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Chapter 19 – Principles of Reactivity: Entropy and Free Energy

Chapter 19 – Principles of Reactivity: Entropy and Free Energy. Objectives: Describe terms: entropy and spontaneity. Predict whether a process will be spontaneous. Describe: free energy. Describe the relationship between D G, K, and product favorability. Thermodynamics.

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Chapter 19 – Principles of Reactivity: Entropy and Free Energy

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  1. Chapter 19 – Principles of Reactivity: Entropy and Free Energy Objectives: Describe terms: entropy and spontaneity. Predict whether a process will be spontaneous. Describe: free energy. Describe the relationship between DG, K, and product favorability.

  2. Thermodynamics • Thermodynamics is _______________ ____________________. • First Law of Thermodynamics • The law of conservation of energy: ______ ________________________________. DE = q + w • The change in internal energy of a system is the sum of the heat transferred to or from the system and the work done on or by the system.

  3. Spontaneous Change • Chemical changes, physical changes • Spontaneous change: occurs _____________________. It leads to ____________. • Example: heat transfers spontaneously from a hotter object to a cooler object. • ____________ is reached in product-favored and in reactant-favored processes.

  4. Spontaneous Chemical Reactions 2 H2 + O2 2 H2O CH4 + 2 O2  CO2 + 2 H2O 2 Na + Cl2  2 NaCl HCl + NaOH  NaCl + H2O • Common feature: _____________ • But many processes are ____________ and spontaneous. • H2 + I2 2 HI (g) _______________ can be approached from either direction.

  5. Spontaneous Processes

  6. Spontaneous Processes • Dissolving NH4NO3 in water: DH = +25.7 KJ/mol • Expansion of a gas into a vacuum: energy neutral, heat is neither evolved nor required. • Phase changes: melting of ice requires ~ 6 kJ/mol; but only occurs if T > 0oC. • ______________ determines whether a process is spontaneous. • Heat transfer: The T of a cold substance in a warm environment will rise until the substance reaches the ambient T. • The required heat comes from the _____________.

  7. Entropy • To predict whether a process will be spontaneous. • Entropy, S is a thermodynamic function • State function: a quantity whose value is determined only by the initial and final states of a system. • Second Law of Thermodynamics • ____________________________________________________________________________. • ____________________________________________________________________________.

  8. Dispersal of Energy • By statistical analysis: • Energy is distributed of a number of particles • Most often case is when energy is distributed over all particles and to a large number of states. • As the number of particles and the number of energy levels grows, one arrangement turns out to be vastly more probable than all others.

  9. Dispersal of Energy • Dispersal of __________ often contributes to energy dispersal.

  10. Boltzmann Equation • Ludwig Boltzmann (1844-1906) • Look at the distribution of energy over different energy states as a way to calculate ____________. S = k log W • K – Boltzmann constant • W – represent the number of different ways that the energy can be distributed over the available energy levels. • A maximum entropy will be achieved at _________________ , a state in which W has the maximum value.

  11. Matter and Energy Dispersal

  12. Matter and Energy Dispersal

  13. Summary: Matter and Energy Dispersal • A final state of a system can be more probable than the initial state if: • The atoms and molecules can be more ____________ and/or • ___________ can be dispersed over a greater number of atoms and molecules. • If energy and matter are both dispersed in a process, it is _______________. • If only matter is dispersed, then quantitative information is needed to decide whether the process is spontaneous. • If energy is not dispersed after a process occurs, then that process will ____________ _____________________.

  14. Entropy • Entropy is used to __________________ ___________ resulting from dispersal of energy and matter. The greater the _______ in a system, the greater the value of S. • Third Law of Thermodynamics • There is no disorder in a perfect crystal at 0K, S=0. • The entropy of a substance at any T can be obtained by measuring the heat required to raise the T from 0K, where the conversion must be carried by a reversible process (very slow addition of heat in small amounts).

  15. Entropy • The entropy of a substance at any T can be obtained by measuring the ________ required to raise the T from 0K, where the conversion must be carried by a reversible process (very slow addition of heat in small amounts). • The entropy added by each incremental change is: DS = • Adding the entropy changes gives the total entropy. • All substances have ___________ entropy values at temperatures above 0K.

  16. Standard Molar Entropy Values

  17. Standard Molar Entropy Values

  18. Thermodynamics • First Law: The total energy of the universe is a constant. • Second Law: The total entropy of the universe is always increasing. • Third Law: The entropy of a pure, perfectly formed crystalline substance at 0K is zero. - A local decrease in entropy (the assembly of large molecules) is offset by an increase in entropy in the rest of the universe -.

