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9 Hydrogen bonds highly labile; lifetime , sec at room temperature Relatively weak: 4.5 * 10 3 calories (4.5 kcal) to break up mole of hydrogen bonds (takes 110 kcal/mol to break O-H bonds within a water molecule)

23 Given a mass of gas in thermal equilibrium we may measure its pressure (p) temperature (T) and volume (V). Boyle demonstrated that pV/T is a constant Volume occupied is proportional to the mass of gas, we can write the above constant as µR where µ is the mass in moles and R is a constant. R = joule/mole K Thus Equation #2 pV=µRT Where the gase expands from an initial volume Vi to a final volume Vf. Using equation #2, we can see that the work done per mole is Equation #4 At a constant temperature this may be written as Equation #5 We can use this result to describe the free energy of diffusion of a particular ion in our cell or outside of the cell as follows: Equation #6 G diff = RTln(c) Now the work done by an expanding gas can be calculated as follows: Equation #3 Diffusional forces acting on an ion

24 According to Faraday the charge on a mole of material is z Coulombs where z is the charge of each atom or ion (the valency of an ion). This is the Faraday constant or F. Electrical forces acting on an ion Thus the electrical energy in a mole of an ion may be expressed as Equation #1 G elect = zFE If z =valency of ion E = electrical potential across the delimiting membrane.

25 The free energy of ions inside or outside of the cell is the sum of these forces So the free energy (G) inside or outside of the cell may be expressed as Equation #7 G = G diff + G elect = RTln(c) + zFE (from #1 and #2) The difference between free energy inside the cell and outside defines the free energy driving movement of the ion across the cell membrane which may be expressed as follows: Equation #8   G)= (RTln(c out ) +zFE out ) —(RTln(c in ) + zFEin) Which simplifies to: Equation #9)  (G) = R.T.ln(cout/cin) + z.F.δ(E) By definition at equilibrium DG = 0 thus the equilibrium potential for any given ion is given by: Equation #10 δ(E) = RT/ZF ln(c in /cout) This is the Nernst-Einstein equation.

31 Current flowing across the giant axon membrane may be represented by the sum of conductive components (What we now identify as ion channels) and capacitance (the cell membrane). The currents are described in the following circuit diagram.