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OBJECTIVES

Determine order of reaction involving asingle reactant using

Initial rate methodThe units of rate constants, kHalf-life based on the graph of concentrationagainst time

Linear graph method based on theintegrated rate equation and rate law

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THE INIT IAL RA TE METHODTHE INIT IAL RA TE METHOD

DETE RMINA T IO N THE O RDE R OF ADETE RMINA T IO N THE O RDE R OF ARE AC T IO NRE AC T IO N

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DETE RMINAT IO N THE O RDE R OF ADETE RMINAT IO N THE O RDE R OF A

RE AC T IO NRE AC T IO N1) THE INIT IAL RA TE METHOD

Compare the initial rate at differenceconcentration

Rate exp 1 =k[A1][B1]Rate exp 2 k [A2 ][ B2 ]

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Example

Exp[ ] mol dm -3 Initial Rate

(Ms -1)[ A ] [ B ]

1 0.50 1.0 2.02 0.50 2.0 8.03 0.50 3.0 18.0

4 1.00 3.0 18.0a) Determine the order of the reaction

b) Write the rate law

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SolutionSolution

To determine the order with respect to AF rom exp . 3 & 4 ( keep [ B ] constant )

Rateexp 3

= k[A3 ] x[B3 ]y

Rate exp 4 k [A4 ]x[B4 ]y

18.0 Ms -1 = k (0. 5 )x(3.0 )y

18.0 Ms -1 k (1 .0 )x(3.0 )yx = 0

ZE RO ZE RO -- O RDE RO RDE R with respect to A

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To determine the order with respect to BF

rom exp 1 & 2 ( keep [ A ] constant )Rate exp 1 = k[A1] x[B1]y

Rate exp 2 k [A2 ]x[B2 ]y

2.0 Ms -1 = k ( 0. 5 )x(1 .0 )y

8.0 Ms -1 k (0. 5 )x(2.0 )y

y = 2

S E CO NDS E CO ND --O RDE RO RDE R with respect to B

b ) Rate law, rate = k [ B ] 2

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UNIT S OF RATE CO NS TANT S, k

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2) UNI T S OF RA TE CO NS T ANT S, k

O RDE R Unit of k

ZE RO Ms -1

F IRS T s -1

S E CO ND M-1s -1

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EXAMPLE

The conversion of cyclopropane to propene inthe gas phase with a rate constant of 6.7 x 10 -4 s -1 at500 0C.a) If the initial concentration of cyclopropane

was 0.25 M what is its concentration after 8.8min

b) H ow long will it take for the concentration of cyclopropane to decrease from 0.2 5 M to 0. 15 M

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Solution

a) first order reaction (from the unit of k = s -1)ln [A]0 = 6.7 x 10 -4 X (8.8 X 60)

[A][A] = 0. 175 M

b) ln 0.2 5 = 6 . 7 x 1 0 -4 s -1 x t0. 15

t = 76 2.42 s

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Iodine atoms combine to form molecular iodine inthe gaseous phaseI (g) + I (g) p I2 (g)

This reaction is a second order reaction , with therate constant of 7 .0 x 1 0 9 M-1 s -1

i) If the initial concentration of iodine was0.08 6

M,calculate its concentration after 2 min .ii) Calculate the half life of the reaction if the

initial concentration of iodine is 0.0 6 M and0.42

M respectively.

E XAMPL E

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i) 1[A]1[A]0

= Kt +

1[A]

1

[0.08 6

]

= (7 .0 x 1 0 9 x 2 x 6 0 ) +

[A] = 1.190 x 10 -12 M

ii) [I2 ]0 = 0.0 6 M

t1/2 = 1k[A]0= 1

(7 .0 x 1 0 9 x 0.0 6 )

= 2.38 x 1 0 -9 s

Solution

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DETE RMINA T IO N THE O RDE R OFDETE RMINA T IO N THE O RDE R OFA R E AC T IO NA R E AC T IO N

GRAP H ICAL M ETHODGRAP H ICAL M ETHODH ALF -LIFE b ased on theGRAP H of [ ] vs time

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3) Graphical method3) Graphical methoda) H ALF -LIFE METHOD

Plot the GRAP H [ ] VS tGRAP H [ ] VS tZE ROZE RO --O RDE R R E AC T IO NO RDE R R E AC T IO N

212

][

t

Ak !

VIDEO 11

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693.0t

k !

210][

1t A

k !VIDEO 1 2 VIDEO 1 3

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The following results were obtained from anexperimental investigation on dissociation of dinitrogen pentoxide at 4 5 oC

N2 O 5 (g)2

NO 2 (g) + O 2 (g)time, t/min 0 10 20 30 40 5 0 6 0

[N2 O 5 ] x 1 0 -4 M 176 124 9 3 7 1 5 3 3 9 2 9

Plot graph of [N 2 O 5 ] vs time , determinei) The order of the reaction

ii) the rate constant k

EXAMPLE

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[N2O 5] x 10 -4 M

Time ( min)

180

16 0

140

80

120

100

80

6 0

40

20

10 20 30 40 5 0 6 0 7 0

Solution

t1/2 t1/2

i. F

irst order because the half lifeis not depending on the initialconcentration

ii.

21

693.0t

k ! = ln 2 / t1/ 2 = 0. 69 3 / 20

= 0.035 min -1

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BAS ED O N THE AB O VE GRAP H ,

Time taken for concentration of N 2 O 5 to change from17 6 x 10 -4 M to 88 x 10 -4 M is 20 min

Time taken for concentration of N 2 O 5 to change from 88x 10 -4 M to 44 x 10 -4 M is also 20 min

T he half life for the reaction is a constant and doesnot depend on the initial concentration of N 2 O 5

Thus , the a

bove reaction is first order

i)

ii) k =ln 2

20 min= 0.03 min -1

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4) GRAP H ICAL M ETHOD4) GRAP H ICAL M ETHODLINE AR GRAP H METHOD based on

INTE GRA TED RATE EQ UAT IO N

ANDRATE LAW

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4 ) LINEAR GRAPH METHOD4 ) LINEAR GRAPH METHODZE ROZE RO --O RDE R R E AC T IO NO RDE R R E AC T IO N

[ A ]

RateRate

Rate = kRate = kk = M sk = M s --11

[A]0 [A]

Based on theBased on the RATE LAWRATE LAW Based on theBased on theINTE GRA TED RATE EQINTE GRA TED RATE EQ

t[A]0 - [A] =k

t

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Rate ln [A]0 /[A]

[ A ] t

F IRS TF IRS T--O RDE R R E AC T IO NO RDE R R E AC T IO NBased on theBased on the RATE LAWRATE LAW

Rate = k [A]

k = s -1

Based on theBased on theINTE GRA TED RATE EQINTE GRA TED RATE EQ

ln ([A]o /[A]) = k t

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rate

[A]2

A product

Rate = k [A]2

k= M -1s -1

Based onBased on thethe RATE LAWRATE LAW

S E CO ND S E CO ND --O RDE R R E AC T IO NO RDE R R E AC T IO N

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1/[A]

t

Based on theBased on theS E CO ND O RDE RS E CO ND O RDE R INTE GRA TED RATE EQINTE GRA TED RATE EQ

1 =1 = k k t +t + 11[A] [A][A] [A] 00

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t

1/ [A] 1/ [A]0

Based on theBased on theS E CO ND O RDE R IN TE GRA TED RATE EQS E CO ND O RDE R IN TE GRA TED RATE EQ

11 -- 11 == k k tt[A] [A][A] [A]

00

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