Sciencemadness Discussion Board - How do you find the temperature at equilibrium?

If you specifically want K to be equal to 1, then deltaG will be zero. (Since deltaG = -RTlnK)

deltaG = deltaH - TdeltaS. DeltaH and deltaS don't change much with temperature, so solve for T.

I'm getting like -6 Kelvin. But this is similar to what I got in my original equations. Ah well.

DraconicAcid - 4-8-2023 at 08:24

If you specifically want K to be equal to 1, then deltaG will be zero. (Since deltaG = -RTlnK)

deltaG = deltaH - TdeltaS. DeltaH and deltaS don't change much with temperature, so solve for T.

I'm getting like -6 Kelvin. But this is similar to what I got in my original equations. Ah well.

And that means that the reaction is never spontaneous. A lot of reactions are like that.

Neal - 4-8-2023 at 08:28

If you specifically want K to be equal to 1, then deltaG will be zero. (Since deltaG = -RTlnK)

deltaG = deltaH - TdeltaS. DeltaH and deltaS don't change much with temperature, so solve for T.

I'm getting like -6 Kelvin. But this is similar to what I got in my original equations. Ah well.

And that means that the reaction is never spontaneous. A lot of reactions are like that.

Even though the delta G values are negative?

DraconicAcid - 4-8-2023 at 09:23

Sorry- it's either always spontaneous, or never spontaneous.

2 Na + Cl2 -> 2 NaCl will happen at all temperatures.

2 NaCl -> 2 Na + Cl2 will not happen at any temperature.

unionised - 4-8-2023 at 09:47

2 NaCl -> 2 Na + Cl2 will not happen at any temperature.

Yes it will.

Even the salt you are sprinkling onto your food is, to an unimaginably small extent, dissociating into Na and Cl2.

The extent of the decomposition can be calculated from the equations early in the thread and it might be " a 1 in a billion chance of a single Na atom in a block of salt the size of Jupiter", but it's not zero.

And, at a high enough temperature there's quite a lot of Na present.
That's how it can make a Bunsen flame yellow.

That's the whole point of equilibria.

DraconicAcid - 4-8-2023 at 09:50

Quote: Originally posted by unionised

2 NaCl -> 2 Na + Cl2 will not happen at any temperature.

But no matter the temperature, you'll never get the equilibrium constant up to 1. Which is what the OP was talking about.

Neal - 4-8-2023 at 10:21

So then how does NaCl work with Le Chatelier's principle? You can't put too much product or reactant to get it to reverse itself?

SplendidAcylation - 5-8-2023 at 03:53

This is the sort of stuff I am trying to understand at the moment as well.

The formula "DeltaG(standard) = -RT ln K" relates the equilibrium constant to the standard Gibbs Free Energy for the forward reaction.

I initially thought that this could be used to find the equilibrium constant at different temperatures, but I since discovered that this is incorrect.

For instance, if you are given the equilbrium constant (K) for a reaction at a temperature of 400K, then you can find the standard Gibb's Free Energy at 400K using this formula:

DeltaG = -R*400 ln K

You can't use it to find the standard Gibbs Free Energy at 300K if you only know the equilibrium constant at 400K.

Alternatively, if you have the standard Gibb's Free Energy for the reaction at 400K, then of course you can work out the equilibrium constant.

If you wish to find the equilibrium constant at a different temperature from one you have been given, you need to use the Van 't Hoff equation.

Of course, you can work out the standard Gibb's Free Energy at any temperature, if you know the standard reaction enthalpy and the standard entropy, using the relation:

DeltaG = DeltaH - T*DeltaS

However, if you've only been given the DeltaG, and the temperature, then you can't find DeltaH or DeltaS from this alone, and therefore you can't work out DeltaG at a different temperature.

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

DeltaG = DeltaG(standard) + RT ln Q

This tells you the Gibbs Free Energy for the reaction if you know the standard Gibbs Free Energy at that temperature, AND the reaction quotient, Q.

For a reaction of the form:

wA + xB <--> yC + zD

Q= [yC][zD] / [wA][xB]

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

DeltaG = DeltaG(standard) + RT ln Q

Substituting Q=K and DeltaG = 0:

0 = DeltaG(standard) + RT ln K

DeltaG(standard) = -RT ln K

-----------------------------

When you have a system as equilibrium, and then you add one of the reactants or products, or change something else, that disturbs the equilibrium, then the system is no longer at equilibrium, and Q is no longer equal to K, and DeltaG is no longer 0.

The system will then move towards equilibrium; If DeltaG is >0 then the reaction will move towards equilibrium in the backwards direction, if DeltaG <0, then it will move in the forward direction.
Eventually, equilibrium will again be attained.

