Solved Problems (Subjective) Problem 1: A (half life = 12.8 h) transforms to B (35%), C (19%) and D (43%) each by 1st order

kinetics as shown below.

Solution: Let the rate constants of the above processes be k1, k2 and k3, respectively and the

overall rate constant be k. Then

Also, k1 = 0.38 k =

Similarly,

= 67.36h

Where t1, t2 and t3 are the partial half-lives for the conversion of A and B, C and D, respectively, by 1st order kinetics.

Problem 2: The half-life of Pb212 is 10.6 h. It undergoes decay to its daughter (radioactive)

element Bi212 of half-life 60.5 minute. Calculate the times at which daughter element will have the maximum activity?

Solution:

= 227.1 min

Problem 3: t1/2 of a first order reaction at 27°C is 100 min. 10–6 % of the reactant molecules exists

in the activated state at 27°C calculate maximum rate constant of the reaction.

Solution:

=

k = Ae–Ea/RT

A = = 6.93 107 min–1

When Thus A is the maximum rate constant of the reaction. Problem 4: The half life of P32 is 14.3 days. Calculate the specific activity of a phosphorus

containing specimen having1.0 ppm P32 (Atomic weight of P = 31) Solution: Activity of 1g sample is the specific activity of the sample 106g contains 1g P32

1g sample will contain 10–6 P32 i.e. atoms of P32

=

= 1.089 1010 dps Since this is activity i.e. rate of decay of 1g sample Specific activity = 1.089 1010 dps per gram

= = 0.295 CI per g

Problem 5: The gas phase decomposition of dimethyl ether follows first order kinetics. CH3OCH3(g) CH4(g_ + H2(g) + CO(g) The reaction is carried out in a constant volume container at 500°C. Initially only

dimethyl ether is present at a pressure of 0.40 atm. The total pressure of the gaseous reaction mixture after 12 minute is 0.75 atm. Calculate half-life of the reaction.

Solution: CH3OCH3(g) CH4(g_ + H2(g) + CO(g) P0 0 0 0 …t = 0 P0 – x x x x …t = 12 minute From question P0 = 0.40 atm Total pressure after 12 minute = P0 = x + x + x + x = 0.75 atm or P0 + 2x = 0.75 x = 0.175 atm At constant volume mole pressure

So,

=

= 0.0475 min-1

Problem 6: The rate of a firs order reaction is 0.8 mol L–1 min–1 after 10 min and 0.1 after 30 min

from the start of the reaction. Find half time of the reaction. Solution: Rate Cn, where n is the order of reaction For a first order reaction: N = 1 so

Problem 7: The half-life period of C14 is 5760year. An ancient piece of wood has -activity of

4.4 atm per gram carbon. A living tree has a -activity of 17.6 dpm per gram carbon. A dead coal when examined in the same Gniger-Muller instrument, it was found to have an activity of 1.1 compounds per gram carbon. Calculate how long before the tree was cut.

Solution:

= 13380.13 years

Problem 8: The isotpe U238- and U235 occur in nature in the ratio of 140 : 1. Assuming that at the

time of earth formation, they were present in equal ratio, make an estimation of the age of the earth. The half life period of U238 and U235 as 4.5 109 and 7.13 108 year respectively.

Solution: In nature

At the time of earth formation

If and be the decay constants of U238 and U235 respectively, then

Putting the value of and , we get T = 6.04 109 year

Problem 9: Rate law for the following reaction; is Ester + H+ Acid + Alcohol; is

[ester]1 [H+]0

What would be the effect on the rate if i) Concentration of ester is doubled ? ii) concentration of H+ ion is doubled ?

Solution: The rate law expression in this question, suggests that concentration of acid is nothing to play with velocity.

i) When concentration of ester is doubled; velocity of the reaction will become double.

ii) When concentration of H+ ion is doubled velocity will be unaffected.

Problem 10: The reaction 2A + B + C D + 2E; is found to be first order in A; second order in B and zero order in C.

i) Give the rate law for the above reaction in the form of a differential equation. ii) What is the effect on the rate of increasing the concentration of A, B and C two

times?

