Chemical kinetics

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CHEMICAL KINETICS

A.K.GUPTA, PGT CHEMISTRY, KVS ZIET BHUBANESWAR

A.K.GUPTAPGT CHEMISTRY

KVS ZIET BHUBANESWAR

2

Rate and Order Rate and Order of Reactionsof Reactions


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For the reaction aA + bB cC + dD

Rate k[A]n[B]m

Rate law or rate equation: Experimentally derived algebraic equation which relates the rate of reaction with the concentration of the reactants


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Rate k[A]n[B]m

where n and m are the orders of reaction with respect to A and Bn and m can be integers or fractional n m is the overall order of reaction.


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Rate k[A]n[B]m

For multi-step reactions, n & y have no direct relation to the stoichiometric coefficients and can ONLY be determined experimentally.For single-step reactions (elementary reactions),n = a and m= b


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Rate k[A]n[B]m

n = 0 zero order w.r.t. An = 1 first order w.r.t. An = 2 second order w.r.t. A

m = 0 zero order w.r.t. Bm = 1 first order w.r.t. Bm = 2 second order w.r.t. B


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For the reaction aA + bB cC + dDRate k[B]2

Describe the reaction with the following rate law.

The reaction is zero order w.r.t. A andsecond order w.r.t. B.


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Rate k[A]n[B]m

Where k is the rate constant (specific rate) of the reaction


• Temperature-dependent• Can only be determined from

experimentsA.K.GUPTA, PGT CHEMISTRY, KVS

ZIET BHUBANESWAR

k is

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Rate k[A]n[B]m

units of k : -mol L1 s1/(mol L1)n+m or,mol L1 min1 /(mol L1)n+m


m1n1

11

mn )L (mol)L (mols L mol

[B][A]ratek


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Rate k[A]0[B]0

units of k= mol L1 s1/(mol L1)0+0

= mol L1 s1

= units of rate



11

Rate k[A][B]0


= s1



12

Rate k[A][B]


= mol1 L1 s1


The overall order of reaction can be deduced from the units of k


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Rate k[A]n[B]m[C]p…

For the reaction aA + bB + cC + … products

units of k : -mol L1 s1/(mol L1)n+m+p+…


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Determination of rate equations

To determine a rate equation is to find n, m, p, z,…

Rate k[A]n[B]m[C]p…

Two approaches : -1. Initial rate method 2. Graphical method


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Determination of Determination of Rate LawRate Law

by by Initial Rate Initial Rate MethodsMethods


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5Cl(aq) + ClO3(aq) + 6H+(aq) 3Cl2(aq) +

3H2O(l)

Expt [Cl(aq)] (mol L1)

[ClO3(aq)]

(mol L1)[H+(aq)] (mol L1)

Initial rate (mol L1 s1 )

1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

Suppose the rate law for the reaction is rate k[Cl(aq)]n[ClO3

(aq)]m[H+

(aq)]p


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[ClO3(aq)]



1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

pmn

pmn

5

5

(0.20)(0.08)(0.15)(0.40)(0.08)(0.15)

101.0104.0

From experiments 1 and 2,

4 = 2p

p = 2

= 2p


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[ClO3(aq)]



1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

pmn

pmn

5

5

(0.40)(0.08)(0.15)(0.40)(0.16)(0.15)

104.0108.0


2 = 2m

m = 1

= 2m


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[ClO3(aq)]



1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

pmn

pmn

5

5

(0.20)(0.08)(0.15)(0.20)(0.08)(0.30)

101.0102.0


2 = 2n

n = 1

= 2n


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rate k[Cl(aq)][ClO3(aq)][H+

(aq)]2


[ClO3(aq)]



1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

From experiment 1, 1.0105 k(0.15)(0.08)(0.20)2 k = 0.02 mol3 L3 s1


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rate k[Cl(aq)][ClO3(aq)][H+

(aq)]2


[ClO3(aq)]



1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

From experiment 2, 4.0105 k(0.15)(0.08)(0.40)2 k = 0.02 mol3 L3 s1


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Question 2C + 3D + E P + 2Q

Exp.[C]

