# Gibbs Free energy change

1. Aug 31, 2013

### sgstudent

In reaction there is a Gibbs Free energy change. I'm still not sure what this means as in my secondary school years we just learned of one component of it which is the enthalpy change. That was simple as it just showed how much heat was released or absorbed. But with Gibbs Free energy, i don't really know what's going on.

My teacher said that it represents the free energy to do non expansion work. But in a reaction we would definitely get the ΔH from it and no more or less than that. So how can energy be "taken away" for the TΔS component? So I don't quite understand the sentence that it is the free energy to do non expansion work actually.

We are still in a basic chem course so we do not know or understand any mathematical derivatives so I'm hoping to get a answer in a more physical sense. Thanks in advance for the help.

2. Aug 31, 2013

### Staff: Mentor

The change in enthalpy takes into account the fact that if there is a change in volume (at constant pressure), you can get extra compression work from the atmosphere (if $\Delta V < 0$) or there is less energy available as some was used to expand against the external pressure (if $\Delta V > 0$).

The Gibbs free energy extends that concept to the case of a constant temperature. You can get extra energy in the form of heat flowing into the system, or have to manage the extra entropy created when heat will leave the system.

A simple way to think about these things is to consider creating a system out of nothing (ex nihilo). Enthalpy $H$ is the total energy needed to create the system itself plus the energy expanded in pushing the atmosphere away to make place for the system. The Gibbs free energy takes into account the fact that you actually need less energy than that ($G = H -TS$), as you can get some of that energy for "free" by letting the environment furnish $TS$ in energy in the form of heat entering spontaneously the system.

3. Sep 1, 2013

### sgstudent

Hi thanks for the reply :)

hmm.. I'm still quite puzzled by that. ΔH=ΔU+PΔV=Q and this would be for constant pressure. And for ΔG it tells us of the free energy at constant temperature. So would this mean the actual heat evolved would be lesser or more from the TΔS component?

If so, where would this energy TΔS come from? And why would it 1) Take away or 2) Bring in more energy into the system?

Sorry about my basic concept. In my course we've only learned that a negative ΔG means a spontaneous reaction and vice versa and not so much on how these 2 terms come together.

Thanks so much :)

4. Sep 4, 2013

### Staff: Mentor

The key point is the fact that temperature is maintained constant. You can thus see the environment, which is responsible for that temperature being constant, as a heat source or heat sink as necessary.

Entripy has a lot to do with that. But I tried to give you answers without envoking entropy too much as I guessed that you weren't yet completely familiar with it.

5. Sep 4, 2013

### sgstudent

Hi thanks for the great reply. Regarding the entropy part, i understand that entropy is a measure of disorder within the system. I watched this youtube lecture: and there are some parts that i don't understand.

He said that the number of moles n is constant as the mass of the closed system remains constant. I don't quite understand this because if a reaction proceeds like this A+B->C then even though the mass remains constant, the number of moles decreases. Also, even if the number of moles remain constant, like in this equation A+B->C+D if some of the products or reactants are not gaseous, then the ideal gas law can't be used. He explained using PV=nRT then since P, V and n are constants, V must also be a constant. But assuming my 2 possible misconceptions are true, V might actually change.

According to the youtube lecture, it seems that the volume must remain constant for ΔH=Q but I thought the only requirement for ΔH to be equal to the heat transfer is for the external pressure to be constant (I found the derivation for that here http://answers.yahoo.com/question/index?qid=20070719190446AArMS7b) which is the case for most reactions as they occur under atmospheric conditions. So again I don't understand the need the volume to be constant for ΔH=Q.

Also, when deriving the Gibbs Free Energy formula, he said that ΔSsurr=-ΔHsurr/T. Would the ΔHsurr=ΔHsys because for the system if 10J of energy goes in then the ΔH would be 10J but for the surrounding, 10J of energy leaves the system so ΔH=-10J?

Lastly, at the final part of the video it was taught that if the -TΔSuni is negative then the reaction would be spontaneous. This makes sense as if -TΔSuni<0, ΔSuni>0 making the reaction spontaneous as stated by the 2nd Law of Thermodynamics. So for this I have 2 parts that i don't quite understand. 1) I don't understand if the entropy of the universe increases, why would it indicate a spontaneous change and 2) How would these variables that determine the entropy change of the universe tell us how much Free Energy there is. Because the way i see this derivation is to determine whether ΔSuni is positive or negative and nothing else. I don't really see how these variables tell us how much energy is available.

