1. From the rate expression for the following reactions ... 12TH/CLASS 12TH CHEMIS…1. From the...

1

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

(a) Rate = K [NO]2

(b) Rate =

(c) Rate = K

(d) Rate =

(e) Rate = K

• Solution: Order of the reaction is the sum of powers of each concentration term raised in the rate equation.

(a) The experimentally determine rate law = K[NO]2

The order with respect to (NO) = 2 Rate = K[NO]2

K =

Dimension of K =

= . (b) Rate = K[H2O2][I

-] order of reaction = 2

K =

Dimension of K =

= .

(c) Rate = K ( Order of reaction = 1.5

K =

Dimension of K = = mol-1/2L1/2sec-1

(d) Rate = K (CH Cl3 ) (Cl2)1/2

Order of reaction = 1.5

MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315

CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA, DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi EMAIL:1 [email protected] web site www.mathematic.in

1

www.mathem

atic.in

09810144315

2

K =

Dimension of K =

= (e) Rate = K (C2 H5 Cl) order of reaction=1

K =

Dimension of K =

= sec –1

2. For the reaction

2 A + B + C A2 B Rate = K A [B]2 with K = 2.0 × 10 –6 M-2 S –1 Calculate the initial rate of the reaction when [A] = 0.1 M, [B] = 0.2 M. If the rate of reverse reaction is negligible then calculate the rate of reaction after [A] is reduced to 0.06 M

• Solution: Rate = K [A] [B]2

Rate = 2.0 × 10-6 × 0.1 × (0.2)2

Rate = 8 × 10-9 M-2 S-1

When [A] is reduced to 0.06M the rate: 2moles of A react with one mole of B When [A] is reduced to 0.06 mole , 0.04 mole of A has reacted. It would have reacted with 0.2 mole of B Remaining concentration of B is 2-0.02=1.8 mole Rate=2.0 × 10-6 × 0.06 × (1.8)2 3.88 × 10-7

3. The rate law for the decomposition of NH3 on Pt surface is zero order. What are the rates of

production of N2 and H2 if K = 2.5 × 10-4

M S-1.

• Solution:

The reaction is According to rate law expression. For zero order reaction.

Rate = Rate = K

i.e., Rate = 2.5 × 10-4 MS-1

The rate of disappearance of ammonia is expressed as rate = = = Rate = 2.5 × 10-4 MS-1

Rate of productivity of NH3 is twice the rate of decomposition of NH3



2

www.mathem

atic.in

09810144315

3

€Rate of production of N2 = =1.25 10-4 mol/L/s.

(ii) Rate of production of H2 = 3/2 (rate of decomposition of NH3),

Rate of production of H2 = 2.5 × 10-4 × 3/2 Rate of production of H2 = 3.75× 10-4 M/L S-1.

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

H2 and CO and the reactionrate is given by Rate = K [CH3 O CH3]3 / 2

Rate of the reaction is followed by increase in pressure in a closed vessel and the rate can be expressed in terms of the partial pressure of dimethyl ether i.e.,| Rate = K (P CH3 O CH3)

3 / 2 . If the pressure is measured in bar and time in minutes then what are the units of rate and rate constant.

• Solution: Rate = K (P CH3 O CH3)

3 / 2

Partial pressure is expressed in bar

K = = K = bar-1/2 min-1 rate = bar.min-1

5. Mention the factors that affect the rate of a chemical reaction.

• Solution: The rate of any particular reaction depends upon the following factors. (1) Nature of the reactants: Consider the following two reactions:

fast

slow These reactions appear to be similar but the first one is fast while the second one is slow. This is because different amounts of energy are required for breaking of different bonds and different amounts of energy are released in the formation of different bonds. (2) Concentration of the reactants: Greater the concentration of the reactants, faster is the reaction conversely as the concentration of the reactants decreases the rate of reaction also decreases.

(3) Temperature: The rate of reaction increases with increase in temperature. In most of the cases the rate of reaction becomes nearly doubled for 10K rise of temperature. In some cases, reactions do not take place at room temperature but take place at higher temperatures.

(4) Presence of Catalyst: A catalyst generally increases the speed of a reaction without itself being consumed in the reaction. In case of reversible reactions, a catalyst helps to attain equilibrium quickly without disturbing the state of equilibrium.

(5) Surface area of the reactants: For a reaction involving a solid reactant or catalyst, smaller the particle size greater is the surface area and hence faster is the reaction.