Solution: i) The rate law according to given information may be given as,

ii) When concentration of A, B and C are doubled then rate will be

i.e., rate becomes 8 fold, the original rate.

Problem 11: At 27°C it was observed, during a reaction of hydrogenation that the pressure of H2 gas decreases from 2 atm to 1.1 atm in 75 min. Calculate the rate of reaction (molarity/sec). Given (R = 0.082/litre atom K–1 mole–1)

Solution: Rate =

atm/min

Rate in atm/sec atm/sec.

Answer is required in molarity per second PV = nRT

sec = molarity per second

. Rate in molarity/sec = (n/V)/sec

Problem 12: In presence of an acid N-chloro acetanilide changes slowly into p-chloro acetanilide. Former substance liberated iodine from KI and not the later and hence progress of reaction can be measured by titrating iodine liberated with Na2S2O3 solution, the results obtained were as follows :

Time (hours) 0 1 2 4 6 8 (a–x)i.e., hypo 45 32 22.5 11.3 3.7 2.9 show that reaction is uniomlecular and find out the fraction of N-chloroacetanilide

decomposed after three hours.

Solution: The present reaction is —

Let us apply the kinetics of first order reaction.

After 1 hours = 0.34098 hour–1

After 2 hours = 0.34660 hour–1

After 4 hours = 0.3455 hour–1

After 6 hours = 0.4164 hour–1

After 8 hours = 0.3428 hour–1

Average value of constant = 0.3584 hours–1

Since, on applying first order kinetics, we get almost same values of rate constant after different time intervals, hence, the reaction is of first order.

Let us see the fraction decomposed after 3 hours.

0.3584 =

[fraction decomposed]

Problem 13: For the decomposition of dimethyl ether, A in the Arrhenious equation K = Ae–E/RT has a value of 1.26 1013 and Ea value of 58.5 kcal. Calculate half life period for first order decomposition at 527°C.

Solution: Taking logarithm of Arrheniuos equation K = Ae–E/RT we get

log K = log A – … (i)

Given A = 1.26 1013 E = 58.5 kcal T = 527 + 273 = 800 K Substituting these value in Eq. (i), we get

log K = log (1.26 1013) –

= 13.1003 – 15.9799 = –2.8796 K = 1.3194 10–3 sec–1

sec = 525 sec

Problem 14: For the reaction — 2NO(g) + H2(g) N2O(g) + H2O(g) at 900 K, the following data are obtained : Initial pressure Initial pressure Rate of NO (atm) of H2(atm) (atm min–1) 0.150 0.400 0.020 0.075 0.400 0.005 0.150 0.200 0.010 Find the rate law and the value of rate constant.

Solution: Let order with respect to NO(g) is `m’ and order with respect to H2(g) is `n’ Then, Rate = … (1) 0.020 = K [0.15]m [0.40]n … (2) 0.005 = K [0.075]m [0.40]n … (3) 0.010 = K [0.15]m [0.2]n … (4) Dividing Eq. (2) by (3), we get

m = 2 Dividing Eq. (2) by (4), we get

n = 1 Rate = Substituting the values of m, n in Eq. (2), we get 0.020 = K (0.15)2 [0.40] K = 2.22 atm–2 min–1

Problem 15: 10 gram atoms of an -active radio isotope are disintegrating in a sealed container. In one hour the helium gas collected at STP is 11.2 cm3. Calculate the half-life of the radio-isotope.

Solution: No. of atoms of helium is 11.2 cc at NTP

= 6.02 1023 = 3.01 1020 atoms

Since, helium atom corresponds to -particle. Thus, Rate of disintegration = 3.01 1020 per hour. We know, Rate = Rate constant concentration in atom 3.01 1020 = K 10 6.02 1023 K = 0.05 10–3 hour–1

= 13860 hours

years

Problem 16: A carbon radio isotope ZXA (half life 10 days) decays to give Z–2YA–4. If 1.00 gm atom of ZXA is kept in a sealed tube, how much helium will accumulate in 20 days ? Express the result in cm3 at STP.

