April 29, 2024, 03:31:42 AM
Forum Rules: Read This Before Posting


Topic: Gibbs' vs Standard gibbs' energies  (Read 6230 times)

0 Members and 1 Guest are viewing this topic.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Gibbs' vs Standard gibbs' energies
« on: May 05, 2013, 02:04:17 PM »
ΔG=ΔH-T·ΔS
ΔG°=ΔH°-T·ΔS°

If ΔH and ΔS are considered to be temperature independent, then as far as I can see ΔH=ΔH° and ΔS=ΔS°. But this means ΔG=ΔG° which would mean Q=K=1 - obviously wrong. Where does the logic go wrong? I have a feeling the definition of ΔH and ΔS need to be clarified for me ...

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3484
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' vs Standard gibbs' energies
« Reply #1 on: May 05, 2013, 06:15:25 PM »
Why would you think that if ΔH° is temperature independent, this would mean that ΔH°=ΔH?  Maybe a better question is - what is ΔH°?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' vs Standard gibbs' energies
« Reply #2 on: May 05, 2013, 06:21:44 PM »
Why would you think that if ΔH° is temperature independent, this would mean that ΔH°=ΔH?  Maybe a better question is - what is ΔH°?

The standard enthalpy change, as determined at a temperature of 298.15 K and a pressure of 1 bar (105 Pa). So I suppose there is both temperature and pressure independency. But that doesn't change the principle of my issue.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3484
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' vs Standard gibbs' energies
« Reply #3 on: May 05, 2013, 07:02:02 PM »
No, I mean, what IS it?  I think you and I have been down this road before, but do not be discouraged.  It is very confusing, I know.  I still get confused about it. :)

The standard heat of formation of any substance is the amount of energy it takes to form 1 mole of the substance from its pure elemental starting materials in their standard states.  Note that temperature isn't what makes the heat of formation "standardized".  The heat of formation is specified for a specific temperature, and it is USUALLY 298 K, but it doesn't have to be.  Many students get confused on this point, so we should be clear about that up front. 

The standard enthalpy change for a reaction (formation of products from reactants), as you know, is found by adding up all the standard heats of formation for the products and subtracting from them all the heats of formation for the reactants (scaled by stoichiometry).  So what does the standard enthalpy change represent?  Essentially it is the amount of energy that would hypothetically be gained or lost if you broke down all the reactants into their pure elemental starting materials in their standard states, and then formed the products from those starting materials.  Importantly, there is a specific quantity of reacting substances implicit to this definition - since heats of formation are specified for 1 mole of substance, so to is the standard enthalpy change for a reaction specified for a certain stoichiometric amount of reactants being converted COMPLETELY into the stoichiometric amount of products.

Note that this kind of reaction never actually happens.  There is never total conversion of reactants to products and it's unlikely you're going to have exactly one mole of a starting material.  So what's the point?  The point is that it serves as a reference point for real reactions of like kind.  Because enthalpies are difficult to measure exactly.

So that's ΔH° for a reaction.  And typically we can assume that it is temperature independent.  Which means that the amount of energy it takes to form 1 mole of molecule AB from its pure elemental starting materials A and B at 298 K is roughly the same amount of energy it takes to do the same process at 288 K or 308 K.  For small temperature differences, it's not a bad approximation, because.... well, I'll let you think about that. :)

What, then, is ΔH for a reaction?  ΔH for a reaction is the amount of enthalpy gained or lost for a process between its starting point and its ending point.  Note (and this is the most important thing to realize) that the ending point for a reaction process is NOT total conversion of reactants to products.  It's going to equilibrium, which will almost always involve some reactants and products in solution at the same time.  ΔH may also be temperature independent, but this does not mean that ΔH is the same as ΔH°.  This common error is due to the equally common misconception that the "standard" in "standard enthalpy change" has anything to do with temperature, per se.  Probably this error arises due to "standard" being used in other contexts like "standard temperature and pressure" for a lot of other thermodynamical topics.  In that it's probably an unfortunate choice of words, but it is what it is.  "Standard" here means that it's a reference point, a standardized state of a substance - it is a way to standardize enthalpy changes (or entropy or Gibbs energy changes) for a wide range of reactions performed under many different conditions.  Remember, thermodynamical quantities like enthalpy, entropy and so forth are (almost) always measured as relative values.  You have to establish a common reference point; otherwise reported values have no context.  This is the purpose of using "standardized" values.  Frankly this is an important point that most textbooks just don't explain very well, and it's no wonder so many students are confused by it.

