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##### Spontaneity Entropy and Free Energy

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**Spontaneity Entropy and Free Energy**WHAT DRIVES A REACTION TO BE SPONTANEOUS?**ENTHALPY (H)**heat content (exothermic reactions are generally favored)**1st Law of Thermodynamics**The total energy in the universe is constant. DE = q (heat) + w (work) -heat out -work on surroundings + heat in + work on system Calculating DH: • stoichiometry—using info given with equation • H rxn = ∑ n DHf° products -- ∑ n DHf° reactants • Hess’s Law—adding up equations and DH’s • calorimetry q = mCDT or q = CDT • bond energies ∑ bonds broken -- ∑ bonds formed • Note: pure elements have DHf° values of 0**ENTROPY ‘S’The Second Law of Thermodynamics**The universe is constantly increasing disorder. DSuniv = DSsystem + DSsurroundings ΔSsurroundingsbased on heat flow exothermic + DSsurr endothermic -DSsurr**ENTROPY (S)**Disorder of a system (more disorder is favored) Nature tends toward chaos! Think about your room at the end of the week! Factors that can indicate entropy changes: • Phase changes • Temperature changes • Volume changes • Mixing substances • Change in number of particles • Change in moles of gas**Spontaneous reactions are those that occur without outside**intervention. They may occur fast OR slow (that is kinetics). Some reactions are very fast (like combustion of hydrogen) other reactions are very slow (like graphite turning to diamond)**Predicting the entropy of a system based on physical**evidence: • The greater the disorder or randomness in a system, the larger the entropy. • The entropy of a substance always increases as it changes from solid to liquid to gas. • When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases (carbonates are an exception! --they interact with water and actually bring MORE order to the system)**Predicting the entropy of a system based on physical**evidence: • When a gas molecule escapes from a solvent, the entropy increases. • Entropy generally increases with increasing molecular complexity (crystal structure: KCl vs CaCl2) since there are more MOVING electrons! • Reactions increasing the number of moles of particles often increase entropy.**Exercise 2 Predicting Entropy Changes**Predict the sign of the entropy change for each of the following processes. A: Solid sugar is added to water to form a solution. B: Iodine vapor condenses on a cold surface to form crystals.**Solution:**A: +∆S Sugar molecules have been randomly dispersed in the water. Greater entropy + DS B: -∆S Gaseous iodine is forming a solid. Less entropy - DS**Sample Problem A**Which of the following has the largest increase in entropy? a) CO2(s) CO2(g) b) H2(g) + Cl2(g) 2 HCl(g) c) KNO3(s) KNO3(l) d) C(diamond) C(graphite)**Answer:**• the substance changes from a highly organized state to a more disorganized state.**Choose the substance expected to have the greater absolute**entropy. • Pb(s) or Cgraphite • He(g) at 1 atmosphere or He(g) at 0.5 atmosphere • H2O(l) or CH3CH2OH(l) at the same temperature • Mg(s) at 0°C or Mg(s) at 150 °C both at the same pressure**Pb has greater molar entropy. Pb, with metallic bonding,**forms soft crystals with high amplitudes of vibration; graphite has stronger (covalent) bonds is more rigid and thus more ordered. • He(g) at 0.05 has greater molar entropy. At lower pressure (greater volume) He atoms have more space in which to move so are more random. • CH3CH2OH has greater molar entropy. Ethanol molecules have more atoms and thus more vibrations; water exhibits stronger hydrogen bonding • Mg(s) at 150 °C has greater molar entropy. At the higher temperature the atoms have more kinetic energy and vibrate faster and, thus, show greater randomness.**ENTROPYThe Third Law of Thermodynamics**• The entropy of a perfect crystal at 0 K is zero. • not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even elements**This means the absolute entropy of a substance can then**be determined at any temperature higher than 0 K. (Handy to know if you ever need to defend why G & H for elements = 0. . . BUT S does not!) So what?**BIG MAMMA, verse 2**S°rxn = S°(products) - S°(reactants) S is + when disorder increases (favored) S is – when disorder decreases Units are usually J/K mol (not kJ ---tricky!)**Sample Problem B**Calculate the entropy change at 25C, in J/K for: 2 SO2(g) + O2(g) 2 SO3(g) Given the following data: SO2(g) 248.1 J/K mol O2(g) 205.3 J/Kmol SO3(g) 256.6 J/K mol Entropy change = -188.3 J/K**ENTROPY CHANGES FOR REVERSIBLE PHASE CHANGES**(that’s a phase change at constant temperature) ΔS = heat transferred = q temperature at which change occurs T **where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvins**EX: water (l @ 100 → water (g @ 100)**the entropy will increase. Taking favored conditions into consideration, the equation above rearranges into: S = - H T Give signs to ΔH following exo/endo guidelines! (If reaction is exo.; entropy of system increases—makes sense!)