Solution: Initial concentration (N0) of radio-isotope is 1 gm atom. Concentration remained after 20 days may be calculated as

Nt = N0

where n = n0 of half lives = 20/10 = 2 =

Concentration decayed to -particles = 1 – gm atom. An -particle takes 2

electron from air and from helium gas. Thus,

Helium formed gm atom cc cc

Problem 17: Radioactive decay is a first order process. Radioactive carbon wood sample decays with a half life of 5770 years. What is the rate constant in (years)–1 for the decay? What fraction would remain after 11540 years?

Solution:

= 1.201 10–4 year–1

1.201 10–4

4.002 =

(Remaining fraction) =

Problem 18: A sample of 53I131, as iodide ion, was administered to a patient in a carrier consisting of 0.10 mg of stable iodide ion. After 4.00 days, 67.7% of the initial radioactivity was detected in the thyroid gland of the patient. What mass of the stable iodide ion had migrated to the thyroid gland? Given T1/2 I131 = 8 days.

Solution: We know,

70.7% of initial activity is present. Given that 67.7% activity is migrated to thyroid gland.

Thus, weight of I– migrated to thyroid gland may be calculated as

100 = 95.75%

i.e, 0.1 = 0.09575 mg

Solved Problems (Objective) Problem 1: The rate constant of a reaction: 3P, is for the rate of

consumption of A at 27°C. The initial rate of formation of P when concentration of A is 1.5 2/3 M,is:

(a) 1.67 10–3 mol L–1 min–1 (b) 2.5 10–3 mole L–1 min–1 (c) 2.5 10–3 mole L—1 min–1 (d) None of these Solution: The unit of K indicates that it is a 1st order reaction

=

(b) Problem 2: A catalyst lowers Ea of a reaction by 25%. The temperature at which rate of

uncatalysed reaction will be equal to that of the catalysed reaction at 27°C under the same condition of concentration of reactant is equal to

(a) 127°C (b) –48°C (c) 27°C (d) None of these Solution: , Let k at T1 is equal to kcat at T2, then

(a)

Problem 3: If a reaction A + B → C is exothermic to the extent of 30 kJ/mol and the forward reaction has an activation energy 70 kJ/mol, the activation energy for the reverse reaction is

(a) 30 kJ/mol (b) 40kJ/mol (c) 70 kJ/mol (d) 100 kJ/mol

Solution:

By seeing the curve, activation energy for backward reaction = 100 kJ ∴ (d) Problem 4: The progress of a 1st order reaction:

3B carried out at 27°C is shown below graphically. Half time of the reaction at 27° is nearly

(a) 20 min (b) 40 min (c) 49 min (d) None of these

Solution: a 0 t = 0 a – x 3x t = t From curve: a – x – 3x when t = 20 min a = 4x

Upon solving: t1/2 – 48.94 min 49 min (c) Problem 5: When a reaction is carried out taking 4M concentration of the reactant, the half-time

of the reaction is 20 min. When the same reaction is carried out taking 1 M concentration of the reactant, in concentration of the reactant is reduced to 0.25 M in 4 hours. Hence order of reaction is

(a) one (b) two (c) half (d) zero

Solution: For a second order reaction:

(b)

Problem 6: A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to 10 kJ mole–1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C is

(a) –123°C (b) 327°C (c) 327°C (d) + 23°C

Solution:

∴T2 = 600 K = 327° C ∴ (b) Problem 7: Another first order reaction B is 75% complete in 40 min at the same temperature. If

these two reactions are carried out simultaneously at 27°C taking equal initial concentration of the reactants A and B in each reaction, the ratio of concentration of to that of B after 60 minute will be

(a) 1:2 (b) 1:1 (c) 1:3 (d) 2:1 Solution: For the reaction:

For the reaction:

