R = k [A]m [B]n If A + B  C DETERMINING THE RATE LAW Then the rate law would ultimately be written as R = k [A]m [B]n

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

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

R = k [NH4+]m [NO2]n exp. var control Dep. var 2x

R = k [NH4+]1 [NO2]n exp. var control Dep. var 2x

R = k [NH4+]1 [NO2]n exp. var control Dep. var 2x

R = k [NH4+]1 [NO2] exp. var control Dep. var 2x 1

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

R = k [0.02M]1 [NO2-]1

R = k [0.02M]1[0.200M]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

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) 1st order in Br2 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 2 2 (0.0320)x 1 = .5x 4 x = 2 R = k[NO]2 [Br2]1

H3C N C: H3C C N: CH3NC1 Time (sec) 150 900 140 2,500 130 5,000 118 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:

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 R = - [A] = k[A] t ln[A]t - ln[A]0 = ln [A]t [A]0 = - kt ln [A]t = - k t + ln [A]0 y = mx + b

H3C N C: H3C C N:

H3C N C: H3C C N: CH3NC1 1 mm Hg 150 140 130 118 90 70 55 35 1/ CH3NC3 .0067 .0071 .0076 .0084 .0111 .0142 .0182 .0286 R = - [A] = k[A]2 t 1 = kt + [A]t [A]0 y = mx + b 3 r = 0.92

H3C N C: H3C C N: CH3NC1 Time (sec) 1 mm Hg 1/ CH3NC3 .0067 .0071 .0076 .0084 .0111 .0142 .0182 .0286 ln CH3NC2 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55 2 r = 0.98 3 r = 0.92 150 140 130 118 90 70 55 35 900 2,500 5,000 10,000 15,000 20,000 30,000

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] t ln[A]t - ln[A]0 = ln [A]t = - kt [A]0 ln [A]t = - k t + ln [A]0 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 = - kt Let’s begin with ln[A]t - ln[A]0 = ln ln [A]t [A]0 = - kt, given let [A]t = ½ [A]0 ½ [A]0 then ln = - kt, and ln ½ = -kt ½ [A]0 t ½ = -ln ½ = 0.693 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

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

...And What Do the Mathematicians Have to Say About Collision Theory? k = Ae-Ea/RT

Rewriting the Arrhenius Equation k = Ae-Ea/RT ln k1 = -Ea/RT1 + ln Ae ln k2 = -Ea/RT2 + ln Ae lnk1 - ln k2 = (-Ea/RT1 + ln Ae) - (Ea/RT2 + ln Ae) ln k1 k2 Ea R 1 T2 T1 - =

But there is a simpler way Getting a Line on the Arrhenius Equation But there is a simpler way to think about the Arrhenius Equation Slope = -Ea/R* k = Ae-Ea/RT ln k ln k1 = -Ea/RT1 + ln Ae y = m x + b 1/T * R (ideal gas constant) = 8.31 J/K-mol

Mastering the Arrhenius Equation with your TI The following table shows the rate constants for the rearrangement of CH3CN (methyl isonitrile) at various temperatures Temperature (°C) k (s-1) 189.7 198.9 230.3 251.2 2.52 x 10-5 5.25 x 10-5 6.30 x 10-4 3.16 x 10-3 From these data calculate the activation energy of this reaction

Mastering the Arrhenius Equation with your TI k = Ae-Ea/RT ln k1 = -Ea/RT1 + ln Ae L1 L2 L3 L4 L5 T(°C) L1 + 273 1/L2 k ln L4 y = m x + b Slope = -Ea/R* It’s as easy as L1, L2, L3 ln k (L5) 1/T (L3)

Mastering the Arrhenius Equation with your TI The following table shows the rate constants for the following reaction at varying temperatures: CO(g) + NO2(g)  CO2(g) + NO(g) Temperature (°C) k (M-1 s-1) 600 650 700 750 800 0.028 0.22 1.3 6 23 From these data calculate the activation energy for this reaction Now, you try it!

Reaction Mechanisms: The Shortest Distance Between Two Points Often Requires More Steps 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: The Shortest Distance Between Two Points Often Requires More Steps NO2(g) +NO2(g) NO3(g) + NO(g) Rate Determining step 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 Yea, but what if the second step is the rate determining step?

Reaction Mechanisms: The Shortest Distance Between Two Points Often Requires More Steps OK. Let’s try this one experimentally: 2NO(g) + Br2(g)  2NOBr(g) what we get is R= k[NO]2[Br2]

Reaction Mechanisms: The Shortest Distance Between Two Points Often Requires More Steps k1 NO(g) +Br2(g) NOBr2(g) Fast step k-1 NOBr2(g) + NO(g)  2NOBr(g) k2 Rate Determining step Slow step 2NO(g) + Br2(g)  2NOBr(g) 1. We assume that reaction 1 is at equilibrium and that Rf = Rr 2. If the above is true, then k1[NO][Br2] = k-1 [NOBr2] 3. 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. 4. We can substitute for NOBr2 using step two, however. Doing so would allow us to use NO(g) and Br2(g) which are not intermediates

Reaction Mechanisms: The Shortest Distance Between Two Points Often Requires More Steps k1 k-1 Let [NOBr2] = [NO][Br2] If R = k [NOBr2][Br2]. Then by substitution, R = k2 [NO][Br2] [Br2]. k1 k-1 k1 k-1 We then let k = k2 Finally, we get the rate law R = k[NOBr2][Br2].

Reaction Mechanisms: The Shortest Distance Between Two Points Often Requires More Steps Prove that the following mechanism is consistent with R = k[NO]2[Br2], the rate law which was derived experimentally: k1 NO(g) +NO(g)  N2O2(g) k-1 N2O2(g) + Br2(g)  2NOBr(g) k2 2NO(g) + Br2(g)  2NOBr(g) Now, you try it

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

Catalyst:a substance that changes the speed of a chemical reaction without it self undergoing a permanent chemical change in the process Ethylene Ethane This metallic substrate is a heterogeneous catalyst