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For example, if a macromolecule M has three binding sites, K′ 1 describes a ligand being bound to any of the three binding sites. In this example, K′ 2 describes two molecules being bound and K′ 3 three molecules being bound to the macromolecule. The microscopic or individual dissociation constant describes the equilibrium of ligands ...
An often considered quantity is the dissociation constant K d ≡ 1 / K a , which has the unit of concentration, despite the fact that strictly speaking, all association constants are unitless values. The inclusion of units arises from the simplification that such constants are calculated solely from concentrations, which is not the case.
The Boltzmann constant (k B or k) is the proportionality factor that relates the average relative thermal energy of particles in a gas with the thermodynamic temperature of the gas. [2] It occurs in the definitions of the kelvin (K) and the gas constant , in Planck's law of black-body radiation and Boltzmann's entropy formula , and is used in ...
= , where k B is the Boltzmann constant, and Ω denotes the volume of macrostate in the phase space or otherwise called thermodynamic probability. d S = δ Q T {\displaystyle dS={\frac {\delta Q}{T}}} , for reversible processes only
[c] [2] For example, a hypothetical weak acid having K a = 10 −5, the value of log K a is the exponent (−5), giving pK a = 5. For acetic acid, K a = 1.8 x 10 −5, so pK a is 4.7. A higher K a corresponds to a stronger acid (an acid that is more dissociated at equilibrium
kT (also written as k B T) is the product of the Boltzmann constant, k (or k B), and the temperature, T.This product is used in physics as a scale factor for energy values in molecular-scale systems (sometimes it is used as a unit of energy), as the rates and frequencies of many processes and phenomena depend not on their energy alone, but on the ratio of that energy and kT, that is, on E ...
where k B is the Boltzmann constant, S is the classical thermodynamic entropy, and Ω is the number of microstates. So =. Substituting into the definition of β from the statistical definition above gives
This would thus allow the calculation of K −1. By plotting a graph of ε HG versus K −1, the result would be a linear relationship. When the procedure is repeated for a series of concentrations and plotted on the same graph, the lines intersect at a point giving the optimum value of ε HG and K −1.