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## Chapter 19 – Principles of Reactivity: Entropy and Free Energy

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**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.**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.**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.**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.**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 _____________.**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 • ____________________________________________________________________________. • ____________________________________________________________________________.**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.**Dispersal of Energy**• Dispersal of __________ often contributes to energy dispersal.**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.**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 ____________ _____________________.**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).**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.**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 -.**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.**Entropy**• The entropy of liquid water is ___________ than the entropy of solid water (ice) at 0˚ C. S˚(H2O sol) < S˚(H2O liq)**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.**Which substance has the higher entropy, why?**• O2 (g) or 03 (g) • SnCl4 (l) or SnCl4 (g)**Arrange the substances in order of increasing entropy.**Assume 1 mole of each at standard conditions. HCOOH(l) CO2(g) Al(s) CH3COOH(l)**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)**S increases slightly with T**S increases a large amount with phase changes Entropy Change**Entropy Change**• Entropy usually increases when a pure liquid or solid ______________ in a solvent. • Entropy of a substance ____________ with temperature.**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.**Calculate the standard entropy changes for the evaporation**of 1.0 mol of liquid ethanol to ethanol vapor. C2H5OH(l) C2H5OH(g)**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)**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)**Calculate the standard entropy change for the oxidation of**ethanol vapor (CH2H5OH (g)).**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)**Show thatDS0univis positive (>0) for dissolving NaCl in**water DS0univ = DS0sys + DS0surr 1) Determine DS0sys 2) Determine DS0surr NaCl(s) NaCl (aq)**Show thatDS0univis positive (>0) for dissolving NaCl in**water**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.**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**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?**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 ____________**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.**“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 _______________.**Calculate DGo for the reaction below at 25.0 C.**P4(s) + 6 H2O(l) → 4 H3PO4(l) DG0rx = DH0 – TDS0**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 _________.**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)**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**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**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**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)**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**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