( mol L1)

[D] ( mol L1)

[E] ( mol L1)

Initial rate ( mol L1 s-1)

1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(a) rate k[C]n[D]m[E]p


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Exp. [C] ( mol L1)

[D] ( mol L1)

[E] ( mol L1)


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102


pmn

pmn

3

-2

(0.10)(0.10)(0.10)(0.10)(0.10)(0.20)

103.0102.4


8 = 2n

n = 3= 2n


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Exp. [C] ( mol L1)

[D] ( mol L1)

[E] ( mol L1)


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102


pmn

pmn

3

-3

(0.10)(0.10)(0.10)(0.10)(0.20)(0.10)

103.0103.0


1 = 2m

m = 0

= 2m


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Exp. [C] ( mol L1)

[D] ( mol L1)

[E] ( mol L1)


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(a) rate k[C]x[D]y[E]z

pmn

pmn

3

-2

(0.10)(0.10)(0.10)(0.30)(0.10)(0.10)

103.0102.7


9 = 3p

p = 2

= 3p


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Exp. [C] ( mol L1)

[D] ( mol L1)

[E] ( mol L1)


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(a) rate k[C]3[D]0[E]2

= k[C]3[E]2


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Exp. [C] ( mol L1)

[D] ( mol L1)

[E] ( mol L1)


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(b) rate k[C]3[E]2

From experiment 1, 3.0103 k(0.10)3(0.10)2 k = 300 mol4 L4 s1


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QuestionH+

CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + H+(aq) + I(aq)

Initial rate(mol L1 s1)

Initial concentration(mol L1)

[I2(aq)] [CH3COCH3(aq)] [H+(aq)]3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a)Suppose, rate k[I2(aq)]n[CH3COCH3(aq)]m[H+

(aq)]p


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[I2(aq)] [CH3COCH3(aq)] [H+(aq)]3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a) rate k[I2(aq)]n[CH3COCH3(aq)]m[H+(aq)]p

p3-m1-n4-

p-3m-1n-4

5

-5

)10(5.0)10(2.0)10(1.5)10(5.0)10(2.0)10(2.5

103.5103.5


1 = 1.67n

n = 0

= 1.67n


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[I2(aq)] [CH3COCH3(aq)] [H+(aq)]3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a) rate k[I2(aq)]n[CH3COCH3(aq)]m[H+

(aq)]p

p3-m1-n4-

p-3m-1n-4

5

-5

)10(5.0)10(2.0)10(2.5)10(5.0)10(4.0)10(2.5

103.5107.0


2 = 2mm= 1= 2m


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[I2(aq)] [CH3COCH3(aq)] [H+(aq)]3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a) rate k[I2(aq)]n[CH3COCH3(aq)]m[H+

(aq)]p

p3-m1-n4-

p-2m-1n-4

5

-4

)10(5.0)10(4.0)10(2.5)10(1.0)10(4.0)10(2.5

107.0101.4


2 = 2p p = 1

= 2p


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[I2(aq)] [CH3COCH3(aq)] [H+(aq)]3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a) Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)]

= k[CH3COCH3(aq)][H+(aq)]


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[I2(aq)] [CH3COCH3(aq)] [H+(aq)]3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(b) Rate = k[CH3COCH3(aq)][H+(aq)]From experiment 1, 3.5105 k(2.0101)(5.0103) k = 0.035 mol1 L1 s1A.K.GUPTA, PGT CHEMISTRY, KVS

ZIET BHUBANESWAR

34

Determination of Determination of Rate LawRate Law

by by Graphical Graphical MethodsMethods


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The rate of a reaction may be expressed as : -

(1) Differential rate equation(2) Integrated rate equation


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A products

nk[A]dtd[A] Rate

(Differential rate equation)

shows the variation of rate with [A]Two types of plots to determine k and n


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[A]

ratenk[A]dt

d[A] rate

n = 0

krate = k


Concentration of reactant A

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Examples of zero-order reactions : -

2NH3(g) N2(g) + 3H2(g)

Fe or W as catalyst

Decomposition of NH3/HI can take place only on the surface of the catalyst. Once the surface is covered completely (saturated) with NH3/HI molecules at a given concentration of NH3/HI, further increase in [NH3]/[HI] has no effect on the rate of reaction.

