Chemical Kinetics How fast can a reaction go Four things that affect rate a. concentration b. temperature c. catalyst d. surface area

Average rate Instantaneous rate

Rate Laws NH4+ + NO2-  N2 + 2H2O R = k [NH4+]m [NO2-]n k = rate proportionality constant m = reaction order for the ammonium ion n = reaction order for the nitrite ion

How do the three chemicals compare? Rate N2 + 3H2 → 2NH3 R or rate = -Δ[N2] = M/s Δt How do the three chemicals compare?

NH4+ + NO2-  N2 + 2H2O indep. var control Dep. var

R = k [NH4+]m [NO2]n Skeleton rate law Dep. var indep. var control 2x

R = k [NH4+]1 [NO2]1

10.8 x 10-7M/s = k [0.02M]1[0.200M]1

10.8 x 10-7M/s = k[0.02M]1 [0.200M]1 k= 10.8 x 10-7 M/s /[0.02M]1[0.200M]1 k = 10.8 x 10-7 M/s / 0.004M2 k = 2.7 x 10-4 1/M • s M = ? M1 M1 s ?

Suppose the concentrations for NH4+ and NO2- are 0.15 M and 0.20 M, respectively Since R = k[NH4+]1 [NO2]1 where k = 2.7 x 10-4 1/M - s at 25 C then R = 2.7 x 10-4 1/M - s [0.15 M] [0.20 M] R = 8.1 x 10-6 M/s

2NO(g) + Br2(g)  2NOBr 1 0.0160 0.0120 3.24 x 10-4 2 0.0160 0.0240 6.38 x 10-4 3 0.0320 0.0060 6.42 x 10-4 Initial [NO] Initial [Br2] Rate of appearance NOBr (M/s) (M/L) R = k[NO]m [Br2]n 3.24 x 10-4 k(0.0160)x (0.0120)1 = 6.42 x 10-4 k(0.0320)x (0.0060)1 1 (0.0160)x 1 2 = .5x x = 2 = 2 (0.0320)x 4 R = k[NO]2 [Br2]1

Using the calculator to find order H3C N C: H3C C N: Using the calculator to find order CH3NC1 Time (sec) 1 mm Hg 150 140 130 118 90 70 55 35 900 2,500 5,000 10,000 15,000 20,000 30,000

H3C N C: H3C C N:

y = m x + b ln [A]t = - k t + ln [A]0 H3C N C: H3C C N: CH3NC1 1 mm Hg 150 140 130 118 90 70 55 35 Ln CH3NC2 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55 2 r = 0.98 ln [A]t = - k t + ln [A]0 y = m x + b

H3C N C: H3C C N: 1 = kt + [A]t y = mx + b [A]0 CH3NC1 1/ CH3NC3 150 140 130 118 90 70 55 35 .0067 .0071 .0076 .0084 .0111 .0142 .0182 .0286 1 = kt + [A]t [A]0 y = mx + b 3 r = 0.92 1 mm Hg

H3C N C: H3C C N: CH3NC1 ln CH3NC2 1/ CH3NC3 Time (sec) 150 140 130 118 90 70 55 35 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55 .0067 .0071 .0076 .0084 .0111 .0142 .0182 .0286 900 2,500 5,000 10,000 15,000 20,000 30,000

zero order Rate = k[A]o 1st order Rate = k[A]1 t plot concentration [A] vs. time (t) – linear, negative slope integrated rate law - [A]t = -kt + [A]0 half life – t1/2 = [A]0/2k [A] 1st order Rate = k[A]1 t plot of ln conc. vs. time is linear, negative slope integrated rate law - ln[A]t = -kt + ln[A]0 half life – t1/2 = .693/k ln[A] 2nd order Rate = k[A]2 t plot inverse of conc. vs. time is linear, positive slope integrated rate law - 1/[A]t = kt + 1/[A]0 half life – t1/2 = 1/k[A]0 1/[A] t

