Chemistry A Molecular Approach 1 st Ed Nivaldo

Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 17 Free Energy and Thermodynamics 2008, Prentice Hall

First Law of Thermodynamics • you can’t win! • First Law of Thermodynamics: Energy cannot be Created or Destroyed üthe total energy of the universe cannot change üthough you can transfer it from one place to another · DEuniverse = 0 = DEsystem + DEsurroundings Tro, Chemistry: A Molecular Approach 2

First Law of Thermodynamics • Conservation of Energy • For an exothermic reaction, “lost” heat from the system • • • goes into the surroundings two ways energy “lost” from a system, üconverted to heat, q üused to do work, w Energy conservation requires that the energy change in the system equal the heat released + work done ü DE = q + w ü DE = DH + PDV DE is a state function üinternal energy change independent of how done Tro, Chemistry: A Molecular Approach 3

Energy Tax • you can’t break even! • to recharge a battery with 100 k. J of • useful energy will require more than 100 k. J every energy transition results in a “loss” of energy ü conversion of energy to heat which is “lost” by heating up the surroundings Tro, Chemistry: A Molecular Approach 4

Heat Tax fewer steps generally results in a lower total heat tax Tro, Chemistry: A Molecular Approach 5

Thermodynamics and Spontaneity • thermodynamics predicts whether a process will proceed under the given conditions üspontaneous process Ønonspontaneous processes require energy input to go • spontaneity is determined by comparing the free energy of the system before the reaction with the free energy of the system after reaction. üif the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable. • spontaneity ≠ fast or slow Tro, Chemistry: A Molecular Approach 6

Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end. Tro, Chemistry: A Molecular Approach 7

Reversibility of Process • any spontaneous process is irreversible ü it will proceed in only one direction • a reversible process will proceed back and forth between the two end conditions ü equilibrium ü results in no change in free energy • if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction Tro, Chemistry: A Molecular Approach 8

Thermodynamics vs. Kinetics Tro, Chemistry: A Molecular Approach 9

Diamond → Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations). Tro, Chemistry: A Molecular Approach 10

Factors Affecting Whether a Reaction Is Spontaneous • The two factors that determine thermodynamic • • • favorability are the enthalpy and the entropy. The enthalpy is a comparison of the bond energy of the reactants to the products. übond energy = amount needed to break a bond. ü DH The entropy factors relates to the randomness/orderliness of a system ü DS The enthalpy factor is generally more important than the entropy factor Tro, Chemistry: A Molecular Approach 11

• • Enthalpy related to the internal energy DH generally k. J/mol stronger bonds = more stable molecules if products more stable than reactants, energy released üexothermic ü DH = negative if reactants more stable than products, energy absorbed üendothermic ü DH = positive The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law DH°rxn = S(DH°prod) - S(DH°react) Tro, Chemistry: A Molecular Approach 12

Entropy • entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S · S generally J/mol • S = k ln W ü k = Boltzmann Constant = 1. 38 x 10 -23 J/K ü W is the number of energetically equivalent ways, unitless • Random systems require less energy than ordered systems Tro, Chemistry: A Molecular Approach 14

W Energetically Equivalent States for the Expansion of a Gas Tro, Chemistry: A Molecular Approach 15

Macrostates → Microstates These microstates all have the same macrostate So there are 6 macrostate can be achieved through different. This particle several different arrangements of the particles arrangements that result in the same macrostate Tro, Chemistry: A Molecular Approach 16

Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State B There are 6 possible arrangements that give State C Therefore State C has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy Tro, Chemistry: A Molecular Approach 17

Changes in Entropy, DS • entropy change is favorable when the result is a more • random system. ü DS is positive Some changes that increase the entropy are: üreactions whose products are in a more disordered state. Ø(solid > liquid > gas) üreactions which have larger numbers of product molecules than reactant molecules. üincrease in temperature üsolids dissociating into ions upon dissolving Tro, Chemistry: A Molecular Approach 18

Increases in Entropy Tro, Chemistry: A Molecular Approach 19

The 2 nd Law of Thermodynamics • the total entropy change of the universe must be positive for a process to be spontaneous ü for reversible process DSuniv = 0, ü for irreversible (spontaneous) process DSuniv > 0 • DSuniverse = DSsystem + DSsurroundings • if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount ü when DSsystem is negative, DSsurroundings is positive • the increase in DSsurroundings often comes from the heat released in an exothermic reaction Tro, Chemistry: A Molecular Approach 20

