unit 7 (chp 14): chemical kinetics (rates) john d. bookstaver st. charles community college st....

Unit 7 (Chp 14):

Chemical Kinetics(rates)

John D. Bookstaver

St. Charles Community College

St. Peters, MO

2006, Prentice Hall, Inc.

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

Chemical Kinetics is the study of:

1) Rates at which reactants are consumed or products are produced in a chemical rxn.

2) Factors that affect the rate of a reaction according to Collision Theory (temperature, concentration, surface area, & catalyst).

3) Mechanisms, or sequences of steps, for how a reaction actually occurs.

4) Rate Laws (equations) used to calculate rates, rate constants, concentrations, & time.

Chemical Kinetics

Reaction Rates

Reaction rates are described by the change in concentration (DM) of reactants (consumed) or of products (produced) per change in time.

Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

[C4H9Cl] measured at various times

recall:[ ] brackets represent concentration in molarity (M)


[C4H9Cl]t

“change in”

Reaction Rates

fewer reactant collisions

average rate slows

WHY?

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

• The slope of a line tangent to the curve at any point is the instantaneous rate at that time.


[C4H9Cl]t

(rise)(run)

• All reactions slow down over time.

• The initial rate of reaction is commonly chosen for analysis and comparison.


rate of consumptionof reactant

=rate of production

of product.

(IF 1:1 mol ratio)

–[C4H9Cl]t

= [C4H9OH]t

↓ reactant = ↑ product

Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Rate:

Reaction Rates:

3. production of products per time

2. consumption of reactants per time

4. Are equalized to a stoich. coefficient of 1

2 NO

1 O2

2 NO2 2 NO + O2

–[NO2]2 t

=[NO]2 t =

[O2]t

–[NO2]t

+[NO]t

[NO2]t

1. slope =

5. change during rxn time (t)

change in conc.change in time

conc

entr

atio

n (M

)

Initial rate of production of H2 is 0.050 M∙s–1.What is the rate of consumption of HI?

aA + bB cC + dD

Rate Ratios

Rate 1a

[A]t =

1b

[B]t =

1c

[C]t

1d

[D]t=

2 HI(g) H2(g) + I2(g)

− −=

0.050 M∙s–1 H2 xmol H2

mol HI1

=2 0.10 M∙s–1 HI

–0.10 M∙s–1 HImolL∙s mol ratio

=?

HW p. 619 #20

Rate equations (or rate laws) have the form:

Rate Laws

rate = k[A]x[B]y[C]z

rate constant

order with respect to reactantsA, B, & C

…or…number of each particle involved

in collision that affects the rate

overall order of reaction = x + y +…

rate = k[BrO3–][Br–][H+]2

Example: overall order= ___ order4th

(4 particles in collision)

• reactant bonds break, then product bonds form.• Reaction rates depend on collisions

between reactant particles with:

1) greater frequency

2) enough energy

3) proper orientation

The Collision Model

reactants productsCollision

activation energy:minimum E required

to start reaction(Eact )

successful

unsuccessful

Reaction Coordinate Diagramtransition state

∆E

Eact

…aka…EnergyProfile

Pot

entia

l Ene

rgy

Reaction progress

demo

4 Factors that Affect Reaction Rates:

1) Concentration

2) Temperature

3) (exposed) Surface Area (particle size)

4) Catalyst

Factors That Affect Reaction Rates

1) Concentration of Reactants ↑ concentration, ↑ collision frequency increase pressure of gases


Fe(s) + O2(g) Fe2O3(s) 20% of air

is O2(g)

100% O2(g)

Eact at higher Temp more particles

over Eact

collisions of…greater frequencygreater energy

unsuccessful collisions(bounce off)

Factors That Affect Reaction Rates2) Temperature

successful collisions(react)

↑ Temp, ↑ rate

↑ Temp, ↑ ratek is temp. dependent

(k changes with temp)

Temperature and k (rate constant)

rate = k [A]x

3) Surface Area (particle size) smaller pieces, more exposed

surface area for collision.


consumed, then produced (not used up)

4) Catalyst


Uncatalyzed …lowering the Eact .

reaction progress

pote

ntia

l ene

rgy

2 H2O2 2 H2O + O2

2 H2O2

2 H2O + O2

+ 2 Br– + 2 H+

+ 2 Br– + 2 H+

Br2(intermediate)

Catalyzed

↑ rate by changing the reaction mechanism by…

Br–, H+

demo

Surface CatalystsCatalysts can orient reactants to help bonds break and form.

