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Neal
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How do you find the temperature at equilibrium?
Say 2 Na + Cl2 -> 2 NaCl.
There is a formula, where K = e ^ -(delta)G / RT
And just plug in 1 for K? Well, problem is, Delta G, and R, are all defined for 298 K. So this formula can only work for T = 298 K. So this formula is
not what I'm looking for...
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Antigua
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Firstly, the Gas Constant is not defined for 298K. It's a constant. It's value depends only on the Boltzmann constant and Avogadro constant (R =
Kb*NA), which both do not depend on temperature. They're constant.
Secondly ∆G (change of Gibbs free energy) depends on temperature. Standard change of Gibbs free energy (denoted as ∆G°) is defined for some set
of standard conditions. But G depends on the temperature according to it's definition: dG = dH - TdS - SdT for the differential form.
You're gonna have to read up a bit on where these equations come from and what they mean to use them correctly. Take a look at the Van't Hoff
equation.
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Neal
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Oh, this 1?
In this equation K1 is the equilibrium constant at absolute temperature T1, and K2 is the equilibrium constant at absolute temperature T2.
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DraconicAcid
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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.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Neal
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Quote: Originally posted by DraconicAcid  | 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.
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DraconicAcid
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| Quote: Originally posted by Neal | | Quote: Originally posted by DraconicAcid | 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.
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Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Neal
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| Quote: Originally posted by DraconicAcid | | Quote: Originally posted by Neal | | Quote: Originally posted by DraconicAcid | 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?
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DraconicAcid
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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.
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unionised
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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.
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DraconicAcid
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| Quote: Originally posted by unionised |
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. |
But no matter the temperature, you'll never get the equilibrium constant up to 1. Which is what the OP was talking about.
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Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Neal
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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?
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SplendidAcylation
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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]
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unionised
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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)
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unionised
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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)?
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SplendidAcylation
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| Quote: Originally posted by DraconicAcid | 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.
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Neal
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| 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
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]
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SplendidAcylation
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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?
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DraconicAcid
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| Quote: Originally posted by Neal |
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] |
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.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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unionised
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| Quote: Originally posted by SplendidAcylation | 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.
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Neal
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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]
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Rainwater
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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.
"You can't do that" - challenge accepted
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DraconicAcid
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| 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
...
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.
Please remember: "Filtrate" is not a verb.
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DraconicAcid
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| Quote: Originally posted by Neal |
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 DeltaGo = 0, then K will be equal to 1. At any other temperature, K will not be 1, and
DeltaGo will not be zero.
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Rainwater
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https://youtu.be/4ZcTEiJru84
A video explanation i found
"You can't do that" - challenge accepted
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Neal
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| Quote: Originally posted by DraconicAcid | | Quote: Originally posted by Neal |
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 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.
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