kinetics: rates and mechanisms of reactions

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Kinetics: Rates and Mechanisms of Reactions

Chemical Kinetics tells us: …how fast a reaction will occur …how molecules react (MECHANISM)

a mechanism is a sequence of steps that lead to the product

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Factors Affecting the rate:

1. concentration:

2. temperature:

generally a 10oC increase will double the rate

3. nature of the reactant: i.e. surface area

4. catalyst: (two types: homogenous and heterogeneous)

5. mechanism: (orientation, shape, & order)

COLLISION THEORY = CAPPING A MARKER

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Rate of RXN =

The increase in concentration of a product per unit time.or

The decrease in concentration of a reactant per unit time.

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Conc. is usually measured in M (Molarity= mol/L) for solutions.

Rate = M time

= mol L • s

or mol•L-1•s-1 or M•s -1

Since many reactions involve gases, P is often used for concentration.

Square brackets [ ] are often used to express molarity (i.e. [HCl] means Molarity of HCl)

... of units has Vn and RT

Vn P nRT, PV from

moles/L

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3ClO- (aq) 2Cl-(aq) + ClO3-(aq)

Consider the reaction (net ionic eq.):

Rate could be defined in at least 3 ways: (3 coefficients and ions)

1. appearance of ClO3-

time][ClO rate 3

2. appearance of Cl-

time][Cl rate

-

3. disappearance of ClO-

time][ClO rate

-

Disappearance

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Consider the reaction (net ionic):

1. appearance of ClO3-

time][ClO rate 3

2. appearance of Cl-

time][Cl rate

-

3. disappearance of ClO-

time][ClO rate

-

Question: Are these three rates equal?

Disappearance

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Consider the reaction (net ionic):

Let’s make these three rates equal.

time3][ClO

time][Cl

time]ClO[ rate

---3

2

Note the use of coefficients and the -

sign

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General Form:

aA + bB cC + dD

rate = [C] ctime

= [D] dtime

= - [A] atime

= - [B] btime

“PRODUCTS” “REACTANTS”

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timeconc rate

Average rate = slope (over time period)

0.014710-400.26-0.70 slope 0.0146

10-400.74-0.30 slope -

negative sign

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timeconc rate

instantaneous rate = tangent slope (changing)

WHY?

Collision Theory!

11interest ofpoint at tangent

xy Slope

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IF we can now somehow get a linear plot in the form of:

y = mx + b.

The slope would be a constant independent

of concentration!

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We could call the slope the rate constant and assign it the letter

k!

rate constant = k instantaneous rate or rate

“call in the mathematicians”

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rate constant: k is conc. independent

Is still temperature dependent and mechanism dependent!

General form of rate law :

for RXN: A products

mk[A] rate m = RXN order according to A.Determined by experiment only!

conc. of Arate constant

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General form of rate law :

zyxw ]C[]D[]B[k[A] rate

RXN orders (w, x, y, and z)must be determined by exp. only!

cC D bB aA

Total (overall) order = individual orders

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General Equation Forms:

0 order: rate = k 1st order: rate = k[A]2nd order: rate = k[A]2 or rate = k[A][B]3rd order: rate = k[A]3 or ........

Simple experiments are done by doubling 1 conc. at a timeand looking at the effect.

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General Equation Forms:

0 order: rate = k 1st order: rate = k[A]2nd order: rate = k[A]2 or rate = k[A][B]3rd order: rate = k[A]3 or ........

Simple experiments are done by doubling 1 conc. at a timeand looking at the effect.

order doubling effect on rate 0 [2]0 = 1 none 1/2 [2]1/2 = 1.41.. increase by 1.41.. 1 [2]1 = 2 doubles 2 [2]2 = 4 quadruples

Question: suppose rate = k[A]2[B] what is the effect of doubling both A and B?

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Let’s look a some rate data for the RXN:

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

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NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

doubles

doubles

double

double

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NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

2

2

for NH4+ : [2]x = 2

x = 1 (1st Order)for NO2

- : [2]y = 2 y = 1 (1st Order)

2

2

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1 ]NO[]k[NH rate :law rate 4

first order with respect to each reactant, 2nd order overall.

orders usually have integer values, but can be fractional.Can also be (-) (inhibitors).

Since rate has units, k must also have units.

timeLmol

timeM

rate

so units of k must work with [ ] to match units.

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Determine the units of k in each of the following:

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14 ]NO[]k[NH rate 3.

1. rate = k[A]

2. rate = k[A]2

Since rate has units, k must also have units.

timeM rate (so units of k must work with [ ] to match units.)

timeM M

? kunits must = time-1 tM

tM :that so

t

M M2

? kunits must = time-1 M-1 tM

tM :that so

MM k t

M so k must have units of M-1t-1

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4. rate = k[A][B]2[C]

5. rate = k[A]0

kunits=M-3time-1

timeMkunits

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Ebbing 4th ed. P 490

14.38 In a kinetic study of the reaction:

2NO(g) + O2(g) 2NO2(g)the following data were obtained for the initial rates disap-pearance of NO:

INITIALCONC.