Entropy Change in State Change • when materials change state, the number of macrostates it can have changes as well üfor entropy: solid < liquid < gas übecause the degrees of freedom of motion increases solid → liquid → gas Tro, Chemistry: A Molecular Approach 21

Entropy Change and State Change Tro, Chemistry: A Molecular Approach 22

Heat Flow, Entropy, and the nd 2 Law Heat must flow from water to ice in order for the entropy of the universe to increase Tro, Chemistry: A Molecular Approach 23

Temperature Dependence of DSsurroundings • when a system process is exothermic, it adds heat to the • • surroundings, increasing the entropy of the surroundings when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings the amount the entropy of the surroundings changes depends on the temperature it is at originally ü the higher the original temperature, the less effect addition or removal of heat has Tro, Chemistry: A Molecular Approach 24

Gibbs Free Energy, DG • maximum amount of energy from the system • available to do work on the surroundings G = H – T∙S DGsys = DHsys – TDSsys DGsys = – TDSuniverse DGreaction = S n. DGprod – S n. DGreact when DG < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when DG is negative Tro, Chemistry: A Molecular Approach 25

Ex. 17. 2 a – The reaction C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(g) has DHrxn = -2044 k. J at 25°C. Calculate the entropy change of the surroundings. Given: Find: Concept Plan: DHsystem = -2044 k. J, T = 298 K DSsurroundings, J/K T, DH DS Relationships: Solution: Check: combustion is largely exothermic, so the entropy of the surrounding should increase significantly

Free Energy Change and Spontaneity Tro, Chemistry: A Molecular Approach 27

Gibbs Free Energy, DG • process will be spontaneous when DG is negative · DG will be negative when ü DH is negative and DS is positive Øexothermic and more random ü DH is negative and large and DS is negative but small ü DH is positive but small and DS is positive and large Øor high temperature • DG will be positive when DH is + and DS is − ünever spontaneous at any temperature • when DG = 0 the reaction is at equilibrium Tro, Chemistry: A Molecular Approach 28

DG, DH, and DS Tro, Chemistry: A Molecular Approach 29

Ex. 17. 3 a – The reaction CCl 4(g) C(s, graphite) + 2 Cl 2(g) has DH = +95. 7 k. J and DS = +142. 2 J/K at 25°C. Calculate DG and determine if it is spontaneous. Given: Find: Concept Plan: DH = +95. 7 k. J, DS = 142. 2 J/K, T = 298 K DG, k. J T, DH, DS DG Relationships: Solution: Answer: Since DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.

Ex. 17. 3 a – The reaction CCl 4(g) C(s, graphite) + 2 Cl 2(g) has DH = +95. 7 k. J and DS = +142. 2 J/K. Calculate the minimum temperature it will be spontaneous. Given: Find: Concept Plan: DH = +95. 7 k. J, DS = 142. 2 J/K, DG < 0 T, K DG, DH, DS T Relationships: Solution: Answer: The temperature must be higher than 673 K for the reaction to be spontaneous

The 3 rd Law of Thermodynamics Absolute Entropy • the absolute entropy of a substance • is the amount of energy it has due to dispersion of energy through its particles the 3 rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K ü therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy ü therefore, the absolute entropy of substances is always + Tro, Chemistry: A Molecular Approach 32

Standard Entropies • S° • extensive • entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution Tro, Chemistry: A Molecular Approach 33

Relative Standard Entropies States • the gas state has a larger entropy than the liquid • state at a particular temperature the liquid state has a larger entropy than the solid state at a particular temperature Substance S°, (J/mol∙K) H 2 O (g) 70. 0 H 2 O (l) 188. 8 Tro, Chemistry: A Molecular Approach 35

Relative Standard Entropies Molar Mass • the larger the molar mass, • the larger the entropy available energy states more closely spaced, allowing more dispersal of energy through the states Tro, Chemistry: A Molecular Approach 36

Relative Standard Entropies Allotropes • the less constrained the structure of an allotrope is, the larger its entropy Tro, Chemistry: A Molecular Approach 37

Relative Standard Entropies Molecular Complexity • larger, more complex • molecules generally have larger entropy more available energy states, allowing more dispersal of energy through the states Tro, Chemistry: A Molecular Approach Molar S°, Substance Mass (J/mol∙K) Ar (g) 39. 948 154. 8 NO (g) 30. 006 210. 8 38

Relative Standard Entropies Dissolution • dissolved solids • generally have larger entropy distributing particles throughout the mixture Tro, Chemistry: A Molecular Approach Substance S°, (J/mol∙K) KCl. O 3(s) 143. 1 KCl. O 3(aq) 265. 7 39

