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[*] posted on 7-8-2023 at 05:04


Quote: Originally posted by Rainwater  

2) delta H, and delta S are really only valid at the pressure and temperature at which they were recorded. They do change.

delta H, and delta S also change as the phase of matter changes.
For most calculations having one value for solid, another for liquid, one for gas, and another for aquaus solution
is sufficient to determine the basic requirements for a reaction to take place. Im going to go ahead and toss disassociation into cations and anions into this category

What matters to me is, whether they both increase or decrease together, without the ratio changing much. -TdeltaS.

As well as whether they both postive/negative and change together.

If that happened, then it would surely be a problem.

But for using room temp calculations, I'm getting temps of below absolutely 0. But these are questions I may ask at the Chemistry Stack Exchange.
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[*] posted on 7-8-2023 at 05:19


Quote: Originally posted by Neal  
Quote: Originally posted by DraconicAcid  


At whatever particular temperature at which DeltaGo = 0, then K will be equal to 1. At any other temperature, K will not be 1, and DeltaGo will not be zero.

Right, so DeltaG-standard is not 0. So somethingThatIsNot0 = -RT * 0.

A contradiction still.
Uh, no. It’s not a contradiction. DA literally said deltaG is 0 when K is 1 and in any other case it will not be. When deltaH equals T(deltaS) deltaG equals 0. Change T and deltaH no longer equals T(deltaS)… deltaG no longer equals 0.

I’m not sure what you’re not understanding.




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[*] posted on 7-8-2023 at 05:29


Quote: Originally posted by SplendidAcylation  

As for Le Chatelier's principle, well, there is another formula, related to the first one:

DeltaG = DeltaG(standard) + RT ln Q

When Q = 1, then it doesn't matter whether he meant Delta G or Delta G standard = 0, the whole thing collapses.

You either get.

0 = somethingNot0 + 0

or SomethingNot0 = 0 + 0.

[Edited on 7-8-2023 by Neal]
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[*] posted on 7-8-2023 at 07:29


Quote: Originally posted by Neal  
Quote: Originally posted by SplendidAcylation  

As for Le Chatelier's principle, well, there is another formula, related to the first one:

DeltaG = DeltaG(standard) + RT ln Q

When Q = 1, then it doesn't matter whether he meant Delta G or Delta G standard = 0, the whole thing collapses.

You either get.

0 = somethingNot0 + 0

or SomethingNot0 = 0 + 0.

[Edited on 7-8-2023 by Neal]


No. When Q =1 , then deltaG = deltaGo, because when Q =1 , then you're at standard conditions of pressure and concentration.




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[*] posted on 7-8-2023 at 10:26


Quote: Originally posted by DraconicAcid  
Quote: Originally posted by Neal  
Quote: Originally posted by SplendidAcylation  

As for Le Chatelier's principle, well, there is another formula, related to the first one:

DeltaG = DeltaG(standard) + RT ln Q

When Q = 1, then it doesn't matter whether he meant Delta G or Delta G standard = 0, the whole thing collapses.

You either get.

0 = somethingNot0 + 0

or SomethingNot0 = 0 + 0.

[Edited on 7-8-2023 by Neal]


No. When Q =1 , then deltaG = deltaGo, because when Q =1 , then you're at standard conditions of pressure and concentration.

That's weird, he originally said

At equilibrium, Q=K, and DeltaG is 0, which is how the first equation is derived, because:

That's like saying all equations are at equilibrium at 298 K.
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[*] posted on 7-8-2023 at 12:24


Quote: Originally posted by Neal  
Quote: Originally posted by DraconicAcid  
Quote: Originally posted by Neal  
Quote: Originally posted by SplendidAcylation  

As for Le Chatelier's principle, well, there is another formula, related to the first one:

DeltaG = DeltaG(standard) + RT ln Q

When Q = 1, then it doesn't matter whether he meant Delta G or Delta G standard = 0, the whole thing collapses.

You either get.