So, to recap: ΔH° is the amount of enthalpy gained or lost when taking a specified amount of reactants and converting them completely to products.  ΔH is the amount of enthalpy gained or lost between the starting (non-equilibrium) point and the equilibrium point.  ΔH depends on the specific conditions and quantities of reactants/products at your starting point; ΔH° does not, because it's a standardized reference value.

This line of discussion applies equally to standard enthalpies and standard Gibbs energies.  If you know the standard Gibbs energy change for a reaction, then you can easily predict which way the reaction will proceed from a nonequilibrium position.  This is the whole point of having the standardized values.

You may want to check out Wikipedia's article on Standard State.  It actually does a good job of explaining what "standard" means, for better than most textbooks I've come across.

http://en.wikipedia.org/wiki/Standard_state
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' vs Standard gibbs' energies
« Reply #4 on: May 05, 2013, 07:30:49 PM »
No, I mean, what IS it?  I think you and I have been down this road before, but do not be discouraged.  It is very confusing, I know.  I still get confused about it. :)

The standard heat of formation of any substance is the amount of energy it takes to form 1 mole of the substance from its pure elemental starting materials in their standard states.  Note that temperature isn't what makes the heat of formation "standardized".  The heat of formation is specified for a specific temperature, and it is USUALLY 298 K, but it doesn't have to be.  Many students get confused on this point, so we should be clear about that up front. 

The standard enthalpy change for a reaction (formation of products from reactants), as you know, is found by adding up all the standard heats of formation for the products and subtracting from them all the heats of formation for the reactants (scaled by stoichiometry).  So what does the standard enthalpy change represent?  Essentially it is the amount of energy that would hypothetically be gained or lost if you broke down all the reactants into their pure elemental starting materials in their standard states, and then formed the products from those starting materials.  Importantly, there is a specific quantity of reacting substances implicit to this definition - since heats of formation are specified for 1 mole of substance, so to is the standard enthalpy change for a reaction specified for a certain stoichiometric amount of reactants being converted COMPLETELY into the stoichiometric amount of products.

Note that this kind of reaction never actually happens.  There is never total conversion of reactants to products and it's unlikely you're going to have exactly one mole of a starting material.  So what's the point?  The point is that it serves as a reference point for real reactions of like kind.  Because enthalpies are difficult to measure exactly.

So that's ΔH° for a reaction.  And typically we can assume that it is temperature independent.  Which means that the amount of energy it takes to form 1 mole of molecule AB from its pure elemental starting materials A and B at 298 K is roughly the same amount of energy it takes to do the same process at 288 K or 308 K.  For small temperature differences, it's not a bad approximation, because.... well, I'll let you think about that. :)

What, then, is ΔH for a reaction?  ΔH for a reaction is the amount of enthalpy gained or lost for a process between its starting point and its ending point.  Note (and this is the most important thing to realize) that the ending point for a reaction process is NOT total conversion of reactants to products.  It's going to equilibrium, which will almost always involve some reactants and products in solution at the same time.  ΔH may also be temperature independent, but this does not mean that ΔH is the same as ΔH°.  This common error is due to the equally common misconception that the "standard" in "standard enthalpy change" has anything to do with temperature, per se.  Probably this error arises due to "standard" being used in other contexts like "standard temperature and pressure" for a lot of other thermodynamical topics.  In that it's probably an unfortunate choice of words, but it is what it is.  "Standard" here means that it's a reference point, a standardized state of a substance - it is a way to standardize enthalpy changes (or entropy or Gibbs energy changes) for a wide range of reactions performed under many different conditions.  Remember, thermodynamical quantities like enthalpy, entropy and so forth are (almost) always measured as relative values.  You have to establish a common reference point; otherwise reported values have no context.  This is the purpose of using "standardized" values.  Frankly this is an important point that most textbooks just don't explain very well, and it's no wonder so many students are confused by it.