NO

INITIALCONC.

O2

Initial rateof RXN of

NO

Exp. 1 0.0125 M 0.0253 M 0.0281 M/sExp. 2 0.0250 M 0.0253 M 0.112 M/sExp. 3 0.0125 M 0.0506 M 0.0561 M/s

Obtain the rate law. What is the value of the rate constant?

.

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Ebbing 4th ed. P 49014.38 In a kinetic study of the reaction:

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

the following data were obtained for the initial rates disap-pearance of NO:

INITIALCONC.

NO

INITIALCONC.

O2


NO



Write overall rate equation: rate = k[NO]x[O2]y

For NO: select a pair of experiments in which the conc. of [NO] is changed, but other concentrations are unchanged.

Let’s use Exp. 1 and 2

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2NO(g) + O2(g) 2NO(g)INITIALCONC.

NO

INITIALCONC.

O2


NO



overall rate equation: rate = k[NO]x[O2]y

y

y

kk

raterate

22x2

12x1

]O[[NO]]O[[NO]

2 1

2 Exp.1 Exp.

“Let’s divide Exp.1 by Exp.2 to allow us to cancel terms.

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2NO(g) + O2(g) 2NO(g)INITIALCONC.

NO

INITIALCONC.

O2


NO



overall rate equation: rate = k[NO]x[O2]y

y

y

kk

raterate

22x2

12x1

]O[[NO]]O[[NO]

2 1

2 Exp.1 Exp.

x

x

[0.0250]

[0.0125] 0.1120.0281

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2NO(g) + O2(g) 2NO(g)overall rate equation: rate = k[NO]x[O2]y

y

y

kk

raterate

22x2

12x1

]O[[NO]]O[[NO]

2 1

2 Exp.1 Exp.

x

x

[0.0250]

[0.0125] 0.1120.0281

x

0.02500.0125

0.1120.0281

How do we solve for x?

0.02500.0125lnx

0.1120.0281lnUse logarithms

0.5lnx 0.250ln

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2NO(g) + O2(g) 2NO(g)overall rate equation: rate = k[NO]x[O2]y

x

0.02500.0125

0.1120.0281

How do we solve for x?

0.02500.0125lnx

0.1120.0281lnUse logarithms

0.5lnx 0.250ln

2 ln[0.5]

ln[0.250] x

Therefore the rate equation becomes: rate = k[NO]2[O2]y

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Now let’s determine the reaction order with respect to [O2]

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


INITIALCONC.

NO

INITIALCONC.

O2


NO


Write overall rate equation: rate = k[NO]2[O2]y

For O2: select a pair of experiments in which the conc. of [O2] is changed, but all other concentrations are unchanged.


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


INITIALCONC.

NO

INITIALCONC.

O2


NO




y12

x1

y32

x3

][Ok[NO]

][Ok[NO]1 rate3 rate

1 Exp.3 Exp.

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2NO(g) + O2(g) 2NO2(g)INITIALCONC.

NO

INITIALCONC.

O2


NO



y12

x1

y32

x3

][Ok[NO]


1 Exp.3 Exp.

y

0.02530.0506

0.02810.0561

y[2] 2

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


y12

x1

y32

x3

][Ok[NO]


1 Exp.3 Exp.

y

0.02530.0506

0.02810.0561

y[2] 2 y = 1

Therefore: rate = k[NO]2[O2]1

Now solve for k.

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2NO(g) + O2(g) 2NO2(g)INITIALCONC.

NO

INITIALCONC.

O2


NO


Overall rate equation: rate = k[NO]2[O2]1

Choose any Exp. and substitute in experimental to obtain k.

i.e. Exp1. : rate = k[NO]2[O2]1

so : 0.0281 = k[0.0125]2[0.0253]

k = 7100 M-2s-1

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Ebbing 4th ed. P 490

14.40 Iodine ion is oxidized to hypoiodite ion, IO-, by hypochlorite ion, ClO-, in basic solution.

the following initial rate experiments were run:

INITIALCONC.

I-

INITIALCONC.

ClO-

INITIALCONC.

OH-

Initial rateof RXN ofNO (M/s)

Exp. 1 0.010 M 0.020 M 0.010 M 12.2 x 10-2

Exp. 2 0.020 M 0.010 M 0.010 M 12.2 x 10-2

Exp. 3 0.010 M 0.010 M 0.010 M 6.1 x 10-2

Exp. 4 0.010 M 0.010 M 0.020 M 3.0 x 10-2


)aq(Cl )aq(IO )aq(ClO (aq)I -- -OH--

.