0 = somethingNot0 + 0

or SomethingNot0 = 0 + 0.

[Edited on 7-8-2023 by Neal]


No. When Q =1 , then deltaG = deltaGo, because when Q =1 , then you're at standard conditions of pressure and concentration.

That's weird, he originally said

At equilibrium, Q=K, and DeltaG is 0, which is how the first equation is derived, because:

That's like saying all equations are at equilibrium at 298 K.


No, because deltaGo is not zero at equilibrium.




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[*] posted on 7-8-2023 at 12:33


Quote: Originally posted by DraconicAcid  
Quote: Originally posted by Neal  
Quote: Originally posted by DraconicAcid  
Quote: Originally posted by Neal  
Quote: Originally posted by SplendidAcylation  

As for Le Chatelier's principle, well, there is another formula, related to the first one:

DeltaG = DeltaG(standard) + RT ln Q

When Q = 1, then it doesn't matter whether he meant Delta G or Delta G standard = 0, the whole thing collapses.

You either get.

0 = somethingNot0 + 0

or SomethingNot0 = 0 + 0.

[Edited on 7-8-2023 by Neal]


No. When Q =1 , then deltaG = deltaGo, because when Q =1 , then you're at standard conditions of pressure and concentration.

That's weird, he originally said

At equilibrium, Q=K, and DeltaG is 0, which is how the first equation is derived, because:

That's like saying all equations are at equilibrium at 298 K.


No, because deltaGo is not zero at equilibrium.

It is if the 2 Delta-Gs equal each other, and 1 of them equals 0. See the bolded quote.


[Edited on 7-8-2023 by Neal]

[Edited on 7-8-2023 by Neal]
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[*] posted on 7-8-2023 at 12:56


They will only equal each other if the system is at standard conditions (1 atm pressure for all gases, 1 mol/L for all solutes). This is not usually the case for equilibrium, where K = 1 and deltaG = 0.



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[*] posted on 7-8-2023 at 19:53


Okay, I guess I got so messed up when you guys threw in Q. I'm looking for T, when K = 1. What do I care to know if Q = K if K doesn't equal 1. I would only care about Q = K if they = 1.
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[*] posted on 7-8-2023 at 23:30


When K = 1, DeltaG(standard) is zero. T = deltaH/deltaS.



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[*] posted on 8-8-2023 at 11:39


I spent a lot of time working through these types of problems before I got a good hold on what I was doing.
Many beers later it became clear.
Hope this helps

Quote:
But for using room temp calculations, I'm getting temps of below absolutely 0. But these are questions I may ask at the Chemistry Stack Exchange.

What this basicly means is the reaction will be non spontaneous.
For example taking CO2 and producing C and O2
DeltaG = 0 at -133215.3 K
Far below absolute zero, currently out of the amateurs ability.

Im missing what your not understanding. So here is an example very similar to your origional post, from start to finish
Lets look at making ammonia at 1 atm and no catalyst
( and no kinetics, just thermodynamics )

$$N_2 + 3H_2 \leftrightarrows 2NH_3$$
Code:
Thermodynamic values reactant state ΔH(KJ/mol) ΔS(J/mol*K) N2       (g)       0        191.50168 H2       (g)       0        130.586824 Nh3      (g)  -46.10768     192.33848

First we need the sum of the deltaH and deltaS of the reagents and products

Reagents
0N2 ΔH + 3 * 0H2 ΔH = 0ΔH reagents
We have 3 hydrogen molecules so that value has to be multiplied by 3.