(d) Problem 8: In a first order reaction 10–6% of the reactant molecule are able to cross-over the

potential energy barrier at 27°C. The Ea (kJmol–1) of the reaction will be nearly (a) 100 (b) 45 (c) 50 (d) 20 Solution: = fraction of the reactant molecules crossing over the barrier at temperature T

So,

(b) Problem 9: The incorrect statement among the following is: (a) Order of reaction is an experimental property (b) Order of reaction is a fixed property i.e. independent of experimental conditions

of the reaction. (c) Moleculartiy is a theoretical property and it concerns with mechanism. (d) Order of reaction may be any number integral, fractional and zero. Solution: The reaction

is a first order at low gas and 2nd order at high gas pressure. (b) Problem 10: The conc. vs time at curve of a reaction

is an given below. The order of reaction is

(a) zero (b) one (c) two (d) half

Solution: The curve shows that

Thus,

i.e. order = 2 (c)

Problem 11: The reaction A(g) + 2B(g) C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to the initial rate is

(a) 1/48 (b) 1/24 (c) 9/16 (d)1/6

Solution: After reaction A + 2B C + D 0.6 – 0.2 0.8 – 0.4 0.2 0.2 0.4 0.4 0.2 0.2

∴ (d)

Problem 12: Thermal decomposition of a compound is of first order. If 50% of a sample of the compound is decomposed in 120 minute show long will it take for 90% of the compound to decompose?

(a) 399 min (b) 410 min (c) 250 min (d) 120 min

Solution: …(1)

= …(2)

Equating (1) and (2)

t = 399 minutes ∴ (a)

Problem 13: If concentration are measured in mole/litre and time in minutes, the unit for the rate constant of a 3rd order reaction are

(a) mol lit–1min–1 (b) lit2 mol–2 min–1

(c) lit mol–1min–1 (d) min–1

Solution: K = [conc.]1–n min–1 For 3rd order reaction = [mole/litre]1–3 min–1 = lit2.mole–2min–1 ∴ (b)

Problem 14: What is the activation energy for the decomposition of N2O5 as

N2O5 2NO2 + O2

If the values of the rate constants are 3.45 × 10–5 and 6.9 × 10–3 at 27°C and 67°C respectively.

(a) 102 × 102 kJ (b) 488.5 kJ (c) 112 kJ (d) 112.5 kJ

Solution:

Ea = 112.5 kJ ∴ (d) Problem 15: The kinetic data for the reaction: 2A + B2 2AB are as given below: [A] [B2] Rate mol L–1 molL–1 molL–1 min–1 0.5 1.0 2.5 10–3 1.0 1.0 5.1 10–3 0.5 2.0 1 10–2 Hence the order of reaction w.r.t A and B2 an, respectively (a) 1 and 2 (b) 2 and 1 (c) 1 and 1 (d) 2 and 2 Solution: Data suggest that when the conc. of A is doubled keeping the conc. of B2 constant,

the rate is just doubled and when conc. of B2 is doubled keeping the conc. of A constant, the rate increases 4 times i.e.

Rate [A], when [B2] is constant and Rate [B2]2, when [A] is constant Hence order w.r.t. A and B2 are 1 and 2, respectively

Problem 16: For the first order reaction: A B, the conc. vs time curve is as

given below: Hence, the rate of constant of the reaction

is as follows: (a) 0.0231 min–1 (b) 0.033 min–1 (c) 0.0693 min–1 (d) None of these Solution: When t = 30 min, CA = CB, so t1/2 = 80 minutes

= 0.0231 min–1

(a) Problem 17: Lead remains always present in uranium ore due to radio active decay of U238 (t1/2 =

109 years) to lead – 2906 (the enol product). In an uranium mineral, U238 to Pb206 mole rate was found to be 1:3. Hence age of the mineral is:

(a) 2.25 109 years (b) 1.5 109 years (c) 9.0 109 years (d) 1.25 109 years Solution: U238 : Pb206 :: 1:3 Massof U238 remains (1/4th) of original compound

