GC Chemical Kinetics

Chemical Kinetics

• Study of speed with which a chemical reaction occurs and the factors affecting that speed

• Provides information about the feasibility of a chemical reaction

• Provides information about the time it takes for a chemical reaction to occur

• Provides information about the series of elementary steps which lead to the formation of product

time (seconds) Concentration A mol/L

Concentration B, mol/L

Concentration C, mol/L

0 0.76 0.38 01 0.31 0.16 0.202 0.13 6.5 x 10-2 0.403 5.2 x 10-2 2.6 x 10-2 0.584 2.1 x 10-2 1.1 x 10-2 0.735 8.8 x 10-3 4.4 x 10-3 0.866 3.6 x 10-3 1.8 x 10-3 0.957 1.4 x 10-3 7.0 x 10-4 1.028 6.1 x 10-4 3.1 x 10-4 1.079 2.5 x 10-4 1.3 x 10-4 1.07

10 1.0 x 10-4 5.0 x 10-5 1.07

Rate Data for A + B → C

A + B → C

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

time (seconds)

Conc

entr

ation

(mol

/L)

A

B

C

The Rate of a Chemical Reaction

• The speed of a reaction can be examined by the decrease in reactants or the increase in products.

• a A + b B → c C + d D

m nRate = k A B

Where m and n are determined experimentally, and not necessarilyEqual to the stiochiometry of the reaction

Reaction A → 2 B

A A A

A A

A A A

A

A

B B

B B B

B B

B B

B B

B

B B

B B

B B

BB

1 mol/L 2 mol/L

A = 6.022 x 1022 molecules B 6.022 x 1022 molecules=

in a 1.00 L container in a 1.00 Liter container

Δ A Δ B1- =

t 2 t

Average Rate

• Rate of A disappearing is• Let’s suppose that after 20 seconds ½ half of A

disappears.• Then

• And Rate of B appearing is • Then

Δ A-

t

Δ B1

2 t

-2 -2f i

f i

Δ B B -B1 1 1.00 mol/L - 0.00 mol/L mol = = x = 2.5 x 10 M/s or 2.5 x 10

2 t t - t 2 20 s - 0 s L-s

-2 -2f i

f i

Δ A A A 0.50 mol/L - 1.00 mol/L mol- = - = - = - 2.5 x 10 M/s or - 2.5 x 10

t t t 20 s - 0 s L-s

Average Rate Law for the General Equationa A + b B → c C + d D

Δ A Δ B Δ C Δ D1 1 1 1- x = - x = x = x

a Δt b Δt c Δt d Δt

For Example:N2O5 (g) → 2 NO2 (g) + ½ O2 (g)

2 5 (g) 2 (g) 2 (g)Δ N O Δ NO Δ O1- = x = 2 x

Δt 2 Δt Δt

Determination of the Rate Equation

• Determined Experimentally• Can be obtained by examining the initial rate

after about 1% or 2% of the limiting reagent has been consumed.

Consider the Reaction:CH3CH2CH2CH2Cl (aq) + H2O (l) → CH3CH2CH2CH2OH (aq) + HCl (aq)

time (seconds) Concentration n-butyl chloride

mol/L

0 0.1050 9.05 x 10-2

100 8.2 x 10-2

150 7.41 x 10-2

200 6.71 x 10-2

300 5.49 x 10-2

400 4.48 x 10-2

500 3.68 x 10-2

800 2.00 x 10-2

Average Rates, mol

L-s

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

0.0 0.1000 1.90 x 10-4

50.0 0.0905

mol

L-s

4 9[C H Cl]

t

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

50.0 0.0905 1.70 x 10-4

100.0 0.0820

mol

L-s

4 9[C H Cl]

t

Average Rates, mol

L-s

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

100.0 0.0820 1.58 x 10-4

150.0 0.0741

mol

L-s

4 9[C H Cl]

t

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

150.0 0.0741 1.74 x 10-4

200.0 0.0671

mol

L-s

4 9[C H Cl]

t

Average Rates, mol

L-s

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

200.0 0.0671 1.22 x 10-4

300.0 0.0549

mol

L-s

4 9[C H Cl]

