Integrated rate laws rate kA lnA kt lnA0
![Integrated rate laws rate = k[A] ln[A] = -kt + ln[A]0 rate = k[A]2 Integrated rate laws rate = k[A] ln[A] = -kt + ln[A]0 rate = k[A]2](https://slidetodoc.com/presentation_image_h/793d61482135bdd5e2f7ea4abff68980/image-1.jpg)





![Reaction Mechanism H 2(g) + I 2(g) 2 HI(g) rate = k [H 2] Reaction Mechanism H 2(g) + I 2(g) 2 HI(g) rate = k [H 2]](https://slidetodoc.com/presentation_image_h/793d61482135bdd5e2f7ea4abff68980/image-7.jpg)



![Reaction Mechanism 2 NO + O 2 2 NO 2 rate = k[NO]2 2 Reaction Mechanism 2 NO + O 2 2 NO 2 rate = k[NO]2 2](https://slidetodoc.com/presentation_image_h/793d61482135bdd5e2f7ea4abff68980/image-11.jpg)
![2 NO + O 2 2 NO 2 rate = k[NO]2 step 2 2 2 NO + O 2 2 NO 2 rate = k[NO]2 step 2 2](https://slidetodoc.com/presentation_image_h/793d61482135bdd5e2f7ea4abff68980/image-12.jpg)
- Slides: 12
Integrated rate laws rate = k[A] ln[A] = -kt + ln[A]0 rate = k[A]2 1/[A] = kt + 1/[A]0
Reaction mechanism reaction = Elementary steps Molecularity Elementary step rate law unimolecular A product bimolecular A+ B product termolecular A+B+C product A+A+A product k[A][B][C] k[A]3 chemical reaction = sum of elementary steps rate law and stoichiometry
Reaction mechanism 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2(g) Gorxn = [ 2(-237. 9)] - [2(-131. 67)] = -212. 46 k. J spontaneous reaction experimental rate law: rate = k[H 2 O 2] [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 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2(g) rate = k[H 2 O 2] [I-] k 1 step 1 H 2 O 2 + I- H 2 O + OIk 2 step 2 H 2 O 2 + OI- H 2 O + O 2 +I 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2(g) step 1 rate = k 1 [H 2 O 2] [I-]
Reaction mechanism k 1 step 1 H 2 O 2 + I- H 2 O + OIk 2 step 2 H 2 O 2 + OI- H 2 O + O 2 +Iwhat about step 2 ? step 1 assume k 2 >> k 1 rate determining step I- catalyst consumed in rate determining 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 H 2(g) + I 2(g) 2 HI(g) rate = k [H 2] [I 2] reaction faster in light free radical unpaired electron
Reaction Mechanism H 2(g) + I 2(g) 2 HI(g) step 1 rate = k [H 2] [I 2] I 2 2 I. forward rate = kf [I 2]. reverse rate = kr [I ]2. 2 k [I ] = k [I ] equilibrium f 2 r Keq= . [I ]2 = kf [I 2] kr
Reaction Mechanism H 2(g) + I 2(g) 2 HI(g) step 2 step 1 rate = k [H 2] [I 2]. I 2 2 I. H 2 + 2 I 2 HI H 2 + I 2 2 HI need H 2 in the rate determining step from step 2 rate = k [H 2] [I. ]2. I = intermediate
Reaction Mechanism H 2(g) + I 2(g) 2 HI(g) step 2 step 1 rate = k [H 2] [I 2]. I 2 2 I. H 2 + 2 I 2 HI . rate = k [H 2] [I ]2 rate = k [H 2] Keq [I 2] rate = k’ [H 2] [I 2] . Keq = [I ]2 = kf [I 2] kr . [I ]2 = Keq [I 2]
Reaction Mechanism 2 NO + O 2 2 NO 2 rate = k[NO]2 2 NO N 2 O 2 step 1 N 2 O 2 + O 2 2 NO 2 overall reaction: 2 NO + O 2 2 NO 2 step 2 intermediates: N 2 O 2 relative rates of steps 1 and 2
2 NO + O 2 2 NO 2 rate = k[NO]2 step 2 2 NO N 2 O 2 + O 2 2 NO 2 step 1












