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NCERT Textual Exercise (Solved)

166

Chemical Kinetics

1. From the rate expressions for the following reactions, determine their order of reaction and the dimensions of the rate constants :

(a) 3 NO (g) →N2O (g) + NO2 (g); Rate = k [NO]2

(b) H2O2 (aq) + 3I– (aq) + 2H+→2H2O (l) + I3–; Rate = k [H2O2] [I

–] (c) CHCl3 (g) + Cl2 (g) → CCl4 (g) + HCl (g); Rate = k [CHCl3] [Cl2]

1/2

(d) C2H5Cl (g) → C2 H4 (g) + HCl (g); Rate = k [C2 H5 Cl].

Sol. (a) Order = 2, dimension of k = Rate[NO]

mol L smol L

L mol s2 = =− −

−− −

1 1

1 21 1

( )

(b) Order = 2, dimension of

k = Rate

[H O ] [Imol L s

mol L mol LLmol s

2 2

] ( ) ( )

1 1

1 1 11 1

(c) Order =32

, dimension of

k = Rate

[CHCl ] [Clmol L s

mol L mol LL mol

3 2

1/2] ( ) ( )/ /1 2

1 1

1 1 1 21

// 2 1s

(d) Order = 1, dimension of k = Rate

[C H Cl]mol L smol L

s2 5

1 1

11

( ) .

2. For the reaction, 2A + B → A2 B, rate = k [A] [B]2 with k = 2⋅0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0⋅1 mol L–1 and [B] = 0⋅2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0⋅06 mol L–1.

Sol. Initial rate of reaction = k [A] [B]2

= (2⋅0 × 10–6 mol–2 s–1) (0⋅1 mol L–1) (0⋅2 mol L–1)2

= 8 × 10–9 mol–1 L s–1. When [A] is reduced from 0⋅10 mol L–1 to 0⋅06 mol L–1 (i.e. 0⋅04 mol L–1 of

A has reacted), the concentration of B reacted, is =12

× 0⋅04 mol L–1 = 0⋅02 mol L–1

Concentration of B, remained after reaction with A = 0⋅2 – 0⋅02 = 0⋅18 mol L–1. Now, rate = (2⋅0 × 10–6 mol–2 L2 s–1) (0⋅06 mol L–1) (0⋅18 mol L–1)2

= 3⋅89 × 10–9 mol–1 L s–1 . 3. The decomposition of NH3 on platinum surface is zero order. What are the

rates of production of N2 and H2 if k = 2⋅5 × 10–4 mol–1 L s–1?


167

Chemical Kinetics

Sol. 2NH3 → N2 + 3H2

rate = 12

13

3 2 2ddt

ddt

ddt

[ ] [ ] [ ]NH N H

Rate of zero order = k = 2.5 × 10–4 M s–1

Rate of production of N2 = ddt[ ] .N Ms2 4 12 5 10= × − −

Rate of production of H2 = ddt[ ] .H2 43 2 5 10 = 7.5 × 10–4 M s–1.

4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and

CO and the reaction rate is given by: Rate = k p[ ] .CH3OCH3

32 If the pressure

is measured in bar and time in minutes, then what are the units of the rate

and rate constants? Sol. As the concentration in the rate law equation is given in terms of pressure, ∴ unit of rate = bar min–1

∴ Unit of k = Rate

[ ]bar minbarCH3OCH3

3/2

1

3/2p

= bar–1/2 min–1.

5. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactants is (i) doubled (ii) reduced to half ?

Sol. Rate = k [A]2 = ka2

(i) If [A] = 2a, rate = k (2a)2 = 4ka2 = 4 times.

(ii) If [A] = a/2, rate = k a ka2

14

14

22

= = th .

6. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on the rate constant be represented quantitatively?