t

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

300.0 0.0549 1.01 x 10-4

400.0 0.0448

mol

L-s

4 9[C H Cl]

t

Average Rates, mol

L-s

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

400.0 0.0448 8.00 x 10-5

500.0 0.0368

mol

L-s

4 9[C H Cl]

t

time, seconds [CH3CH2CH2CH2Cl] Average Rate,

500.0 0.0368 5.60 x 10-5

800.0 0.0200

mol

L-s

4 9[C H Cl]

t

0 100 200 300 400 500 600 700 800 9000

0.02

0.04

0.06

0.08

0.1

0.12

time (seconds)

Conc

entr

ation

(mol

/L)

Instantaneous Rate or initial rate at t=0 s

Instantaneous Rate at t = 500 s

at 0 s

-4at 0 s

0.10 M - 0.060 MInstaneous Rate =

190 s - 0 s0.040 M

Instaneous Rate = = 2.1 x 10 M s190 s

at 500 s

-5at 500 s

0.042 M - 0.020 MInstaneous Rate =

800 s - 400 s0.022 M

Instaneous Rate = = 5.5 x 10 M s400 s

Order of Reaction

• Zero order –independent of the concentration of the reactants, e.g, depends on light

• First order - depends on a step in the mechanism that is unimolecular

• Pseudo first order reaction – one of the reactants in the rate determining step is the solvent

• Second order – depends on a step in the mechanism that is bimolecular

• Rarely third order – depends on the step in the mechanism that is termolecular

time (seconds) Concentration n-butyl chloride

mol/L

0 0.10

50 9.05 x 10-2

100 8.2 x 10-2

150 7.41 x 10-2

200 6.71 x 10-2

300 5.49 x 10-2

400 4.48 x 10-2

500 3.68 x 10-2

800 2.00 x 10-2

Data from the hydrolysis of n-butyl chloride

time (seconds) [C4H9Cl]

0 0.10

50 9.05 x 10-2

100 8.2 x 10-2

150 7.41 x 10-2

200 6.71 x 10-2

300 5.49 x 10-2

400 4.48 x 10-2

500 3.68 x 10-2

800 2.00 x 10-2

IF Zero Order

0 100 200 300 400 500 600 700 800 9000

10

20

30

40

50

60

time (seconds)

[n-b

utyl

chlo

ride]

Therefore, the reaction is not zero order

If Second Order

time (seconds) 1/[C4H9Cl]

0 10

50 11.0

100 12.2

150 13.5

200 14.9

300 18.2

400 22.3

500 27.2

800 50

0 100 200 300 400 500 600 700 800 9000

10

20

30

40

50

60

time (seconds)

1/[n

-but

yl ch

lorid

e]

Therefore, the reaction is not second order

time (seconds)

log [C4H9Cl] ln[C4H9Cl]

0 -1 -2.350 -1.04 -2.4

100 -1.09 -2.51150 -1.13 -2.60200 -1.17 -2.69300 -1.26 -2.90400 -1.35 -3.11500 -1.43 -3.29800 -1.7 -3.92

IF First Order Reaction

First Order Plot

0 100 200 300 400 500 600 700 800 900

-1.8

-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

f(x) = − 0.000873532550693703 x − 0.99846318036286

time (seconds)

log

[n-b

utyl

chlo

ride]

First Order Plot

0 100 200 300 400 500 600 700 800 900

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

f(x) = − 0.00201664887940235 x − 2.29870864461046

time (seconds)

ln[C

4H9C

l]

Slope

kslope = -

2.303

-2.303 x slope = k

-4

-3

-2.303 x - 9.0 x 10 = k

2.1 x 10 = k

Slope

slope = - k

3 1

3 1

- (- 2 x 10 s ) = k

2 x 10 s = k

Rate of the Reaction

Rate = k [n-butylchloride]

For the ReactionN2O5 (g) → 2 NO2 (g) + ½ O2 (g)

2 5Rate = k [N O ]

The rate can be used to explain the mechanism

(1) Slow Step

(2) Fast Step

NO

N

OO

O O....