2HI(g) H2(g) + I2(g) Au as catalyst


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[A]

ratenk

dtd[A] rate [A]

n = 1

slope = k

linear



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[A]

rate2k[A]dt

d[A] rate

n = 2

k cannot be determined directly from the graph

parabola



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nk[A]dtd[A] rate

[A]

rate n = 2 n =

1

n = 0



42 log10[A]

log10rate

nk[A] rate n

1010 k[A]log ratelog

slope

y-intercept

n = 0

n = 1

log10k

n = 2

[A]nlogklog 1010

slope = 1

slope = 2

slope = 0


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nk[A]dtd[A]


kdtd[A]

t

0 0

A

A

t

tdtkd[A]

kt[A][A] 0t

[A]t = [A]0 – kt (Integrated rate equation)

If n = 0Derivation not required


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[A]t = [A]0 – kt (Integrated rate equation)

shows variation of [A] with time

time

[A]t

rate kdtd[A]slope

[A]0

constant rate


Conc

entra

tion

of re

acta

nt A

tAA

k t][][ 0

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nk[A]dtd[A]


(Integrated rate equation)

If n = 1, k[A]dtd[A]

kdt[A]d[A]

t

0 0

[A]

[A]

t

tdtkd[A][A]

1

loge[A]t – loge[A]0 = kt

Or [A]t [A]0 ekt loge[A]t = loge[A]0 kt

ln


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Two types of plots to determine k and n


time

log[A]t

log[A]0

slope = k/2.303 k= -2.303xslpoe

linear n = 1


][][

log303.2 0

tAA

tk

intercept = log[A]0

[A]0=antilog (intercept)

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Two types of plots to determine k and n


time

[A]t

[A]t varies exponentially with time

constant half life n = 1


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seconds 100tlife, half21


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21tt when 0t [A]

21[A]

693.0301.0303.2k

log2303.2t21 x


tAA

tk

][][log303.2 0

50

seconds 100tlife, half21

s 1002.303log2k = 6.9103

s1


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QuestionFor hydrolysis sucrose fructose + glucoseRate = k[sucrose] k = 0.208 h1 at 298 Ka. Determine the rate constant of the reaction.b. Calculate the time in which 87.5%of sucrose has decomposed

(a) h 3.33h 0.208

12.303x0.30k

2.303log2t 121


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Q.17sucrose fructose + glucose

Rate = k[sucrose] k = 0.208 h1 at 298 K

(b)

87.5% decomposed [A]t = 0.125[A]0

On solving we get = 9.99 h


tAA

tk

][][log303.2 0

0

0

][125.0][log303.2A

Ak

t

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mol1 L1 s1k against t2

s1k/2.303log[A]t against t1

mol L1 s1k[A]t against t[A]t [A]0 – kt0

Units of kSlopeStraight line plot

Integrated rate

equationOrder

t[A]1

kt[A]

1[A]

1

0t

kt/2.303[A][A]

log0

t

Summary : - For reactions of the typeA Products


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2H2O2(aq) 2H2O(l) + O2(g)

Rate = k[H2O2(aq)]

Examples of First Order ReactionsExamples of First Order Reactions


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Examples of First Order ReactionsExamples of First Order Reactions

Reaction Rate equation2N2O5(g) 4NO2(g) + O2(g) Rate = k[N2O5(g)]SO2Cl2(l) SO2(g) + Cl2(g) Rate = k[SO2Cl2(l)]

(CH3)3CCl(l) + OH-(aq) (CH3)3COH(l) + Cl-(aq)

Rate = k[(CH3)3CCl(l)](SN1)

All radioactive decays e.g. Rate = k[Ra]