3

www.mathem

atic.in

09810144315

4

6. A reaction is second order with respect to a reaction. How is the rate

of the reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

• Solution: The rate law expression to the above reaction is represented as Rate = K (A)2 as it is a second order reaction, where K is rate constant. For E.g. consider rate α (A)2

Let us assume that concentration of [A] is unity rate α(1)2 = 1 rate is unity.

Consider a situation where [A] = ½ [A]

i.e., rate

rate = original rate i. Consider another situation when [A] = [2A]

i.e., rate rate = 4 original rate Therefore on increasing the concentration by 2 times the rate becomes 4 times and by decreasing the concentration by half rate of the reaction is reduced by 4 times.

7. What is the effect of temperature on the rate constant of reaction ?

How can this temperature effect on rate constant be represented quantitatively ?

• Solution: Quantitatively the effect of temperature on the rate of a reaction and hence on the rate constant K was proposed by Arrhenius (1889). The equation, called Arrhenius equation is usually written in the form.

K = Where the pre-exponential factor A is a constant and is called frequency factor. Ea is energy of

activation R gas constant and T is the absolute temperature. The factor gives the fraction of molecules having energy equal to or greater than the activation energy Ea. The energy of activation Ea is an important quantity as it is characteristic of a reaction. Using the above equation, its value can be predicted whether the reaction do take place or not. On increasing the temperature, for most of the reactions the rate increases by 100% for a 10° C rise near the room temperature. This is explained by the fact that certain threshold energy is required for the molecules to react. The fact can be explained by the plot of fraction of molecules against energy.



4

www.mathem

atic.in

09810144315

5

increases considerably when temperature increases by 10 K.ΣFrom the graph it is seen that the average relative kinetic energy increases as expected but populations of colliding molecules (given by the shaded area under the curve) possessing the threshold relative kinetic energy

8. In a pseudo first order hydrolysis of ester in water the following

results were obtained: t/s 0 30 60 90

[Ester]/mol L-1 0.55 0.31 0.17 0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

• Solution:

(i) Average rate = =

Average rate = = .

(Rate constant ,K=

=

=1.90 10-2 s-1

9. A reaction is first order in A and second order in B.

(i) Write the differential rate equation. (ii) How is the rate affected when the concentration of B is tripled ?



5

www.mathem

atic.in

09810144315

6

(iii) How is the rate affected when the concentrations of both A and B are doubled ?

• Solution:

(i) A + B reaction.

Rate = = Rate law = k [A] [B]2

(ii) When the concentration of B is tripled the rate of the reaction increased by 9 times. (iii) When the concentration of A and B are doubled the rate increases 8 times. rate α [A] [B]2

rate α [2] [2]2 where (A) (B) is doubled. rate α€8. The above expression shows that the rate increases 8 times.

10. In a reaction between A and B the initial rate of reaction was

measured for different initial concentrations of A and B as given below: A/M 0.20 0.20 0.40 B/M 0.30 0.10 0.05 Rate/MS-1 5.07 × 10-5 5.07 × 10-5 1.43 × 10-4

What is the order of the reaction with respect to A & B ?

• Solution: Consider the second experiment in which ion concentration of B is changed therefore the order with respect to it is calculated by comparing with the first experiment. rate i α (A)x (B)y

rate (i) = k (0.20)x (0.30)y - (i) rate (ii) = k (0.20)x (0.10)y - (ii) dividing eq.(i) by eq. (ii)

=

= 1 = 3y

y = zero. Order w.r.t to (B) = zero |||rly the order w.r.t to A Considering the 3rd and 2nd experiment. rate 3 α K (A)x (B) rate 2 α€K (A)x (B)0

rate 3 = 1.43 × 10-4 = k (0.40)x -- (iii) rate 2 = 5.07 × 10-5 = k (0.20)x -- (iv) Dividing eq. 3 by 4

= 2.8 = 2x

x=1.52 order with respect A almost equal to 1 rate = R [A][B]0.



6

www.mathem

atic.in

09810144315

7

11. Reaction between NO2 and F2 to give NO2 F takes place by the following mechanism.

• Solution: The rate Law equation involves one molecule of NO2 and mole of F2 as the first step is slow step. The slow step is the rate determining step. Rate of reaction = K [NO2] [F2]

12. The following results have been obtained during the kinetic studies

of the reaction.