: :.. ..

: : : :.. ..- -

slow

--.... ::::

....::

..

.. OO

O O

NO

N .+ +

++

+

N2O5NO2

NO3

-..

::

..:O

O

N .++

NO2NO3

+ON

O

O....

:..

: :..-

fast 1/2 O2

Sum of the two steps:

NO

N

OO

O O....

: :.. ..

: : : :.. ..- - -

..::

..:O

O

N .+ + ++

N2O5NO2

1/2 O2

+.N

O

O:

: :..-

NO2

+

N2O5 → 2 NO2 + ½ O2

or2 N2O5 → 4 NO2 + O2

ApplicationMechanism of a Chemical Reaction

Suggest a possible mechanism forNO2 (g) + CO (g) → NO (g) + CO2 (g)

Given that 2

2(g)Rate = k [NO ]

(a)

(b)

Suggest a possible mechanism for2 NO2 (g) + F2 (g) → 2 NO2F (g)

Given that

2 (g) 2 (g)Rate = k [NO ] [F ]

Factors Affecting the Rate of a Chemical Reaction

• The Physical State of Matter• The Concentration of the Reactants• Temperature• Catalyst

For A Reaction to Occur

• Molecules Must Collide• Molecules must have the Appropriate

Orientation• Molecules must have sufficient energy to

overcome the energy barrier to the reaction-• Bonds must break and bonds must form

A Second Order Reaction

-2 2 (aq) (aq) 2 (l) 2 (g)H O + I H O + O

-2 2(aq) (aq)Rate = k [H O ] [I ]

Rate Constant “k”

• Must be determined experimentally• Its value allows one to find the reaction rate

for a new set of concentrations

The following data were collected for the rate of the reaction Between A and B, A + B → C , at 25oC. Determine the rate law for the reaction and calculate k.

Experiment [A], moles/L [B], moles/L Initial Rate, M/s

1 0.1000 0.1000 5.500 x 10-6

2 0.2000 0.1000 2.200 x 10-5

3 0.4000 0.1000 8.800 x 10-5

4 0.1000 0.3000 1.650 x 10-5

5 0.1000 0.6000 3.300 x 10-5

Solution A:

From Experiments 1 and 2

m nRate = k A B

m n-6(1) 5.5 x 10 M/s = k 0.1000 M 0.1000 M

m n-5(2) 2.2 x 10 M/s = k 0.2000 M 0.1000 M

Divide equation (1) into equation (2)

m n-5

m n-6

m

k 0.2000 M 0.1000 M2.2 x 10 M/s =

5.5 x 10 M/s k 0.1000 M 0.1000 M

4 2

2 = m

Solution B:

Subtract equation (2) from equation (1)

-6 -5log (5.5 x 10 ) -log (2.2 x 10 ) = m [log 0.1000 -log 0.2000 ]

-5.3 - (-4.7) = m [-1 - (-0.7)]

-0.6 = m [-0.3]

-0.6 = m

-0.32 = m

-6(1) log (5.5 x 10 ) = log k + m log 0.1000 + n log 0.1000

-5(2) log (2.2 x 10 ) = log k + m log 0.2000 + n log 0.1000

Solution A:

From Experiments 4 and 5

Divide equation (1) into equation (2)

m n-5(1) 1.65 x 10 M/s = k 0.1000 M 0.3000 M

m n-5(2) 3.3 x 10 M/s = k 0.1000 M 0.6000 M

m n-5

m n-6

n

k 0.1000 M 0.6000 M3.3 x 10 M/s =

1.65 x 10 M/s k 0.1000 M 0.3000 M

2 2

1 = n

Solution B:

Subtract equation (2) from equation (1)

-5(1) log (1.65 x 10 ) = log k + m log 0.1000 + n log 0.3000

-5(2) log (3.3 x 10 ) = log k + m log 0.1000 + n log 0.6000

-5 -5log (1.65 x 10 ) -log (3.3 x 10 ) = m [log 0.3000 -log 0.6000 ]