SN1 : 1st order Nucleophilic Substitution ReactionA.K.GUPTA, PGT CHEMISTRY, KVS

ZIET BHUBANESWAR

56

1. For a reaction involving one reactant only:

2NOCl(g) 2NO(g) + Cl2(g) Rate = k[NOCl(g)]2

2NO2(g) 2NO(g) + O2(g) Rate = k[NO2(g)]2

Examples of Second Order Examples of Second Order ReactionsReactions


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Examples of Second Order Examples of Second Order ReactionsReactions

Reaction Rate equation

H2(g) + I2(g) 2HI(g) Rate = k[H2(g)][I2(g)]

CH3Br(l) + OH(aq) CH3OH(l) + Br(aq)

Rate = k[CH3Br(l)][OH(aq)] (SN2)

CH3COOC2H5(l) + OH(aq) CH3COO(aq) + C2H5OH(l)

Rate = k[CH3COOC2H5(l)][OH(aq)]

SN2 : 2nd order Nucleophilic Substitution Reaction

2. For a reaction involving one reactant only:


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2. For a reaction involving two reactants:A + B products

Rate = k[A][B]To determine the rate equation, the concentration of one of the reactants must be kept constant (in large excess) such that the order of reaction w.r.t. the other reactant can be determined.


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2. For a reaction involving two reactants:A + B products

Rate = k[A][B]

When [B] is kept constant, excess

rate = k’[A] (where k’ = k[B]excess)


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Rate = k[A][B]excess = k’[A]k can be determined from k’ if [B]excess is known

Linear first order


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2. For a reaction involving two reactants: A + B products Rate = k[B][A]

• When [A] is kept constant,rate = k”[B] (where k” = k[A]excess)

excess


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Rate = k[A]excess[B] = k’’[B]k can be determined from k’’ if [A]excess is known

Linear first order


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Rate Equations and Order of Reactions

(a)The reaction between tyrosine (an amino acid) and iodine obeys the rate law: rate = k [Tyr] [I2].Write the orders of the reaction with respect to tyrosine and iodine respectively, and hence the overall order. Answer(a) The order of the reaction with respect to

tyrosine is 1, and the order of the reaction with respect to iodine is also 1. Therefore, the overall order of the reaction is 2.


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Rate Equations and Order of Reactions

(b) Determine the unit of the rate constant (k) of the following rate equation:

Rate = k [A] [B]3 [C]2

(Assume that all concentrations are measured in mol dm–3 and time is measured in minutes.)Answer

(b) k =

Unit of k =

= mol-5 dm15 min-1

23 ]C[]B][A[Rate

63-

-1-3

)dm (molmin dm mol


65

The initial rate of a second order reaction is 8.0 × 10–3 mol dm–3 s–1. The initial concentrations of the two reactants,A and B, are 0.20 mol dm–3. Calculate the rate constant of the reaction and state its unit.

Zeroth, First and Second Order Reactions

Answer 8.0 10-3 = k (0.20)2

k = 0.2 mol-1 dm3 s-1


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Determination of Simple Rate Equations from Initial Rate Method

For a reaction between two substances A and B, experiments with different initial concentrations of A and B were carried out. The results were shown as follows:Expt Initial conc.

of A (mol L-1)Initial conc. of B (mol L-

1)

Initial rate (mol L-1 s-1)

1 0.01 0.02 0.0005

2 0.02 0.02 0.001 0

3 0.01 0.04 0.002 0


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(a)Calculate the order of reaction with respect to A and that with respect to B.

Answer


68


(a) Let x be the order of reaction with respect to A, and y be the order of reaction with respect to B. Then, the rate equation for the reaction can be expressed as:Rate = k [A]x [B]y

Therefore,0.0005 = k (0.01)x (0.02)y .......................... (1)0.0010 = k (0.02)x (0.02)y .......................... (2)0.002 0 = k (0.01)x (0.04)y .......................... (3)Dividing (1) by (2),

x = 1

x)02.001.0(

0010.05 0.000


69


(a) Dividing (1) by (3),

y = 2

y)04.002.0(

0010.05 0.000


70


(b) Using the result of experiment (1),Rate = k [A] [B]2

0.000 5 = k 0.01 0.022

k = 125 mol-2 dm6 s-1

(c) Rate = 125 [A] [B]2

(b) Calculate the rate constant using the result of experiment 1.