For 2) if i were to base the Gibbs Free Energy as hoe it is defined of the amount of energy available to do non expansion work, I'm unsure that if ΔH=-100J and TΔSsys=-10J would that mean in the conditions for Gibbs Free energy where P and T are constant, would there only be 90J of heat being evolved from the system or would there still be 100J of energy evolved. Or would there be 90J of heat evolved and 10J of expansion within the system? If the 10J of expansion is true then would it mean that if my volume, pressure and temperature are kept constant, then there would actually be 100J of energy given out as heat? Again, I still don't really understand why those variables come together and tell me this numbers actually as mentioned in the previous paragraph before this.

Thanks so much for helping me out here and sorry for the long post.

Last edited by a moderator: Sep 25, 2014
6. Sep 9, 2013

### Staff: Mentor

This is a simplification of what entropy really is. It is not a bad image to have in your mind when thinking about entropy, but be aware that this is not the full picture.

You are correct. I think the video presents a very bad explanation of the Gibbs free energy. G is very useful because it works for constant T and P, which are the conditions you usually work under in real life, where the system is in contact with the atmosphere, which can be seen most of the time at being at SATP (298 K, 1 atm). The idea that the system most be closed still holds, so you only have to be careful if gases are involved, as the container must be closed while the pressure must still be held constant.

You are again correct. The entire point of H is to not have to bother with work being done by/on the atmosphere. So long as there is no other source of work (e.g., electrical), then $\Delta H = q$.

He actually wrote
$$\Delta S_\mathrm{surr} = \frac{\Delta H_\mathrm{surr}}{T}$$
and then used $\Delta H_\mathrm{surr} = -\Delta H_\mathrm{sys}$, based on the first law.

One way to formulate the 2nd law is that spontaneous processes increase the entropy of the universe. But you have to be careful that a process that decreases the entropy will only be spontaneous if there is enough energy to initiate the process.

This is the reason why the Gibbs free energy is so wonderful: it implicitly takes into account what is happening in the environment. The only way to get $\Delta G < 0$ is if $\Delta S_\mathrm{total} > 0$, where $\Delta S_\mathrm{total} = \Delta S_\mathrm{sys} + \Delta S_\mathrm{env}$. The key here is that since $T_\mathrm{env} = T_\mathrm{sys} = T =\mathrm{const.}$, you can know exactly how much entropy is created in the environment when $q$ heat leaks into the environment.

Lets start by examining $\Delta H=-100\ \mathrm{J}$. It means that, after taking into account any expansion/contraction of the system at constant $P$, 100 J can be extracted to perform other work. Now, at constant $T$, some heat will leak out to the environment, since the entropy of the system has decreased ($\Delta S_\mathrm{sys}<0$), but the heat leaking into the environment will compensate for that by producing entropy in the environment. There is then 90 J left to perform other work. That could be, for instance, electrical work if the reaction you are considering is taking place inside a battery.

Hope this helps.

Last edited by a moderator: Sep 25, 2014
7. Sep 9, 2013

### DrDu

I would not look at it this way. What you were told is that Delta G is the maximal amount of non-volume work which can be done by a system while Delta H is the heat released by the system if no work is done.
But according to the second law of thermodynamics it is not possible to convert heat completely into work, hence the two quantities are different.

8. Sep 27, 2013

### sgstudent

Thanks so much for the insightful post. I mostly cleared out my misconceptions but I have a couple questions left about this.

I don't quite understand the last paragraph where after taking into account any volume changes 100J is available to do work. I thought the 10J was the work done to do the expansion work? And the remaining 90J for other things?

Also, you explained that at constant T the entropy of system decreases and so the surrounding increases. I don't quit get why the entropy of the system would decrease and for the surrounding it increases.

9. Sep 27, 2013

### Staff: Mentor

I started from $\Delta H = -100\ \mathrm{J}$, so the expansion/contraction work has already been taken into account. The -100 J is the energy left for other work. When you consider the Gibbs free energy, and therefore take a constant temperature, some of those 100 J will leak out to the environment, $T \Delta S_\mathrm{sys} = -10\ \mathrm{J}$, and you will be left with 90 J.

