Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Enthalpy and Spontaneous Rxns Early on in the development of thermodynamics, it was believed that if a reaction was exothermic, it was spontaneous. But this can’t be the whole story---consider an ice cube on a warm day. Consider the following reaction: H 2 O(s) H 2 O(l)  H° rxn = kJ Endothermic (ice feels cold) …..yet spontaneous (and it melts) !

Enthalpy and Spontaneous Rxns Consider the following problem: Mixing of a gas inside a bulb adiabatically (q = 0). q = 0, w = 0,  E = 0, and  H = 0 This process is NOT exothermic, but it still happens.

Statistical Interpretation of Entropy Imagine that we have a collection of 3 distinguishable particles who have a total energy of 3 . Let’s ask the question, “How will this fixed amount of energy distribute itself over the particles?”

Statistics and Entropy (cont.) Our system consists of three distinguishable particles. There are three “quanta” of energy (  ) available for a total of energy of “3  ”

First Arrangement: All on one The first possible arrangement we consider is one in which all energy resides on one particle There are three ways to do this

Second Arrangement: 2, 1, 0 Next arrangement: 2  on 1, 1  on another, and the third has 0  Six ways to do this

Third Arrangement The final possible arrangement is 1  on each particle. Only one way to do this.

Which Arrangement? Which arrangement is most probable? Ans: The arrangement which the greatest number of possibilities In this case: “2, 1, 0”

The Dominant Configuration Configuration: a type of energy distribution. Microstate: a specific arrangement of energy corresponding to a configuration. Which configuration will you see? The one with the largest # of microstates. This is called the dominant configuration (Why does a rope tangle?)

Determining Weight Weight (  ): the number of microstates associated with a given configuration. We need to determine , without having to write down all the microstates. A = the number of particles in your system. a i is the number of particles with the same amount of energy. ! = factorial, and  means take the product. 

Determining Weight (cont.) Consider 300 students where 3 students have 1  of energy, and the other 297 have none. A = 300 a 1 = 3 a 0 = 297 = 4.5 x 10 6

Weight and Entropy The connection between weight (  ) and entropy (S) is given by Boltzmann’s Formula: S = k(ln  k = Boltzmann’s constant = R/N a = 1.38 x J/K The dominant configuration will have the largest  ; therefore, S is greatest for this configuration

Young Ludvig Boltzmann

A devoted father And husband. His wife called him “My sweet, fat darling”

Boltzmann at 58 Troubled by severe bouts of depression, and criticism of his scientific ideas

Boltzmann took his own life while on a family vacation in Switzerland.

Example: Crystal of CO Consider the depiction of crystalline CO. There are two possible arrangements for each CO molecule. Each arrangement of CO is possible. For a mole of CO:  = N a !/(N a /2!) 2 = 2 Na

Example: Crystal of CO For a mole of CO:  = N a !/(N a /2!) 2 = 2 Na Then, S = k ln(  ) = k ln (2 Na ) = N a k ln(2) = R ln(2) = 5.64 J/mol.K

Another Example: Expansion What is  S for the expansion of an ideal gas from V 1 to 2V 1 ? Focus on an individual particle. After expansion, each particle will have twice the number of positions available.

Expansion (cont.) Original Weight =  Final Weight =  Then  S = S 2 -S 1 = k ln(2  ) - kln(  ) = k ln(2  /  ) = k ln(2)

Expansion (cont.) Therefore, the  S per particle = k ln (2) For a mole of particles:  S = k ln (2 Na ) = N a k ln(2) = R ln(2) = 5.64 J/mol.K

Expansion (cont.) Note in the previous example that weight was directly proportional to volume. Generalizing:  S = k ln (  final ) - kln(  initial ) = k ln(  final /  initial ) = Nk ln(  final /  initial ) for N molec. = Nkln(V final /V initial )