**Calculate ∆Ssurr for each of these reactions at 25C and**1 atm. Sb2S3(s)+3 Fe(s) 2 Sb(s)+3 FeS(s) ∆H = -125kJ Sb4O6(s)+6 C(s)4 Sb(s)+6 CO(g) ∆H = 778kJ**Solution:**∆Ssurr= 419 J/K ∆Ssurr = -2.61 × 103 J/K**SUMMARY**ENTROPY: S = + MORE DISORDER (FAVORED CONDITION) S = - MORE ORDER**Whether a reaction will occur spontaneously may be**determined by looking at the S of the universe. ΔS system + ΔS surroundings = ΔS universe • IF ΔS universe is +, then reaction is spontaneous • IF ΔS universe is -, then reaction is NONspontaneous**Consider**2 H2 (g) + O2 (g) → H2O (g) ignite & rxn is fast! ΔSsystem = -88.9J/K Entropy declines (due mainly to 3 → 2 moles of gas!)**First law of thermodynamics demands that this energy is**transferred from the system to the surroundings so... -ΔHsystem = ΔHsurroundings OR - (- 483.6 kJ) = + 483.6 kJ**ΔS°surroundings =**ΔH°surroundings = + 483.6 kJ T 298 K = 1620 J/K**Now we can find ΔS°universe**ΔS system + ΔS surroundings = ΔS universe (-88.9 J/K) + (1620 J/K) = 1530 J/K Even though the entropy of the system declines, the entropy change for the surroundings is SOOO large that the overall change for the universe is positive.**Bottom line:**A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering.**FREE ENERGY**• Calculation of Gibb’s free energy is what ultimately decides whether a reaction is spontaneous or not. • NEGATIVE G’s are spontaneous.**G can be calculated one of several ways:**Gºrxn = Go(products) - Go(reactants) ΔG°f = 0 (for elements in standard state) G = H - TS**G = Go + RT ln (Q)**Define terms: G = free energy not at standard conditions Go = free energy at standard conditions R = universal gas constant 8.3145 J/molK T = temp. in Kelvin ln = natural log Q = reaction quotient: (for gases this is the partial pressures of the products divided by the partial pressures of the reactants—all raised to the power of their coefficients) Q = [products] [reactants]**“RatLink”:**G = -RTlnK Terms: Basically the same as above --- however, here the system is at equilibrium, so G = 0 and K represents the equilibrium constant under standard conditions. K = [products] still raised to power of coefficients [reactants]**“nFe”: G = - nFE Remember this!!**Terms: Go = just like above—standard free energy n = number of moles of electrons transferred (look at ½ reactions) F = Faraday’s constant 96,485 Coulombs/mole electrons Eo= standard voltage ** one volt = joule/coulomb****Sample Problem C:**Find the free energy of formation for the oxidation of water to produce hydrogen peroxide. 2 H2O(l) + O2(g) 2 H2O2(l) Given the following information: G°f H2O(l) -56.7 kcal/mol O2(g) 0 kcal/mol H2O2(l) -27.2 kcal/mol Free energy of formation = 59.0 kcal/mol**Calculate ∆H, ∆S, and ∆G**2 SO2 (g) + O2 (g) 2 SO3 (g) carried out at 25C and 1 atm.**Solution:**∆H = -198 kJ ∆S = -187 J/K ∆G = -142 kJ**Hess’s law of summation**Using the following data @ 25C, calculate Go Cdiamond (s) Cgraphite (s) Cdiamond (s) +O2 (g)CO2 (g) ∆G= -397 kJ Cgraphite (s) +O2 (g) CO2 (g) ∆G= -394 kJ ∆G = -3 kJ**G = -RTlnK )**The overall reaction for the corrosion (rusting) of iron by oxygen is 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) Calculate the equilibrium constant for this reaction at 25C. K = .548**Gibb’s equation can also be used to calculate the phase**change temperature of a substance. During the phase change, there is an equilibrium between phases so the value of G is zero. Really just like what we did earlier in this unit with enthalpy and entropy!**Sample Problem:**Find the thermodynamic boiling point of H2O(l) H2O(g) Given the following information: Hvap = +44 kJ Svap = 118.8 J/K Thermodynamic BP = 370K**SUMMARY OF FREE ENERGY:**G = + NOT SPONTANEOUS G = - SPONTANEOUS**Conditions of G:**H S Result neg pos spontaneous at all temps pos pos spontaneous at high temps neg neg spontaneous at low temps pos neg not spontaneous, ever**Relationship to K and E:**G K E 0 at equilibrium; K = 1 0 negative >1, products favored positive positive <1, reactants favored negative**Summary**Signs are important! DS + increased disorder - decreased disorder DH + endothermic -exothermic DG + nonspontaneous - spontaneous 0 @ equilibrium**3 equations**D?rxn = ∑ ?f products - ∑ ?f reactants DG = DH - TDS DG = -RTlnK**Enthalpy change**ΔH o = ∑ Δ Hof,products - ∑ ΔHof, reactants Δ Hrxn = ∑ bonds broken - ∑ bonds formed**Enthalpy**ΔS = ∑ Sproducts - ∑ Sreactants ΔG = ΔH - TΔS At equilibrium ΔG = 0, therefore ΔS = ΔH / T