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Chemical Thermodynamics

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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

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Reversible Processes Reversible Process When 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.

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Irreversible Processes Irreversible Process A process that cannot simply be reversed to restore the system and surroundings. Must take alternative pathway Only system is returned to its original state

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Entropy Processes where the disorder of the system increases tend to occur spontaneously Ice melting Salts dissolving Change in disorder and change in energy determine spontaneity Entropy (S) is a state function (ΔS) that measures of disorder Units – J/K

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Determining Entropy Change Predict whether ΔS is positive or negative for each of the following processes. H 2 O(l) → H 2 O(g) Ag + (aq) + Cl - (aq) → AgCl(s) 4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s) CO 2 (s) → CO 2 (g) CaO(s) + CO 2 (g) → CaCO 3 (s)

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Calculating Entropy In a reversible process there is only one heat for both processes (q r e v ) When a process occurs at constant temperature entropy is related to both heat and absolute temperature – ΔS = q r e v / T When 1mol of water is converted to 1mol of steam at 1 atm, ΔH v a p = 40.67kJ/mol, what is the change in entropy?

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Calculating Entropy The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9°C, and its molar enthalpy of fusion is ΔH f 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

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Calculating Entropy The normal boiling point of ethanol is 78.3°C 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

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Second Law of Thermodynamics Second law = Entropy of the universe increases in any spontaneous process. ΔS univ = ΔS sys + ΔS surr = 0 → reversible process ΔS univ = ΔS sys + ΔS surr > 0 → irreversible process Unlike energy, entropy is not conserved Examples Straightening up your room 4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s) Exceptions – Isolated Systems ΔS sys = 0 → reversible process ΔS sys > 0 → irreversible process

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Calculating Entropy Consider the reversible melting of 1 mol of ice in a large isothermal water bath at 0°C 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

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Molecular Interpretation of Entropy Why does the entropy increase when a gas expands? Why does entropy decrease in the following reaction - 2NO(g) + O 2 (g) → 2NO 2 (g)? Degrees of Freedom Translational Motion Vibrational Motion Rotational Motion

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Molecular Interpretation of Entropy Lowering 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.

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Entropy Changes in Reactions Entropy 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 25°C 1 mol of HCl(g) or 1 mol Ar (g) at 25°C 1 mol of H 2 O(s) at 0°C or 1 mol of H 2 O(l) at 25°C

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Predicting Entropy Changes Predict whether the entropy change of the system in each of the following isothermal reactions if positive or negative. CaCO 3 (s) → CaO(s) + CO 2 (g) N 2 (g) + 3H 2 (g) → 2NH 3 (g) N 2 (g) + O 2 (g) → 2NO(g) HCl(g) + NH 3 (g) → NH 4 Cl(s) 2SO 2 (g) + O 2 (g) → 2SO 3 (g)

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

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Entropy Changes in Chemical Reactions Entropy Change in a reaction can be calculated using a table of values ΔS°= ΣnS°(prod)-ΣmS°(react) Calculate ΔS° for the synthesis of ammonia from nitrogen and hydrogen at 298K. N2(g) + 3H 2 (g) → 2NH 3 (g) Answer: -198.3 J/K

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Entropy Changes in the Surroundings Surroundings 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. ΔS s u r r = -q s y s /T If the reaction happens at constant P, what is q? Calculate the ΔS u n i v given the heat of formation of ammonia (-46.19kJ/mol)

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Gibbs Free Energy Spontaneity depends on enthalpy and entropy. Gibbs Free Energy – G = H – TS In a chemical reaction at constant T ΔG = ΔH – TΔS Algebra fun When both T and P are constant If ΔG is negative, the reaction is spontaneous in the forward direction If ΔG is zero, the reaction is at equilibrium If ΔG is positive, the forward reaction is nonspontaneous, work must be done to make it occur. The reverse reaction is spontaneous.

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Standard Free Energy Changes Gibbs free energy is a state function. For Standard free energies Free energies for standard states are set to zero. Gases should be at 1 atm Solutions should be 1M in concetration Solids and Liquids should be in their pure forms ΔG°= ΣnG°(prod)-ΣmG°(react)

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Calculating Standard Free Energies Using the data from Appendix C, calculate the standard free-energy change for the following reaction at 298K: P 4 (g) + 6Cl 2 (g) → 4PCl 3 (g) What about the reverse reaction? Answer: -1054.0kJ

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Calculating Standard Free Energies Without using data from appendix C, predict whether ΔG° for this reaction is more or less negative than ΔH° C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ΔH°=- 2220kJ Using the data from appendix C, calculate the standard free energy change for this reaction. What your prediction correct? Answer: -2108kJ

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Calculating Standard Free Energies Consider the combustion of propane to form CO 2 (g) and H 2 O(g) at 298K. Would you expect ΔG° to be more negative or less negative than ΔH°?

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Free Energy and Temperature ΔG = ΔH – TΔS Ice melting ΔG° = ΔH° – TΔS° Use to estimate at other temperatures

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

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