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S = k ln W A vignette….. Let’s consider a simpler case first Thought experiment: Consider a beaker with a partition right in the middle that starts out.

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Presentation on theme: "S = k ln W A vignette….. Let’s consider a simpler case first Thought experiment: Consider a beaker with a partition right in the middle that starts out."— Presentation transcript:

1 S = k ln W A vignette….

2 Let’s consider a simpler case first Thought experiment: Consider a beaker with a partition right in the middle that starts out holding molecules in only one half. What happens when you remove the partition? What is  H for this process? Zero (at least if molecules don’t interact, “ideal solution”) So why does it happen? How about the reverse process? [Relevance to drug distribution ]

3 How about the reverse process? Thought experiment: Molecules spontaneously moving to only one half of beaker?  H is still zero This seems unlikely, except maybe if there are very few particles Let’s calculate the probabilities quantitatively

4 Probability of being on one side 1 mol N = 1 p = 1/2 N = 2 p = 1/4 N p = (1/2) N N ~ 6 x p = (1/2) N = very very small This fits with our intuition (and the second law) which says that things tend toward disorder, i.e., higher entropy [unless there is an input of energy to counteract …]

5 Statistical thermodynamics tells us how to compute entropy There are multiple ways to define entropy, including the one on Boltzmanns tombstone In modern notation, this is typically written S = k B ln W W is the number of configurations available to a system, and is related to the probabilities we just calculated. Note that S is a state function, and positive values of  S are favorable, not negative like  H. Units on S are J/K (more on this later).

6 Let’s compute  S for this process This is not just a thought experiment. This is equivalent to entropy change upon dilution, or mixing two solutions. State 1State 2 S 1 = k B ln W 1 S 2 = k B ln W 2

7 N = 1  S = k B ln 2 = 1.38x J/K * ~ J/K We can define the number of “configurations” of the system in different ways, but regardless, W clearly doubles for a single molecule when we double the volume. To get units of energy, multiply by temperature (300 K), which gives 3x J. TINY amount of energy.

8 N = 1  S = k B ln 2 N = 2  S = k B ln 4 We assume that the 2 molecules are uncorrelated, i.e., the configuration of one is unrelated to the other. (This is again the definition of an “ideal solution”.) This then implies that the total number of configurations is the product (not sum) of the configurations for each molecule.

9 N = 1  S = k B ln 2 N = 2  S = k B ln 4 N molecules  S = k B ln 2 N =Nk B ln 2

10 N = 1  S = k B ln 2 N = 2  S = k B ln 4 N = N A = 1 mole  S = N A k B ln 2 = R ln 2 N = # of molecules N A = Avogadro’s number k B = Boltzmann’s constant R = N A k B = gas constant

11 N = 1  S = k B ln 2 N = 2  S = k B ln 4 N = N A = 1 mole  S = N A k B ln 2 = R ln 2 Generalizing to any volume change, Molar entropy: N = # of molecules N A = Avogadro’s number k B = Boltzmann’s constant R = N A k B = gas constant

12 V = volume N = # of molecules n = # of moles R = N A k B = gas constant We can also think of this in terms of concentrations State 1State 2 Convert to concentrations, in molarity units: M = n/V, so V = n/M

13 There are other flavors of entropy We focused here on “translational” entropy, which is equivalent to entropy of dilution, or entropy of mixing There is also “rotational” entropy and “vibrational entropy”, very important in drug binding as well

14 Free Energy At constant pressure, the correct criterion is  G < 0 where G = H – TS If we additionally assume constant T, you get the famous equation  G =  H – T  S Key point: note the sign on the entropy term When  G = 0, this is called “equilibrium” Molar free energy for a substance also referred to as “chemical potential”:

15 An intuitive feel for free energy? Enthalpy = energy Entropy = disorder Free energy = ?? Best understood as a statement of the 2 nd law of thermodynamics

16 Free energy of dilution/mixing State 1State 2 Recall that …

17 Derivation of  G=-RTln K for a simple A  B reaction

18 At equilibrium (constant T, P): Assume ideal solution, Mº = 1 molar (to make argument of the logarithm unitless) Equating these two:

19 Constant that depends only on nature of reactants and products DEFINE K M = (M B /M A ) eq

20 Extension to more general equilibria: aA + bB  cC + dD At equilibrium (constant T, P; closed system): Alternate Notation

21 Other concentration units Derivations are extremely similar. For molality (mol/kg solvent), the chemical potential can be written as where the standard state is now defined by mº = 1 molal. The equilibrium constant winds up looking like Expression for gasses is also very similar, with partial pressures replacing concentrations.


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