Substance S , J/mol K NH 3(g) 192. 8 O 2(g) 205. 2 NO(g) 210. 8 H 2 O(g) standard entropies from Appendix IIB 188. 8 Ex. 17. 4 –Calculate DS for the reaction 4 NH 3(g) + 5 O 2(g) 4 NO(g) + 6 H 2 O(l) Given: Find: DS, J/K Concept Plan: S NH 3, S O 2, S NO, S H 2 O, DS Relationships: Solution: Check: DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules

Calculating DG • at 25 C: • DGoreaction = Sn. Gof(products) - Sn. Gof(reactants) at temperatures other than 25 C: üassuming the change in DHoreaction and DSoreaction is negligible DG reaction = DH reaction – TDS reaction Tro, Chemistry: A Molecular Approach 41

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Ex. 17. 7 –Calculate DG at 25 C for the reaction CH 4(g) + 8 O 2(g) CO 2(g) + 2 H 2 O(g) + 4 O 3(g) Substance CH 4(g) O 2(g) CO 2(g) H 2 O(g) O 3(g) DG f, k. J/mol -50. 5 0. 0 -394. 4 -228. 6 163. 2 Given: standard free energies of formation from Appendix IIB Find: DG , k. J Concept Plan: Relationships: Solution: DG f of prod & react DG

Ex. 17. 6 – The reaction SO 2(g) + ½ O 2(g) SO 3(g) has DH = -98. 9 k. J and DS = -94. 0 J/K at 25°C. Calculate DG at 125 C and determine if it is spontaneous. Given: Find: Concept Plan: DH = -98. 9 k. J, DS = -94. 0 J/K, T = 398 K DG , k. J T, DH , DS DG Relationships: Solution: Answer: Since DG is -, the reaction is spontaneous at this temperature, though less so than at 25 C

DG Relationships • if a reaction can be expressed as a series of reactions, the sum of the DG values of the individual reaction is the DG of the total reaction ü DG is a state function • if a reaction is reversed, the sign of its DG value • reverses if the amounts of materials is multiplied by a factor, the value of the DG is multiplied by the same factor ü the value of DG of a reaction is extensive Tro, Chemistry: A Molecular Approach 45

Free Energy and Reversible Reactions • the change in free energy is a theoretical limit as • to the amount of work that can be done if the reaction achieves its theoretical limit, it is a reversible reaction Tro, Chemistry: A Molecular Approach 46

Real Reactions • in a real reaction, some of the free energy is “lost” as heat üif not most • therefore, real reactions are irreversible Tro, Chemistry: A Molecular Approach 47

DG under Nonstandard Conditions DG = DG only when the reactants and products are in their standard states there normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M under nonstandard conditions, DG = DG + RTln. Q Q is the reaction quotient at equilibrium DG = 0 DG = ─RTln. K Tro, Chemistry: A Molecular Approach 48

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Example - DG • Calculate DG at 427°C for the reaction below if the PN 2 = 33. 0 atm, PH 2= 99. 0 atm, and PNH 3= 2. 0 atm N 2(g) + 3 H 2(g) 2 NH 3(g) Q= PNH 32 PN 2 x PH 2 1 3 = (2. 0 atm)2 (33. 0 atm)1 (99. 0)3 = 1. 2 x 10 -7 DH° = [ 2(-46. 19)] - [0 +3( 0)] = -92. 38 k. J = -92380 J DS° = [2 (192. 5)] - [(191. 50) + 3(130. 58)] = -198. 2 J/K DG° = -92380 J - (700 K)(-198. 2 J/K) DG° = +46400 J DG = DG° + RTln. Q DG = +46400 J + (8. 314 J/K)(700 K)(ln 1. 2 x 10 -7) DG = -46300 J = -46 k. J 50

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Example - K • Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2(g) + 3 H 2(g) Û 2 NH 3(g) DH° = [ 2(-46. 19)] - [0 +3( 0)] = -92. 38 k. J = -92380 J DS° = [2 (192. 5)] - [(191. 50) + 3(130. 58)] = -198. 2 J/K DG° = -92380 J - (700 K)(-198. 2 J/K) DG° = +46400 J DG° = -RT ln. K +46400 J = -(8. 314 J/K)(700 K) ln. K = -7. 97 K = e-7. 97 = 3. 45 x 10 -4 since K is << 1, the position of equilibrium favors reactants 52

Temperature Dependence of K • for an exothermic reaction, increasing the • temperature decreases the value of the equilibrium constant for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant Tro, Chemistry: A Molecular Approach 53