Sol. The rate constant of a reaction increases with increase of temperature and becomes nearly double for every 10° rise of temperature. Quantitatively, this temperature effect on the rate constant is represented as,

K =A E RTe a− / where Ea is the activation energy of the reaction and A represents the

frequency factor. 7. Inapseudofirstorderhydrolysisofanesterinwater,thefollowingresults

were obtained: t/s 0 30 60 90 [Ester] mol L–1 0⋅55 0⋅31 0⋅17 0⋅085


168

Chemical Kinetics

(a) Calculate the average rate of reaction between the time interval 30–60 s.

(b) Calculatethepseudofirstorderrateconstantforthehydrolysisofester.

Sol. (a) Average rate during the interval 30 – 60 s = C C2 1

2 1

−−t t

= 031 01760 30

01430

= 4.66 × 10–3 s–1.

(b) k′ =2 303⋅

tlog

[A ][A]0

where initial concentration of A, [A] = 0⋅55 M

when t = 30 s k'1 = 230330

055031

⋅ ⋅⋅

log = 1.91 × 10–2 s–1

when t = 60 s k'2 =230360

055017

⋅ ⋅⋅

log = 1.95 × 10–2 s–1

when t = 90 s k'3 = 230390

0550085

⋅ ⋅⋅

log = 2.075 × 10–2 s–1

∴ Average k′ =

k k k1 2 32

3191 196 207 10

3( ) = 1.978 s–1.

8. AreactionisfirstorderinAandsecondorderinB. (a) Write differential rate equation. (b) How is the rate affected on increasing the concentration of B three

times? (c) How is the rate affected when concentration of both A and B is

doubled?

Sol. (a) dxdt

= k [A] [B]2

(b) Rate = kab2

when [B] becomes three times, rate = ka (3b)2 = 9 kab2 = 9 times. (c) When both [A] and [B] are doubled, rate = k (2b)2 = 8 kab2 = 8 times. 9. In a reaction between A and B, the initial rate of reaction was measured for

different initial concentrations of A and B are given below: A/mol L–1 0⋅20 00⋅20 0⋅04 B/mol L–1 0⋅30 0⋅10 0⋅05


169

Chemical Kinetics

ro/mol L–1 s–1 5⋅07 ×10–5 5⋅07×10–5 7⋅16×10–5

What is the order of reaction with respect to A and B ? Sol. Rate = [A]α [B]β

r1 = 5⋅07 × 10–5 = (0⋅20)α (0⋅30)β (i) r2 = 5⋅07 × 10–5 = (0⋅20)α (0⋅10)β (ii) r3 = 7⋅16 × 10–5 = (0⋅40)α (0⋅50)β (iii)

rr1

2

5

55 07 105 07 10

0 200 20

0 30010

=⋅ ×⋅ ×

=⋅⋅

⋅⋅

−

−( )( )

( )( )

α

α

β

β

= 1 = (3)β ⇒ ∴ β = 0

rr3

2

4

50 40 0 050 00 0 10

1 43 105 07 10

= =××

−

−( . ) ( . )( . ) ( . )

.

.

α β

α β

2 12

2 8 .

Since, q = 0 α = log .log

..

2 82

0 450 3010

=

2 12

2 80 . α = 1.5

2α = 2.8 order w.r.t. α = 1.5 α log 2 = log 2.8 order w.r.t. β = 0. 10. The following rate data were obtained at 303 K for the following reaction: 2 A + B → C + D

Experiment [A], mol L–1 [B], mol L–1 Initial Rate of Formation of D, mol

L–1 min–1

I

II

III

IV

0.1

0.3

0.3

0.4

0.1

0.2

0.4

0.1

6.0 × 10–3

7.2 × 10–2

2.88 × 10–1

2.4 × 10–2

What is the rate law ? What is the order with respect to each reactant and the overall order ? Also calculate the rate constant and write its units:

Sol. Rate = k [A]α [B]β

r1 = k [0⋅1]α [0⋅1]β = 6⋅6 × 10–3


170

Chemical Kinetics

r2 = k [0⋅3]α [0⋅2]α = 7⋅2 × 10–2

r3 = k [0⋅3]α [0⋅4]α = 2⋅88 × 10–1

r4 = k [0⋅4]α [0⋅1]α = 2⋅4 × 10–2

rr1

4

3

260 102 4 10

010 4

14

( )( )

