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Thermodynamics Chapter 19
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19.1 Spontaneous Processes – process that occurs without any outside intervention, the internal energy alone determines if a reaction will occur
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Spontaneous Processes example:
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Spontaneous Processes example: when T>0 o C, ice melts spontanously; when T<0 o C water turns to ice spontaneously; when T= 0 o C, the system is at equilibrium and neither process is preferred
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Reversible process - when a system and surroundings is changed in such a way that they can be restored by reversing the change exactly
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Reversible process Whenever a chemical system is in equilibrium, reactants and products can interconvert reversibly
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Irreversible Process - a change to a system and surroundings that can be restored by taking a new path to get back and changing the surroundings
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Sum Up 1. Whenever a chemical system is in equilibrium, reactants and products can interconvert reversibly
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Sum Up 2. In any spontaneous process, the path between reactants and products is irreversible
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19.2 Entropy disorder of a system The more disorder of a system, the larger the entropy S = S final -S initial
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Entropy When S > 0, the system is disordered When S < 0, the system is more ordered or less random
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Second Law of Thermodynamics states the entropy of the universe increases in any spontaneous process (entropy is not conserved)
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19.3 The Molecular Interpretation of Entropy Entropy changes are related to the way the particles in a system can be arranged
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19.3 The Molecular Interpretation of Entropy Particles change through translational motion, vibrational motion, and rotational motion
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19.3 The Molecular Interpretation of Entropy Entropy and movement decrease with decreasing T
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Third Law of Thermodynamics
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entropy of a pure crystalline substance at absolute zero (0 Kelvin) is zero
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Entropy Generalizations: Entropy is expected to increase for processes in which: 1. Liquids or solutions are formed from solids
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Entropy Generalizations: 2. Gases are formed from either solids or liquids
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Entropy Generalizations: 3. The number of molecules of gas increases during a chemical rxn
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Sample Exercise 19.3 Predicting the Sign of ΔS Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature:
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Sample Exercise 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a)1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b)2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c)1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.
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19.4 Entropy Changes in Rxns Standard Molar Entropies (S o ) - the molar entropy values of substances in their standard states, pure substance at 1 atm; found on appendix C
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Standard Molar Entropies (S o ) Characteristics: 1. Molar entropies of elements are not 0 2. S o (g) > S o (l) >S o (s) 3. S o increases w/ increasing MM 4. Increase w/ increasing atoms
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Entropy Change S o = nS o (products) - mS o (reactants), where m and n are coefficients of the balanced chemical eq
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Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies Calculate ΔSº for the synthesis of ammonia from N 2 (g) and H 2 (g) at 298 K: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)
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19.5 Gibbs Free Energy G=H-TS G = H-T S at constant temperature
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Relationship between G and spontaneity 1. If G is negative, the rxn is spontaneous in the forward direction
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Relationship between G and spontaneity 2. If G is zero, the rxn is at equilibrium
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Relationship between G and spontaneity 3. If G is positive, the rxn in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. The reverse rxn is spontaneous
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Sample Exercise 19.6 Calculating Free-Energy Change from ΔH°, T, ΔS° Calculate the standard free energy change for the formation of NO(g) from N 2 (g) and O 2 (g) at 298 K: N 2 (g) + O 2 (g) → 2 NO(g) given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these circumstances?
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Standard Free-Energy Changes G o = nG f o (products) - mG f o (reactants)
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Standard Free-Energy Changes
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when G>0 (nonspontaneous), it is the measure of the minimum amount of work that must be done to cause the process to occur
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Sample Exercise 19.7 Calculating Standard Free-Energy Change from Free Energies of Formation (a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K: P 4 (g) + 6 Cl 2 (g) → 4 PCl 3 (g)
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C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l) ΔH° = –2220 kJ (a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct? Sample Exercise 19.8 Estimating and Calculating ΔG°
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19.6 Free Energy and Temperature The value of H and S do not change much with temperature The value of T plays a major role in determining G
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19.6 Free Energy and Temperature
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Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity The Haber process for the production of ammonia involves the equilibrium Assume that ΔH°(-92.38kJ) and ΔS°(-198.3J/K) for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.
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19.7 Free Energy and the Equilibrium Constant Calculating G at nonstandard conditions G= G o + RTlnQ, where R is the ideal gas constant 8.314J/mol-K, T is abs T, and Q is the reaction quotient
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Free Energy and the Equilibrium Constant G is related to Keq, at equilibrium; G=0 and Q = Keq G o = - RT lnKeq Keq= e - G/RT
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Sample Exercise 19.12 Calculating an Equilibrium Constant for ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25 °C for the reaction involved in the Haber process: (see 19.9)
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