How fast can a reaction go

Four things that affect ratea. concentration b. temperaturec. catalystd. surface area

Chemical KineticsChemical Kinetics

Average rateAverage rate

Instantaneous rateInstantaneous rate

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

Rate Laws

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

indep. var control Dep. var

R = k [NH4+]m [NO2]n

2x

Skeleton rate law

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, respectivelySince 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

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)(M/L)

2NO(g) + Br2(g) 2NOBr

R = k[NO]m [Br2]n

3.24 x 10-4

6.42 x 10-4=

k(0.0160)x (0.0120)1

k(0.0320)x (0.0060)1

12

=(0.0160)x

(0.0320)x

2 1

4= .5x x = 2x = 2

R = k[NO]2 [Br2]1

CH3NC1 Time (sec)

1 mm Hg

15014013011890705535

0900

2,5005,000

10,00015,00020,00030,000

H3C N C: H3C C N:

Using the calculator to find order

H3C N C: H3C C N:

CH3NC1

1 mm Hg

15014013011890705535

Ln CH3NC2

5.024.944.864.774.494.244.003.55

2 r = 0.98

ln [A]t = - k t + ln [A]0

y = m x + b

H3C N C: H3C C N:

CH3NC1

1 mm Hg

15014013011890705535

1/ CH3NC3

.0067

.0071

.0076

.0084

.0111

.0142

.0182

.0286

3 r = 0.92

y = mx + b

1

= kt + [A]t

1

[A]0

H3C N C: H3C C N:

CH3NC1 Time (sec)

15014013011890705535

0900

2,5005,000

10,00015,00020,00030,000

H3C N C: H3C C N:

1/ CH3NC3

.0067

.0071

.0076

.0084

.0111

.0142

.0182

.0286

ln CH3NC2

5.024.944.864.774.494.244.003.55

zero order Rate = k[A]o

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 tplot 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

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 = - 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

Let’s begin with ln[A]t - ln[A]0 = ln[A]t

[A]0

= - kt

ln[A]t

[A]0

= - kt,

let [A]t = ½ [A]0given

ln½ [A]0

[A]0

= - kt,then and ln ½ = -kt ½

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.

ln 1.2 x 10-7 g/cm3

5.0 x 10-7 g/cm3= - k (1.00 yr.)

k = 1.43 yr-1 t ½ = 0.693 = 0.484 yr.1.43 yr-1

Examining the Relationship Between the Rate Constantand Temperature

Why does Temperature Have an Effect on Rate?

..it’s because ofCollision Theory

It’s All About Activation Energy and Getting overthe HUMP!!

1010 collisions/sec occur 1 in 1010 collisions succeeds

...And What Do the Mathematicians Have to Say AboutCollision Theory?

k = Ae-Ea/RT

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

R1

T2

1T1

-

Rewriting the Arrhenius Equation

ln k1 = -Ea/RT1 + ln Ae

y = m x + bln

k

1/T

Slope = -Ea/R

R (ideal gas constant) 8.31 J/K-mol

Getting a Line on the Arrhenius Equation

ln k1 = -Ea/RT1 + ln Ae

y = m x + b

L1 L2 L3 L4 L5

T(°C) L1 + 273 1/L2 k ln L4

ln k

(L

5)

1/T (L3)

Slope = -Ea/R*

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)

189.7198.9230.3251.2

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

The following table shows the rate constants for the following reaction at varyingtemperatures: CO(g) + NO2(g) CO2(g) + NO(g)

Temperature (°C) k (M-1 s-1)600650700750800

0.0280.221.3623

From these data calculate the activation energy for this reaction

Reaction Mechanisms 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 orientationsNO(g) + O3(g) NO2(g) + O2(g)

A single reaction event is called an elementary reactionelementary reactionNO2(g) + CO(g) NO(g) + CO2(g)

While this reaction looks like an elementary reaction, 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 MechanismsNO2(g) +NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

NO2(g) + CO(g) NO(g) + CO2(g)

Slow step

Fast step

Rate Determining step

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

NO(g) +Br2(g) NOBr2(g)

NOBr2(g) + NO(g) 2NOBr(g)

2NO(g) + Br2(g) 2NOBr(g)

Slow step

Fast stepRate Determining step

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]

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

This metallic substrate is a heterogeneous catalyst

Heterogeneous Catalysts

Ethylene

Ethane