How Does ΔS(total) = R*lnK Align with Entropy Being Zero at Equilibrium?

[jsmith613]

Given that at equilibrium total entropy change = 0
how does the following equation make sense
ΔS(total) = R*lnK
if ΔS(total) is 0?
thanks

 

[Borek]

how does the following equation make sense
ΔS(total) = R*lnK

Units don't match, so it doesn't make sense. Units of entropy are JK^-1, units of ideal gas constant are JK^-1mol^-1.

 

[jsmith613]

no units of entropy are J K^-1 mol^-1

 

[jsmith613]

oh Standard entropy = R lnK
my bad

nonetheless how can a standard entropy change of zero have an equation R*lnK

 

[Mike H]

Questions to ask yourself:

1.) What is K in your equation?

2.) What does K equal when the system is at equilibrium?

If you do this, the answer will fall right into your lap.

 

[jsmith613]

1) K is the equilibrium constant
2) when the system is at equilibrium K = 1

...so your saying that ONLY at the point of equilibrium does total entropy change = zero
...i am still confused
sorry :(

 

[jsmith613]

Questions to ask yourself:

1.) What is K in your equation?

2.) What does K equal when the system is at equilibrium?

If you do this, the answer will fall right into your lap.

in fact K does NOT nessecaryily = 1 at equilibrium
...therefore how can the two be related

 

[jsmith613]

In fact:
http://www.edexcel.com/migrationdoc...0/January 2011 - QP/6CH04_01_que_20110126.pdf

How can q17 work
surely as its an equilibrium the total entropy change should be 0

I am totally confused :S please can someone put me right

 

[Mike H]

I have to say I've never seen this particular twist in presentation of thermo before, which is why my earlier "shot-from-the-hip" answer isn't right. And I should probably read more carefully...

It's as if your text/reference material is trying to rework what is normally seen via Gibbs free energy statements in terms of entropy for whatever inexplicable reason - if you'll remember, we have the well-known equality

ΔG = ΔG° + RT*ln(Q)

where ΔG = 0 at equilibrium, and as Q → K at equilibrium, we have

ΔG° = -RT*ln(K).

It seems as if your text is trying to do something similar in terms of ΔS and ΔS° for whatever reason, such that

ΔS = ΔS° - R*ln(K),

so when ΔS = 0,

ΔS° = R*ln(K).

I suppose it's valid, although I've never seen it presented this particular way, at least that I can recall.

 

[jsmith613]

I have to say I've never seen this particular twist in presentation of thermo before, which is why my earlier "shot-from-the-hip" answer isn't right. And I should probably read more carefully...

It's as if your text/reference material is trying to rework what is normally seen via Gibbs free energy statements in terms of entropy for whatever inexplicable reason - if you'll remember, we have the well-known equality

ΔG = ΔG° + RT*ln(Q)

where ΔG = 0 at equilibrium, and as Q → K at equilibrium, we have

ΔG° = -RT*ln(K).

It seems as if your text is trying to do something similar in terms of ΔS and ΔS° for whatever reason, such that

ΔS = ΔS° - R*ln(K),

so when ΔS = 0,

ΔS° = R*ln(K).

I suppose it's valid, although I've never seen it presented this particular way, at least that I can recall.

what is the difference between ΔS° and ΔS
and are they both calculated in the same way