ln [A]t = - (1.45 yr.-1)(1.00 yr.) + ln( 5.0 x 10-7 g/cm3) Using Rate Law Equations to Determine Change in Concentration Over Time The first order rate constant for the decomposition of a certain insecticide in water at 12 C° is 1.45 yr1-. A quantity of insecticide is washed into a lake in June leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the effective temperature of the lake is 12 C°. (a)What is the concentration of the insecticide in June of the following year? (b) How long will it take for the concentration of the insecticide to drop to 3.0 x 10-7 g/cm3? R = - [A] = k[A] ln [A]t = - k t + ln [A]0 t ln [A]t = - (1.45 yr.-1)(1.00 yr.) + ln( 5.0 x 10-7 g/cm3) ln [A]t = - 15.96, [A]t = e -15.96 [A]t = 1.2 x 10-7 g/cm3

Using Rate Law Equations to Determine the Half-Life of a Reaction Let’s begin with ln[A]t - ln[A]0 = ln = - kt [A]0 [A]t given ln let [A]t = ½ [A]0 [A]0 ½ [A]0 then ln = - kt, and ln ½ = -kt ½ [A]0 t ½ = -ln ½ = 0.693 = - kt, k k

Using Rate Law Equations to Determine the Half-Life of a Reaction The concentration of an insecticide accidentally spilled into a lake was measured at 1.2 x 10-7 g/cm3. Records from the initial accident show that the concentration of the insecticide was 5.0 x 10-7 g/cm3. Calculate the 1/2 life of the insecticide. 1.2 x 10-7 g/cm3 ln = - k (1.00 yr.) 5.0 x 10-7 g/cm3 k = 1.43 yr-1 t ½ = 0.693 = 0.484 yr. 1.43 yr-1

Examining the Relationship Between the Rate Constant and Temperature

..it’s because of Collision Theory Why does Temperature Have an Effect on Rate? ..it’s because of Collision Theory

It’s All About Activation Energy and Getting over the HUMP!!

1010 collisions/sec occur 1 in 1010 collisions succeeds

Reaction Mechanisms NO(g) + O3(g)  NO2(g) + O2(g) A Reaction Mechanism is the process which describes in great detail the order in which bonds are broken and reformed, changes in orientation and the energies involved during those rebondings, and changes in orientations NO(g) + O3(g)  NO2(g) + O2(g) A single reaction event is called an elementary reaction NO2(g) + CO(g)  NO(g) + CO2(g) While this reaction looks like an elementary reaction, it actually takes place in a series of steps NO2(g) +NO2(g) NO3(g) + NO(g) NO3(g) + CO(g)  NO2(g) + CO2(g) NO2(g) + CO(g)  NO(g) + CO2(g)

Reaction Mechanisms NO2(g) +NO2(g) NO3(g) + NO(g) Rate Determining step NO2(g) +NO2(g) NO3(g) + NO(g) Slow step NO3(g) + CO(g)  NO2(g) + CO2(g) Fast step NO2(g) + CO(g)  NO(g) + CO2(g) The rate law is constructed from the rate determining step Rate = k1[NO2]2

Reaction Mechanisms: Catalysts NO + O3 → NO2 + O2 O + NO2 → NO + O2 O + O3 → 2O2

Fast step NO(g) +Br2(g) NOBr2(g) NOBr2(g) + NO(g)  2NOBr(g) Rate Determining step Slow step 2NO(g) + Br2(g)  2NOBr(g) If step 2 is the rate determining step, then R = k[NOBr2][Br2]. Keep in mind, however, that one cannot include an intermediate in the rate determining step. We can substitute for NOBr2 using the previous equation NO(g) and Br2(g) which are not intermediates rate = k [NO]2[Br2]

rate law which was derived experimentally: Prove that the following mechanism is consistent with R = k[NO]2[Br2], the rate law which was derived experimentally: NO(g) +NO(g)  N2O2(g) N2O2(g) + Br2(g)  2NOBr(g) 2NO(g) + Br2(g)  2NOBr(g)

Catalyst: a substance that changes the speed of a chemical reaction without it self undergoing a permanent chemical change in the process or it is reconstituted at the end. Br - is a homogeneous catalyst

Heterogeneous Catalysts Ethylene Ethane This metallic substrate is a heterogeneous catalyst