N =

n = 2 age T =

(c) Problem 18: During the fission of U235, energy of the order of 180 MeV is generated per nucleus

fissioned. The amount of energy released by the fission of 0.235g of U235 is: (a) 1.73 107 kJ (b) 1.08 1025 kJ (c) 1.73 1016 kJ (d) 1.08 107 kJ

Solution: No. of nuclei of U235 undergoing fission =

= Energy released =

= =

(a)

MCQ for practice Level 1 1. For the 1st order reaction A(g) 2B(g)

- + C(s) t1/2 = 24 minute the reaction is carried out taking certain mass of A enclosed in a vessel in which it exerts a pressure of 400 mm Hg. The pressure of the reaction mixture after expiry of 48 minutes will be

(a) 700mm (b) 600mm (c) 800mm (d) 1000mm 2. The ratio of the rate constant of a reaction at any temperature T to the rate constant is

equal to (a) Energy of activation of the reaction (b) Fraction of molecules in the activated state (c) Average life of the reaction (d) Pre-exponential factor in the Arrhenius equation 3. The rate constant of a reaction: at 27°C is at 27°C and at this

temperature per cent of the reactant molecules are able to cross-over the P.E. barrier. The maximum rate constant of the reaction is:

(a) (b) (c) 0.2 s–1 (d) 20s–1 4. For a first order reaction: A B, whose

concentration vs. time curve is as shown in the figure. The rate constant is equal to

(a) 41.58 h–1 (b) 4.158 s–1 (c) (d) 6.93 min–1

5. The rate of reaction: is doubled when concentration of A is increased 4-fold.

If half-time of the reaction with 0.2 M concentration of A is 25 minutes then that for 0.8 M concentration of A the half time of the reaction will be

(a) 12.50 min (b) 50 min (c) 5 min (d) 625 min 6. The incorrect statement amongst the following about order of reaction is: (a) It is an experimental property (b) It concerned with kinetics (c) It may be any number including zero (d) It is a fixed property which does not change in any case.

7. If of a reaction be positive and k1 and k2 be the rate constants of forward reaction and backward reaction, respectively, at temperature t°C and be the respective rate constants at (t + 10)°C then

(a) (b)

(c) (d) None of these

8. For which of the following reactions, the average life will be the minimum? (a) (b)

(c) (d) PCl5 PCl3 + Cl2 9. A decomposes following two parallel first order mechanisms

the rate of disappearance of A i.e. taking 1 M concentration of A is equal to

(a) (b) (c) (d) None of these 10. Among equilibrium constant (K), rate constant and Ea the one or more than one that is

(are) not affected by catalyst is (one) (a) Rate constant and (b) K and Ea (c) and K (d) Only K Answers 1. a 2. b

3. d 4. c

5. b 6. d

7. c 8. b

9. a 10. c

Level 2

1. The decomposition of nitrogen pentoxide can be represented as

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

The rate of the reaction can be expressed as

(a)

(b)

(c)

(d)

2. The instantaneous rate of disappearance of the Mn ion in the following reaction is 4.56 10–3 Ms–1

2Mn + 10I– + 16H+ 2Mn2+ + 5I2 + 8H2O The rate of appearance of I2 is (a) 1.14 10–3 M s–1 (b) 5.7 10–3 M s–1 (c) 4.56 10–4 M s–1 (d) 1.14 10–2 M s–1

3. The rate constant for the reaction 2N2O5 → 4NO2 + O2 is 3 × 10–5s–1. If the rate at a given time is 2.40 × 10–5 mol L–1 s–1, then concentration of N2O5

at that time is: (a) 1.4 (b) 1.2 (c) 0.04 (d) 0.8

4. The rate expression for the reaction A(g) + B(g) C(g) is rate = . What changes in the initial concentrations of A and B will cause the rate of reaction to increase by a factor of eight ?

(a) CA 2; CB 2 (b) CA 2; CB 4 (c) CA 1; CB 4 (d) CA 4; CB 1

5. Calculate the half-life of the first-order reaction C2H4O(g) CH4(g) + CO(g)

if the initial pressure of C2H4O(g) is 80 mm and the total pressure at the end of 20 minutes is 120 mm.

