Integrated rate laws zero order t1/2 = [A]0 2k rate = k

Integrated rate laws 1st order ln[A] = -kt + ln[A]0 rate = k[A] t1/2 =

2nd order 1/[A] = kt + 1/[A]0 rate = k[A]2 t1/2 = 1 k [A]0 can also get negative or fractional orders

Reaction mechanism reaction =  Elementary steps Molecularity rate law unimolecular A  product k[A] bimolecular A+ B  product k[A][B] termolecular A+B+C  product k[A][B][C] A+A+A  product k[A]3 chemical reaction = sum of elementary steps rate law and stoichiometry

Reaction mechanism increase rate of reaction 2H2O2 (aq)  2H2O(l) + O2(g) Gorxn = [ 2(-237.9)] - [2(-131.67)] = -212.46 kJ spontaneous reaction experimental rate law: rate = k[H2O2] [I-] I- = catalyst increase rate of reaction not consumed in the overall reaction reactant in early elementary step product in later elementary step

Reaction mechanism 2H2O2 (aq)  2H2O(l) + O2(g) rate = k[H2O2] [I-] step 1 H2O2 + I-  H2O + OI- k2 step 2 H2O2 + OI-  H2O + O2 +I- 2H2O2 (aq)  2H2O(l) + O2(g) step 1 rate = k1 [H2O2] [I-]

Reaction mechanism step 1 H2O2 + I-  H2O + OI- step 2 k1 step 1 H2O2 + I-  H2O + OI- k2 step 2 H2O2 + OI-  H2O + O2 +I- what about step 2 ? assume k2 >> k1 step 1 rate determining step I- catalyst consumed in early elementary step regenerated in later elementary step OI- intermediate formed in early step, consumed in later step

Rate determining step on Labor Day weekend “Big Mac” bridge

Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2] reaction faster in light free radical unpaired electron

Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2]  I2 2I. step 1  forward rate = kf [I2] reverse rate = kr [I.]2 equilibrium kf[I2] = kr[I.]2 Keq= [I.]2 = kf [I2] kr

Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2] I2 2I.  step 1 step 2 H2 + 2I. 2HI H2 + I2  2HI need H2 in the rate determining step from step 2 rate = k [H2] [I.]2 I. = intermediate

Reaction Mechanism H2(g) + I2(g)  2HI(g) rate = k [H2] [I2] I2 2I.  step 1 step 2 H2 + 2I. 2HI rate = k [H2] [I.]2 Keq = [I.]2 = kf [I2] kr rate = k [H2] Keq [I2] [I.]2 = Keq [I2] rate = k’ [H2] [I2]

Reaction Mechanism 2NO + O2 2NO2 step 1 2NO  N2O2 step 2 overall reaction: 2NO + O2  2NO2 intermediates: N2O2 [N2O2] = constant d[N2O4] = 0 rate = k2 [N2O2] [O2] dt steady state approximation

2NO + O2 2NO2 step 1 2NO  N2O2 step 2 N2O2 + O2 2NO2 rate = k2 [N2O2] [O2] = k2[O2] k1[NO]2 k2[O2] + k-1 k1[NO]2 = k2 [N2O2][O2] + k-1[N2O2] produce N2O2 consume N2O2 [N2O2] = k1 [NO]2 k2[O2] + k-1

2NO + O2 2NO2 step 1 2NO  N2O2 step 2 N2O2 + O2 2NO2 2NO + O2  2NO2 low [O2] rate = k2[O2] k1[NO]2 rate = k'[O2][NO]2 k2[O2] + k-1 high [O2] rate = k1[NO]2