So, to recap: ΔH° is the amount of enthalpy gained or lost when taking a specified amount of reactants and converting them completely to products.  ΔH is the amount of enthalpy gained or lost between the starting (non-equilibrium) point and the equilibrium point.  ΔH depends on the specific conditions and quantities of reactants/products at your starting point; ΔH° does not, because it's a standardized reference value.

This line of discussion applies equally to standard enthalpies and standard Gibbs energies.  If you know the standard Gibbs energy change for a reaction, then you can easily predict which way the reaction will proceed from a nonequilibrium position.  This is the whole point of having the standardized values.

You may want to check out Wikipedia's article on Standard State.  It actually does a good job of explaining what "standard" means, for better than most textbooks I've come across.

http://en.wikipedia.org/wiki/Standard_state

Thank you so much! This makes things far clearer.

So ΔH is scaled not only for conditions, but also for extent of reaction (i.e. equilibrium) - and that is why it is very rare to say ΔH=ΔH° or ΔG=ΔG°, certainly not based on purely physical conditions. And this relationship between them is held by the extent-defining formula ΔG=ΔG°+RT·loge(Q). Right?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Gibbs' vs Standard gibbs' energies
« Reply #5 on: May 06, 2013, 02:02:43 AM »
So that's ΔH° for a reaction.  And typically we can assume that it is temperature independent.  Which means that the amount of energy it takes to form 1 mole of molecule AB from its pure elemental starting materials A and B at 298 K is roughly the same amount of energy it takes to do the same process at 288 K or 308 K.

From what I remember ΔH° was at std. states of pressure etc. but at the reaction temperature.

Doesn't one have to correct using Cp0 for both reactants and products to account for temperature changes?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3484
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' vs Standard gibbs' energies
« Reply #6 on: May 06, 2013, 10:28:54 AM »
Doesn't one have to correct using Cp0 for both reactants and products to account for temperature changes?
Technically, yes.  However for small changes in temperature, the variation of ΔH° is small enough that you can neglect it.

For example, for the reaction CO (g) + H2O (g)  :rarrow: CO2 (g) + H2 (g), ΔH° @ 298.15 K is -41.173 kJ/mol and at 308 K it is -41.139 kJ/mol.  At 1500 K however it is -30.676 kJ/mol.  These values were determined using the heat capacity method you mention.  So you can see it's not really necessary to use for small temperature changes but if you are going to be comparing very large temperature changes, then you need to do it.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Gibbs' vs Standard gibbs' energies
« Reply #7 on: May 06, 2013, 10:41:16 AM »
Doesn't one have to correct using Cp0 for both reactants and products to account for temperature changes?
Technically, yes.  However for small changes in temperature, the variation of ΔH° is small enough that you can neglect it.

Fair enough. OTOH for small enough changes in temperature you can ignore most things. :)

Coming from a Chemical Engineering background it was fairly standard to include the Cp0 correction or at least explicitly mention it was not done in cases it wasn't.

In fact I've almost grown with the habit of writing not ΔG° but ΔG°298 etc. which is a explicit reminder each time as to what temperature the standard quantity was evaluated at.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' vs Standard gibbs' energies
« Reply #8 on: May 06, 2013, 10:45:51 AM »
Doesn't one have to correct using Cp0 for both reactants and products to account for temperature changes?
Technically, yes.  However for small changes in temperature, the variation of ΔH° is small enough that you can neglect it.

For example, for the reaction CO (g) + H2O (g)  :rarrow: CO2 (g) + H2 (g), ΔH° @ 298.15 K is -41.173 kJ/mol and at 308 K it is -41.139 kJ/mol.  At 1500 K however it is -30.676 kJ/mol.  These values were determined using the heat capacity method you mention.  So you can see it's not really necessary to use for small temperature changes but if you are going to be comparing very large temperature changes, then you need to do it.