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]OH[]ClO][k[I rate

-

1- ]OH][ClO][k[I rate

k = 6.1 s-1

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This data isn’t linear! What can we do?

Integrated Rate Laws

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IF we can now somehow get a linear plot in the form of:

y = mx + b.

The slope would be a constant, independent

of concentration!

40

We could call the slope the rate constant and assign it the letter

k!

rate constant = k rate

“call in the mathematicians”

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OOrrddeerr iinn [[AA]]

RRaattee LLaaww** IInntteeggrraatteedd RRaattee LLaaww((iinn yy == mmxx ++ bb ffoorrmm))

LLiinneeaarrGGrraapphh ?? vvss tt

SSllooppee ooffLLiinnee

EEqquuaallss

HHaallff lliiffeeEEqquuaattiioonnss

00 rraattee == kk [[AA]]tt == --kktt ++ [[AA]]00 [[AA]]tt --kk tt11//22 == [[AA]]00//22kk11 rraattee == kk[[AA]] llnn[[AA]]tt == --kktt ++ llnn[[AA]]00 llnn[[AA]]tt --kk tt11//22 == 00..669933//kk22 rraattee == kk[[AA]]22 11//[[AA]]tt == kktt ++ 11//[[AA]]00 11//[[AA]]tt kk tt11//22 == 11//kk[[AA]]00

Key Equations:

**Since the units of rate are Since the units of rate are concentration/time, the units of k (the rate concentration/time, the units of k (the rate constant) must dimensionally agree. So for constant) must dimensionally agree. So for each order, k will have different units and each order, k will have different units and those units can tell one which equation to those units can tell one which equation to use. [ ] means the concentration of the use. [ ] means the concentration of the enclosed species in Molarity (M).enclosed species in Molarity (M).

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The data below was collected for the reaction:

NOCl(g) NO(g) + 1/2Cl2(g)

Time (s) [NOCl] (M) 0 0.100 30 0.064 60 0.047 100 0.035 200 0.021 300 0.015 400 0.012

Prepare THREE graphs to determine if the RXN is ZERO, 1st, or 2nd order. Then determine the value and units of the rate constant k.

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Zero Order Plot

[A]t vs. time

rate = k

[A]t = -kt + [A]0

y = mx + b

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First Order Plot

ln[A]t vs. time

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

y = mx + b

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2nd Order Plot

1/[A]t vs. time

Plot is linear so 2nd Order

k = slope = 0.185 M-1s-1

rate = k[A]2

1/[A]t = kt + 1/[A]0

y = mx + b

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Zero Order:rate = k[A]0 = k

rate = k integrated gives: [A]t = -kt + [A]0

y = mx + b

slope = -k

If a RXN is zero order, a plot of [A] vs. time should be linear and the slope = -k.

Integrated rate laws:

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Integrated rate laws:

1st order rate laws:

rate = k[A] integrated gives: kt- ]A[]A[ln t 0

rearranged to : y = mx + b gives:ln[A]t = -kt + ln[A]0

slope = -k

If reaction data is 1st order, a plot of ln[A] vs. time should be

linear.

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2nd Order Integrated Rate Equations:

rate = k[A]2 integrated gives: 0t [A]

1 kt ]A[

1

y = mx + b

slope = k

If a RXN is 2nd order, a plot of 1/[A] vs. time should be linear and the slope = k.

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slope = -k

RXN is first order with respect to CH3NC

ln[CH3NC]t = -kt + ln[CH3NC]0

Zero Order Plot 1st Order Plot

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2nd order Plot

Slope = k

First order plot

RXN is 2nd order with respect to [NO2]

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General form 1st order: ln[A]t = -kt + ln[A]0

Note: This is a formula that can be used to solve (1st order) problems. (If all but one of the variables are given)

1. Given the RXN: C3H6 CH2=CHCH3

Where k = 6.0 x 10-4 s-1 @500oC.Looking at the units of k, determine the order.

Problem: if [C3H6]0 = 0.0226 M, find [C3H6] @ 955 s.

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

ln[0.0226] s

(955s)10x 6.0- ln[A]-4

t

ln[A]t = -4.362[A]t = 0.0127 M

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The other equations can be used in a similar fashion.

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Half-life: the time it takes to decrease the concentration to 1/2its initial value.

“fold paper to view subsequent half-lives”

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Formulas:

1st order: t1/2 = 0.693/k

Derivation: kt]A[]A[

ln t 0

@t1/2 :212

1/

ktln

and: 212

/ktln

so: k

.k

lnt/

6930221

Half-lives:

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Half-life formulas:

Zero Order First Order 2nd Order

k]A[t

/ 20

1 2

k.

klnt

/

6930221

01

12 ]A[k

t/

These 2 are conc. dependent (and not very useful).

