure the result is -57.1 kJ/mole

All the energy of this reaction has come from the difference between the bond energies in the new and old arrangements.
In this case all the reactants and products remain in solution so there is no change in volume and no work is done.

$$\Delta E = {q_v} = \Delta H = {q_p}$$

Now consider the reaction of an acid on zinc granules dropped into solution.

$${\rm{Zn(s) + 2}}{{\rm{H}}^ + }{\rm{(aq) = Z}}{{\rm{n}}^{2 + }}{\rm{(aq) + H(g)}}$$

If we perform this in an open beaker, we will measure an evolved heat of -152 kJ/mole

If however we perform the same experiment in a sealed vessel, as before so no hydrogen gas is liberated we measure
-154.47 kJ/mole

So what is the difference?

Well in the in open beaker is the heat evolved is the enthalpy change

$${q_p} = \Delta H = - 152$$

In the second the system is at constant volume. Since $$\Delta V = 0$$, no work is done and all the internal energy changes appear as heat.

$$\Delta E = {q_v} = \Delta H - 2.47 = -154.47$$

What of this extra -2.47?

Well in the open beaker $$\Delta V$$ is not zero so using

$$\Delta E = \Delta H - P\Delta V = \Delta H - \Delta nRT$$

Since $$P\Delta V = \Delta nRT$$ from the gas law and $$\Delta n$$ is the change in the number of moles of gas.

So this is work done by the system expending some of its change in internal energy on work expanding a mole of hydrogen against atmospheric pressure in the open beaker.

Calorimeter measurement at constant volume measure the change in internal energy

Calorimeter measurements at constant pressure measure the change in enthalpy

We cannot measure at both contant volume and pressure at the same time.
P: 60
 Quote by Studiot Not necessarily. Work can still be done by change of volume. It is worth getting these ideas clear before proceeding further. So let us look at a few real world examples. Suppose we have a calorimeter and we measure the 'heat of a chemical reaction'. What does this mean i.e. what have we measured? Well as I keep stressing it depends upon the conditions. Consider the neutralisation of hydrchloric acid by sodium hydroxide. When we take 1 mole of each in solution and allow them to react we measure the same heat whether we do this in an open beaker or in a vessel full to the brim and sealed. $${\rm{HCl(aq) + NaOH(aq) = NaCl(aq) + }}{{\rm{H}}_2}{\rm{O(l)}}$$ Whichever way we measure the result is -57.1 kJ/mole All the energy of this reaction has come from the difference between the bond energies in the new and old arrangements. In this case all the reactants and products remain in solution so there is no change in volume and no work is done. $$\Delta E = {q_v} = \Delta H = {q_p}$$ Now consider the reaction of an acid on zinc granules dropped into solution. $${\rm{Zn(s) + 2}}{{\rm{H}}^ + }{\rm{(aq) = Z}}{{\rm{n}}^{2 + }}{\rm{(aq) + H(g)}}$$ If we perform this in an open beaker, we will measure an evolved heat of -152 kJ/mole If however we perform the same experiment in a sealed vessel, as before so no hydrogen gas is liberated we measure -154.47 kJ/mole So what is the difference? Well in the in open beaker is the heat evolved is the enthalpy change $${q_p} = \Delta H = - 152$$ In the second the system is at constant volume. Since $$\Delta V = 0$$, no work is done and all the internal energy changes appear as heat. $$\Delta E = {q_v} = \Delta H - 2.47 = -154.47$$ What of this extra -2.47? Well in the open beaker $$\Delta V$$ is not zero so using $$\Delta E = \Delta H - P\Delta V = \Delta H - \Delta nRT$$ Since $$P\Delta V = \Delta nRT$$ from the gas law and $$\Delta n$$ is the change in the number of moles of gas. So this is work done by the system expending some of its change in internal energy on work expanding a mole of hydrogen against atmospheric pressure in the open beaker. Calorimeter measurement at constant volume measure the change in internal energy Calorimeter measurements at constant pressure measure the change in enthalpy We cannot measure at both contant volume and pressure at the same time.
Great I understand everything in that! please continue. Sorry for my late reply as I have midsemester exams drawing near. Thanks for your reply.

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