3-d plotContour Plot2 of 3 coordinatesSymmetric StretchrEnergySimplest possible “reaction”D + H2→→→→ HD + HrHHEnergyV(rHH) →→→→H2as H - D distance increases.Potential Energy SurfacesPotential Energy CurvesEnergyQEnergyQReac. Prod.EndothermicEnergyQReac. Prod.ExothermicEa∆∆∆∆E or ∆∆∆∆H∆∆∆∆E or ∆∆∆∆HEaActivation Energy, EaActivation Energy, EaReaction CoordinateReaction CoordinateReac. Prod.Temperature Dependence of a Reaction Rate Constant:Temperature Dependence of a Reaction Rate Constant:Velocity distribution in gas phase at temperature TNv|v|Critical velocity for reaction, vc|v| < voNo Reaction|v| > voReactionT = 300 KT = 500 KT = 1000 KT = 5000 KThe minimum energy that must be supplied by a collision is the Activation Energy.KE = ½mv2≥≥≥≥ Eak = A∞∞∞∞e-Ea/kBT|v|NvRate constantBoltzmann ConstantPre-exponential factorPopulation with KE sufficient to drive the reaction.k = A∞∞∞∞e-Ea/kBTTemperaturekk = 0.9 A∞∞∞∞kBT = 10 EakBT = Eak = 0.36 A∞∞∞∞Ea= Joules/moleculefor Ea= Joules/molek = A∞∞∞∞e-Ea/RTRT = 10 EaRT = Eak = A∞∞∞∞A∞∞∞∞= maximum possible rate at infinite temperatureln k = -Ea1 + ln A∞∞∞∞RTA plot of ln k vs 1/T will be linear. lnk1/TIntercept = ln A∞∞∞∞Slope = -Ea/RArrhenius EquationArrhenius EquationArrhenius PlotArrhenius PlotActivation barriers are determined experimentally by measuring the rate over as large a range of T as possible.But What About Entropy?But What About Entropy?An exothermic reaction will occur rapidly if Ea< RT and only very slowly if Ea>> RT. Potential Energy∆∆∆∆EEa∆∆∆∆HReaction CoordinateAn endothermic reaction will have a large activation energy, Ea.Reaction Coordinate∆∆∆∆HEaPotential EnergyEntropy accounts for the number of different states that contribute all along the reaction path.We plot Gibbs energy along reaction coordinate to account for entropy.Gibbs Energy∆∆∆∆Gŧ∆∆∆∆rGReaction Coordinate∆∆∆∆Gŧis the free energy or Gibbs energy of activation. = kBTe∆∆∆∆Sŧ/R e-∆∆∆∆Hŧ/RTh= kBTe-(∆∆∆∆Hŧ-T∆∆∆∆Sŧ)/RThk = kBTe-∆∆∆∆Gŧ/RThPre-exponential factor A∞∞∞∞kh = Planck’s Constant∆∆∆∆Sŧ= Entropy of Activation∆∆∆∆Hŧ= Enthalpy of Activation ⇒⇒⇒⇒ microscopic EaSample Problem:Sample Problem:Sample Problem:If a reaction doubles its rate when the temperature is increasedfrom 305 K to 315 K, what is the activation energy?k = A∞∞∞∞e-Ea/RTk(315 K) = A∞∞∞∞e-Ea/R(315 K)= 2k(305 K) A∞∞∞∞e-Ea/R(305 K)2 = e-Ea(1/305 – 1/315)/RA∞∞∞∞is approximately T independentln2 = -Ea(1/305 – 1/315)/REa= ln2 (8.3145 J mol-1K-1) 10-3kJ0.0001041 K-1J= 55.4 kJ/molCatalysis:Catalysis:Catalysis:Consider: 2 H2O2(l) →→→→ 2 H2O (l) + O2(g)∆∆∆∆rG∅∅∅∅= 2(-237.13) + 0 - 2(-120.35) = -233.56 kJ∆∆∆∆rH∅∅∅∅= 2(-285.83) + 0 - 2(-187.78) = -196.10 kJIn fact the equilibrium constant is huge. K = e-∆∆∆∆rG /RT∅∅∅∅K = exp = 1.15 ×××× 1041233.56 ×××× 103J8.31451 J K-1298EnergyQReac. Prod.HO – OH→→→→HO – H+ O••••∆∆∆∆rH∅∅∅∅Very high activation energyThis reaction is exothermic and spontaneous!2 H2O2(l) 2 H2O (l) + O2(g)MnO2Add MnO2MnO2acts as a catalyst – that is, it provides a mechanism with a much smaller activation energy.Old PathEnergy Along the Catalyzed PathEnergyQH2O + O2H2O2Initial and final states are unaffected by catalyst. The nature of the reaction coordinate is changed. “Q”EnergyQCAT“Q”EnergyQCATCatalyst is in a different phase than reactants.Heterogeneous Catalysis:Heterogeneous Catalysis:Usually a solid surface catalyzing gas or liquid phase reactions.Homogeneous Catalysis:Homogeneous Catalysis:Catalyst is in the same phase as the reactants.Usually in liquid solution.Rhodium, Platinum, Zeolites, Metal Oxides, …Complexes of Rhodium and Platinum, Enzymes, Strong Acids, …Examples:Examples:2 H2O2(l) 2 H2O (l) + O2(g)CatalaseIn the presence of CatalaseManganese Catalase in certain Lactobacillus type