Then the same for ΔS
191.50168N2 ΔS + 3 * 130.586824H2 ΔS = 583.262152ΔS reagents

Products
2 * -46.10768NH3 ΔH = -92.21536ΔH product
2 * 192.33848 NH3 ΔS = 384.67696 NH3 ΔS

Subtract the ΔH of the reactants from the products for -92.2154  kJ/mol
Subtract the ΔS of the reactants from the products for -198.5852 J mol/K

Now all the variables for the formula ΔG = ΔH - TΔS are available.
To solve for T when ΔG=0 we have to rearrange the formula like so

0 = ΔH - TΔS ... add "TΔS" to both sides
TΔS = ΔH      ... divide both sides deltaS
T = ΔH ÷ ΔS

Now we have to convert our units to match, im just making everything kJ
T is in kelvin
ΔH ÷ (ΔS÷1000) = T
-92.2154  ÷ ( -198.5852 ÷ 1000 ) = 464.36 K
We have solved all varables, for deltaG = 0

now lets see what the equilibrium equation says.
ΔG = -RT ln(Keq)
There is no ΔH or ΔS here. So ignore their existence for a moment.

Keq is basicly a value which was recorded experimently and some genuine genius used it to discover a mathematical relationship to deltaG for reactions taking place in the gas phase
(need verification here, i could be wrong about the gas phase bit)
Thanks to this relationship, we can rearrange the equation to solve for unknown values.
If deltaG is = 0, because of this relationship, Keq = 1
0 = -RTwe calculated above*ln(Keq)
Given that Keq is 1 at equilibrium
And
deltaG is 0 at equilibrium, for any given reaction at the calculated T.
This equation holds true, simply because the natural log of 1equilibrium is 0, and anything multiplied by 0 is 0
If any conditions of the reaction change, the ΔG = ΔH - TΔS needs to be recalculated to produce the correct results.


Back to the orginal question
At first glance, I would solve this by plugging in the values for sodium and chlorine out of the appendix of my text book, then solving for T when deltaG = 0

[Edited on 9-8-2023 by Rainwater]




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[*] posted on 9-8-2023 at 07:09


Quote: Originally posted by DraconicAcid  
When K = 1, DeltaG(standard) is zero. T = deltaH/deltaS.

Okay, so I'm getting -411/.072 = -5,708 K.

Or at least, 2 * that, for 2 mol.

[Edited on 9-8-2023 by Neal]
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[*] posted on 9-8-2023 at 08:48


Quote: Originally posted by Neal  
Quote: Originally posted by DraconicAcid  
When K = 1, DeltaG(standard) is zero. T = deltaH/deltaS.

Okay, so I'm getting -411/.072 = -5,708 K.

Or at least, 2 * that, for 2 mol.

[Edited on 9-8-2023 by Neal]


If you get a negative temperature in this calculation, then the reaction never has K = 1. either K is always larger, or K is always smaller.

If you change how you write the equation (such as Na + 1/2 Cl2 --> NaCl vs 2 Na + Cl2 --> 2 NaCl), you'll double both deltaH and deltaS, but the ratio (and thus T) will remain unchanged.




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[*] posted on 9-8-2023 at 09:17


Quote: Originally posted by DraconicAcid  
Quote: Originally posted by Neal  
Quote: Originally posted by DraconicAcid  
When K = 1, DeltaG(standard) is zero. T = deltaH/deltaS.

Okay, so I'm getting -411/.072 = -5,708 K.

Or at least, 2 * that, for 2 mol.

[Edited on 9-8-2023 by Neal]


If you get a negative temperature in this calculation, then the reaction never has K = 1. either K is always larger, or K is always smaller.

If you change how you write the equation (such as Na + 1/2 Cl2 --> NaCl vs 2 Na + Cl2 --> 2 NaCl), you'll double both deltaH and deltaS, but the ratio (and thus T) will remain unchanged.

What about this equation.

H2O(l) <-> H2O(s)

Equilibrium should be at 0 C, right? Water and ice.

My chem textbook doesn't have the values for ice, only liquid water and steam.

But I'll use liquid water.

But even getting -285/.07 = -4071 K. Something doesn't seem right here.

[Edited on 9-8-2023 by Neal]
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[*] posted on 9-8-2023 at 10:21


It looks like you're dividing the enthalpy of formation of liquid water by its absolute entropy. You have to use reaction enthalpies and reaction entropies, or you're calculating nonsense.

