CHEMICAL KINETICS

Solution to Subjective Problems 

 

LEVEL – I

1. The one existing for long time will cause serious effect.

2. Use concepts of complex reactions.
3. Arrhenius equation.
4. Calculate‘t’.
5. Use first order kinetics concepts.
6. Calculate‘t’.
7. Use pressure concepts in first order kinetics.

LEVEL – II

1. Calculate no. of He particles and proceed.

4. Write the expression and then calculate (a – x).
5. Compare both the isotopes’ rate and proceed.
6. Activity α vol. of blood.
7. Find a relation between the total rotation and individual rotations.

LEVEL – III

4. Calculate initial nuclei of and proceed.

5. Calculate initial moles of Cl238 and proceed.
6. Find the relation between total rotation and individual rotations from the graph & find log A.
7. Use concepts of complex reactions.
8. Each Po nucleus gives one mole of He.
9. Use concepts of complex reactions.
10. Volume α activity.

Solution to Subjective Problems 

 

LEVEL – I

1. 2.303 log =

2.303 log =

K2 = 9.38 × 10–3 sec–1

For I131 time required to fall its activity to 1%

t = = 53.17 days

For Sr90, time required to fall its activity to 1% be, = = 59.8 days

Thus 90Sr which takes longer time to fall its activity to 1% is to be serious

2. Time of maximum activity tmax =

λd ⇒ disintegration constant of daughter element

λp ⇒ disintegration constant of parent elements

λd = = 0.01145 min–1

λp = = 0.00109 min–1

tmax =

tmax = 227 min

 

3. Since the plot of logarithms of partial pressure versus time is linear, hence the reaction is of first order

The equation for first reaction is

or t = = –

This is a straight line equation (y = mx + c) hence its slope is equal to – 2.303 / k from which the specific rate reaction can be calculated. The intercept of this equation log a: hence the value of ‘a’ can also be calculated by knowing the value of k.

4. i) A nB

The initial concentration of A = 0.6 M

The equilibrium concentration of A = 0.3 M

Hence concentration of A lost = 0.3 M

If x is the concentration of A lost, then more is the concentration, B obtained

Hence, B = nx = n × 0.3 M

B is 0.6 M (from the graph)

0.6 M = n × 0.3 M

∴ n = 2

1. ii) K = = 1.2

iii) When t = 0, [A] = 0.6 M

t = 1 hr, [A] = 0.5 M

∴ Rate of conversion = 0.5 – 0.6 = – 0.1 M hr–1

5. Given t1/2 = 28.1 years N0 = 10-6 g, t = 20 yrs

t =   =

N  = 6.1 x 10-7g 

6. t = = = 491.16 year

∴ the painting was done in the year (1999-491) = 1508 which was within the life span of Raphael (1483-1520)

∴ It was not a forgery

7. a) Rate =
8. b) Rate =

Change in molarity =

8. Rate = k [NO]x [Cl2]y where x = order w.r.t. NO

i.e. R = k [NO]x [Cl2]y y = Order w.r.t. Cl2

Case (1) 8R = k [2NO]x [2Cl2]y

Case (2) 2R = k [NO] x [2Cl2]y

Dividing the equation we have

4 = 2x ∴ x = 2

Similarly 2R = k [NO] x [2Cl2]y

  R = k [NO] x [Cl2]y

∴ 2 = 2y ⇒ y =1

∴Overall order = x + y = 2 + 1 = 3

9. Pb present  =

U present =

No = U present + U decayed

= 0.1 + 0.1= 0.2

and N = 0.1

∴ t = =

  t = 4.5 × 109  years

10. i) k = =

= 6.2 × 10–3 s–1

1. ii) Again t = 100 s, Pt = ?