-4.78 - (-4.5) = m [-0.5227 - (-0.2218)]

-0.3 = m [-0.3]

-0.3 = m

-0.31 = m

Rate Constant k

m nRate = k A B

2 1Rate = k A B

2

Rate = k

A B

Rate Constant kFrom Experiment 3

2

Rate = k

A B

-5

2

M8.800 x 10

s = k 0.4000 M 0.1000 M

-32

2-3

2

15.500 x 10 = k

M s

L5.500 x 10 = k

mol s

Rate Constant kFrom Experiment 1

2

Rate = k

A B

-32

2-3

2

15.500 x 10 = k

M s

L5.500 x 10 = k

mol s

-5

2

M5.500 x 10

s = k 0.1000 M 0.1000 M

Your Understanding of this ProcessConsider the Data for the Following Reaction:

Experiment [CH3CO2CH3]M

[-OH]M

Initial Rate, M/s

1 0.050 0.050 0.000342 0.050 0.100 0.000693 0.100 0.100 0.00137

Determine the Rate Law Expression and the value of k consistentWith these data.

CH3 C

O

OCH3

+ OH_

CH3 C

O_

O

+ CH3OH

Solution :

From Experiments 1 and 2

Divide equation (1) into equation (2)

nm -3 2 3Rate = k [CH CO CH ] OH

m n-4(1) 3.4 x 10 M/s = k 0.050 M 0.050 M

m n-4(2) 6.9 x 10 M/s = k 0.50 M 0.100 M

m n-4

m n-5

n

k 0.050 M 0.100 M6.9 x 10 M/s =

3.4 x 10 M/s k 0.050 M 0.050 M

2 2

1 = n

Solution :

From Experiments 2 and 3

Divide equation (1) into equation (2)

m n-4(1) 6.9 x 10 M/s = k 0.050 M 0.050 M

m n-3(2) 1.37 x 10 M/s = k 0.100 M 0.100 M

m n-3

m n-5

m

k 0.100 M 0.100 M1.37 x 10 M/s =

6.9 x 10 M/s k 0.050 M 0.100 M

2 2

1 = m

Rate Expression

-3 2 3Rate = k [CH CO CH ] [ OH]

-3 2 3

Rate = k

[CH CO CH ] [ OH]

M0.00137

s = k[0.100 M] [0.100 M]

10.137 = k

M sL

0.137 = kmol s

AssignmentDetermine the Rate Law for the following reaction from the given data:

Experiment [NO (g)]M

[O2 (g)]M

Initial Rate, M/s

1 0.020 0.010 0.0282 0.020 0.020 0.0573 0.020 0.040 0.1144 0.040 0.020 0.2275 0.010 0.020 0.014

2 NO (g) + O2 (g) → 2 NO2 (g)

Relationship Between Concentration and Time

First Order Reactiono

A product

a xx

o

o

o o

o o

o

o

o

o

-ln (a -x) = kt + C

at x = o and t = o, C = -ln a

-ln (a -x) = kt - ln a

ln a - ln (a -x) = kt

aln = kt

a -x

or

a ktlog =

a -x 2.303

o

o

o

o

dx = k (a -x)

dtdx

= k dta -x

dx = k dt

a -x

let s = a -x

ds = -dx

ds = k dt

s-ln s = kt + C

A plot of o

o

aln versus t

a -x

Gives a Straight line

The Following Reaction is a First Order Reaction:

C

C C

H

H

H H

H

H

C

CC

H

H

H

H

H

H

Plot the linear graph for concentration versus timeand obtain the rate constant for the reaction.