Note: This is all based on my tenuous understanding of the topic, I'd appreciate it if anyone would correct any mistakes I've probably made!

[Edited on 5-8-2023 by SplendidAcylation]

unionised - 5-8-2023 at 04:12

There are two effects to consider.
One is the energy change that goes into making and breaking bonds.
That's not strongly dependent on temperature. It's Delta H

And the other is the energy change due to changes in entropy- if a material turns from a solid to a gas, for example.
That is strongly dependent on temperature. It's T times Delta S

The equilibrium depends on balancing those two terms.

At the point where delta H = T delta S the system is at equilibrium

(Because delta G at that point is zero and the log of 1 is zero)

unionised - 5-8-2023 at 04:17

But no matter the temperature, you'll never get the equilibrium constant up to 1. Which is what the OP was talking about.

You will if it's hot enough.
As you raise the temperature, more material will dissociate. The equilibrium constant for dissociation rises.

What stops it reaching 1 (or more)?

SplendidAcylation - 5-8-2023 at 04:35

Sorry- it's either always spontaneous, or never spontaneous.

2 Na + Cl2 -> 2 NaCl will happen at all temperatures.

2 NaCl -> 2 Na + Cl2 will not happen at any temperature.

NaCl decomposing to form sodium and chlorine is endothermic, with an increase in entropy.

DeltaH is therefore positive, and DeltaS is also positive

DeltaG=DeltaH-T*deltaS

In order for DeltaG to be below zero, T has to be high enough so that T*DeltaS is bigger than DeltaH.

So this reaction will happen at a high enough temperature.

Neal - 5-8-2023 at 05:06

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

DeltaG = DeltaG(standard) + RT ln Q

This tells you the Gibbs Free Energy for the reaction if you know the standard Gibbs Free Energy at that temperature, AND the reaction quotient, Q.

For a reaction of the form:

wA + xB <--> yC + zD

Q= [yC][zD] / [wA][xB]

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

DeltaG = DeltaG(standard) + RT ln Q

Substituting Q=K and DeltaG = 0:

0 = DeltaG(standard) + RT ln K

DeltaG(standard) = -RT ln K

Yea this is another catch-22.

So at equilibrium, Delta G equals 0 and K = 1. Ln (1) = 0.

So you get.

0 = T*0

T can be anything. It can be absolute zero, room temp, oven temp, etc.

So this goes back to what DraconicAcid said?

[Edited on 5-8-2023 by Neal]

SplendidAcylation - 5-8-2023 at 05:35

Oh I see what you mean... So you are referring to the particular circumstances when K=1...

I have no idea why this is, I'd be interested to find out the explanation!

Is there something special about K=1?

DraconicAcid - 5-8-2023 at 09:04

No. At equilibrium, deltaG = 0. deltaG(standard) is a function of the reaction, and does not depend on whether the reaction is at equilibrium or not.

K depends on deltaG(standard), not deltaG.

unionised - 6-8-2023 at 00:45

Oh I see what you mean... So you are referring to the particular circumstances when K=1...

I have no idea why this is, I'd be interested to find out the explanation!

Is there something special about K=1?

Yes.
K=1 is the point where you get as much reactants as products (At least in simple cases).
It's the situation where the forward reactions and backward reactions go at the same rate.

Neal - 6-8-2023 at 05:49

Substituting Q=K and DeltaG = 0:

0 = DeltaG(standard) + RT ln K

DeltaG(standard) = -RT ln K

Okay, with what DraconicAcid said, the math doesn't add up now.

You now got.

<SomethingThatisNot0> = -RT ln (1)
<SomethingThatisNot0> = -RT * 0

Weird.

[Edited on 6-8-2023 by Neal]

Rainwater - 6-8-2023 at 15:26

Several things that I feel should be pointed out and underlined

1) the gibbs free energy only calculates work as heat and pressure

Heat in this context being thermal radiation. There are other forms of energy that are within the system but not being accounted for. Most of the time, their contributions are negligible.

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

3) In an extreme example
It was suggested above, a flame test with salt
2NaCl = 2Na + Cl₂
Gibbs perdicted at tempatures above 2400c this reaction would be spontaneous.
At these temperatures, plasma and gas would be the states of matter.
Essentially turning the NaCl into Na⁺ and Cl^- (this is probably incorrect in some way, but hopefully you get the idea im trying to convey)

But the flame test occurs at lower temperatures than this.
So you must also consider alternative reactions at test conditions.
Products of combustion and such.

Quote:

So you get.

0 = T*0

T can be anything.

Thats simply the frustration required to reach understanding.
It means your really trying.

