ReviewIntegrated rate lawsSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Pseudo order reactionsSlide 12Slide 13ReviewDifferential Rate Lawsrate (M s-1) =A + 3B  2Crate =Assume that A is easily detectedk [A]a[B]b ...-[A] t = - [B] tinitial rate =what is-[B] t1x10-3 M s-1a) 1x10-3Ms-1b) 3x10-3Ms-1c) 0.33x10-3Ms-11/3Integrated rate lawsdifferential rate laws rate = k[A]rate = k[A]2are differential equationsdifferential rate eqn integrated rate eqn= -d[A]/dt= -d[A]/dtt = 0t = trate = k = -d[A]/dtzero order reactionfirst order reactionsecond order reactiont = 0t = 0t = tt = tIntegrated rate lawsdifferential rate laws are differential equationsdifferential rate eqn integrated rate eqn[A] =-kt+ [A]0y = mx + bslope = -kintercept = [A]0rate = k = -d[A]/dtzero order reactionrate = kzero order in ethanolk = - .037 - .06560 - 0= 4.7 x 10-4rate = 4.7 x 10-4M min-1CH3CH2OH + NAD+  CH3CHO + NADH + H+ t (min) [ethanol] M0 0.06515 0.05830 0.05145 0.04460 0.037[CH2CH2OH] = -kt + [CH3CH2OH]0[CH3CH2OH]02 k= 70 mint1/2 =Integrated rate lawsdifferential rate laws are differential equationsdifferential rate eqn integrated rate eqnln [A] =-kt+ ln [A]0y = mx + bslope = -kintercept = ln [A]0rate = k[A] = -d[A]/dtfirst order reactionrate = k [cis-platin]first order in cis-platink = - (-6.32)-(-5.12)800 - 0= 1.5x10-3 rate = 1.5 x 10-3 min-1cis-[Pt(NH3)2Cl2] + H2O  [Pt(NH3)2Cl(H2O)]Cl- t (min) [cis-platin] M0 0.0060200 0.0044400 0.0033600 0.0024800 0.0018ln [cis-platin] = -kt +ln [cis-platin]0 k= 462 mint1/2 =ln 2PtClClNH3NH3+H2O PtClOH2NH3NH3+Cl-rate = k [cis-platin]first order in cis-platink = - (-6.32)-(-5.12)800 - 0= 1.5x10-3 rate = 1.5 x 10-3 min-1cis-[Pt(NH3)2Cl2] + H2O  [Pt(NH3)2Cl(H2O)]Cl- t (h) [cis-platin] M0 0.0060200 0.0044400 0.0033600 0.0024800 0.0018ln [cis-platin] = -kt +ln [cis-platin]0 k= 462 mint1/2 =ln 2radioactive decay 1st order14C datingt1/2 = 5730 yearsIntegrated rate lawsdifferential rate laws are differential equationsdifferential rate eqn integrated rate eqn1/[A] =kt+ 1/[A]0y = mx + bslope = kintercept = 1/[A]0rate = k[A]2= -d[A]/dtsecond order reactiont (s) [X] (M)2 0.3924 0.3366 0.2948 0.26110 0.2351/.235 – 1/.39310 - 2= .213 1/.235 =(.213)(10) + 1/[X]0[X]0 = 0.471 = ksecond order in X1/[X] = kt + 1[X]0rate = [X]2M-1s-10.213t1/2 =1k[X]0Integrated rate lawssecond order reactionsrate = k [A]2 1 = kt + 1 [A]t [A]0many second order reactionsA + B  C[A]0 = [B]0if not, A and B consumed stoichiometricallyrate = k[A] [B]no analytical solutionPseudo order reactionshigh order reactionsput in large excess of all but one reagentrate = k [A]a [B]b [A]0 (M) [B]0 (M)[B] rate =difficult to analyze1.0 x 10-31.0 -0.5 x 10-3 mol -0.5x 10-3 mol0.5 x 10-3 0.999 constantk’ [A]ak = k’ [B]bk[A]a[B]b= k’[A]X + 2Y  2Zt (s) [X] (M) [Y] (M)0 0.470 2.02 0.4484 0.4276 0.4098 0.39210 0.376second order in Xt (s) [X] (M) [Y] (M)0 0.470 4.02 0.4274 0.3916 0.3618 0.33510 0.313X + 2Y  2Zsecond order in X[Y] = 2.0 Mk’ =2.70 - 2.1310 - 0= 0.057k’’ =4.26 - 2.1310 - 0= .106.106.057= (4.0)y(2.0)yy =1first order in Yrate = k [X]2[Y] rate = k’ [X]2[Y] = 4.0 Mk = k’[Y]k = .028