If $T \Delta S_\mathrm{sys} = -10\ \mathrm{J}$, then for the process $\Delta S_\mathrm{sys} < 0$, so the entropy of the system decreases. By the second law, for the process to occur spontaneously, you need $\Delta S_\textrm{universe} > 0$, so we need $\Delta S_\mathrm{env} > \left| \Delta S_\mathrm{sys} \right|$, so the entropy of the environment increases.

10. Sep 27, 2013

### sgstudent

Thanks for the quick reply :)

For the first point, how do we tell how much energy is being used for the expansion work actually?

And what happens to the leaked energy to the environment?

Lastly, can I check with you whether or not if we use the ΔH° and ΔS° but at a temperature that's not 298K, like maybe 1000K would the Gibbs free energy of the reaction be ΔG° and can be used in the equation ΔG°=-RTlnKeq, providing if the reaction has a equilibrium and is reversible?

Thanks so much for the help :)

11. Sep 27, 2013

### Staff: Mentor

If it is an ideal gas, you can calculate it using $w = P \Delta V$. If it is not an ideal gas, then you need to measure the volume change or know the change in internal energy $\Delta U$ of the system.

It is lost to the environment.

I'm not sure I understand your question here. If you have ΔH° at a given T, you can calculate it at another T with the help of the heat capacity Cp, and so on. So if you have the adjusted values of ΔH° and ΔS° for the temperature you want, then you can use that to get ΔG° at that temperature, and use that to get the equilibrium constant K.

12. Sep 27, 2013

### sgstudent

Thanks again for the reply :)

Besides measuring it directly by using PΔV, how do we use the ΔU to determine the expansion work? Do we use ΔU=ΔH+PΔV and since we have ΔU and ΔH we can tell the PΔV? If so, since temperature is constant can we assume ΔU to be constant? I'm not sure about this because for a substance that doesn't change, if temperature remains constant then the ΔU would be 0. BT now that a new product is formed I'm not too sure about it.

For the energy (10J) leaked to the environment, how does it get leaked and in what form does it occur? Is it heat? If so if the other 90J is being used as a heat source won't it have to be leaked out to the environment too?

Lastly, for the last question. Sorry for being vague, I meant to say if we used ΔH°-TΔS° where T is a temperature other than 298K, would the ΔG still be considered a standard condition with the naught (ΔG°)? And if so can it be used to find the Keq by just using this formula ΔG°=-RTlnKeq? Because I'm a bit unsure whether the ΔG of the reaction still gets the ° at temperatures other than 298K?

Thanks so much :)

13. Sep 27, 2013

### Staff: Mentor

Yes.

$\Delta U =0$ at constant T only for an ideal gas. In reality, as is the case if you have a reaction, then the interparticle interaction changes and the internal energy can change. Think about it this way: if you have a diatomic molecule that dissociates into two atoms, even considering an ideal gas approximation, you go from $U = \frac{5}{2} n R T$ to $U = \frac{3}{2} 2 n R T$. The internal energy has changed.

Yes. In that example, we were considering a process at constant temperature. If the reaction is exothermic, the only way we can have const. T is to transfer heat to the environment.

Depends what you do with it. If the reaction is taking place inside a battery, that energy is converted to electrical work. You can use it to heat a resistor, where indeed it will end up in the environment, but you can also use it to power a motor to lift a weight. In the later case, you have converted the energy into gravitational potential energy.

You cannot use another T than that for which ΔH° and ΔS° are standard. If you want to use another T, you need to convert ΔH° and ΔS° to that T first. Note that the convention I am used to (and I don't know if there is another one) doesn't fix the value of T for which you put a naught. For example, the standard enthalpy of melting will be given for the temperature at which the substance melts for a pressure of 1 bar.

14. Sep 27, 2013

### sgstudent

Thanks so much for the quick replies :)

Oh yes ΔU is only 0 at constant temperature for an ideal gas because for solids or liquids at a phase change T is constant but still there is a change in U right?