14

14

∴ α = 1

rr k2

3

2

17 2 10288 10

0 3 0 20 3 0 4

=⋅ ×⋅ ×

=⋅ ⋅⋅ ⋅

−

−k ( ) ( )( ) ( )

α β

α β

14

12

12

12

∴ β = 2 ∴ Rate law expression, is, Rate = k [A] [B]2

∴ Order with respect to A = 1 and Order with respect to B = 2 Overall order = 1 + 2 = 3. Calculation of rate constant k Substituting values of experiment I, we get 6⋅0 × 10–3 = k (0⋅1) (0⋅1)2

= k × 1 × 10–3

∴ k = 6 mol–2 L2 s–1. 11. ThereactionbetweenAandBisfirstorderwithrespecttoAandzeroorder

with respect to B. Fill in the blanks in the following table:

Experiment [A], mol L–2 [B], mol L–2 Initial rate, mol L–1 min–1

IIIIIIIV

0.1–

0.4–

0.10.20.40.2

2.0 × 10–2

4.0 × 10–2

–2.0 × 10–2


171

Chemical Kinetics

Sol. Rate law expression : Rate = k [A]1 [B]0 = k [A] R1 = 2⋅0 × 10–2 mol L–1 min–1 = k [0⋅1] mol L–1

∴ k = 0⋅2 min–1

R2 = 4⋅0 × 10–2 mol L–1 min–1 = (0⋅2 min–1) [A] ∴ [A] = 0⋅2 mol L–1

R3 = Rate = k [A] = (0⋅2 min–1) (0⋅4 mol L–1) = 0⋅08 mol L–1 min–1

R4 = 2⋅0 × 10–2 mol L–1 min–1

= k [A] = 0⋅2 min–1 [A] ∴ [A] = 0⋅1 mol L–1. 12. Calculatethehalf-lifeofafirstorderreactionfromtheirrateconstantsgiven

below: (a) 200 s–1 (b) 2 min–1

(c) 4 year–1

Sol. Half-lifeforafirstorderreactionis,t1/2 =0693⋅

k

(a) t1/2 =0693200 1

s= 0⋅346 × 10–2 s = 3⋅46 × 10–3 s

(b) t1/2 =06932 1

min= 0⋅346 min = 3⋅46 × 10–1 min.

(c) t1/2 =06934 1

year = 0⋅713 year = 1⋅73 × 10–1 year.

13. The half-life for radio active decay of 14C is 5730 year. An archaeological artefact contained wood that had only 80% of the 14C found in living tree. Estimate the age of the sample.

Sol. Radioactivedecayfollowsfirstorderkinetics.

Decay constant (k) = 0693 0693

57301 2

t /

year–1.

t = 2 303 2 303

0 693 573010080

01

⋅=

⋅⋅ −k

log[ ][ ] ( / )

logAA years


172

Chemical Kinetics

= 2303 57300693

00969

= 1845 years.

14. The experimental data for the decomposition of N2O5 [2 N2 O5→4 NO2 + O2] in gas phase at 318 K is given below:

t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ mol L–1 1⋅63 1⋅36 1⋅14 0⋅93 0⋅78 0⋅64 0⋅53 0⋅43

0⋅35 (a) Plot log [N2O5] against time. (b) Find the half-life period for the reaction. (c) Draw a graph between log [N2O5] and t. (d) What is rate law? (e) Calculate the rate constant. (f) Calculate the half-life period from k and compare it with (b). Sol. t/s 0 400 800 1200 1600 2000 2400 2800 3200 10–2×[N2O5](mol L–1) 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 log[N2O5] –1.79 –1.87 –1.94 –2.03 –2.11 –2.19 –2.28 –2.37 –2.46 (a) Plot of [N2 O5] vs. time

(b) Initial concentration of N2O5 = 1⋅63 × 10–2 M

Half of this concentration of N2 O5 = 163 102

2

= 0⋅815 × 10–2 M From the plot, time corresponding to this half concentration,

t2 = 1440 s.