(c)Write the rate equation for the reaction.Answer


71

In the kinetic study of the reaction,CO(g) + NO2(g) CO2(g) + NO(g)

four experiments were carried out to determine the initial reaction rates using different initial concentrations of reactants. The results were as follows:


Expt Initial conc. of CO(g)

(mol dm-3)

Initial conc. of NO2(g)

(mol dm-3)

Initial rate (mol dm-3 s-1)

1 0.1 0.1 0.0152 0.2 0.1 0.0303 0.1 0.2 0.0304 0.4 0.1 0.060A.K.GUPTA, PGT CHEMISTRY, KVS

ZIET BHUBANESWAR

72

(a) Calculate the rate constant of the reaction, and hence write the rate equation for the reaction.

14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31)

Answer


73

14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31)

(a) Let m be the order of reaction with respect to CO, and n be the order of reaction with respect to NO2. Then, the rate equation for

the reaction can be expressed as:Rate = k [CO]m [NO2]n

Therefore,0.015 = k (0.1)m (0.1)n .......................... (1)0.030 = k (0.2)m (0.1)n .......................... (2)0.030 = k (0.1)m (0.2)n .......................... (3)Dividing (1) by (2),

m = 1

m)2.01.0(

030.00.015


74


(a) Dividing (1) by (3),

n = 1 Rate = k [CO] [NO2]

Using the result of experiment (1),0.015 = k (0.1)2

k = 1.5 mol-1 dm3 s-1

Rate = 1.5 [CO] [NO2]

n)2.01.0(

030.00.015


75


(b) Determine the initial rate of the reaction when the initial concentrations of both CO( g) and NO2( g) are 0.3 mol dm–3.

Answer(b) Initial rate = 1.5 0.3 0.3 = 0.135 mol dm-3 s-1


76

(a) Write a chemical equation for the decomposition of hydrogen peroxide solution.

Determination of Simple Rate Equations from Differential Rate Equations

Answer(a) 2H2O2(aq) 2H2O(l) + O2(g)


77

(b) Explain how you could find the rate of decomposition of hydrogen peroxide solution in the presence of a solid catalyst using suitable apparatus.


Answer


78


(b) In the presence of a suitable catalyst such as manganese(IV) oxide, hydrogen peroxide decomposes readily to give oxygen gas which is hardly soluble in water. A gas syringe can be used to collect the gas evolved. To minimize any gas leakage, all apparatus should be sealed properly. A stopwatch is used to measure the time. The volume of gas evolved per unit time (i.e. the rate of evolution of the gas) can then be determined.


79

(c) The table below shows the initial rates of decomposition of hydrogen peroxide solution of different concentrations. Plot a graph of the initial rate against [H2O2(aq)].


Answer

[H2O2(aq)] (mol L-1)

0.100 0.175 0.250 0.300

Initial rate (10-4 mol L-1

s-1)

0.59 1.04 1.50 1.80


80

14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)

(c)


81

(d) From the graph in (c), determine the order and rate constant of the reaction.


Answer


82


(d) There are two methods to determine the order and rate constant of the reaction.Method 1:When the concentration of hydrogen peroxide solution increases from 0.1 mol dm–3 to 0.2 mol dm–3, the reaction rate increases from 0.59 × 10–4 mol dm–3 s–1 to about 1.20 × 10–4 mol dm–3 s–1.

∴ Rate [H2O2(aq)]

Therefore, the reaction is of first order.The rate constant (k) is equal to the slope of the graph.

k =

= 6.0 10-4 s-1

3-

-1-3

dm mol 0) - .3000(s dm mol 0) - 4-10 (1.8


83


(d) Method 2:The rate equation can be expressed as:Rate = k [H2O2(aq)]x

where k is the rate constant and x is the order of reaction.Taking logarithms on both sides of the rate equation,log (rate) = log k + x log [H2O2(aq)] ................. (1)

-3.74-3.82-3.98-4.23log (rate)

-0.523-0.602-0.757-1.000log [H2O2(aq)]


84


(d) A graph of log (rate) against log [H2O2(aq)] gives a straight line

of slope x and y-intercept log k.