H2 + H2C=CH2 H3C–CH3

H2 + H2C=CH2

CH3CH3

• biological catalysts in living systems.• A substrate fits into the active site of the

enzyme much like a key fits into a lock.

(IMAFs work here)

Enzymes

HW p. 621#50,51,64


Rate Laws


rate constant








recall…

Reaction MechanismsThe sequence of molecular collisions and changes by which reactants become products is called the reaction mechanism.

• Rxns may occur in separate elementary steps.

• The overall reaction occurs only as fast as the slowest, rate-determining step. (RDS)

Slow First Step

• A proposed mechanism for this reaction is:

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: NO3 + CO NO2 + CO2 (fast)

• NO3 intermediate is produced then consumed.

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

• CO is not involved in the slow RDS,

so it does not appear in the rate law.

(OR…the order w.r.t. CO is ___)

• Rate law depends on the slow 1st step:

rate = k [NO2]2

0th

Slow Second Step

• A proposed mechanism is:

Step 2: NOBr2 + NO 2 NOBr (slow)

Step 1: NO + Br2 NOBr2 (fast)

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

• NOBr2 intermediate is produced then consumed.

• Rate law depends on the slow 2nd step:

rate = k2 [NOBr2] [NO]

• But cannot include intermediate [NOBr2] (b/c

it’s difficult to control conc.’s of intermediates).

Slow Second Step

b/c step 1 is in equilibrium



From Step 1: rateforward = ratereverse

k1 [NO] [Br2] = k−1 [NOBr2]

k1

k−1

[NO] [Br2] = [NOBr2]

From RDS Step 2: rate = k2 [NOBr2] [NO]

solve for [NOBr2]

substitute for [NOBr2]

Slow Second Step



k1

k−1

[NO] [Br2] = [NOBr2]

rate = k2 [NOBr2] [NO]

substitute for [NOBr2]

k2k1

k−1

rate = [NO] [Br2] [NO]

rate = k [NO]2 [Br2]

HW p. 625 #92, 94

WS Reaction Mechanisms


Rate Laws


rate constant








recall…

Orders from Experimental Data: Rate varies with Concentration

Initial Concentrations Rate in M per unit

timeExperiment [BrO3

–] , M [Br–] , M [H+] , M

1 0.0050 0.25 0.30 10

2 0.010 0.25 0.30 20

3 0.010 0.50 0.30 40

4 0.010 0.50 0.60 160



–] , M [Br–] , M [H+] , M

1 0.0050 0.25 0.30 10

2 0.010 0.25 0.30 20

3 0.010 0.50 0.30 40

4 0.010 0.50 0.60 160In experiments A & B:• doubling [BrO3

–] doubles the rate.(other conc.’s kept constant)

• rate is 1st order w.r.t. [BrO3–]


rate = k [BrO3–]x

2 = [2]x

“w.r.t.” (with respect to)

1 = x



–] , M [Br–] , M [H+] , M

1 0.0050 0.25 0.30 10

2 0.010 0.25 0.30 20

3 0.010 0.50 0.30 40

4 0.010 0.50 0.60 160In experiments B & C:• doubling [Br–] doubles the rate

(other conc.’s kept constant)• rate is 1st order w.r.t. [Br–]


rate = k [Br–]y

2 = [2]y

1 = y



–] , M [Br–] , M [H+] , M

1 0.0050 0.25 0.30 10

2 0.010 0.25 0.30 20

3 0.010 0.50 0.30 40

4 0.010 0.50 0.60 160In experiments C & D:• doubling [H+] quadruples the rate

(other conc.’s kept constant)• rate is 2nd order w.r.t. [H+]


rate = k [H+]z

4 = [2]z

2 = z



–] , M [Br–] , M [H+] , M

1 0.0050 0.25 0.30 10

2 0.010 0.25 0.30 20

3 0.010 0.50 0.30 40

4 0.010 0.50 0.60 160

Rate law is:rate = k [BrO3

–] [Br–] [H+]2


Orders found by experiment are:x = 1 , y = 1 , z = 2

1)…only found experimentally (from data).

