Integrated rate lawsSlide 2Slide 3Reaction mechanismSlide 5Slide 6Slide 7Slide 8Reaction MechanismSlide 10Slide 11Slide 12Slide 13Slide 14Slide 15Integrated rate lawszero orderrate = k[A]0 2kt1/2 =Integrated rate laws1st orderrate = k[A]ln[A] = -kt + ln[A]0ln 2 kt1/2 =2nd orderrate = k[A]21/[A] = kt + 1/[A]01k [A]0t1/2 =can also get negative or fractional ordersReaction mechanismreaction =Elementary stepchemical reaction Elementary stepsunimolecularA  productk[A]bimolecularA+ B  productk[A][B]termolecularA+B+C  product k[A][B][C]A+A+A  product k[A]3rate law and stoichiometry= sum of elementary stepsMolecularity rate lawReaction mechanism2H2O2 (aq)  2H2O(l) + O2(g)spontaneous reactionexperimental rate law: rate = k[H2O2]I- =increase rate of reaction not consumed in the overall reactionreactant in early elementary stepproduct in later elementary stepGorxn =[ 2(-237.9)] - [2(-131.67)] = -212.46 kJ[I-]catalystReaction mechanism2H2O2 (aq)  2H2O(l) + O2(g)rate = k[H2O2] [I-]step 1 H2O2k1step 2 H2O2 +k22H2O2 (aq)step 1+ I- H2O + OI-OI-  H2O + O2+I-rate = k1[H2O2] [I-] 2H2O(l) + O2(g)Reaction mechanismstep 1H2O2 + I-  H2O + OI-step 2H2O2 + OI-  H2O + O2 +I-k2what about step 2 ?assume k2step 1I- catalystrate determining stepk1consumed in early elementary stepregenerated in later elementary stepOI- formed in early step, consumed in later step>> k1 intermediateRate determining step on Labor Day weekend“Big Mac” bridgeReaction MechanismH2(g) + I2(g)  2HI(g)rate = k [H2] [I2]reaction faster in lightfree radical unpaired electronReaction Mechanismstep 1H2(g) + I2(g)  2HI(g)rate = k [H2] [I2]forward rateequilibriumI2= kf [I2]reverse rate =kr [I.]2kf[I2] =kr[I.]2kf Keq=[I2]kr2I.Reaction Mechanismstep 1H2(g) + I2(g)  2HI(g)rate = k [H2] [I2]step 2I. =need H2 in the rate determining steprate =H2 +k [H2] [I.]2intermediate2I. 2HIfrom step 2H2+ I2 2HII2 2I.Reaction MechanismH2(g) + I2(g)  2HI(g)rate = k [H2] [I2]step 2 H2 + 2I. 2HIrate = k [H2] [I.]2 Keq =rate =rate = k’ [H2] [I2]eq [I2]k [H2]eq [I2][I2] kfkrstep 1I2 2I.Reaction Mechanism2NO + O2 2NO2step 1step 2overall reaction:intermediates:2NO + O2  2NO2N2O2N2O2 + O2 2NO22NO  N2O2rate =[N2O2] [O2] k2steady state approximation[N2O2] = constantd[N2O4]dt= 02NO + O2 2NO2step 1step 22NO + O2  2NO2N2O2 + O2 2NO22NO  N2O2rate =[N2O2] [O2] k2k1[NO]2k2 [N2O2][O2]produce N2O2consume N2O2+ k-1[N2O2] =[N2O2] =k1 [NO]2k2[O2] + k-1 = k2[O2]k1[NO]2k2[O2] + k-12NO + O2 2NO22NO + O2  2NO2N2O2 + O2 2NO22NO  N2O2rate = k2[O2]k1[NO]2k2[O2] + k-1 low [O2]rate = k'[O2][NO]2high [O2]rate =k1[NO]2step 1step