To be more on point, delta g(standard) cannot be at any temperature. Thats why its called standard, to specify what temperature it implies. 273.15 kelvin i think.

It may be helpful for you simplifie those equations again, but replace all the standard delta g values with the function ΔG = ΔH - TΔS for each particular compound.

Spreadsheets are the way to go, as graphing result is helpful in understanding the processes effect as things change.

DraconicAcid - 6-8-2023 at 17:06

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
...
To be more on point, delta g(standard) cannot be at any temperature. Thats why its called standard, to specify what temperature it implies. 273.15 kelvin i think.

Enthalpies and entropies of reaction don't change greatly with temperature. They do change VERY significantly with phase changes, though.

DeltaG(standard) can actually be at different temperatures. Most phys chem books will quote the definition of standard as 1 atm for all gases, all solutes are 1 mol/L, and T = 298.15 K unless otherwise specified. You can use all the equations for deltaG(standard) at a different temperature, provided you're using the thermodynamic measurements at that temperature. You cannot, however, change the concentrations or pressures of the species involved without making it nonstandard.

DraconicAcid - 6-8-2023 at 17:12

Substituting Q=K and DeltaG = 0:

0 = DeltaG(standard) + RT ln K

DeltaG(standard) = -RT ln K

Okay, with what DraconicAcid said, the math doesn't add up now.

You now got.

<SomethingThatisNot0> = -RT ln (1)
<SomethingThatisNot0> = -RT * 0

Weird.

[Edited on 6-8-2023 by Neal]

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

Rainwater - 7-8-2023 at 04:48

https://youtu.be/4ZcTEiJru84
A video explanation i found

Neal - 7-8-2023 at 04:51

Substituting Q=K and DeltaG = 0:

0 = DeltaG(standard) + RT ln K

DeltaG(standard) = -RT ln K

Okay, with what DraconicAcid said, the math doesn't add up now.

You now got.

<SomethingThatisNot0> = -RT ln (1)
<SomethingThatisNot0> = -RT * 0

Weird.

[Edited on 6-8-2023 by Neal]

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

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

A contradiction still.

Neal - 7-8-2023 at 05:04

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.

Texium - 7-8-2023 at 05:19

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

Neal - 7-8-2023 at 05:29

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]

DraconicAcid - 7-8-2023 at 07:29

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

DeltaG = DeltaG(standard) + RT ln Q

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

Neal - 7-8-2023 at 10:26

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

DeltaG = DeltaG(standard) + RT ln Q

No. When Q =1 , then deltaG = deltaG^o, 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.

DraconicAcid - 7-8-2023 at 12:24

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

DeltaG = DeltaG(standard) + RT ln Q

No. When Q =1 , then deltaG = deltaG^o, 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 deltaG^o is not zero at equilibrium.

Neal - 7-8-2023 at 12:33

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

DeltaG = DeltaG(standard) + RT ln Q

No. When Q =1 , then deltaG = deltaG^o, 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 deltaG^o 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]

DraconicAcid - 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.

Neal - 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.

DraconicAcid - 7-8-2023 at 23:30

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

Rainwater - 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
0_{N2 ΔH} + 3 * 0_{H2 ΔH} = 0_{ΔH reagents}
We have 3 hydrogen molecules so that value has to be multiplied by 3.

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

Products
2 * -46.10768_{NH3 ΔH} = -92.21536_{ΔH 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(K_eq)
There is no ΔH or ΔS here. So ignore their existence for a moment.

K_eq 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, K_eq = 1
0 = -RT_{we calculated above}*ln(K_eq)
Given that K_eq 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 1_equilibrium 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]

Neal - 9-8-2023 at 07:09

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]

DraconicAcid - 9-8-2023 at 08:48

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.

Neal - 9-8-2023 at 09:17

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]

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]

DraconicAcid - 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?

Neal - 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.

DraconicAcid - 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.

Neal - 9-8-2023 at 16:13

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.

Texium - 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.

DraconicAcid - 9-8-2023 at 17:17

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.

Neal - 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.

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...

DraconicAcid - 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.

metalresearcher - 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.

Neal - 13-8-2023 at 08:37

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]

Rainwater - 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

H₂O_solid = H₂O_liquid

Code:

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

H₂O_liquid = H₂O_Gas

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]

Neal - 14-8-2023 at 06:08

Code:

Formula State  Enthalpy (kJ/mol) Entropy (J mol/K) 

H2O	(s)	-291.83996	47.91464

So replace -293 with -292 and .041 with .048.

So -6/-.022 instead of -7/-.029

T = 273 K or 0 C instead of -32 C. So Wikipedia's values sucked.

[Edited on 14-8-2023 by Neal]

Neal - 14-8-2023 at 06:19