My handy textbook doesn't have ice in it, but it does have steam. So let's look at H2O(l) <==> H2O(g)

Enthalpy of rxn = products - reactants = -241.83 kJ/molk - -285.83 kJ/mol = +44.00 kJ/mol (deltaHrxn)
Entropy of rxn = products - reactants = 188.84 J/molK - 69.95 J/moK = + 118.89 J/molK (deltaSrxn)

T = 44 000 J/mol / 118.89 J/molK = 370.1 K (96.9 C, because enthalpy and entropy of rxn isn't entirely independent of temperature)

Does that make more sense?




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[*] posted on 9-8-2023 at 14:47


Okay, so 2NaCl should be.

Delta H = 2* -410 - (2*0 + 0) = -820
Delta S = 2*.072 - (2*.051 + .2223) = .144 - .3243 = -.1803

-820/-.1803 = 4547 K sheesh.

I will try Na + 1/2 Cl2 -> NaCl

Delta H = -410

Delta S = .072 - .2733 = -.2013

-410/-.2013 = 2036 K

Ah, so ratio is not the same.

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[*] posted on 9-8-2023 at 15:58


Using your numbers, deltaS = 0.072 - (0.051 + 0.2223/2) = -0.09015 kJ/molK

-410 kJ/mol / -0.09015 kJ/molK = 4547 K

Same thing.




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[*] posted on 9-8-2023 at 16:13


Quote: Originally posted by DraconicAcid  
Using your numbers, deltaS = 0.072 - (0.051 + 0.2223/2) = -0.09015 kJ/molK

-410 kJ/mol / -0.09015 kJ/molK = 4547 K

Same thing.

Ah. 1/2 Cl2 I forgot to / by 2.

Cool thanks.

Okay so we're at 4547 K, or 4274 C.

Here's something to make it weirder. NaCl has a melting point, and bp.

This temperature is above both.

So, at this temp NaCl wouldn't exist as a solid. Making it quite contradictory to be at equilibrium?

Ironically, my text does have values for NaCl liquid and gas. As well as Na for liquid and gas. And this is above Na's mp and bp too. So I could run 2 versions of this in liquid and gas phase. Weird.
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[*] posted on 9-8-2023 at 16:39


Well yeah, that’s fully logical. When you heat NaCl up to its melting point it melts, but it doesn’t start spontaneously splitting into its constituent elements… that reaction is still very far from its equilibrium. There is no instance that you would have Na(s) Cl2(g) and NaCl(s) hanging out together in a state of equilibrium. But you can imagine that at an extremely high temperature, at which NaCl is a gas, under certain conditions, the drive of entropy could make it favorable for it to split. Then it would be at equilibrium.



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[*] posted on 9-8-2023 at 17:17


Quote: Originally posted by Neal  


Here's something to make it weirder. NaCl has a melting point, and bp.

This temperature is above both.

So, at this temp NaCl wouldn't exist as a solid. Making it quite contradictory to be at equilibrium?

Ironically, my text does have values for NaCl liquid and gas. As well as Na for liquid and gas. And this is above Na's mp and bp too. So I could run 2 versions of this in liquid and gas phase. Weird.


That's not really weird. The values we use are for 298 K, so our calculations are only good for temperatures close to that.

Above the melting point of sodium, our calculations are useless, because we don't have solid sodium. We'd have to recalculate the things with liquid sodium instead (and at higher temperatures, gaseous sodium, and liquid NaCl).

Now, this doesn't mean that this reaction can never be at equilibrium at reasonable temperatures. You would just have infinitesimal quantities of sodium and chlorine coexisting with the salt.




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[*] posted on 9-8-2023 at 17:36


Does Na "liquid or gas" can react with Cl2 to get NaCl liquid or gas?

Otherwise, I feel no point in doing those calculations with liquid/gas.

Quote: Originally posted by DraconicAcid  

If you get a negative temperature in this calculation, then the reaction never has K = 1.