2)…do NOT come from the coefficients of reactants of an overall reaction.

3)…represent the number of reactant particles (coefficients) in the RDS of the mechanism.

4)…zero order reactants have no effect on rate b/c they do not appear in the RDS of the mechanism (coefficient of 0 in RDS).

5)…typically 0, 1, 2, but can be any # or fraction

Orders in Rate Laws

OrderRate Law

k Units

M = ?∙M3

sM = ?∙M2

sM = ?∙M s

M = ? s

Units of k (rate constant)Units of k give info about order.

Rate is usually (M∙s–1), or (M∙min–1), etc.

M∙s–1 s–1 M–1∙s–1 M–2∙s–1

1 M2∙s

1M∙s

1 s

Ms

0th 1st 2nd 3rd

rate =k[A]0

rate =k[A]

rate =k[A]2

rate =k[A]2[B]

HW p. 618 #22,24,28

Determine the rate law for the reaction (from experimental data).

HW p. 619 #28

rate = k [ClO2]x [OH–]y

rate1

rate2

= (0.0248)= (0.00276)

= k (0.060)x(0.030)y

= k (0.020)x(0.030)y

0.060 x 0.020

0.030 y 0.030

(3)x (1)y

9 = (3)x

(0.0248) =(0.00276) =

9 =

x = 2

(a)

Determine the rate law for the reaction (from experimental data).

HW p. 619 #28

rate = k [ClO2]2 [OH–]y

rate3

rate2

= (0.00828)= (0.00276)

= k (0.020)2(0.090)y

= k (0.020)2(0.030)y

3 = (3)y y = 1

rate = k [ClO2]2 [OH–]1

OR rate = k [ClO2]2 [OH–]

(a)

Calculate the rate constant (with units).

HW p. 619 #28

rate = k [ClO2]2 [OH–]

Exp 1: (0.0248) = k (0.060)2(0.030)

(b)

(0.0248)(0.060)2(0.030)

k = 230

k =

M–2∙s–1

M = ?∙M2∙M s

Calculate the rate when [ClO2] = 0.010 M and [OH–] = 0.025 M.

HW p. 619 #28

rate = k [ClO2]2 [OH–]

rate = (230)(0.010)2(0.025)

(c)

rate = 0.00058 M∙s–1

Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.

Integrated rate laws express (reveal) the relationship between

concentration of reactants and time.

The differential rate law is usually just called “the rate law.”

Rate Laws

Integrated Rate Laws

ln[A]t

[A]0

= −kt

ln [A]t – ln [A]0 = –kt

ln [A]t = –kt + ln [A]0

y = mx + b

given on exam

Using calculus to integrate a first-order rate law gives us:

[A]0 = initial conc. of A at t = 0 .[A]t = conc. of A at any time, t .

If a reaction is first-order, a plot of ln [A] vs. t is a straight line, and the slope of the line will be –k.

First-Order Processes

ln [A]t = –kt + ln [A]0

y = mx + b

ln [A]

t

m = –k

first-order


CH3NC CH3CN

at 198.9 oC


• When ln P is plotted as a function of time, a straight line results.

• Therefore,The process is first-order.k is the negative slope: 5.1 10–5 s−1.

Second-Order Processes

Similarly, integrating the rate law for a process that is second-order in reactant A, we get…

1[A]t

= kt +1

[A]0

y = mx + b

1[A]t

–1

[A]0

= ktgiven

on exam


1[A]t

= kt +1

[A]0

y = mx + b

If a reaction is second-order, a plot of 1/[A] vs. t is a straight line, and the slope of the line is k.