• Solution: To determine the rate law the order of the reaction is calculated. To find the order w. r. to A consider the expt (i) and (iv) respectively. Rate (iv) = K [A]x [B]y

2.40 x 10-2 = K [0.4]x [0.1]y …………………… (i) rate (i) = K [A]x [B]y

6.0 x 10-3 = K (0.1)x (0.1)y ……………………. (ii) divide eq. (i) by eq. (ii)

= 4 = 4x

x = 1 order with respect [A] = 1 To calculate order w.r.t B. Consider expt 3 and 2 respectively. Rate (iii) = K [A]x [B]y

2.88 x 10-1 = K (0.3)1 (0.4)y ……………….. (3) rate (ii) = K [A]x [B]y

7.2 x 10-2 = K [0.3]1 [0.2]y ………………. (4) Dividing eq (3) by (2) we get

=

4 = 2y

y = 2 The order w. r to B is 2 Rate Law = K [A]1 [B]2

Rate = K [A] [B]2



7

www.mathem

atic.in

09810144315

8

6.0 x 10-3 = K [0.1] [0.1]2

K = K = 6 M-2 S-1

13. The reaction between A and B is first order with respect to A and

zero order with respect to B. Fill in the blanks in the following table.

• Solution: I. Rate = K (A)1 (B)0

2 x 10-2 = R (0.1)1 (o.1)0

2 x 10-2 = R 0.1 x 1

K = K = 0.2 Sec-1

II. Experiment Concentration A = x Concentration B = 0.2 Initial Rate = 4 x 10-2 M min-1

Rate = K (A)1 (b)0

4 x 10-2 = 0.2 x x 1

x = = 0.2 The concentration of (A) = 0.2 III. Experiment Rate = R (A)1 (B)0

Rate = 0.2 x 0.4 x (0.4)0

= 0.2 x 0.4 x 1 = 0.08 M min-1

IV. Experiment Rate = K (A) (B)0

2 x 10-2 = 0.2 X x X 1 2 x 10-2 = 0.2 x

x = = 0.1 x = 1 Concentration (A) = 0.1 Experiment (A) / M B / M Initial rate / M min-1 I 0.1 0.1 2 x 10-2

II 0.2 0.2 4 x 10-2

III 0.4 0.4 8 X 10-2

IV 0.1 0.2 2 X 10-2



8

www.mathem

atic.in

09810144315

9

14. Calculate the half life of a first order reaction from their rate

constants given below. (a) 200s-1 (b) 2 min-1 (c) 4 year-1

• Solution:

t1/2 =

(a) t1/2 =

(b) t1/2 = = 0.3465 min

(c) t1/2 = = 0.17325 years.

15. The experimental data for decomposition of N205 ( 2N205 4NO2 +

O2) in gas phase at 318K are given below.

• Solution: (a) To plot (N2O5) against t. The plot of (N2O5) against time give a parabolic curve.

Initial rate = -d

= rate = 3.4 x 10-3 M s-1

rate constant K =

= K = 2.5 x 10-3

(b) Half life period of first order reaction

t1/2 =

t1/2 = (from the plot)



9

www.mathem

atic.in

09810144315

10

(c) The plot of log [N2O5] against t gives a straight line.

The reaction is first order as it gives a straight line.

t1/2 = = 4620 sec t1/2 = 4620 sec.

(d) Since the above graph is a straight line it is a first order reaction.

Hence the rate law is : rate =k[N2O5]

(e) k=

=4.53 10-4 s-1

(f) t1/2 = =1529 s.

16. The rate constant for first order reaction is 60 s-1. How much time

will it take to reduce the initial concentration of the reactant to thof its value.

• Solution: The first order rate equation

K

=

K = 60 sec

-1

t = ? a = 1



10

www.mathem

atic.in

09810144315

11

a – x = substituting the values

t =

=

=

= t = 4.62 X 10-2 sec.

17. The rates of most reactions double when their temperature is raised

from 298 K to 308 K.Calculate their activation energy.

• Solution: The relationship between rates at two different temperatures with activation energy is

=

log 2 =

0.3010 =

0.3010 =

Ea = = 52922.3 J Ea = 52.922 KJ.

18. The half life for radio active decay of C is 5730 Y. An

archaeological artifact contained wood had only 80% of the C found in a living tree. Estimate the age of the sample.

• Solution:

=

K =

=

K = 1.2 x 10-4

t= log

Let = 100%

Let = 80%

log



11

www.mathem

atic.in

09810144315

12

= log 1.25

= 0.0969 = 19191.6 0.0969 t = 1859.6 years.

19. During nuclear explosion, one of the products is 90Sr with half life of

28.1 y. If 1 µµµµ g of 90Sr was absorbed in the 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.