85


(d) Slope of the graph =

=1.0 The reaction is of first order.Substitute the slope and one set of value into equation (1):-4.23 = log k + (1.0) (-1.000)log k = -3.23 k = 5.89 10-4 s-1

)8.0(5.0)02.4(71.3


86


(a)Decide which curve in the following graph corresponds to(i) a zeroth order reaction;(ii) a first order reaction.

(a) (i)(3)(ii)(2)

Answer


87


(b) The following results were obtained for the decomposition of nitrogen(V) oxide.

2N2O5(g) 4NO2(g) + O2(g)

Concentration of N2O5 (mol dm-3)

Initial rate (mol dm-

3 s-1)1.6 10-3 0.122.4 10-3 0.183.2 10-3 0.24


88


(i) Write the rate equation for the reaction.Answer

(i) The rate equation for the reaction can be expressed as:Rate = k [N2O5(g)]m

where k is the rate constant and m is the order of reaction.


89


(ii) Determine the order of the reaction.Answer


90


(ii) Method 1:A graph of the initial rates against [N2O5(g)] is shown as follows:


91


As shown in the graph, when the concentration of N2O5 increases

from 1.0 10–3 mol dm–3 to 2.0 10–3 mol dm–3, the rate of the reaction increases from 0.075 mol dm–3 s–1 to 0.15 mol dm–3 s–1. Rate [N2O5(g)] The reaction is of first order.Then, the rate constant k is equal to the slope of the graph.

k =

= 75 s-1 The rate equation for the reaction is:

Rate = 75 [N2O5(g)]

1-3-3-

-1-3

s dm mol 0) - 10 .61(s dm mol 0)(0.12


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(iii) Determine the initial rate of reaction when the concentration of nitrogen(V) oxide is:(1) 2.0 × 10–3 mol dm–3.(2) 2.4 × 10–2 mol dm–3. Answer


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(iii) The rate equation, rate = 75 [N2O5(g)], is used for the

following calculation.(1) Rate = 75 [N2O5(g)]

= 75 s–1 2.0 10–3 mol dm–3

= 0.15 mol dm–3 s–1

(2) Rate = 75 [N2O5(g)]

= 75 s–1 2.4 10–2 mol dm–3

= 1.8 mol dm–3 s–1


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The half-life of a radioactive isotope A is 1 997 years. How long does it take for the radioactivity of a sample of A to drop to 20% of its original level?

Determination of Simple Rate Equations from Integrated Rate Equations

Answer


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As radioactive decay is a first order reaction,

= 3.47 10-4 year-1

t = 4638 years It takes 4638 years for the radioactivity of a sample of A to

dropt to 20 % of its original level.

kt 693.0

21

1997693.0

k

kt)[A][A]( ln 0

t410473.)% 20% 100( ln


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(a) At 298 K, the rate constant for the first order decomposition of nitrogen(V) oxide is 0.47 × 10–4 s–1. Determine the half-life of nitrogen(V) oxide at 298 K.

N2O5 2NO2 + O221

Answer


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(a) Let the half-life of nitrogen(V) oxide be .

The half-life of nitrogen(V) oxide is 14 745 s.

21t

21

14 693.0s1047.0t

s 745 1421 t


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(b) The decomposition of CH3N = NCH3 to form N2 and C2H6 follows first order kinetics and has a half-life of 0.017 minute at 573 K. Determine the amount of CH3N = NCH3 left if 1.5 g of CH3N = NCH3 was decomposed for 0.068 minute at 573 K.


Answer


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(b)

Let m be the amount of CH3N=NCH3 left after 0.068

minute.

m = 0.094 g

1

21

min76.40min017.0

693.0693.0 t

k

0.068 40.76)1.5( ln m


Chemical kinetics

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