In any reaction if the TΔS component is negative, this means the ΔSsurr must increase so would this mean heat is always given out to the surroundings. If the reaction is endothermic, then this doesn't really make sense for heat to leave into the surroundings though. If the ΔH and TΔS are positive and negative respectively that means energy is being absorbed. So does it mean this drives heat out from the system to the surrounding? What does the energy from outside the system do for this positive Gibbs Free energy case actually?

Sorry my english is not very good. if we change the ΔH° and ΔS° for another temperature then they won't be ΔH° or ΔS° anymore. So we cannot use the resulting ΔG to solve for Keq? What if we assumed that the ΔH° or ΔS° has a negligible change. Then could the ΔG we get be used to solve for Keq directly using ΔG°(of the new temperature)=-RTlnKeq?

I interpreted that if T is other than 298K then the ΔG for the reaction wouldn't have the naught but I'm not sure if that's what you meant. Could you clarify that with me again? Sorry my English is very good.

Thanks so much I really appreciate all the replies :)

15. Sep 30, 2013

### Staff: Mentor

The key word here is ideal, not gas. In a real gas, particle-particle interactions will change the energy of the gas by changing for instance pressure, even at constant T.

If the reaction is endothermic, you will get heat from the surroundings. But if TΔS is positive and ΔG is negative, then you have a spontaneous reaction where the entropy is increasing, the process is already increasing the entropy of the universe, so you don't necessarily need heat flow. It could be a constant T process.

Well, it depends at what you declare "standard" to be. To quote from Atkin's Physical Chemistry:
So changing the temperature doesn't necessarily mean that the label ° goes away.

Yes, if you can neglect the change in ΔH and ΔS, but that is a big if.

16. Sep 30, 2013

### sgstudent

Hi thanks for the great reply :)

if our ΔG is negative and our ΔH(reaction) and TΔS(system) are negative and positive respectively. So in this case ΔS(total)>0 and ΔS(surrounding)=-ΔH(reaction)/T and since our ΔH(reaction) is negative it means the ΔS(surrounding) increases. and since TΔS(system) is positive, ΔS(system) increases. So does it mean heat goes to the surrounding and system? Where does the heat come from here?

Then is ΔG is negative and our ΔH(reaction) is positive while our TΔS(system) is also positive, this means ΔS(surrounding) is negative while ΔS(system) is positive. So heat goes to the system from the surroundings such that the temperature remains constant?

If ΔG is negative and our ΔH(reaction) is negative while our TΔS(system) is also negative, this means ΔS(surrounding) is positive while ΔS(system) is negative. So heat is loss from the system to the surrounding so that the temperature remains constant?

Then when ΔG is positive and our ΔH(reaction) and TΔS(system) are positive and negative respectively. So both our ΔS(surrounding) and ΔS(system) decreases. So heat leaves them? But I'm not sure where the heat goes. I think its similar to the first paragraph where I'm not sure where the heat comes from.

Then when ΔG is positive and our ΔH(reaction) is positive while the TΔS(system) is also positive, this means the ΔS(surrounding) is negative while ΔS(system) increases. So does this mean heat from the surrounding enters the system so that temperature remains constant?

when ΔG is positive and our ΔH(reaction) is negative while the TΔS(system) is also negative this would mean ΔS(surrounding) is positive while ΔS(system) is negative. So heat is loss from the system to the surrounding to maintain the same temperature?

Going on to the other point, at standard state the ° also represents the stoichiometric amount in the equation right? Like if we have 2B-->A then ΔG°, ΔH° and ΔS° would be for 2 moles of B to 1 mole of A. While for the actual ΔG, ΔH and ΔS they might not be for 2 moles of B to 1 mole of A because there might be an equilibrium? And in that case not all 2 moles of B would be reacted as the reaction is reversible so there is a ratio of B to A?

Lastly, if we neglect the change to ΔH° and ΔS° then for the new ΔG° we can simply use ΔG°=-RTlnKeq. But if we were to calculate the actual new ΔH and ΔS, then when we substitute to get ΔG can we just use the formula? I think we can because for the new ΔG found, it is still just the calculated Gibbs Free energy change from reactants to products while the actual Gibbs Free energy change that is going on in the reaction would be zero right? So we can still use the ΔG from the altered ΔH and ΔS to find the Keq?

Thanks so much for the help :)