  19. Standard Entropy • So, is the entropy gained by converting it from a perfect crystal at 0K to standard state conditions (1 bar, 1 molal solution). • Units: J/Kmol • Entropies of gases are ____________than those for liquids, entropies of liquids are ____________ than those for solids. • Larger molecules have a _________ entropy than smaller molecules, molecules with more complex structures have ________entropies than simpler molecules.

  20. Entropy • The entropy of liquid water is ___________ than the entropy of solid water (ice) at 0˚ C. S˚(H2O sol) < S˚(H2O liq)

  21. Entropy Entropies of ionic solids depend on ___________________________. So (J/K•mol) MgO 26.9 NaF 51.5 Mg2+ & O2- Na+ & F- The larger coulombic attraction on MgO than NaF leads to a lower entropy.

  22. Which substance has the higher entropy, why? • O2 (g) or 03 (g) • SnCl4 (l) or SnCl4 (g)

  23. Arrange the substances in order of increasing entropy. Assume 1 mole of each at standard conditions. HCOOH(l) CO2(g) Al(s) CH3COOH(l)

  24. Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. CO(g) + 3 H2(g)  CH4(g) + H2O(g) 2 H2O(l)  2 H2(g) + O2(g) I2(g) + Cl2(g)  2 ICl(g)

  25. S increases slightly with T S increases a large amount with phase changes Entropy Change

  26. Entropy Change • Entropy usually increases when a pure liquid or solid ______________ in a solvent. • Entropy of a substance ____________ with temperature.

  27. Entropy Change • The entropy change is the sum of the entropies of the products minus the sum of the entropies of reactants: DS0system = S S0(products) – S S0(reactants) You will find DSo values in the Appendix L of your book.

  28. Calculate the standard entropy changes for the evaporation of 1.0 mol of liquid ethanol to ethanol vapor. C2H5OH(l)  C2H5OH(g)

  29. Calculate the standard entropy change for forming 2.0 mol of NH3(g) from N2(g) and H2(g) N2(g) + 3 H2(g)  2 NH3 (g)

  30. Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.35 moles of NO(g) react at standard conditions. 2 NO(g) + O2(g)  2 NO2(g)

  31. Calculate the standard entropy change for the oxidation of ethanol vapor (CH2H5OH (g)).

  32. Entropy in the Universe DS0univ = DS0sys + DS0surr • 2nd Law of Thermodynamics: DSuniv is positive for a spontaneous process. • For a nonspontaneous process: DS0univ < 0 (negative) • If DSuniv = 0 the system is at equilibrium. • Calculate first the DS0sys, then DS0surr. DS0surr = qsurr/T = -DH0sys/T DH0sys = S H0(products) – S H0(reactants)

  33. Show thatDS0univis positive (>0) for dissolving NaCl in water DS0univ = DS0sys + DS0surr 1) Determine DS0sys 2) Determine DS0surr NaCl(s) NaCl (aq)

  34. Show thatDS0univis positive (>0) for dissolving NaCl in water

  35. Predicting whether a Process will be Spontaneous – Table 19.2 Based on the values of DH0sys and DS0sys there are 4 types: 1) DH0sys < 0 Exothermic & DS0sys > 0 Less order DS0univ > 0 ________________ under all conditions. 2) DH0sys < 0 Exothermic & DS0sys < 0 More order Depends on values, more favorable at __________ temperatures. 3) DH0sys > 0 Endothermic & DS0sys > 0 Less order Depends on values, more favorable at __________ temperatures. 4) DH0sys > 0 Endothermic & DS0sys < 0 More order DS0univ < 0 ___________________ under any conditions. Remember that –∆H˚sys is proportional to ∆S˚surr An exothermic process has ∆S˚surr > 0.

  36. Classify the following as one of the four types of Table 19.2 DH0 (kJ) DS0 (J/K) CH4 (g) + 2 O2(g) 2 H2O (l) + CO2(g) -890 -242.8 2 FeO3(s) + 3 C (graphite) 4 Fe(s) + 3 CO2(g) +467 +560.7

  37. Calculate the entropy change of the UNIVERSE when 1.890 moles of CO2(g) react under standard conditions at 298.15 K. Consider the reaction6 CO2(g) + 6 H2O(l)  C6H12O6 + 6O2(g)for which DHo = 2801 kJ and DSo = -259.0 J/K at 298.15 K. • Is this reaction reactant or product favored under standard conditions?

