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Topic: Difference between K_eq, K_p, K_c, and K  (Read 3575 times)

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Offline burkeas

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Difference between K_eq, K_p, K_c, and K
« on: December 21, 2014, 01:13:18 AM »
Hi,

I have a question that was sparked by a recent exam...
We are given the formula delta G standard=-RT*ln(K). However, I do not understand exactly what K to use there.

Generally, I have seen that K be referred to as K_eq. How does that relate to K_c and K_p? Or is there another value that is just K? My teacher seems to think that K_eq can be K_p or K_c, but I do not understand how that could be possible... since K_p=K_c (RT)^ΔN, it does not seem that K_p and K_c would be interchangeable in the above equation.

Hopefully my question is somewhat coherent. The exact question on the exam gave a delta G standard value and asked for the value of K_p... I used the ΔG standard=-RT*ln(K) equation, but I then used K_p=K_c (RT)^ΔN because I thought K in the first equation was K_c... I got it wrong because apparently K in the ΔG standard=-RT*ln(K) can be K_c OR K_p, according to my teacher. Help please?
Thanks in advance

Offline mjc123

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Re: Difference between K_eq, K_p, K_c, and K
« Reply #1 on: December 22, 2014, 06:49:59 PM »
Keq means the equilibrium constant and is the same as K. For gas phase reactions, K or Keq can be expressed in terms of either the pressures (Kp) or concentrations (Kc), which are related, as you say, by Kp = Kc(RT)Δn. The equation ΔG° = -RTlnK applies to both forms of K, but the reference standard states are different in each case. When you use pressures, the standard state is gases at 1 atm pressure. When you use concentrations, the standard state is a concentration of 1M. This means that ΔG° is different in the two cases. From the above relation, ΔG°(c) = ΔG°(p) + RTΔnln(RT). (Try confirming this by using ΔS = nRln(V2/V1) for the entropy change of a gas on expansion/contraction.)
But you will find that tabulated thermodynamic data for gases, e.g. ΔG°f(CO2), is usually referenced to a standard pressure of 1 atm, so if you are given a value of ΔG° for a gas reaction, you should assume unless specified otherwise that the standard state is 1 atm, and the corresponding value of K is Kp. If you want Kc you then work it out from Kp using the above relation.

Offline mjc123

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Re: Difference between K_eq, K_p, K_c, and K
« Reply #2 on: December 23, 2014, 06:37:02 PM »
Just a caution about units. In  Kp = Kc(RT)Δn, to convert atm to mol/L, you need to use a value of R in L atm/mol/K. In ΔG° = -RTlnK, in SI units R should be in J/mol/K. In the equation ΔG°(c) = ΔG°(p) + RTΔnln(RT), the two Rs must be in different units, so it's probably not a very good equation to use.

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