Chapter 19 Chemical Thermodynamics HW:

Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously effuse

Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C it is spontaneous for ice to melt. Below 0  C the reverse process is spontaneous.

Enthalpy-based Spontaneity OFTEN, spontaneous reactions are exothermic (  H is -) NOT ALWAYS, though! Ex – Ice melting – It is ENDOTHERMIC, but occurs spontaneously when T>0 o C at standard pressure

Entropy-based Spontaneity Entropy (S) is a term coined by Rudolph Clausius in the 19th century.

Entropy Entropy can be thought of as a measure of the randomness of a system. It is related to the various modes of motion in molecules.

Entropy Like total energy (E) and enthalpy (H) entropy is a state function. Therefore,  S = S final  S initial The more disorder the higher the S value. If  S is +, the reaction increases in disorder If  S is -, the reaction decreases in disorder

Solutions Generally, when a solid is dissolved in a solvent, entropy increases.

Entropy Changes In general, entropy increases when – Gases are formed from liquids and solids. – Liquids or solutions are formed from solids. – The number of gas molecules increases. – The number of moles increases.

Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.

Second Law of Thermodynamics In other words: For equilibrium processes:  S univ =  S system +  S surroundings = 0 For spontaneous processes:  S univ =  S system +  S surroundings > 0

Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.

Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. Appendix in the back of the book.

Standard Entropies Larger and more complex molecules have greater entropies.

Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which  H is estimated:  S° =  n  S° (products) -  m  S° (reactants) where n and m are the coefficients in the balanced chemical equation.

Example: 3 O 2 (g) -> 2 O 3 (g) Does this make sense??

Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

Gibbs Free Energy Gibbs free energy,  G.  G is negative, a process is spontaneous.  G is positive, it is not spontaneous (it is spontaneous in the reverse direction)  G = 0, it is at equilibrium

Standard Free Energy Changes Standard free energies of formation,  G  is the free energy change when 1 mole of compound is synthe f  G  =  n  G  (products)   m  G  (reactants) ff where n and m are the stoichiometric coefficients.

Free Energy Changes At temperatures other than 25°C,  G° =  H   T  S  How does  G  change with temperature?

Free Energy and Temperature There are two parts to the free energy equation: –  H  — the enthalpy term – T  S  — the entropy term The temperature dependence of free energy, then comes from the entropy term.

Free Energy and Temperature

Temperature and Chemical Reactions It is possible to calculate the temperature at which a reaction becomes favored. CaCO 3 (s) CaO(s) + CO 2 (g)  H = kJ  S = J/K At what temp does the reaction become favored??? Favored or not favored with respect to…?

During Phase Changes  G = 0 =  H-T  S So…  S = +  H / T where  H is the molar heat of vaporization / fusion / etc. Before / After phase changes, heat is found using q = ms 

Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT lnK Rearranging, this becomes  G  =  RT lnK or, K = e  G  /RT

Free Energy and Equilibrium Kln K GG Favored species >1+-Products =100Both <1-+Reactants