And am I right in thinking these are all values of ΔH°, except that when we refer to "the standard enthalpy change for a reaction" it generally means at 298.15 K, but nonetheless we can calculate ΔH° at any temperature by scaling using heat capacities (or through the heats of formation at that temperature, if they are available, or by calculating them)?

How does one scale for ΔH° in terms of pressure dependence?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3484
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' vs Standard gibbs' energies
« Reply #9 on: May 06, 2013, 10:57:42 AM »
Just for the heck of it (you can see, I'm in major procrastination mode today!), here is a plot of the standard reaction enthalpy change vs. temperature for the reaction I mentioned above, calculated from the heat capacity.  You can see that when dealing with temperatures around room temperature, the variation is small (less than 2 kJ/mol between ~100 and ~500 K). 

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3484
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' vs Standard gibbs' energies
« Reply #10 on: May 06, 2013, 11:04:01 AM »
@BD,

There really is no pressure dependence for ΔH° as far as I understand, because the standard states from which these values are calculated implicitly include pressure (e.g., standard state of oxygen gas is determined at 1 atmosphere).  In other words, if you change the pressure, you are no longer dealing with a standard enthalpy change.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Gibbs' vs Standard gibbs' energies
« Reply #11 on: May 06, 2013, 11:09:31 AM »
Doesn't one have to correct using Cp0 for both reactants and products to account for temperature changes?
Technically, yes.  However for small changes in temperature, the variation of ΔH° is small enough that you can neglect it.

For example, for the reaction CO (g) + H2O (g)  :rarrow: CO2 (g) + H2 (g), ΔH° @ 298.15 K is -41.173 kJ/mol and at 308 K it is -41.139 kJ/mol.  At 1500 K however it is -30.676 kJ/mol.  These values were determined using the heat capacity method you mention.  So you can see it's not really necessary to use for small temperature changes but if you are going to be comparing very large temperature changes, then you need to do it.

Can I ask you to provide the heat capacities? Like any devoted student, I tried to arrive at them through differentiation from the values you gave, but I have only two equations and four variables (one cp for each reactant and product) to calculate! You can provide just 2 of them if you want, that might make an interesting exercise for me :P

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Gibbs' vs Standard gibbs' energies
« Reply #12 on: May 06, 2013, 11:13:44 AM »
Just for the heck of it (you can see, I'm in major procrastination mode today!), here is a plot of the standard reaction enthalpy change vs. temperature for the reaction I mentioned above, calculated from the heat capacity.  You can see that when dealing with temperatures around room temperature, the variation is small (less than 2 kJ/mol between ~100 and ~500 K). 



Nice graph.

I recollected more about why it was so crucial to include ΔCp0 correction: Most industrial gas phase reactions happen at fairly high T (in interest of kinetics) and since reference values were at 298 K even a modest 800 C reaction (not really high temperature as things go) would have a fairly large correction.

Ergo, the "low temp. difference" domain was rarely of interest unless it was liquid phase reactions carried out near room temp.  But for liquids anyways ΔCp0  was moot.

For things like combustion modelling these corrections were often crucial.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3484
  • Mole Snacks: +530/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Gibbs' vs Standard gibbs' energies
« Reply #13 on: May 06, 2013, 11:21:21 AM »
@curious

Plus I would imagine that when you are dealing with very large reaction quantities (as you would be in an industrial setting) even small changes in thermodynamic parameters can translate into large energy costs.  I am not a chemical engineer, but I assume that in these kinds of situations efficiency is paramount (unlike in a small scale research lab).

@BD

Here is where I got the data for that reaction.  See problem 1. (Just did a google search for temperature dependence of reaction enthalpy and looked around until I found an example that had some readily available values.)  The plot is just the equation at the top of page 4. http://www.science.fau.edu/chemistry/CHM3410/TemperatureDependenceofE.pdf
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Sponsored Links