So has anyone ever got temps below absolute zero? I'm thinking now it isn't possible...
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[*] posted on 9-8-2023 at 19:39


Yes, liquid sodium will also react with chlorine to give sodium chloride. I suspect gaseous sodium would also react.

You cannot have a temperature below 0 K. That would imply that the particles had negative amounts of kinetic energy.




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[*] posted on 10-8-2023 at 08:55


Well, *any* compound will disintegrate above a few thousand K.
I think, even CO2 gets unstable above 1000 C (decomposing to CO and O2 ?), water above 2000 C.
At coolest (M class, such as Betelgeuse) star temperatures, only simple compounds like CO, TiO still exist.
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[*] posted on 13-8-2023 at 08:37


Quote: Originally posted by DraconicAcid  
It looks like you're dividing the enthalpy of formation of liquid water by its absolute entropy. You have to use reaction enthalpies and reaction entropies, or you're calculating nonsense.

My handy textbook doesn't have ice in it, but it does have steam. So let's look at H2O(l) <==> H2O(g)

Enthalpy of rxn = products - reactants = -241.83 kJ/molk - -285.83 kJ/mol = +44.00 kJ/mol (deltaHrxn)
Entropy of rxn = products - reactants = 188.84 J/molK - 69.95 J/moK = + 118.89 J/molK (deltaSrxn)

T = 44 000 J/mol / 118.89 J/molK = 370.1 K (96.9 C, because enthalpy and entropy of rxn isn't entirely independent of temperature)

Does that make more sense?

97 C, I'm so glad the result was below the bp of water and not above. Since bp is so subjective to me, as you can have steam exist at room temperature far below the bp, due to something called vapor pressure.

So for ice and water, I was wondering if the calculated would actually be above or below the freezing point, if it isn't on there exactly.

Wikipedia has it for ice.

https://en.wikipedia.org/wiki/Water_(data_page)#Thermodynamic_properties

Scroll to solid properties.

H = -293 - -286 = -7
S = .041 - .070 = -.029

T = -7/-.029 = 241 K, = -32 C.

-32 C a lot farther from 0 C as 97 C to 100 C.

Here's the thing. Our text doesn't have the enthalpy of formation for ice, but it does have for steam.

Well ice can't exist at 25 C..., but steam can. So the heat of formation for steam, that we used, are for room temperature, not 100 C, so that affects calculations?

It won't affect any Na + 1/2 Cl -> NaCl since they are all consistent at 25 C, but using any

H2O solid/liquid <-> H2O liquid/gas

is inherently flawed if we're only looking at values at room-temperature?

In any event, Wikipedia didn't cite their sources, having all 3 states of water, but ice can't exist at room temperature unless different pressure.

[Edited on 13-8-2023 by Neal]
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[*] posted on 13-8-2023 at 11:22


Try labeling each value as you write them down, i think your getting your varables mixed up.(see edit)

Code:
Formula State Enthalpy (kJ/mol) Entropy (J mol/K) H2O (g) -241.818464 188.715136 H2O (l) -285.82996 69.91464 H2O (s) -291.83996 47.91464

data taken from https://www.drjez.com/uco/ChemTools/Standard%20Thermodynamic... March 6 2022 
Runnig the math I get these tempatures

H2Osolid = H2Oliquid
Code:
T = 273.15  Kelvin  ΔH  6.0100  kJ/mol ΔS  22.0000  J mol/K ΔG  0.0007  kJ/mol  Keq= 0.999691809


H2Oliquid = H2OGas
Code:
T = 370.47 Kelvin ΔH 44.0115 kJ/mol ΔS 118.8005 J mol/K ΔG -0.0005 kJ/mol Keq= 1.000170059


[Edited on 13-8-2023 by Rainwater]
Running the numbers off Wikipedia i get the same values.
That's interesting, must be a flaw in the data somewhere

[Edited on 13-8-2023 by Rainwater]




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