It is suggested that as a measurement of the collagen stability, the denature temperature Td is comprised by both the hydrogen bond from triple helix and an entropic part of the fibril. The denature enthalpy ΔHd only consider hydrogen bond within the triple helix (measured by calorimetric meter). Why H is independent of fibril entropy?
It is also stated that, when being dried, the fibril are closed to each other and thus the mobility decreased. yet the denature entropy change ΔSd increases, which means denature is more difficult. Why?
In polymer how is the relation of fibril/side chain mobility and entropy?
Thanks!
By definition, the Gibbs free energy change, which relates to the equilibrium constant of the chemical process you analyze, is related to enthalpy and entropy changes by:
delta G= delta H - T*delta S
The more negative delta G is, the higher the equilibrium constant:
delta G= -R*T*ln Keq
Therefore, a large positive delta S favours the reaction. However, delta S is not only dependent on the internal entropy of the chains, but also on the changes in solvent entropy as the area of interface between protein and solvent increases upon denaturation. Therefore the increase of chain entropy may be offset by decreased solvent entropy.
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