-
CycleTrack and SDN are teaming up to make medical school admissions more transparent and accessible! Read the announcement to learn how we’re supporting the future of CycleTrack’s free application-tracking tools .
Kd vs Ksp
Started by DoctorRealtor
Get help with your application
Use all the free resources available to you from SDN: articles, guides, expert advising, forums discussions, and school research.
If I am not mistaken isn't Kd just the dissociation constant between an antibody and and an antigen? While Ksp the equilibrium constant in solution chemistry?
Kd is a dissociation constant. So if A + B ---> AB, then the Kd would be [A]/[AB] at equilibrium. This is just the inverse of the association constant. These are just plain equilibrium constants.
Ksp is different. Ksp is the solubility product. The key difference is that first, it refers only to a saturated solution. Now, the second difference is that say you're dissolving a solid AB into solution. The equilibrium constant would be [A]/[AB] (note the similarity in form here to the dissociation constant). Now, what's [AB]? Well, it's not the concentration of AB in solution because we assume AB dissolves in solution. It's also not the amount of solid AB because solids don't affect solution equilibria much. So what it is is actually the concentration of AB in solid AB. Weird, I know. But all you should know is that that's a constant. So you multiple both sides of the equilibrium expression by [AB] and you get [A] = K x [AB] and we call the right side of this expression the solubility product constant. So it's a very specific application of equilibrium constants.
Ksp is different. Ksp is the solubility product. The key difference is that first, it refers only to a saturated solution. Now, the second difference is that say you're dissolving a solid AB into solution. The equilibrium constant would be [A]/[AB] (note the similarity in form here to the dissociation constant). Now, what's [AB]? Well, it's not the concentration of AB in solution because we assume AB dissolves in solution. It's also not the amount of solid AB because solids don't affect solution equilibria much. So what it is is actually the concentration of AB in solid AB. Weird, I know. But all you should know is that that's a constant. So you multiple both sides of the equilibrium expression by [AB] and you get [A] = K x [AB] and we call the right side of this expression the solubility product constant. So it's a very specific application of equilibrium constants.
Good explanation and prob just a typo, but shouldn't Kd be [A] /[AB]?
Weird! The forum won't even allow A * B in the numerator. It automatically defaults to [A]/[AB]
Advertisement - Members don't see this ad
[A]/[AB]
ah! I had to try it. I was also wondering about that term for Kd
ah! I had to try it. I was also wondering about that term for Kd
Last edited:
I have a similar question. I became a little confused on the concept of Keq. So far my understanding of Keq is as follows:
∙If Keq>1 then ∆G<0 (spontaneous)
∙If Keq<1 then ∆G>0 (nonspontaneous)
So my questions are:
1) When Keq>1, this describes the reaction as having MORE products than reactants, correct, since Keq is products/reactants, (larger numerator)?
But what I don't understand is when they define a reaction as "spontaneous", does it mean that a reaction is regarded as "spontaneous" when a reaction goes proceeds from reactant ---> products and NOT FROM products---->reactants?
2)And wouldn't a spontaneous reaction favor FORMING PRODUCTS, so why do they term a reaction WHICH HAS ALREADY MADE MORE PRODUCTS THAN REACTANTS as spontaneous when it has already completed forming reactants --->products?
I am really trying to put 2 and 2 together but I don't seem to quite understand this concept at the level which I want to. It would be much appreciated if anyone would take the time to explain this confusing topic. Thank you.
∙If Keq>1 then ∆G<0 (spontaneous)
∙If Keq<1 then ∆G>0 (nonspontaneous)
So my questions are:
1) When Keq>1, this describes the reaction as having MORE products than reactants, correct, since Keq is products/reactants, (larger numerator)?
But what I don't understand is when they define a reaction as "spontaneous", does it mean that a reaction is regarded as "spontaneous" when a reaction goes proceeds from reactant ---> products and NOT FROM products---->reactants?
2)And wouldn't a spontaneous reaction favor FORMING PRODUCTS, so why do they term a reaction WHICH HAS ALREADY MADE MORE PRODUCTS THAN REACTANTS as spontaneous when it has already completed forming reactants --->products?
I am really trying to put 2 and 2 together but I don't seem to quite understand this concept at the level which I want to. It would be much appreciated if anyone would take the time to explain this confusing topic. Thank you.
For a reaction A -> B, A is the reactant and B is the product. At equilibrium, K = B (conc) / A (conc). If K > 1, then the reaction is favored in the forward direction (products). If K < 1, the reaction is favored in the reverse direction (reactants).
Say K > 1, then the reaction is favored in the forward direction which must mean that deltaG in the forward direction < 0 (spontaneous).
You can also look at deltaG = -RTln(K) at equilibrium. As K gets bigger, deltaG gets more negative (more spontaneous).
Say K > 1, then the reaction is favored in the forward direction which must mean that deltaG in the forward direction < 0 (spontaneous).
You can also look at deltaG = -RTln(K) at equilibrium. As K gets bigger, deltaG gets more negative (more spontaneous).
Yes, something is clearly wrong with the formatting.
This will be helpful to you: http://www.chemteam.info/Equilibrium/Writing-Ksp-expression.html. The reason you are taught to "ignore" solids and liquids is because their activities are arbitrarily defined to be unity.