Data for the Transformation of cylcpropane to propene

ao M

XM

ao –xM

Ln[ao]/[ ao –x] tseconds

0.050 0 0.050 0 0

0.050 0.0004 0.0496 9.0 x 10-3 600

0.050 0.0009 0.0491 0.0180 1200

0.050 0.0015 0.0485 0.0300 2000

0.050 0.0022 0.0478 0.045 3000

0.050 0.0036 0.0464 0.075 5000

0.050 0.0057 0.0443 0.120 8000

0.050 0.0070 0.0430 0.150 10000

0.050 0.0082 0.0418 0.180 12000

slope = 2 x 10-5 s-1

0 2000 4000 6000 8000 10000 12000 140000

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

f(x) = 0.000015 x

time (seconds)

ln (a

o /(

ao –

x))

Data for the Transformation of cylcpropane to propene

ao M

XM

ao –xM

Ln [ ao –x] tseconds

0.050 0 0.050 -2.996 0

0.050 0.0004 0.0496 -3.0038 600

0.050 0.0009 0.0491 -3.014 1200

0.050 0.0015 0.0485 -3.026 2000

0.050 0.0022 0.0478 -3.0407 3000

0.050 0.0036 0.0464 -3.070 5000

0.050 0.0057 0.0443 -3.1167 8000

0.050 0.0070 0.0430 -3.1466 10000

0.050 0.0082 0.0418 -3.1748 12000

0 2000 4000 6000 8000 10000 12000 14000

-3.2

-3.15

-3.1

-3.05

-3

-2.95

-2.9

f(x) = − 1.50112367675131E-05 x − 2.99568114479088

time (seconds)

Ln (a

o –

x)

-slope = -2 x 10-5s-1

Slope = 2 x 10-5 s-1

Relationship Between Concentration and Time

Second Order Reactiono

A product

a xx

2o

2o

2o

o

2

dx = k (a -x)

dtdx

= k dt(a -x)

dx = k dt

(a -x)

let s = a -x

ds = -dx

ds = k dt

s1

= kt + Cs

o

o

o o

o o

1 = kt + C

a -x

1at x = o and t = o, C =

a

1 1 = kt +

a -x a

1 1 - = kt

a -x a

A plot of

Gives a Straight line

The Following Reaction is a Second Order Reaction:

2 HI (g) → H2 (g) + I2 (g)

Plot the linear graph for concentration versus timeand obtain the rate constant for the reaction.

o

1 versus t

a -x

ao

MXM

ao –xM

1/[ ao –x]M-1

tMinutes

0.0100 0 0.0100 100 0.00

0.0100 0.0060 0.00400 250 5.00

0.0100 0.0075 0.00250 400 10.0

0.0100 0.0086 0.00143 700 20.0

0.0100 0.0090 0.0010 1000 30.0

0.0100 0.0099 0.00077 1300 40.0

0.0100 0.0094 0.00063 1600 50.0

0.0100 0.0095 0.00053 1900 60.0

Data for the Transformation of hydrogen iodide gas to hydrogen and iodine

0 10 20 30 40 50 60 700

200

400

600

800

1000

1200

1400

1600

1800

2000

f(x) = 30 x + 100

time (minutes)

1/(a

o –

x)

slope = 30. L mol-1 min-1

Graphical Method for Determining the Order of a Reaction

• First Order Reaction: y = ln (ao – x); x = t; slope = -k; and the intercept is ln ao

or y = ln (ao /(ao – x)); x = t; slope = k; and the intercept =0

• Second Order Reaction: y = 1/ (ao – x); x = t; slope = k; and the intercept = 1/ ao

• Zero Order Reaction: y = x; x = t; slope = k and the intercept = 0or y = ao – x ; x = t; slope = -k and the intercept = ao

Zero Order Reaction

dx = k

dtdx = k dt

dx = k dt

x = kt + C

at t = 0 and x =0; C = 0

x = kt

o

o

a -x t

a 0

o o

o o

o o

dx = dt

a - (a -x) = kt

- (a -x) = kt - a

(a -x) = - kt + a

Application of the Graphical Method for Determining the Order of a Reaction

N2O5 (g) → 2 NO2 (g) + ½ O2 (g)

[ N2O5 ]M

tminutes

2.08 3.071.67 8.771.36 14.450.72 31.28

Tabulate the data so that each order may be tested

[ N2O5 ]M

(zero order)

tminutes

ln[ N2O5 ](first order)