  38. Gibbs Free Energy DSuniv= DSsurr+ DSsys DSsurr= -DHsys/T DSuniv= -DHsys/T + DSsys Multiply equation by –T -T DSuniv = DHsys –TDSsys J. Willard Gibbs (1839-1903) DGsys = -T DSuniv DGsys = DHsys –TDSsys DGsys< 0, a reaction is ____________ DGsys= 0, a reaction is _____________ DGsys> 0, the reaction is ____________

  39. Gibbs Free Energy and Spontaneity • J. Willard Gibbs (1839-1903) • Gibbs free energy, G, “free energy”, a thermodynamic function associated with the ________________. • G = H –TS • H- Enthalpy • T- Kelvin temperature • S- Entropy • Changes during a process: DG • Use to determine whether a reaction is __________. • DG is ___________related to the value of the _____________________________ , and hence to product favorability.

  40. “Free” Energy • DG = w max • The free energy represents the maximum energy ____________________________. • Example: C(graphite) + 2 H2 (g) CH4 (g) • DH0rx = -74.9 kJ; DS0rx = -80.7 J/K • DG0rx = DH0 – TDS0 • = -74.9 kJ – (298)(-80.7)/1000 kJ • = -74.9 kJ + 24.05 kJ • DG0rx = - 50.85 kJ • Some of the energy liberated by the reaction is needed to “order” the system. The energy left is energy available energy to do_________, “free” energy. • DG < 0, the reaction is _______________.

  41. Calculate DGo for the reaction below at 25.0 C. P4(s) + 6 H2O(l) → 4 H3PO4(l) DG0rx = DH0 – TDS0

  42. Standard Molar Free Energy of Formation • The standard free energy of formation of a compound, DG0f, is the free energy change when forming __________of the compound from the __________________, with products and reactants in their __________________. • Then, DG0f of an element in its standard states is _________.

  43. Gibbs Free Energy DG0rxn is the increase or decrease in free energy as the reactants in their standard states are converted completely to the products in their standard states. * Complete reaction is not always ________________. * Reactions reach an _____________. DG0system = S G0(products) – S G0(reactants)

  44. Calculating DG0rxn from DG0f • DG0system = S G0(products) – S G0(reactants) • Calculate the standard free energy change for the oxidation of 1.0 mol of SO2 (g) to form SO3 (g). • DG0system = DGf0 (kJ/) SO2(g) -300.13 SO3(g) -371.04

  45. Free Energy and Temperature • G = H – TS • G is a function of T, DG will change as T changes. • Entropy-favored and enthalpy-disfavored • Entropy-disfavored and enthalpy-favored

  46. Changes in DG0 with T

  47. Consider the reaction below. What is DG0at 341.4 K and will this reaction be product-favored spontaneously at this T? CaCO3(s)  CaO(s) + CO2(g) Thermodynamic values: DHf0 (kJ/mol) S0 (kJ/Kmol) CaCO3(s) -1206.9 +0.0929 CaO(s) -635.1 + 0.0398 CO2(g) -393.5 + 0.2136

  48. Estimate the temperature required to decompose CaSO4(s) into CaO(s) and SO3(g). Thermodynamic values: DHf0 (kJ/mol) S0 (J/Kmol) CaSO4(s) -1434.52 +106.50 CaO(s) -635.09 + 38.20 SO3(g) -395.77 + 256.77 CaSO4(s) CaO(s) + SO3(g) DH0sys = S H0(products) – S H0(reactants) DH0sys = S S0(products) – S S0(reactants)

  49. For the reaction: 2H2O(l) 2H2(g) + O2(g)DGo = 460.8 kJ and DHo = 571.6 kJ at 339 K and 1 atm. • This reaction is (reactant,product) _____________ favored under standard conditions at 339 K. • The entropy change for the reaction of 2.44 moles of H2O(l) at this temperature would be _________J/K. • DGorxn = DHorxn - DT Sorxn DSo = (DHo - DGo)/T

  50. DG0, K, and Product Favorability • Large K – ____________ favored • Small K – ____________favored • At any point along the reaction, the reactants are not under standard conditions. • To calculate DG at these points: • DG = DG0 + RT ln Q • R – Universal gas constant • T - Temperature (kelvins) • Q - Reaction quotient

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