1[A]

t

m = k

second-order


The decomposition of NO2 at 300°C is described by the equation

NO2 (g) NO (g) + 1/2 O2 (g)

and yields the following data:

Time (s) [NO2], M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380


• Graphing ln [NO2] vs. t yields:

Time (s) [NO2], M ln [NO2]

0.0 0.01000 −4.610

50.0 0.00787 −4.845

100.0 0.00649 −5.038

200.0 0.00481 −5.337

300.0 0.00380 −5.573

• The plot is NOT a straight line, so the process is NOT first-order in [A].


• Graphing 1/[NO2] vs. t, however, gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• Because this IS a straight line, the process is second-order in [A].

If a reaction is zero-order, a plot of [A] vs. t is a straight line, with a slope = rate.

Zero-Order Processes

[A]

t

zero-order

zero-order

[A]

t

ln [A]t = –kt + ln [A]0

ln [A]

t

m = –kfirst-order

1[A]

t

m = ksecond-order

Summary ofIntegrated Rate Laws

andStraight-Line Graphs

1[A]t

= kt +1

[A]0

HW p. 620 #33, 38, 43

Half-life (t1/2):• time at which half of initial amount remains.

Half-life (t1/2) is constantfor 1st order only.

[A]t = 0.5 [A]0concentration

at time, t

initial concentration at time, t = 0

For a 1st order process:

0.5 [A]0

[A]0

ln = −kt1/2

ln 0.5 = −kt1/2

−0.693 = −kt1/2

= t1/20.693

k

ln [A]t – ln [A]0 = –kt

ln 0.5 [A]0 – ln [A]0 = –kt1/2

[A]t1/2 = 0.5 [A]0

given on

exam

Half-life (t1/2) depends on k:

A wooden object from an archeological site is subjected to radiometric dating by carbon-14.

The activity of the sample due to 14C is measured to be 11.6 disintegrations per second (current amount of C-14).

The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14).

The half-life of 14C is 5715 yr.

What is the age of the archeological sample?

Half-life & Radiometric Dating

A wooden object from an archeological site is subjected to radiometric dating by carbon-14.

The activity of the sample due to 14C is measured to be 11.6 disintegrations per second (current amount of C-14).

The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14).

The half-life of 14C is 5715 yr.

What is the age of the archeological sample?


t1/2 = 5715 yr

[A]0 = 15.2 initial dis/s at time, t0 = 0 s

[A]t = 11.6 current dis/s at time, t

t = ? yr

= 5715 yr 0.693

k

= k 0.693

5715 yr

First, determine the rate constant, k.

k = 1.21 10−4 yr−1

ln Nt − ln N0 = –kt

ln 0.5N0 − ln N0 = –k(5715)

0.5 N0

N0

ln = −k(5715)

–0.693–5715 = k

easy way

= t1/20.693

k


0.5ln = −k(5715)

–0.693 = −k(5715)

(given: t1/2 = 5715 yr)

Now we can determine t:

= −(1.21 10−4) t ln(11.6) – ln(15.2)

= −(1.21 10−4) t –0.270

= t 2230 yr

ln Nt − ln N0 = –kt


Summary 0th Order 1st Order 2nd OrderRate Law Rate = k Rate = k[A] Rate = k[A]2

Integrated Rate Law

[A] = –kt + [A]0 ln[A] = –kt + ln[A]0

Linear plot [A] ln[A]

k & slope of line

Slope = –k Slope = –k Slope = k

M∙s–1 s–1 M–1∙s–1

Half-Life depends on [A]0 depends on [A]0

0

1 1

[ ] [ ]kt

A A

1/ 2

0.693t

k

HW p. 620 #34,36,40,32

M = ?∙M2

sM = ?∙M

sM = ?

s

1[A]

Units of k

t t t

y = mx + b y = mx + b

unit 7 (chp 14): chemical kinetics (rates) john d. bookstaver st. charles community college st....

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