A2 Chem: deltaS = RLnK - help with position of dynamic equilibria?? Watch

Coke Or Pepsi
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Hey everyone...

I'm almost 100% sure my textbook has a mistake. This has been driving me crazy.

It says that in dynamic equilibria deltaS total = 0. The forward and the reverse occur both spontaneously and thus both have equal entropy changes.

Then I look at the formula deltaStotal = RLnK

This shows me that if deltaS total IS zero in dynamic equilibria, then the equilibrium constants Kc and Kp are always one? Ie the equilibrium sits in the middle??

Is this correct or has my book made a mistake. The reason why I think it's incorrect is because Kc = [products]/[reactants]

This doesn't mean that just because something is happening at dynamic equilibrium, that the concentration of products and reactants are equal... There are many examples where dynamic equilibria sit to the left or the right....


Thank you! Any help appreciated xx
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EierVonSatan
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Firstly, I'd like to express my confusion as to how this topic is presented to you Secondly, how rusty my thermodynamics is...:p:

Okay, so I don't know if you've come across the concept of Gibbs free energy yet, but it has an important use in equilibria and spontaneity of reactions.

Gibbs free energy \Delta G is a measure of the total energy of a system/reaction. If \Delta G is negative a reaction is possible on it's own (spontaneous), if it's positive then it's not spontaneous. When it is equal to zero, then the system is in equilibrium.

\Delta G is therefore related to the equilibrium constant K in some way.

the full formula is:

\Delta G = \Delta G^o + RT ln K where \Delta G^o is the Gibbs free energy under standard conditions

At equilibrium \Delta G = 0 and so \Delta G^o = -RT ln K

so what has this got to do with your equation?

\Delta G = -T \Delta S_{Total}

which exam board is this?
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Coke Or Pepsi
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(Original post by EierVonSatan)
Firstly, I'd like to express my confusion as to how this topic is presented to you Secondly, how rusty my thermodynamics is...:p:

Okay, so I don't know if you've come across the concept of Gibbs free energy yet, but it has an important use in equilibria and spontaneity of reactions.

Gibbs free energy \Delta G is a measure of the total energy of a system/reaction. If \Delta G is negative a reaction is possible on it's own (spontaneous), if it's positive then it's not spontaneous. When it is equal to zero, then the system is in equilibrium.

\Delta G is therefore related to the equilibrium constant K in some way.

the full formula is:

\Delta G = \Delta G^o + RT ln K where \Delta G^o is the Gibbs free energy under standard conditions

At equilibrium \Delta G = 0 and so \Delta G^o = -RT ln K

so what has this got to do with your equation?

\Delta G = -T \Delta S_{Total}

which exam board is this?
Edexcel. The specification changed. And I know about Gibbs - how it relates to what I'm saying is (well I thought but you've now definitely confirmed thank you )

-TdeltaStotal = Gibbs free energy. That's what the value of G actually is. A reaction is spontaneous when deltaStotal is positive. As temperature is in kelvin Ie always positive, you can see from my equation that G will be negative. They're the same thing just slightly different. We don't do it in terms of G because it's part of the Entropy unit.

Ok so when a system is in equilibrium Gibbs is zero? Meaning that deltaS total is zero..

But I don't get it because I was under the impression that there are dynamic equilibria where the equilibrium doesn't necessarily sit in the middle?? Or was I wrong.
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EierVonSatan
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(Original post by Coke Or Pepsi)
-TdeltaStotal = Gibbs free energy. That's what the value of G actually is. A reaction is spontaneous when deltaStotal is positive. As temperature is in kelvin Ie always positive, you can see from my equation that G will be negative. They're the same thing just slightly different. We don't do it in terms of G because it's part of the Entropy unit.

Ok so when a system is in equilibrium Gibbs is zero? Meaning that deltaS total is zero..
It's just an unusual way of going about doing the topic, I for one have never seen the equation in this form before - it's just a bit bizarre to me :lolwut: but okay whatever...

In an isolated reversible process the total Gibbs free energy doesn't change - there is no net energy in or out. In the same fashion the total entropy will not be increasing or decreasing. This is different to saying that the change in the standard Gibbs free energy (and thus the standard entropy) is zero.

Imagine you have a reversible reaction not yet at equilibrium and instead we had a ratio of [products]/[reactants] which we will call Q ('the reaction quotient') then the Gibbs free energy is:

\Delta G = -RT ln Q + RT ln K

and as equilibrium is established Q tends towards K and so then the Gibbs free energy tends towards zero, not the standard Gibbs free energy. When Q = K then:

\Delta G = 0 and so \Delta G^o = -RT ln K

Rewriting the above to see how it relates to the entropy (using the last equation in my previous post):

\Delta S_{Tot} = R ln Q - R ln K

then when Q = K:

\Delta S_{Tot} = 0 and \Delta S_{Tot}^o = R ln K

But I don't get it because I was under the impression that there are dynamic equilibria where the equilibrium doesn't necessarily sit in the middle?? Or was I wrong.
No you're right, the equilibrium does not sit in the middle in general

Bit of a nightmare isn't it? :p:
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Coke Or Pepsi
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(Original post by EierVonSatan)
It's just an unusual way of going about doing the topic, I for one have never seen the equation in this form before - it's just a bit bizarre to me :lolwut: but okay whatever...

In an isolated reversible process the total Gibbs free energy doesn't change - there is no net energy in or out. In the same fashion the total entropy will not be increasing or decreasing. This is different to saying that the change in the standard Gibbs free energy (and thus the standard entropy) is zero.

Imagine you have a reversible reaction not yet at equilibrium and instead we had a ratio of [products]/[reactants] which we will call Q ('the reaction quotient') then the Gibbs free energy is:

\Delta G = -RT ln Q + RT ln K

and as equilibrium is established Q tends towards K and so then the Gibbs free energy tends towards zero, not the standard Gibbs free energy. When Q = K then:

\Delta G = 0 and so \Delta G^o = -RT ln K

Rewriting the above to see how it relates to the entropy (using the last equation in my previous post):

\Delta S_{Tot} = R ln Q - R ln K

then when Q = K:

\Delta S_{Tot} = 0 and \Delta S_{Tot}^o = R ln K



No you're right, the equilibrium does not sit in the middle in general

Bit of a nightmare isn't it? :p:
So if not every equilibrium sits in the middle... don't the two theories contradict each other?
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thefinker
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Using a calculator ...

deltaS total = R lnK (where ln is natural log).

so deltaStotal/R = lnK

As deltaStotal = 0, deltaStotal/R = 0 so 0 = lnK

So exp(0) = K

1= K

As to why something would have a 1 for K, as long as both deltaHrxn and deltaSsys have different signs, deltaStotal has to be zero at some temperature.

A quick example using the Haber process

deltaSsys298=-199 J K-1
deltaH-92.4 kJ mol-1 so - -92400/T + -199 = deltaStotal
deltaStotal= 0 at 464.3 K which agrees with G Facer Edexcel A2 Chemistry 2nd Edition page 81.

I think that's all you need for Paper 4 which is simpler than above.
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