6.2 × 10–3 = log or Pt = 0.033 atm

 LEVEL- II

1. No. of α-particles (or) He formed = 2.24 × 1013 min–1

∴ No. of He particles formed in 420 days = 2.24 × 1013 × 420 × 1440 = 1.355 × 1019

Also at 27°C and 750 mm; He = 0.5 mL

Using PV = nRT

= n × 0.0821 × 300 ⇒ n = 2.0 × 10–5 moles

2.0 × 10–5 moles of He = 1.355 × 1019 particles of He

⇒ 1 mole of He = = 6.775 × 1023 particles

∴ Avogadro’s number = 6.775 × 1023 particles/mol

2. Let the order of the reaction is ‘n’

∴ = ;     =

or,   = logK + nlogC1; log = log K + n logC2

∴ n =

n = = = 1.128

So order of the reaction is 1.128

3.

Rearranging and taking indefinite integration within the limit, 

When t → 0, then y →0 & t →, t, the y → y

Then this equation

⇒

⇒

When t → t1/2, then y → C0/2

∴

⇒ proved

4. [N2O5] =

K = 1.68 × 10–2

t = 1 min = 60 sec

[A]t = ?

K =

1.62 × 10–2 =

on solving, n = 0.185M

= 0.9215 mol.

decomposed  = (2.5 – 0.9215) mol = 1.5875 mol

decomposed  = mol = 0.7938 mol

5. Now,

At the time of earth formation,

For X193, λ1 =

For X192, λ2 =

If ‘t’ is the age of earth

∴ λ1 – λ2 = = log

or, =

∴ t = 32.56 × 107 years

6. Let volume of the blood in the animal = V mL

total activity before mixing = 1.8 × 106 cps

total activity after mixing in V mL = × V mL

= 0.6 × 104 V mL

Thus     0.6 × 104 V = 1.8 × 106 (assuming activity remains constant)

V = = 300 mL

7. We know,

log

Ea = energy of activation

Given: Ea = 0

∴ = 0

or log = 0

or = 1

Hence K300  = K280

Or K300 = 1.6 × 106 s–1

8. K =

Total moles of sucrose =

Sucrose     ⎯→      Glucose   +        Fructose

t = 0 0.38r10 0 0

t = 10 min 0.38 r10 –αr10 αr20 –αr30

t = ∞ 0.38 r20 –0.38 r30

Total rotation at t∞ = –4.7120

K = =

K = 0.0113 min–1 , t1/2 = = 61.16 min

9. For first order reactions,

t =

At 298 k; t =

At 309 K; t =

Since time is the same hence,

or or = 2.73

According to Arrhenius equation 

2.303 log

or 2.303 log 2.73 = = 76.65 kJ

and  K318 = 9.036 × 10–4 s–1

10. According to kinetic equation of first order 

K =

a = 5.0 mg/ml (a-x) = 4.2 mg/ml, t = 20 months 

K = log =

K = 8.728 × 10–3  months–1

Suppose t months are sufficient for its 30% decomposition, therefore

8.728 × 10–3 =

t =  log = 263.86 × 0.1549  = 40.87 months

≈ 41 months

Half life period  = = 79.39 months

≈ 80 months 

LEVEL – III

1. At t = 20 min, [R] = [P] i.e. half of the concentration of R has got consumed during this time interval. So, t1/2 = 20 min

K = = 0.03465 min–1

i)

= 2.303 log (3.8 × 10–18)

Ea = 2.303 × 8.314 × 300 × 17.420 = 100.0 kJ mol–1

1. ii) k =

when T →∞, k → A. Thus A is the maximum rate constant of the reaction i.e. the rate constant at temperature approaching infinity.

Putting the value of k and we have

0.03465 = A × 3.8 × 10–18

A = = 9.11 × 1015 min–1

iii)

= 0.561

= antilog (0.561) = 3.65

Thus, the rate will increase 3.65 fold

Alternatively, rate constant at 37°C may also be calculated using the equation

K =

And rate constant at 27°C being known, the ratio may thus be found out.

iv)

log 104=

Ea – EaC= 22.96

EaC = 100 – 22.96 = 77.04 kJ mol–1

(Kcat = Rate constant of catalysed reaction, k  = rate constant of uncatalysed reaction, EaC = energy of activation of the catalysed reaction and Ea = energy of activation of the uncatalysed reaction).