• Solution:

t1/2 =

K = K = 0.0246 y-1

First order rate reaction

K = log ………… (i) (Ao) = 1µ g [A] = ? t = 10 years K = 0.0246 eg (i) is rearranged as

= log

= log 1 - log [A] 0.1068 = - log A log A = - 0.1068. log A = T.8932 [A] = 0.7798 µ g 0.7798 is left 10 yrs II case

= log

= log 1 - log A 0.6409 = -logA log A = -0.6409

= .3591

A = 0.2287 ⇐ µ g

0.2287 µ€g is left after 60 years.

20. A first order reaction takes 40 min. for 30% decomposition.

Calculate t1/2.

• Solution:



12

www.mathem

atic.in

09810144315

13

t=40 min. ; [A0]=100; [A]=100-30=70

k=

=8.92 10-3 min-1

= 0.693/8.92 10-3

=77.7min.

21. For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained. Calculate the rate constant. t (sec) P (mm of

Hg)

0 35.0

360 54.0

720 63.0

• Solution:

Initial pressure 0 0

After time t p p

Total pressure after time t

and

substituting the value of p

, i.e.,

As decomposition of azoisopropane is a first order reaction



13

www.mathem

atic.in

09810144315

14

When

k= 2.175 10-3 s-1

22. The following data were obtained during the first order thermal

decomposition of SO2Cl2 at a constant volume.

SO2Cl2 (g) →→→→ SO2 (g) + Cl2(g)

Experiment Time/s-1 Total pressure/atm

1 0 0.5

2 100 0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.

• Solution:

When

When i.e.,

∴ p = 0.65 -

∴ pressure of at time

Rate

23. The rate constant for the decomposition of N2O5 at various

temperatures is given below: T/˚C 0 20 40 60 80



14

www.mathem

atic.in

09810144315

15

105x k/s-1 0.0787 1.70 25.7 178 2140

Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30˚C and 50˚C.

• Solution: To draw the plot of logK-versus 1/T,

T (K) 273 293 313 333 353

1/T 0.003663 0.003413 0.003195 0.003003 0.002833

log K -6.1040 -4.7696 -3.5900 -2.7496 -1.6996

• • Fig: Graph of log k v/s 1/T

• Slope = =

Therefore, activation energy Ea = = 17.689 R J mol-1.We have, log K = log A

+

Or log K = + log A Compare it with y = mx + c which is equation of line intercept form.

24. The rate constant for the decomposition of hydrocarbons is 2.418 x

10-5 s-1at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?



15

www.mathem

atic.in

09810144315

16

• Solution: K = 2.418 x 10-5s-1

Ea = 179.9 kJ/mol K = Ae - Ea/RT

log K = log A +

log A = log 0.0000241 + A = 3.9 x 1012 s-1.

25. Consider a certain reaction A →Products with k = 2.0 x 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.

• Solution: k = 2.0 x 10-2 s-1, t = 100 sec, a = 1.0 mol L-1

K = log

2.0 x 10-2 = x log

0.8684 = log

Antilog 0.8684 = log 1-x = o.135 Hence, concentration of A remaining after 100 S = 0.135M.

26. Sucrose decomposes in acid solution into glucose and fructose

according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

• Solution: t1/2 = 3.00 hours

For the first order reaction, t1/2 =

Or K = = = 0.231 hr-1

We have, K = log

0.231 = log

log = 0.8024



16

www.mathem

atic.in

09810144315

17

= Antilog 0.8024 = 6.345

= = 0.157M.

27. The decomposition of hydrocarbon follows the equation k = (4.5 x

1011s-1) e-28000K/T Calculate Ea.

• Solution: k = (4.5 x 1011s-1) e-28000K/T

log K = log A -

= 28000 K Ea = 28000 K x R = 28000 x 8.314 = 232792 J mol-1 = 232.792 kJ mol-1.

28. The rate constant for the first order decomposition of H2O2 is given

by the following equation: log k = 14.34 - 1.25 x 104K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

• Solution:

log K = log A - -

Comparing this equation with log K = 14.34 - gives

= 1.25 x 104 K Ea = 1.25 x 104 K x 2.303 x 8.314 J mol-1 K-1 = 239339 J mol-1 = 239.339 kJ mol-1.

When =256 min, k=

=4.50 10-5 s

-1

Substituting this value we get,

Log 4.5 10-5 =14.34-

T= 669 K



17

www.mathem

atic.in

09810144315

18

29. The decomposition of A into product has value of k as 4.5 x 103 s-1 at 10°°°° C and energy of activation 60 kJ mol-1.At what temperature would k be 1.5 x 104s-1?