173

Chemical Kinetics

(c) Plot of log [N2 O5] vs. time.

(d) As the plot obtained for log [N2 O5] vs. time is a straight line, hence

itisafirstorderreaction. ∴ Rate law expression = rate = k [N2 O5]

(e) Slope of the line = – k2303

2 46 1793200 0

0673200

( )

∴ k = 067 23033200

= 4⋅82 × 10–4 s–1

(f) t1/2 =0693 0693

482 10 4

k= 1438 s.

15. Therateconstantofafirstorderreactionis60s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16 th value?

Sol. t = 230316

230360 1

k

aa

log/ s

log 16 = 4⋅62 × 10–2.

16. During nuclear explosion, one of the products is 90Sr with half-life of 28⋅1 year. If 1 µg of 90Sr was absorbed in bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?

Sol. Radioactivedisintegrationfollowfirstorderkinetics,

Decay constant of 90Sr, k = 0693 0693

2811 2

t / year = 2⋅466 × 10–2 year–1

Calculation of amount left after 10 years


174

Chemical Kinetics

a = 1 µg, t = 10 years, k = 2⋅466 × 10–2 year–1, (a – x) = ?

∴ k = 2303

ta

a xlog

2⋅466 × 10–2 = 230310

1

log( )a x

log (a – x) = – 0⋅1071 (a – x) = Antilog 1⋅8929 = 0⋅7814 µg Calculation of amount left after 60 years a = 1 µg, t = 60 years, k = 2⋅466 × 10–1 year–1, (a – x) = ?

k = 2303

ta

a xlog

2⋅466 × 10–2 = 230310

1

log( )a x

log (a – x) = – 0⋅6426 (a – x) = Antilog 1⋅3575 = 0⋅2278 µg. 17. Showthatthetimerequiredfor99%completionofafirstorderreactionis

twice the time required for the completion of 90%.

Sol. Forthefirstorderreaction,t = 2303

ka

a xlog

99% completion means that x = 99% of a – x = 1%

∴ t99% =2 303

0 992 303 102⋅

− ⋅=

⋅k k

log logaa a

t99% =2 2303

k 90% completion means that a = 100%, x = 90%

∴ t90% = 23030 90

2303 10 2303

k

aa a k k

log log

∴ tt99

99

2 2 303

2 3032%

%=

× ⋅

⋅

=k

k ∴ t99% = 2 × t99%

Hence proved.


175

Chemical Kinetics

18. Afirstorderreactiontakes40minfor30%decomposition.Calculatet1/2. Sol. 30% decomposition meant that x = 30% of a = 0⋅30 a

Asthereactionisfirstorder,thereforerate=2303

ta

a xlog

k = 2 30340

10070

. log

= 230340⋅

× 0⋅1549 = 8⋅918 × 10–3 min–1.

t1/2 = 0693 0693

8918 10 3 1

k min= 77⋅7 min.

19. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained:

t/s p (mm of Hg)0

360

720

35.0

54.0

6330 Calculate the rate constant. Sol. The decomposition reaction of azoisopropane is (CH3)2CHN = NCH(CH3)2 (g) → N2 (g) + C6 H14 (g) Initial pressure p0 0 0 After certain p0 – p p p time (t) ∴ Total pressure after time t = pT = (p0 – p) + p + p = p0 + p pT = p0 + p ⇒ p = pT – p0

Let the initial concentration of azoisopropane is a a α p0

Concentration of azoisopropane left after time, t is (a – x) (a – x) α p0 – p α p0 – (pT – p0) α 2p0 – pT

k = 2303 2303

360 20

0

t

aa x

pp pT

log log


176

Chemical Kinetics

For t = 360 s, k1 = 2303360

3502 350 540

2303360

3516

log log

= 2303360

03400 = 2.175 × 10–3 s–1.

For t = 720 s, k2 = 2303720

3502 350 630

2303720

5

log log

= 2303720

06990 = 2⋅235 × 10–3 s–1.

Average value of k = k k1 2 3 12

2175 2 2352

10+

=⋅ + ⋅

× − −( ) s = 2⋅20 × 10–3 s–1.