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Energy Diagrams:All chemical and physical changes are accompanied by energy changes.

ener

gy

time

reactants

products

E

Question: What keeps the reactants from rolling down the hill?

Ea = Activation energy

Exothermic

62

rate constant (k) varies with temperature.

but not with concentration

63

The Arrhenius Equation:

/RTEa- Ae k Temp in K

8.31 J/mol•K

activation energy

base e (natural ln)frequency factor (1/time), fraction ofcollisions with correct geometry.

rate constant

-Ea/RT is always <1 and refers to the fraction of molecules having minimum energy for a RXN.

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/RTEa- Ae k How can we make this a linear equation in the form of

y = mx + b?

Take the ln of each side.

ln k = ln A - Ea/RT

or: T1

RE Aln k ln a-

y = b - mx

A plot of ln k vs. 1/T gives a straight line with the slope = -Ea/R (Ea = -8.31 x slope)

65

At two diff. temps. we get:

21

a

1

2T1

T1

RE

kkln -

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Reaction Mechanisms: Reaction is broken into steps with intermediates being formed.

“some RXNS occur in one step, but most occur in in multiple steps.”

Each Step is called an elementary step, and the number of molecules involved in each step defines the molecularity of the step.

uni-molecular: = 1 i.e. O3* O2 + O

bi-molecular: = 2 (these are the most common) i.e. HI + HI activated complex H2 + I2

ter-molecular: = 3 (rare, due to probability of orientation and energy both being correct.) i.e. O(g) + O2(g) + N2(g) O3(g) + “energetic” N2(g)

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The Raschig process for the preparation of hydrazine (N2H4)

Overall RXN: 2NH3(g) + NaOCl(aq) N2H4(aq) + NaCl(aq) + H2O(l)

Proposed Mechanism: (Only from experiment)

Step 1: NH3(aq) + OCl-(aq) NH2Cl(aq) ‡ + OH-‡ (aq)Step 2: NH2Cl(aq) ‡ + NH3(aq) N2H5

+‡ + Cl-(aq)Step 3: N2H5

+(aq) ‡ + OH-‡ (aq) N2H4(aq) + H2O(l)

“Cancel intermediates and “add steps” to give overall RXN.”2NH3(g) + OCl-(aq) N2H4(aq) + Cl-(aq) + H2O(l)

The overall rate law, mechanism, and the total order can’t be predicted from the stoichiometry, only by experiment.

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The following is only true for individual steps:

The rate law of an elementary step is given by the product of a rate constant and the conc. of the reactants in the step.

Step Molecularity rate law

A Product(s) uni rate = k[A]

A + B Product(s) bi rate = k[A][B]

A + A Product(s) bi rate = k[A]2

2A + B Product(s) ter rate = k[A]2[B]

The overall mechanism must match the observed rate law.

Usually one STEP is assumed to be the rate determining step.

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Example:

Overall RXN: 2NO2(g) + F2(g) 2NO2F(g)

Observed Experimental rate law: rate = k[NO2][F2]

Question: Why does this rule out a single step RXN?

Answer:rate law for single step process would be: rate = k[NO2]2[F2]

“Let’s try to work out a Mechanism that matches the observed rate law.”

70

Example:



slow: NO2(g) + F2(g) NO2F(g) + F(g) ‡k1

fast: NO2(g) + F(g) ‡ NO2F(g)k2

The rate law is dependent upon the slow step.

rate = k[NO2][F2]

2NO2(g) + F2(g) 2NO2F(g)

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Let’s try making the 2nd RXN slow and the first fast



fast: NO2(g) + F2(g) NO2F(g) + F(g) ‡k1

slow: NO2(g) + F(g) ‡ NO2F(g)k2

The rate law is dependent upon the slow step.

2NO2(g) + F2(g) 2NO2F(g)

rate = k[NO2][F ‡]

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1. At low temperatures, the rate law for the reaction:CO(g) + NO2(g) CO2(g) + NO(g) is:rate = k[NO2]2Which mechanism is consistent with the rate law?a. CO + NO2 CO2 + NO

b. 2NO2 N2O4‡

(fast) N2O4

‡ + 2CO 2CO2 + 2NO (slow)

c. 2NO2 NO3 ‡ + NO (slow)

NO3 ‡ + CO NO2 + CO2 (fast)

d. 2NO2 2NO + O2 ‡ (slow)

2CO + O2 ‡ 2CO2 (fast)

rate = k[CO][NO2]

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For the following mechanism:

2NO N2O2 (fast)

N2O2 + H2 H2O + N20 (slow)

N2O + H2 N2 + H2O (fast)

a. Determine the overall reaction

b. Does the mechanism agree with the rate law:

rate = k[NO]2[H2]

kinetics: rates and mechanisms of reactions

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