2. i) Let the rate law be

Rate = k[A]m [B]n

Where m and n are the order w.r.t. A and B respectively, and k = rate constant of the reaction, a constant at constant temperature.

From (a): 5.0 × 10–4 = k(2.5 × 10–4)m (3.0 × 10–5)n

From (b): 4.0 × 10–3 = k(5.0 × 10–4)m (6.0 × 10–5)n

Dividing one by another

or 2m⋅2n = 8

From (c): 1.6 × 10–2 = k(1 × 10–3)m (6 × 10–5)n

Dividing this by the rate law from (b), we get

or 2m = 4 ∴ m = 2

Putting the value of m in the above equation

22 ⋅ 2n = 8, or n = 1

1. ii) Rate = k[A]2 [B]

∴ k = = 2.67 4 × 108 L2 mol–2 s–1

iii)

=

upon solving: Ea = 55333 J mol–1 = 55.333 kJ mol–1

1. iv) k =

or 2.67 × 108 = A exp or A = 1.145 × 1018

3. Let T1 = 300 K and T2 = 310 k

we know that,

log

For (i) reaction

log

or log 2 = —— (i)

For (ii) reaction 

log

Since  =

Here k2 = 2

k2 = 0.0462 min–1 and Ea = EA

Therefore, log

From equation  (i) 

log = log2 = log or =

or k1 = = 0.327 min–1

4. 18 g H2O has 2H atom in it =  6.023 × 1023× 2 

18 g H2O has atoms = 8 ×10–18 × 6.023 ×1023 ×2

∴ 10 g H2O has atoms =

i.e No of = 5.354 × 106 atoms 

Now t =

40 =   = 5.624 ×105 atoms

5. 92U238 = 1.667 g = mole

82Pb206 = 0.227g = mole

 All the lead have come from decay of U.

∴ Moles of Pb formed = 

∴ moles of U decayed =

∴ Total moles of Uranium = + , ie N0

Also N for U238 =

 for U238 t =   

= log 

t = 1.143 × 109 years

6. A + H2O ⎯⎯→ 2B + C

a excess 0 0

(a-x) const. 2x 2x

0 const. 2a a

After completion of reaction 50 × 29 + (–80) × a = 10, ∴ a = 0.5

After 40 minute. 60(a-x) + 50 × 2x – 80 × x = 26 ⇒ x = 0.1

K = log 5/4

∴ K = 5.57 × 10–3 min–1

∴t1/2 = = = 124.4 min

From Arrhenius equation, K = Ae–Ea/RT

Log K = + log A; Intercept = log A

When T = 27 + 273 = 300K , log K = log (5.57 × 10–3) = –2.254

Putting the value in above equation

– 2.254 = × + 15.046

∴ Ea = 99.37 kJ /mol

In order that t1/2 of reaction may decrease from 124.4 min, rate constant will have to increase 4 time.

Log (4 × 5.57 × 10–3) = × + 15.046

∴ – 1.652 – 15.046 = ×

∴ T = 310.80 K

 

7. Let [A]o =  initial concentration of A and 

[A], [B], [C] = concentration of A B and C  at any time t

k1 + k2 = k =

Now [B] =    and [A] = 1

∴calculate & get the answer

8. t1/2 = 138.4 days t = 69.2 days

No. of halves

Amount of Po left after ½ halves =

Amount of Po used = 1-0.707 = 0.293g

84Po210 → 82Po206 + 2He4

210g Po on decay will produce 4g the 

0.293 Po will produce  = = 5.581 × 10-3g He

Volume of He the at STP = = 31.25 ml = 31.25 cm3

9. Let initial amount = 20

∴amount left = 20 x = 15

Calculate k at 25°C from the equation k = with t = 20 mins. 

Now from Arrhenius equation

Calculate & from this k40 can be calculated, as k25 is already known

Let m = amount left in 30% solution

∴k40 =

From this calculate m

& % decomposition =

10.

