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. Example: Diamond converting to graphite

Entropy Most spontaneous processes are exothermic, but some are endothermic. Melting of ice above 0oC Dissolution of sodium chloride in water Why? Second Law of Thermodynamics

Entropy (S) Informally—the disorder or randomness of a system Formally—a 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, ΔSuniverse > 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 ΔS 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 ΔSuniv = ΔSsys + ΔSsurr Process is spontaneous as long as entropy of the universe is positive.

Temperature dependence Freezing of water is a nonspontaneous process above 0oC. Why? Entropy is temperature dependent. Impact of heat flow is greater at lower temperatures Give money to a rich man vs. poor man

Entropy and Temperature

Entropy Changes in surroundings Can use qsys to calculate Ssurr. ΔSsurr = −ΔH/ T If ΔH is positive, then ΔS is negative and non-spontaneous.. If ΔH is negative, then ΔS is positive and spontaneous.

Practice Calculate the entropy change in the surroundings associated with the combustion of propane gas occuring at 25oC (ΔHrxn = -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 ΔHrxn = -107 kJ and a ΔSrxn = 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 ΔS? 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)

Gibb’s 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 ΔG proportional to –ΔSuniv ΔG < 0 = spontaneous ΔG > 0 = non-spontaneous ΔG = ΔH −TΔS Units for ΔG are generally kJ

Summary ΔG = ΔH −TΔS At low temperatures, enthalpy is dominant. At high temperatures, entropy is dominant.

Concept Check Which statement is true regarding the sublimation of dry ice (solid CO2)? Justify your reasoning. ΔH is positive, ΔS is positive, and ΔG is positive at low temperature and negative at high temperature. ΔH is negative, ΔS is negative, and ΔG is negative at low temperature and positive at high temperature. ΔH is negative, ΔS is positive, and ΔG is negative at all temperatures ΔH is positive, ΔS is negative, and ΔG is positive at all temperatures.

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 Factors that affect standard entropy: State of substance Molar mass of substance If substance is in a particular allotrope Molecular complexity Extent of dissolution

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 ΔSorxn ΔSorxn = ΣnpSo(products) – ΣnrSo(reactants) Calculate Δsorxn for the reaction between gaseous ammonia and oxygen producing nitrogen monoxide (gas) and water vapor. Reactant or Product So (J/mol K) NH3(g) 192.8 O2(g) 205.2 NO(g) 210.8 H2O(g) 188.8

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) + ½ O2(g)  SO3(g) Calculate the ΔGorxn at 25oC and determine whether the reaction is spontaneous. Reactant or Product ΔHof (kJ/mol) So (J/mol K) SO2(g) -296.8 248.2 O2(g) 205.2 SO3(g) -395.7 256.8

Estimating Free Energy at Temperatures other than 25oC For the same reaction, estimate the value of ΔGorxn at 125oC. Is the reaction more or less spontaneous at this elevated temperature?

Free Energy under non-standard conditions ΔGrxn = ΔGorxn + 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