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

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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

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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#2

Firstly, I'd like to express my confusion as to how this topic is presented to you Secondly, how rusty my thermodynamics is...

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 is a measure of the total energy of a system/reaction. If 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.

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

the full formula is:

where is the Gibbs free energy under

At equilibrium and so

so what has this got to do with your equation?

which exam board is this?

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 is a measure of the total energy of a system/reaction. If 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.

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

the full formula is:

where is the Gibbs free energy under

__standard conditions__At equilibrium and so

so what has this got to do with your equation?

which exam board is this?

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(Original post by

Firstly, I'd like to express my confusion as to how this topic is presented to you Secondly, how rusty my thermodynamics is...

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 is a measure of the total energy of a system/reaction. If 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.

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

the full formula is:

where is the Gibbs free energy under

At equilibrium and so

so what has this got to do with your equation?

which exam board is this?

**EierVonSatan**)Firstly, I'd like to express my confusion as to how this topic is presented to you Secondly, how rusty my thermodynamics is...

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 is a measure of the total energy of a system/reaction. If 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.

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

the full formula is:

where is the Gibbs free energy under

__standard conditions__At equilibrium and so

so what has this got to do with your equation?

which exam board is this?

-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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#4

(Original post by

-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..

**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..

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:

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:

and so

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

then when Q = K:

and

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.

Bit of a nightmare isn't it?

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(Original post by

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 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

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:

and as equilibrium is established Q tends towards K and so then the Gibbs free energy tends towards zero,

and so

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

then when Q = K:

and

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

Bit of a nightmare isn't it?

**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 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:

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:and so

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

then when Q = K:

and

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

Bit of a nightmare isn't it?

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#6

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.

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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