20. Thefollowingdatawereobtainedduringthefirstorderthermaldecompositionof SO2Cl2 at a constant volume:

SO2 Cl2 (g) → SO2 (g) + Cl2 (g)

Experiment Time/s Total pressure/atm1

2

0

100

0.5

0.6 Calculate the rate of reaction when total pressure is 0⋅65 atm. Sol. SO2Cl2 (g) →SO2 (g) + Cl2 (g) Initial pressure p0 0 0 After some time t p0 – p p p Total pressure after some time t, pT = (p0 – p) + p + p = p0 + p p = pT – p0

Let the initial cone. of SO2Cl2 be concentration of SO2 Cl2 after some time, t = a – xforfirstorderreaction,

k = 2303

ta

a xlog ...(i)

a α p0

(a – x) α (p0 – p) α p0 – (pT – p0) (a – x) α 2p0 – pT

Substituting these values in (i), we get

k = 2 3032

0

0

⋅−t t

logpp p


177

Chemical Kinetics

when t = 100 s, t = 2303100

052 05 06

2303100

125

log log ( )

= 2 303100⋅ (0⋅0969) = 2⋅2316 × 10–3 s–1.

When pT = 0⋅65 atm i.e. p0 + p = 0⋅65 atm, ∴ p = 0⋅65 – p0 = 0⋅65 – 0⋅50 = 0⋅15 atm ∴ Pressure of SO2Cl2 at time t = p0 – p = 0⋅50 – 0⋅15 = 0⋅35 atm

Rate = k [ ]pSO Cl2 2 = (2⋅2316 × 10–3 s–1) (0⋅35 atm) = 7⋅8 × 10–4 atm s–1. 21. The rate constant for the decomposition of a hydrocarbon is 2⋅418 × 10–5 s–1

at 546 K. If the energy of activation is 179⋅9 kJ/mol, what will be the value of pre-exponential factor?

Sol. Given k = 2⋅418 × 10–5 s–1, Ea = 179⋅9 kJ mol–1, T = 546 K Using Arrhenius equation, k = A e E RT− a /

n k = n A – ERT

a

log k = log A –ERT

a2303⋅

or log A = log k +ERT

a2303⋅

= log (2⋅418 × 10–5) + 179 9 10

2 303 8314 546

3⋅ ×⋅ × ⋅ ×

= (– 5 + 0⋅3834) + 17⋅2081 log A = 12⋅5924 s–1

A = Antilog (12⋅5924) s–1 = 3⋅912 × 1012 s–1. 22. Consider a certain reaction, A →Product with k = 2⋅0 × 10–2 s–1. Calculate

the concentration of A remaining after 100 s, if the initial concentration of A is 1⋅0 mol L–1.

Sol. k = 2303⋅t

log[A ][A]0

2⋅0 × 10–2 s–1 = 2 303100

10 1⋅ ⋅ −

smol L[A]

log

log [A] = – 0⋅8684 = Antilog ( )11316⋅ = 0⋅1354 mol L–1.


178

Chemical Kinetics

23. Sucrose decomposes in acid solution into glucose and fructose according tothefirstorderratelawwitht1/2 = 3⋅00 h. What fraction of the sample of sucrose remains after 8 h?

Sol. Decompositionofsucroseisfirstorderreaction

k = 2303⋅t

log[A ][A]0 ...(i)

For time, t1/2 = 3⋅0 h

k = 0 693 0 693

31 2

⋅=

⋅t / h = 2⋅231 h–1

Substituting k = 0⋅231 h–1 in (i), we get

2⋅231 =23038⋅ log

[A ][A]0

log [A ][A]0 = 0⋅8024

[A ][A]0 = Antilog 0⋅8024 = 6⋅345

∴ fraction of sample remains =[A ][A]0

1

63450158.

