## K vs Kc

StephanieGrigorian2J
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Joined: Wed Sep 30, 2020 10:01 pm

### K vs Kc

Can someone explain how to convert between K and Kc? This was mentioned in the textbook and I'm confused on how to do this/what this means.

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### Re: K vs Kc

I believe K and Kc are the same thing. K is a general term for the equilibrium constant disregarding units, while Kc is the equilibrium constant when using units of concentration so M for molarity. (Also, K can be in molarity or bars/atm, so when measuring pressure K becomes Kp.) Hope that helps!

Astha Patel 2J
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### Re: K vs Kc

Correct me if I'm wrong, but I believe that Kc is just a more specific way to describe K. However, if you want to convert between Kc and Kp, then I think the equation is Kp= Kc(RT)^delta(n), where delta n is the change in the number of moles (moles of products - moles of reactants).

SashaAnand2J
Posts: 90
Joined: Wed Sep 30, 2020 9:37 pm

### Re: K vs Kc

Hi Stephanie! I am a bit confused by your question, as K is referring to an equilibrium constant and Kc simply specifies that the equation uses concentrations. Assuming you're asking how to convert between Kc and Kp, you use the formula Kp = Kc (RT) ^delta(n), where delta(n) represents moles of gas products - moles of gas reactants. Kp uses partial pressures and Kp uses molar concentrations. Hope that helps!

AlbertGu_2C
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### Re: K vs Kc

Kc is just K that uses the concentrations of the variables involved. Hope this helps!

Lauren Mungo 1K
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### Re: K vs Kc

Kc is just K that uses concentration as opposed to Kp which uses pressure!

Ayesha Aslam-Mir 3C
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Joined: Wed Sep 30, 2020 9:43 pm

### Re: K vs Kc

Can you recall which section it came from? If I recall correctly, Kc refers to K when the reaction has molarities (typically aqueous reactants and products) when K is just generally referred to as the equilibrium constant.

The textbook says (in 5H.3) "You are free to choose either K or KcKc to report the equilibrium constant of a reaction. However, it is important to remember that calculations of an equilibrium constant from thermodynamic tables of data (standard Gibbs free energies of formation, for instance) give K, not Kc. In some cases, you need to know Kc after you have calculated K from thermodynamic data, and so you need to be able to convert between these two constants" (Chemical Principles, p. 412). Also, "The overall strategy for finding the relation between K and KcKc is to replace the partial pressures that appear in K by the molar concentrations and thereby generate Kc" (Chemical Principles, p. 412).

So I'm assuming we'll be concerned about this when considering thermodynamics, but te textbook goes a bit more in depth regarding consideration of converting to gaseous form.

Ansh Patel 2I
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Joined: Wed Sep 30, 2020 9:42 pm

### Re: K vs Kc

Hi! Kc is simply K along with the concentrations of the variables involved in the reactions.

StephanieGrigorian2J
Posts: 90
Joined: Wed Sep 30, 2020 10:01 pm

### Re: K vs Kc

Astha Patel 2J wrote:Correct me if I'm wrong, but I believe that Kc is just a more specific way to describe K. However, if you want to convert between Kc and Kp, then I think the equation is Kp= Kc(RT)^delta(n), where delta n is the change in the number of moles (moles of products - moles of reactants).

So calculating delta n is taking the difference of the stoichiometric coefficients of the products and reactants?

Astha Patel 2J
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### Re: K vs Kc

StephanieGrigorian2J wrote:
Astha Patel 2J wrote:Correct me if I'm wrong, but I believe that Kc is just a more specific way to describe K. However, if you want to convert between Kc and Kp, then I think the equation is Kp= Kc(RT)^delta(n), where delta n is the change in the number of moles (moles of products - moles of reactants).

So calculating delta n is taking the difference of the stoichiometric coefficients of the products and reactants?

Exactly, so let's say the reaction was N2O4 -> 2NO2. Then delta n would be 2 - 1 = 1.