Question

What is an equilibrium constant perturbed by a solute X equal to? (dK/da), a = activity of solute X

Answer 1

@Vocaloid
@aaronq
@Kainui

What do you guys think?

Answer 2

Aha well since the equilibrium constant is a product of all the activities raised to the power of their coefficient in their balanced equation (Whichever it may be) then depending on if it's product or reactant it will be + or - for that exponent.
Balanced chemical equation:
\[\nu_1 X_1 + \nu_2 X_2 +\cdots \rightleftharpoons \nu_1' X_1'+\nu_2' X_2'+\cdots\]
\[P = (\text{Product of other activities not dependent on X})\]\[K = a_X^{\pm \nu_x} P\]\[\frac{dK}{da_X} = \pm \nu_x a_X^{\pm \nu_x -1 }P\]

Answer 3

At first though I was wondering, how do we change an activity of a single particle while leaving the rest unchanged, so I thought you had meant how the rate constant changes as you perturb the extent of reaction.

Haha I almost fell for it and put:
\[\frac{dK}{da} = 0\]but of course, the rate CONSTANT means constant over time haha:
\[\frac{dK}{dt}=0\]

Answer 4

In the case:
For a two state reaction (for proteins) N <-> D with equilibrium constant K, the perturbed equilibrium constant of solute X is equal to:
\[\frac{ d \ln(K) }{ d \ln(a_X) }=\Delta v_X\]
Where \(\Delta v_X \) is the net change in the number of bound \(X\) molecules.
To take into account the effects of high salt concentrations on the binding of water molecules, we can re-write the above to:
\[\frac{ d \ln(K) }{ d \ln(a_X) }=\Delta v_X-\Delta v_W \frac{ N_X }{ N_W }\]
The last term on the right hand side is the net change in water molecules bound multiplied by the ratio of solute to solvent.
Reciprocal we would get:
\[\frac{ d \ln(K) }{ d \ln(a_W) }=\Delta v_W-\Delta v_X \frac{ N_W }{ N_X }\]

Lets go crazy and find the relation to Gibbs free energy!
\[\Large \Delta G=\Delta G_{H_2O}+RT \ln(K)\]
\[\large \Delta G = \Delta G_{H_2O}+RT \left( \Delta v_X-\Delta v_W \frac{ N_X }{ N_W } \right) \ln \left( a_W \right)\]
This must be true as when \(a_W=1, K_{solute}=K_{H_2O}\).
The activity can be written as an exponential function that depends on the molar fraction:
\[\large \Delta G = \Delta G_{H_2O}+RT \left( \Delta v_X-\Delta v_W \frac{ N_X }{ N_W } \right) \left[ \ln(1-X_X)+\alpha X_X^2 + \beta X_X^2\right]\]

Answer 5

This is just getting stupid O_O

Answer 6

If you have time I really like to go other some mathematical detail of this... cause I think there is something to this. I just need to map all the approximations to this!

Answer 7

This looks interesting, I will look at this later, thanks for posting!

Answer 8

My attempt:
From the law of mass action, we have:
\[\sum\,\nu_i\,\ln{c_i}=\ln{K}\] where \(c_i=\frac{N_i}{N}\) and \(N=\sum\,N_i\). Taking derivative on both sides yields:
\[\nu_{X}\,\mathrm{d}\ln{c_{X}}+\sum_{i\neq X}\,\nu_i\mathrm{d}\ln{c_i}=\mathrm{d}\ln{K}\] dividing yields:
\[\frac{\mathrm{d}\ln{K}}{\mathrm{d}\ln{c_{X}}}=\nu_{X}+\sum_{i\neq X}\,\nu_i\frac{\mathrm{d}\ln{c_i}}{\mathrm{d}\ln{c_X}}\] The second term can be expressed as:
\[\sum_{i\neq X}\,\nu_i\frac{\mathrm{d}\ln{\frac{N_i}{V}}/\mathrm{d}N_{X}}{\mathrm{d}\ln{\frac{N_X}{V}}/\mathrm{d}N_{X}}=-\sum_{i\neq X}\,\nu_{i}\]
So the final expression appears to be\[\frac{\mathrm{d}\ln{K}}{\mathrm{d}\ln{c_{X}}}=\nu_{X}-\sum_{i\neq X}\,\nu_{i}\]

Answer 9

Ok one thing I am working out this derivative with logarithms thing on the first line and I see what you're getting it from now, but the thing is I don't think it should be \(\Delta \nu_x\) I think it should just be \(\nu_x\) since that's just the number for balancing the chemical equation. I think changing the activity is more like changing how the equilibrium constant depends on the concentration rather than changing how the chemicals fundamentally combine.

Like for instance, if you are looking at the reaction to make H2O out of H2 and O2 you can't change the relative coefficients like:
\[2 H_2 + 1 O_2 \to 2 H_2O\]
\[1 H_2+1O_2 \to 2 H_2O\]

that would be like,
\[\Delta \nu_{H_2} = 1- 2 = -1\]

At least that's how I'm thinking of it.

Answer 10

Basically I am kinda looking at a reaction like this:
\[\Large \sf F\bullet \nu H_2O\bullet \mu X \rightleftharpoons D\bullet \nu' H_2O\bullet \mu' X ~+a~X + b~ H_2O\]
Does it make sense? :P

Answer 11

Ah so I kinda see, like the number of water molecules and I guess ions sorta like hydrating/surrounding the protein before and after is what's changing ok I see that I think, interesting.

Answer 12

So like in a way we're looking at the change in polar surface area between folded and unfolded states, since the nonpolar part of the protein when unfolded won't be... Well hmmm, I would think there would be more and so am I right in thinking that your reaction arrows show the transition (left to right) from denatured to natural state?

Answer 13

Folded <-> unfolded (denatured)