Initial amount mo = 1g atom

Let us assure [A]o to be the initial concentration of [A] and [B] to be the concentration of Pb214.

This is an eg. of a consecutive reaction

For the number of nuclei to be maximum, should be equated to zero.

or

or

Now k1 =

k2 =

∴ t = 4.12 mins

11. Infected sample activity  = 1 × 10–7 Ci × = 3.7 × 103 d/sec.

In sample withdrawn : = 0.33 dis/sec

The ratio of total activity to activity of sample withdrawn is equal to the ratio of volumes, where V is original volume of blood.

Thus, =

∴ V = 1.1 × 103 ml = 1.1L

12. N0 = 1g atom t1/2 Ra = 1600 years t = 800 y

t =  

800=

N = 0.707g atom

∴ Amount of Ra decayed = 1-0.707

Moles of Ra decayed = 0.293g atom

& Moles of He formed = 0.293

as 80Ra226 → 86Ra222 + 2He4

∴partial pressure of Helium

P = =

P = 1.44 atm

13. Total activity of  a sample is the sum of the  individual activities of all its components. 

Let the total mass  of the sample be 1 gm and the mass of 239Pu be x gm

∴ × 6.023 ×1023 ×

=  6 × 109 × 365 × 24 ×60 ×60

On calculating,  x = 0.3896 

∴ 239Pu =  38.96% and 240Pu = 61.04%

14. Rate = k [CH3COOC2H5]a[H+]b.

[H+] is constant through out the reaction 

k1 = k [H+]b

Hence ,

b = 1 

k1 = k [H+]

1.1 × 10–4 = k (10–3)

k = 1.1 × 10–1 dm3mol–1 sec–1

15. Let  the nuclei of A236 initially be N0

∴ the nuclei of A234 initially will be 

Initial activity of A236 = 106 =

∴ Initial activity of A234 =

Now for A236, At = A0 e–λt

∴ At = 106

For A234, At =

Since the activities are to be equal 

106 

∴ t = 180  min or 3 hrs.

For the number of nuclei to be same,

For A236, Nt = N0

For A234, Nt =

∴ t = 120  min or 2 hrs.

 

Solution to Objective Problems 

LEVEL – II

1. Order = + 3/2 – 1 = 1/2
2. A(g)      + 2B(g) ⎯→ C(g) + D(g)  

t = 0 0.60 0.80 0 0

t = t (0.6–0.2) (0.8 – 2×0.2) 0.2 0.2

(Rate)i = k [A] [B]2 = k[0.6] [0.8]2

(Rate)t = k[0.4] [0.4]2

= = 1/6

3. For a 3rd order reaction 

Rate = k [A]3

=

is k = lit2 mol–2 min–1

4. t½ = 140 days 

Given 1120 days ∴ x = = 8

∴ After 1120 day, = of it will remain 

5. Equation of a first order reaction 

k =

A plot of log a Vs t for a 1st order reaction is 

 

log [A]t = log  [A]o –

y = c +   m x 

c = log [A]o

m =

Which means a negative slope and non zero intercept 

6. k = k[2]–1/2 [2]3/2 = k[2]3/2–1/2 = k = [2]1 = 2k
7. Average energy of reactants + activation energy = threshold energy 
8. For a first order reaction

As we know that

k = log

Where No = original amount

N = amount left after time t 

When t becomes t3/4 or t0.75 

3/4th  of the original amount converts into the product.

i.e. left amount N  = No – ¾ No = ¼ No

or, k = log  

or, k – = 10-2 sec–1

9. No. of half lives needed for change in the concentration of reactant from 0.08 to 0.02m is 3

∴time needed = 10 × 3 = 30 min.

10. Let original amount, NO = 100

Since 75% completed, so,  final amount N = 100-75 = 25

As we know

, where n = no. of half lives 

or,

or, 4 = 2n or, 22 or, 22 = 2n

∴n= 2

Since total time = n x t1/2

32 minutes = 2 x t1/2

∴t1/2 = 16 minutes

 

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