1/[ N2O5 ]M-1

(second order)2.08 3.07 0.732 0.481

1.67 8.77 0.513 0.599

1.36 14.45 0.307 0.735

0.72 31.28 -0.329 1.390

Data tabulation to determine which order will give a linear graph

Test for zero order reaction

Not linear; therefore, the reaction is not zero order

0 5 10 15 20 25 30 350

0.5

1

1.5

2

2.5

time (minutes)

[N2O

5]

Test for first order reaction

Linear; therefore, the reaction is first order

0 5 10 15 20 25 30 35

-0.4

-0.2

0

0.2

0.4

0.6

0.8

f(x) = − 0.0375520327162226 x + 0.846217630868234

time (minutes)

ln[N

2O5]

Test for second order reaction

Non-linear; therefore, the reaction is not second order

0 5 10 15 20 25 30 350

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

time (minutes)

1/[N

2O5]

Half Life for a First Order Reaction

o1

2

o

12

12

12

aln = kt

1 x a

2ln 2= k t

0.693 = k t

0.693 = t

k

Half Life for a Second Order Reaction

12

oo

12

o o

12

o

12

o

1 1 - = kt

1 aa2

2 1 - = kt a a

1 = kt

a

1 = t

k a

Half Life for a Zero Order Reaction

o 1 o2

o o 12

o 12

o1

2

1 a = - kt + a

21

a - a = - kt 2

1 a = - kt

2a

= t 2 k

Application of Half Life

• The rate constant for transforming cyclopropane into propene is 0.054 h-1

• Calculate the half-life of cyclopropane.• Calculate the fraction of cyclopropane

remaining after 18.0 hours.• Calculate the fraction of cyclopropane

remaining after 51.5 hours.

12

12

0.693 = t

0.054/h12.8 h = t

Half-Life

o

o

o

o

- kto

o

- (0.054/h) x 51.5 ho

o

- 2.8o

o

o

o

aln = kt

a -x

a -xln = - kt

a

a -x= e

a

a -x= e

a

a -x= e

a

a -x= 0.061

a

Fraction of cyclopropaneRemaining after 18.0 hours

Fraction of cyclopropaneRemaining after 51.5 hours

o

o

o

o

- kto

o

- (0.054/h) x 18.0 ho

o

- 0.972o

o

o

o

aln = kt

a -x

a -xln = - kt

a

a -x= e

a

a -x= e

a

a -x= e

a

a -x= 0.38

a

Effect of Temperature on the Reaction Rate

Arrhenius Equation

actE- RTk = A e

actEln k = - + ln A

RT

of the form y = mx + b

Use the Following Data to Determine the Eact

for

2 (g) 2 (g) 2 (g)2 N O 2 N + O

TK

kM-1/s

Ln k 104(1/T)

1125 11.5900 2.450 8.890

1053 1.6700 0.510 9.50

1001 0.3800 -0.968 9.99

838 0.0010 -6.810 11.9

8.5 9 9.5 10 10.5 11 11.5 12 12.5

-8

-6

-4

-2

0

2

4

f(x) = − 3.07115033064349 x + 29.7212674570598

104(1/T)

ln k

act255 kJ/mol = E

4 act

4act

4act

5act

Eslope = -3.07 x 10 K = -

R

-3.07 x 10 K x - R = E

J-3.07 x 10 K x - 8.314 = E

K mol

2.55 x 10 J/mol = E

Effect of a Catalyst on the Rate of a Reaction

• Lowers the energy barrier to the reaction via lowering the energy of activation

• Homogeneous catalyst- in the same phase as the reacting molecules

• Herterogeneous catalyst – in a different phase from the reacting molecules

Example of a Homogeneous Catalyst

-2 2 (aq) (aq) 2 (aq) 2 (l) 2 (g)

-2 2 (aq) 2 (aq) (aq) 2 (l) 2 (g)

1 11. H O + Br Br + H O + O

2 21 1

2. H O + Br Br + H O + O2 2

reaction coordinates

PE

intermediate

reactants

products

Example of Heterogeneous Catalyst

H H C C

H H

HH

Finely divided metal

An interesting problem:

The reaction between propionaldehyde and hydrocyanic acid have been observed by Svirbely and Roth and reported in the Journal of the American Chemical Society. Use this data to ascertain the order of the reaction and the value of the rate constant for this reaction.