• Solution: K1 = 4.5 x 103 s-1 T1 = 273 + 10 Ea = 60 kJ mol-1 T2 = ? K2 = 1.5 x 104s-1

We know that, log =

log =

log =

0.5229 = T2 = 297 K.

30. The time required for 10% completion of a first order reaction at

298K is equal to that required for its 25% completion at 308K. If the value of A is 4 x 1010s-1. calculate k at 318K and Ea.

• Solution:

(i) t10% = log = log

(ii) t25% = log = log According to the problem, t10% = t25%

Therefore, log = log

x [log 10 – log 9] = [log 4 – log 3]

x [1– 0.9542] = [0.6021 – 0.4771]

= = 2.7

log =

log 2.7 =



18

www.mathem

atic.in

09810144315

19

0.4314 =

Ea = = 75.847 kJ mol-1. Calculation of k :- log k=logA+eE

a/RT

=log4 1010 -

k =1.042 10-2 s-1

31. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

• Solution: We have, according to Arrhenius equation,

log =

log =

0.6021 =

Ea = = 5.2863 x 104 J mol-1

Ea = 52.863 kJ mol-1.

32. For the reaction R P, the concentration of a reactant changes from

0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

• Solution: Average rate in mins. = { - ∆[P] } = ½ x [-(0.02 – 0.03)] ∆t 25 min.

= 0.01/25 = 0.0004 = 4 x 10-4mol L-1min-1

Average rate in secs. = 0.01 25x60

= 6.66 x 10-6mol L-1sec-1



19

www.mathem

atic.in

09810144315

20

33. In a reaction, 2A Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval?

• Solution: Rate in mins. = ½ x { - ∆[A] } = ½ x [-(0.4 – 0.5)] ∆t 10 min.

= 0.005mol L-1min-1

34. For a reaction, A + B Product; the rate law is given by, r = k [

A]1/2[B]2.What is the order of the eaction?

• Solution: The order of the reaction is = ½ +2 = 2.5

35. The conversion of molecules X to Y follows second order kinetics. If

concentration of X is increased to three times how will it affect the rate of formation of Y ?

• Solution: For this reaction, Rate = k[3X]2 = 9k[X]2 If the concentration of X is increased three times, the rate of the reaction increases nine times the actual rate.

36. A first order reaction has a rate constant 1.15 × 10-3 s-1. How long

will 5 g of this reactant take to reduce to 3 g?

• Solution: k = 1.15 × 10-3 s-1

t1/2 = log

= log

= (log5 - log3)

= (0.6990 - 0.4771) = 444sec. t1/2 = 444sec.

37. Time required to decompose SO2Cl2 to half of its initial amount is 60

minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.



20

www.mathem

atic.in

09810144315

21

• Solution: K = 0.693 = 0.693 = 1.925 x 10-4s-1

t1/2 0x60

38. What will be the effect of temperature on rate constant?

• Solution: The rate constant of a reaction depends on the temperature. It increases with temperature. It can be explained as follows:

Each reaction rate coefficient k has a temperature dependency, which is usually given by the Arrhenius equation:

Ea is the activation energy and R is the gas constant. Since at temperature T the molecules have energies given by a Boltzmann distribution, one can expect the number of collisions with energy

greater than Ea to be proportional to . A is the pre-exponential factor or frequency factor.The values for A and Ea are dependent on the reaction. There are also more complex equations possible, which describe temperature dependence of other rate constants.

39. The rate of the chemical reaction doubles for an increase of 10K in

absolute temperature from 298K. Calculate Ea.

• Solution: We have,

log =

= 2, T1 = 298K,T2 = 298+10 = 308 K, R = 8.314JK-1mol-1

log2 =

0.3010 = x Ea = 52.89 kJ mol-1.

40. The activation energy for the reaction 2 HI(g) H2 + I2 (g) is 209.5 kJ

mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

• Solution: 2 HI(g) → H2 + I2 (g) Ea = 209.5kJ mol-1



21

www.mathem

atic.in

09810144315

22

T = 581K

The fraction of molecules of reactants having energy equal to or greater than activation energy = 1.462 x 10 –19.



22

www.mathem

atic.in

09810144315

Date post:	31-Oct-2019
Category:	Documents
Upload:	others
View:	0 times
Download:	0 times

1. From the rate expression for the following reactions ... 12TH/CLASS 12TH CHEMIS…1. From the...

Documents