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Chemical Thermodynamics. Spontaneous Processes. First Law of Thermodynamics Energy is Conserved – ΔE = q + w Need value other than ΔE to determine if a process if favored (spontaneous) Spontaneous processes have a direction Spontaneity can depend on temperature. Reversible Processes. - PowerPoint PPT Presentation

Chemical Thermodynamics

Spontaneous ProcessesFirst Law of ThermodynamicsEnergy is Conserved E = q + wNeed value other than E to determine if a process if favored (spontaneous)Spontaneous processes have a directionSpontaneity can depend on temperature

Reversible ProcessesReversible ProcessWhen a change in a system is made in such a way that the system can be restored to its original state by exactly reversing the change.

Irreversible ProcessesIrreversible ProcessA process that cannot simply be reversed to restore the system and surroundings.Must take alternative pathwayOnly system is returned to its original state

EntropyProcesses where the disorder of the system increases tend to occur spontaneouslyIce meltingSalts dissolvingChange in disorder and change in energy determine spontaneityEntropy (S) is a state function (S) that measures of disorderUnits J/K

Determining Entropy ChangePredict whether S is positive or negative for each of the following processes.H2O(l) H2O(g)Ag+ (aq) + Cl-(aq) AgCl(s)4Fe(s) + 3O2(g) 2Fe2O3(s)CO2(s) CO2(g)CaO(s) + CO2(g) CaCO3(s)

Calculating EntropyIn a reversible process there is only one heat for both processes (qr e v)When a process occurs at constant temperature entropy is related to both heat and absolute temperature S = qr e v / TWhen 1mol of water is converted to 1mol of steam at 1 atm, Hv a p = 40.67kJ/mol, what is the change in entropy?

Calculating EntropyThe element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9C, and its molar enthalpy of fusion is Hf u s = 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?

Answer: -2.44 J/K

Calculating EntropyThe normal boiling point of ethanol is 78.3C and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 68.3g of ethanol at 1 atm condenses to liquid at the normal boiling point?

Answer: -163 J/K

Second Law of ThermodynamicsSecond law = Entropy of the universe increases in any spontaneous process.Suniv = Ssys+ Ssurr = 0 reversible processSuniv = Ssys+ Ssurr > 0 irreversible processUnlike energy, entropy is not conservedExamplesStraightening up your room4Fe(s) + 3O2(g) 2Fe2O3(s)Exceptions Isolated SystemsSsys = 0 reversible processSsys > 0 irreversible process

Calculating EntropyConsider the reversible melting of 1 mol of ice in a large isothermal water bath at 0C and 1 atm pressure. The enthalpy of fusion of ice is 6.01 kJ/mol. Calculate the entropy change in the system and in the surroundings, and the overall change in entropy of the universe for this process.

Answer = 22.0 J/K, 0 J/K

Molecular Interpretation of EntropyWhy does the entropy increase when a gas expands?Why does entropy decrease in the following reaction - 2NO(g) + O2(g) 2NO2(g)?Degrees of FreedomTranslational MotionVibrational MotionRotational Motion

Molecular Interpretation of EntropyLowering temperature decreases energy which lowers the degrees of freedom.Third law of thermodynamics = entropy of a pure crystalline substance at absolute zero is zero.Entropy generally increases with temperature.

Entropy Changes in ReactionsEntropy increases for processes in which:Liquids or solutions are formed from solids.Gases are formed from either solids of liquids.The number of molecules of gas increases during a chemical reaction.Choose the sample of matter that has greater entropy and explain your choice.1 mol of NaCl(s) or 1 mol HCl(g) at 25C1 mol of HCl(g) or 1 mol Ar (g) at 25C1 mol of H2O(s) at 0C or 1 mol of H2O(l) at 25C

Predicting Entropy ChangesPredict whether the entropy change of the system in each of the following isothermal reactions if positive or negative.CaCO3(s) CaO(s) + CO2(g)N2(g) + 3H2(g) 2NH3(g)N2(g) + O2(g) 2NO(g)HCl(g) + NH3(g) NH4Cl(s)2SO2(g) + O2(g) 2SO3(g)

Entropy Changes in Chemical ReactionsUsing variation of heat capacity with temperature absolute entropy are measured. Molar entropy at standard states (S)Molar entropies of elements are not zeroMolar entropies of gases are greater than liquids and solidsMolar entropies generally increase with increasing molar mass.Molar entropies generally increase with increasing number of atoms

Entropy Changes in Chemical ReactionsEntropy Change in a reaction can be calculated using a table of valuesS= nS(prod)-mS(react)Calculate S for the synthesis of ammonia from nitrogen and hydrogen at 298K.N2(g) + 3H2(g) 2NH3(g)

Answer: -198.3 J/K

Entropy Changes in the SurroundingsSurroundings are essentially a large constant temperature heat source. The change in entropy will then depend on how much heat is absorbed or given off by the system. Ss u r r = -qs y s /TIf the reaction happens at constant P, what is q?Calculate the Su n i v given the heat of formation of ammonia (-46.19kJ/mol)

Gibbs Free EnergySpontaneity depends on enthalpy and entropy. Gibbs Free Energy G = H TSIn a chemical reaction at constant T G = H TSAlgebra funWhen both T and P are constantIf G is negative, the reaction is spontaneous in the forward directionIf G is zero, the reaction is at equilibriumIf G is positive, the forward reaction is nonspontaneous, work must be done to make it occur. The reverse reaction is spontaneous.

Standard Free Energy ChangesGibbs free energy is a state function.For Standard free energiesFree energies for standard states are set to zero.Gases should be at 1 atmSolutions should be 1M in concetrationSolids and Liquids should be in their pure formsG= nG(prod)-mG(react)

Calculating Standard Free EnergiesUsing the data from Appendix C, calculate the standard free-energy change for the following reaction at 298K:P4(g) + 6Cl2(g) 4PCl3(g)What about the reverse reaction?

Answer: -1054.0kJ

Calculating Standard Free EnergiesWithout using data from appendix C, predict whether G for this reaction is more or less negative than H C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) H=-2220kJUsing the data from appendix C, calculate the standard free energy change for this reaction. What your prediction correct?

Answer: -2108kJ

Calculating Standard Free EnergiesConsider the combustion of propane to form CO2(g) and H2O(g) at 298K. Would you expect G to be more negative or less negative than H?

Free Energy and TemperatureG = H TSIce meltingG = H TSUse to estimate at other temperatures

HomeworkPart 1 - 2, 5, 7, 17, 20, 23, 25, 29, 31, 34, 37Part 2 - 41, 43, 47, 49, 53, 55, 56