(a) 40 min (b) 120 min (c) 20 min (d) 80 min 6. Rate constant of a first order reaction is 0.0693 min–1. If we start with 20 mol L–1, it is

reduced to 2.5 mol L–1 in: (a) 10 min (b) 20 min (c) 30 min (d) 40 min 7. The half life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/litre and

concentration at some instant is 0.01 mol/litre, then t = (a) 10 minutes (b) 30 minutes (c) 20 minutes (d) 40 minutes

8. In the Arrhenius equation k = A exp (–EA/RT), the rate constant (a) Decreases with increasing activation energy and increases with temperature (b) Increases with decrease of activation energy and increase of temperature (c) Decreases with decrease of activation energy and increase of temperature (d) Increases with decrease of activation energy and decreasing temperature

9. Rate of formation of SO3 in the following reaction 2SO2 + O2 → 2SO3 is 100 g min–1. Hence, rate of disappearance of O2 is: (a) 50 g min–1 (b) 100 g min–1 (c) 200 g min–1 (d) 40g min–1

10. For a reaction A(g) + 2B(g) → C(g) + D(g)

Initial pressure of A and B are respectively 0.60 atm and 0.80 atm. At a time when pressure of C is 0.20 atm, rate of the reaction, relative to the initial value is:

(a) (b)

(c) (d)

11. In the first order reaction the concentration of reactant decreases from 2 M to 0.50 M in 20 minutes. The value of specific rate is

(a) 69.32 (b) 6.932 (c) 0.6932 (d) 0.06932

12. The rate of a gaseous reaction is given by the expression K[A]2[B]3. If the volume of reaction vessel is suddenly reduced to one half of the initial volume. The reaction rate relative to the original rate will be

(a) 1/24 (b) 1/32 (c) 32 (d) 24

13. The half life period t1/2 is independent of initial concentration of reactant when the order of

reaction is (a) Negative (b) 0 (c) 1 (d) Fractional

14. A reaction X2 + Y2 → 2XY occurs by the following mechanism X2 X + X … (slow) X + Y2 XY + Y … (fast) X + Y XY … (fast) The order of the overall reactions (a) 2 (b) 1

(c) 1 (d) zero

15. The graph between log K and [K is rate constant (s–1) and T the

temp. (K)] is a straight line with OX = 5 and = tan–1 .

Hence Ea will be

(a) 2.303 2 cal (b)

(c) 2 cal (d) None of these

16. For a chemical reaction A B, the rate of reaction increases by a factor of 1.837 when the concentration of A is increased by 1.5 times. The order of reaction with respect to A is

(a) 1 (b) 1.5 (c) 2 (d) –1

17. Rate of the chemical reaction : nA products, is doubled when the concentration of A is increased four times. If the half time of the reaction at any temperature is 16 min. then time required for 75% of the reaction to complete is

(a) 24.0 min. (b) 27.3 min. (c) 48 min. (d) 49.4 min.

18. The mechanism of the reaction A + 2B + C D is Step – 1 A + B X (Fast) equilibrium Step-2 X + C Y (Slow) Step-3 Y + B D (Fast) which rate law is correct (a) R = K[C] (b) R = K[A][B2][C] (c) R = K[A][B][C] (d) R = K[D]

19. For a first order reaction the ratio of t0.75 to t0.50 would be (a) 4 : 3 (b) 3 : 2 (c) 2 : 1 (d) 1 : 2 20. Two substances A(t1/2 = 5 min.) and B(t1/2 = 15 min.) are taken in such away that initially

[A]0 = 4[B]0. The time after which both the concentrations will be equal is (a) 5 min. (b) 15 min. (c) 20 min. (d) Concentrations can never be equal Answers 1. (d) 2. (d)

3. (d) 4. (b)

5. (c) 6. (c)

7. (b) 8. (b)

9. (d) 10. (a)

11. (d) 12. (c)

13. (c) 14. (b)

15. (c) 16. (b)

17. (b) 18. (c)

19. (c) 20. (b)