C

H

O

+ C

OHC N

H

::

..

H C N :

:..

+C

H

HON C:

..:

CH3CH2 CH3CH2 CH2CH3

time, minutes [HCN] [CH3CH2CHO]

2.78 0.0990 0.0566

5.33 0.0906 0.0482

8.17 0.0830 0.0406

15.23 0.0706 0.0282

19.80 0.0653 0.0229

∞ 0.0424 0.0000

Check to determine first order in HCN

0 2 4 6 8 10 12 14 16 18

-2.8

-2.7

-2.6

-2.5

-2.4

-2.3

-2.2

time (minutes)

ln([H

CN]-x

)

Check to determine first order in propionaldehyde

0 2 4 6 8 10 12 14 16 18

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

time (minutes)

ln ([

prop

iona

ldeh

yde]

-x)

So close; therefore, let’s take another approach. Let [HCN] = ao and [propionaldehyde] = bo

Then,

o o

o o

o o

o o

o o o o o o

dx = k (a -x) (b -x)

dtdx

= k dt(a -x) (b -x)

dx = k dt

(a -x) (b -x)

Solution:

(a -x) b1 1 ln = kt - ln

(a -b ) b -x (a -b ) a

o

o

-1 -1o

o

-1 -1o

o

-1 -1o

o

(a -x)1 1 0.0566 ln = kt - ln

0.0424 M b -x 0.0424 M 0.0990

(a -x)23.6 M ln = kt - 23.6 M ln 0.572

b -x

(a -x)23.6 M ln = kt - 23.6 M (-0.559)

b -x

(a -x)23.6 M ln = kt + 13.2 M

b -x

o

o

-1 -1o

o

-1 -1o

o

-1 -1o

o

(a -x)1 1 0.0566 ln = kt - ln

0.0424 M b -x 0.0424 M 0.0990

(a -x)23.6 M ln = kt - 23.6 M ln 0.572

b -x

(a -x)23.6 M ln = kt - 23.6 M (-0.559)

b -x

(a -x)23.6 M ln = kt + 13.2 M

b -x

time, minutes [HCN] - x [CH3CH2CHO]-x

2.55 0.0906 0.0482 14.9

5.39 0.0830 0.0406 16.9

12.76 0.0706 0.0282 21.7

17.02 0.0653 0.0229 24.8

Let’s construct the data in a different format

3 2

([HCN]-x)(23.6) ln

([CH CH CHO]-x)

0 2 4 6 8 10 12 14 16 180

5

10

15

20

25

30

f(x) = 0.67777080987964 x + 13.183621262835

time, minutes

Slope = 0.678; therefore, k = 0.678 M-1min-1

3 2

([HCN] - x)23.6 ln

([CH CH CHO]-x)

3 2Rate = k [HCN] [CH CH CHO]

Mechanism:

HCN + H2Ok

k

1

2

-1

H3O+ + CN

-1.

CH3CH2 C

O

H

2. + H3O+

k

k-2

CH3CH2 C

OH

H

+

+ H2O

3.

CH3CH2 C

H

OH+

+ CN- C

H

HO

CH3CH2 CN

k3

Steps 1 and 2 are fast equilibrium steps; and step 3 is the rate determining step

Rate = CH3CH2 C

OH

H

+

[ CN- ]k3

k 1 [HCN] [H2O] = k -1 [H3O+] [CN

-]

k 1 [HCN] [H2O]=

k-1 [H3O+]

[CN-]

CH3CH2 C

OH

H

+CH3CH2 C

O

H

[H3O+]

k2

-2[H2O]k

=

Rate = k3 k1 [HCN] [H2O]

k -1

[H3O+]

CH3CH2 C

O

H

[H3O+]

k2

-2 [H2O]k

Rate = [HCN] CH3CH2 C

O

H

k

Revisit the kinetics for

2 NO (g) + O2 (g) → 2 NO2 (g)

22Rate = k [NO] [O ]