# Thermodynamics - PC\|MAC Thermodynamics Spontaneous and Non-spontaneous Processes Spontaneity in chemical processes = the direction in which the reaction proceeds. Not speed of reaction (kinetics) A reaction can be thermodynamically favored but still be slow. Entropy

Most spontaneous processes are exothermic, but some are endothermic. Melting of ice above 0oC Dissolution of sodium chloride in water Why? Entropy (S) Informallythe disorder or randomness of a system Formallya measure of the molecular motional

energy (plus any phase change energy) that has been dispersed in a system at a specific temperature. System with highest entropy has the greatest dispersal of energy For any spontaneous process, SS universe > 0 Units = J/K Phase changes and Entropy Ssolid < Sliquid < Sgas In gas, more ways to distribute energy than a

solid. Energy in solid is due to vibrations between molecules. Energy in gas is due to translational, rotational, and vibrational energy. Predict the sign for SS for each process and justify your answer Melting of ice to water Sublimation of carbon dioxide

H2O(g) H2O(l) 2N2O(g) 2N2(g) + O2(g) Heat Transfer Exothermic process increases entropy of surroundings Endothermic process increases entropy of system SSuniv = SSsys + SSsurr Process is spontaneous as long as entropy of the universe is positive.

Temperature dependence Temperature determines the magnitude of heat flow to the surroundings. Impact of heat flow is greater at lower temperatures Give money to a rich man vs. poor man Entropy and Temperature

Heat Transfer Practice Calculate the entropy change in the surroundings associated with the combustion of propane gas occuring at 25oC (SHorxn = -2044 kJ). Determine sign of the entropy change for the system. Determine the sign of the entropy change

for the universe. Will the reaction be spontaneous? Practice A reaction has SHrxn = -107 kJ and a SSrxn = 285 J/K. At what temperature is the change in entropy for the reaction equal to the change in entropy of the surroundings? Practice Which reaction is most likely to have a

positive SS? Justify your reasoning. a. SiO2(s) + 3C(s) SiC(s) + 2CO(g) b. 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) c. CO(g) + Cl2(g) COCl2(g) d. 3NO2(g) + H2O(l) 2HNO3(l) + NO(g) Practice Elemental mercury is a liquid at room temperature. Its normal freezing point is -38.9oC, and the molar enthalpy of fusion is 2.29 kJ/mol. What is the

entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? Calculating Entropy Third Law of Thermodynamics In a perfect crystal (diamond), the entropy at 0K is 0 J/K. Can use third law to develop standard molar entropy values with spectroscopy techniques.

Factors that affect standard entropy: State of substance Molar mass of substance If substance is in a particular allotrope Molecular complexity Practice--Whiteboard Get into groups of three. On your whiteboard, answer the following question. Arrange these gases in order of increasing standard molar entropy: SO3,

Kr, Cl2. Justify your reasoning. Calculating SSorxn SSorxn = nnpSo(products) nnrSo(reactants) (same as Calculate SSorxn for the reaction between ammonia and oxygen producing nitrogen monoxide and water vapor. Substance So (J/mol K) NH3(g)

192.8 O2(g) 205.2 NO(g) 210.8 H2O(g)

188.8 H) Practice Determine the change in the standard entropy, SS, for the synthesis of carbon dioxide from graphite and oxygen: C(s,graphite) + O2(g) CO2(g) Substance

So (J/mol K) C(s, graphite) 192.8 O2(g) 205.2 CO2(g) 213.8

Practice For the previous example, the change in the standard entropy, SS, for the synthesis of carbon dioxide from graphite and oxygen, use the previously calculated SSsys and standard enthalpy of formation values toSubstance determine Ssurr and SS .

universe o H (kJ/mol) f C(s, graphite) 0 O2(g) 0

CO2(g) -393.5 Gibbs Free Energy (G) A measure of whether or not a process will occur without the input of outside energy. Criterion of spontaneity (thermodynamically favored or unfavored) AKA chemical potential

Entropy and Free Energy Summary SG proportional to SSuniv SG < 0 SG > 0 = spontaneous = non-spontaneous SG = SH TSS Units for SG are generally kJ

Summary SG = SH TSS At low temperatures, enthalpy is dominant. At high temperatures, entropy is dominant. H S + +

+ + G Spontaneous at high temperature

Spontaneous at low temperature Never spontaneous Spontaneous at any temperature Concept Check Which statement is true regarding the sublimation of dry ice (solid CO2)? Justify your reasoning. a. SH is positive, SS is positive, and SG is positive at low temperature and negative at high temperature.

b. SH is negative, SS is negative, and SG is negative at low temperature and positive at high temperature. c. SH is negative, SS is positive, and SG is negative at all temperatures d. SH is positive, SS is negative, and SG is positive at all temperatures. Calculating Standard Change in Free Energy for a reaction One of the possible initial steps in the formation of acid rain is the oxidation of SO2 to SO3 via the following reaction:

SO2(g) + O2(g) SO3(g) Calculate the SGorxn at 25oC and determine Reactantthe or reaction HH (kJ/mol) S (J/mol K) whether is spontaneous. o Product o

f SO2(g) -296.8 248.2 O2(g) 0

205.2 SO3(g) -395.7 256.8 Hesss law of Summation Find Horxn for the following reaction: C + 2H2 CH4 C + O2 CO2

SH = -393 kJ H2 + 1/2O2 H2O SH = -286 kJ CH4 + 2O2 CO2 + 2H2O SH = -892 kJ Hesss Law for G Bond Energies Can utilize bond energies to calculate H. H = n Bond Energies broken n Bond Energies

formed Calculate the change in energy that accompanies the following reaction given the data below. H2(g) + F2(g) 2 HF(g) Bond Type Bond Energy HH 432 kJ/mol FF 154 kJ/mol HF 565 kJ/mol

Potential Energy Diagrams Free Energy under non-standard conditions SGrxn = SGorxn + R T lnQ Q = reaction quotient (equilibrium concept) T = Temperature in K R = 8.314 J/mol K For aA + bB cC + dD Q = [C]c[D]d [A]a[B]b