24. The decomposition of a hydrocarbon follows the equation, k = (4⋅5 × 1011 s–1) e– 2800 K/T. Calculate Ea.

Sol. According to Arrhenius equation,

k a= −A E /RTe

∴ ERT

28000KT

a

Ea = 28000 K × R = 28000 K × 8⋅314 JK–1 mol = 232⋅79 KJ mol–1. 25. TherateconstantforthefirstorderdecompositionofH2 O2 is given by the

following equation: Log k = 14⋅34 – 1⋅25 × 104 K / T Calculate Ea for this reaction and at what temperature will its half period be

256 min? Sol. Using Arrhenius equation, k = Ae a−E /RT

ln k = ln A –ERT

a


179

Chemical Kinetics


a2303⋅ = 14⋅34 – 1⋅25 × 104 K/T {given} (i)

⇒ ER

a2303⋅ = 1⋅25 × 104

Ea = 2.303 R × 1.25 × 104

= 2.303 × (8.314 Jk–1 mol–1) × 1.25 × 104 K = 239.34 kJ mol–1

For t1/2 = 256 min,

k =0693 0693

2500693250 60

t min

k = 4.62 × 10–5 s–1

= 4.62 × 10–5

Putting this value of k in equation (i) we get

log (4⋅51 × 10–5) = 14.34 – 125 104 KT

(–5 + 0⋅6542) = 14.34 – 125 104 KT

125 104 K

T = 18.675

T = 125 104 471.665

T = 669 K. 26. The decomposition of A into product has value of k as 4⋅5 × 103 s–1 at 10°C

and energy of activation 60 kJ mol–1. At what temperature would k be 1⋅5 × 104 s–1?

Sol. Given k1 = 4⋅5 × 103 s–1, T1 = 10 + 273 = 283 K k2 = 1⋅5 × 104 s–1, T2 = ? Ea = 60 kJ mol–1

Using Arrhenius equation,

log kk

a2

1

1

2 303=

⋅

−

ER

T TT T2

1 2

log 15 104 5 10

600002 303 8314

4

3

1

1 1⋅ ×⋅ ×

=⋅ × ⋅

−−

− −J molJK mol

T 283283 T2

2


180

Chemical Kinetics

log 3⋅333 = 3133⋅63T 283283 T2

2

0 5228313363⋅

⋅=

−T 283283 T2

2

0⋅0472 T2 = T2 – 283 0⋅9528 T2 = 283

T2 = 28309528⋅ = 297 K = 24°C.

27. Thetimerequiredfor10%completionofafirstorderreactionat298Kisequal to that required for its 25% completion at 308 K. If the value of a is 4 × 1010 s–1, calculate k at 318 K and Ea.

Sol. k = 2303

ta

a xlog

∴ k298 = 2303

0101

ta

a alog

= 2303 1091

⋅t

log

= 2303

1

⋅t (0⋅0458)

= 01055

1

⋅t

t1 = 01055

298

⋅k ...(i)

k308 = 2303

0 252

ta

a alog

= 2303 4

32

⋅t

log

= 2303

2

⋅t (0⋅125)

k308 = 0 2879

2

⋅t


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Chemical Kinetics

∴ t2 =0 2879

308

⋅k ...(ii)

As given t1 = t2, ∴ comparing equation (i) and equation (ii), we get

01055 0 2879

298 308

⋅=

⋅k k

kk308

2982 728= .

Using Arrhenius equation,

log kk

a308

298 2303

ER

T TT T2 1

1 2

log (2⋅7289) = Ea

2303 8314308 298298 308

0 ⋅ 4360 = Ea

10

2303 8314 298 308

∴ Ea =0 4360 2303 8314 298 308

10 = 76620.79 J

= 76.62 KJ mol–1. Calculation of k318:


a2303⋅

= log (4 × 1010) – 766 23303 8314 318

1 1

1 1

JK mol

2 JK mol K log k = 10⋅6021 – 12⋅5843 = – 1⋅9822 k = Antilog (– 1⋅9822) = Antilog ( )2 0178⋅ = 1⋅042 × 10–2 s–1.

β øãç ø Ù ò - StudyMate · Rate of production of N 2 = d dt []. N 2 =×25 10−−41Ms Rate...

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