Entropy, Free Energy and Equilibrium

Learning Objectives (LO’s) In this lesson you will learn…. 1) What spontaneity is in reactions 2) The likelihood of exothermic reactions to be spontaneous compared to endothermic reactions 3) What entropy is and the 3rd Law of Thermodynamics 4) How to calculate entropy values

Entropy (7 min) https://www.youtube.com/watch?v=LzvkO5ai9YQ

Spontaneous Processes Often spontaneous processes are exothermic (which is ok a ball rolls downhill) Water is made of hydrogen and oxygen, but it does not explode. This is NOT spontaneous… 2H2O(l) + 241 kJ →  2H2 (g) + O2 (g) So how do we get a ball to roll up hill? How can we make ENDOTHERMIC reactions spontaenous? (1 minute Ba(OH)2) https://www.youtube.com/watch?v=tTzNkr06Ap8

Spontaneous vs Nonspontaneous

ENTROPY SPONTANEOUS CHANGES - take place without the need for work Exothermic reactions are usually spontaneous - this is because they go from higher to lower enthalpy However ... Why should reactions with a positive DH value take place spontaneously ? (some salts dissolve readily in water and the solution gets colder, not hotter) ENDOTHERMIC Basically, how does a ball roll UPHILL????????????

The Concept of Entropy (S) Entropy refers to the state of order. A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process. more order less order solid liquid gas more order less order crystal + liquid ions in solution more order less order crystal + crystal gases + ions in solution

Probability- what is the chance that the order of the cards can be restored by reshuffling?

Laws of Thermodynamics 1st Energy can neither be created nor destroyed. It can only change forms. 2nd: Entropy will increase over time; or entropy tends towards the maximum. 3rd The entropy of a perfect crystal at absolute zero is zero https://www.youtube.com/watch?v=kswiDQ2aAKA

Trends in Entropy Entropy for gas phase is greater than that of liquid or solid of same substance I2 (g) has greater entropy than I2 (s) Dissolve a sugar cube creates disorder What are some other examples???

Entropy Changes in a System Qualitative Ssolid < Sliquid

Entropy Changes in a System Qualitative Sliquid < Svapor

Entropy Changes in a System Qualitative Ssolid < Saqueous

Entropy Changes in a System Qualitative Slower temp < Shigher temp

Entropy Changes in a System Qualitative Sfewer moles < Smore moles

Standard Entropies Larger and more complex molecules have greater entropies.

Qualitative Entropy Changes Determine the sign of S for the following, qualitatively 1. Liquid nitrogen evaporates 2. Two clear liquids are mixed and a solid yellow precipitate forms 3. Liquid water is heated from 22.5 C to 55.8 C

Standard Entropies These are molar entropy values of substances in their standard states. These are in Joules not kJ We can use these values to get a number for entropy

Entropy Changes in the System Entropy can be calculated from the table of standard values just as enthalpy change was calculated. Srxn = nS products  nS reactants

Standard Entropy Calculate the standard entropy change for the following using the table of standard values. (first, predict the sign for S qualitatively) 2 NH3(g)  N2(g) + 3 H2(g) oS NH3(g) = + 191.5 J/K oS H2(g) = + 131.0 J/K oS N2(g) = + 193.0 J/K

Standard Entropy 2NH3(g)  N2(g) + 3H2(g) Answer: qualitatively (Entropy increases) (2 mol gas  4 mol gas) Srxn = nS products  nS reactants = [(1)(191.5 J/K · mol) + (3)(131.0 J/K · mol)] - [(2)(193.0 J/K · mol)] = 584.5 J/K · mol - 386 J/K · mol Srxn = 198.5 J/K · mol (Entropy increases)

Your Turn! Calculate the standard entropy change for the following using the table of standard values. (first, predict the sign for S qualitatively) 2H2(g) + O2(g)  2H2O (g) S H2O(g) = +188 J/K.mol S H2 = +131 J/K.mol S O2 = + 205 J/K.mol

Standard Entropy 2H2(g) + O2(g)  2H2O (g) Answer: qualitatively (Entropy decreases, more organized; 3 mol gas  2 mol gas) Srxn = nS products  nS reactants = [(2)188 J/K·mol) – [(2)131 J/K·mol)+ (205 J/K·mol)] = 376 J/K · mol - 467 J/K · mol Srx = -91 J/K·mol (Entropy decreases, more organized; 3 mol gas  2 mol gas)