**CHEMICAL KINETICS**

**Solution to Subjective Problems **

**LEVEL – I**

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

- Use concepts of complex reactions.
- Arrhenius equation.
- Calculate‘t’.
- Use first order kinetics concepts.
- Calculate‘t’.
- Use pressure concepts in first order kinetics.

**LEVEL – II**

# 1. Calculate no. of He particles and proceed.

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

**LEVEL – III**

# 4. Calculate initial nuclei of and proceed.

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

**Solution to Subjective Problems **

**LEVEL – I**

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

- 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**

- 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.

- 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

- 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**

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

t = =

**N = 6.1 x 10****-7****g **

- 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**

- a) Rate =
- b) Rate =

Change in molarity =

- 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**

- Pb present =

U present =

No = U present + U decayed

= 0.1 + 0.1= 0.2

and N = 0.1

∴ t = =

t = **4.5 ****×**** 10****9**** years**

- i) k = =

= 6.2 × 10–3 s–1

- ii) Again t = 100 s, Pt = ?

6.2 × 10–3 = log or Pt = **0.033 atm**

** LEVEL- II**

- 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 ****×**** 10****23**** particles/mol**

- 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

- [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**

- 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 ****×**** 10****7**** years**

- 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**

- We know,

log

Ea = energy of activation

Given: Ea = 0

∴ = 0

or log = 0

or = 1

Hence K300 = K280

Or K300 = **1.6 ****×**** 10****6**** s****–1**

- 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**

- 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**

- 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**

- 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**

- 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 ****×**** 10****15**** 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).

- 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

- 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

- iv) k =

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

- 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**

- 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 ****×****10****5**** atoms**

- 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 ****×**** 10****9**** years**

- 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**

- 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

- 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 cm****3**

- 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**

- 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 ****×**** 10****3**** ml = 1.1L**

- 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**

- 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%

- 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**** dm****3****mol****–1**** sec****–1**

- 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**

- Order = + 3/2 – 1 = 1/2
- 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

- For a 3rd order reaction

Rate = k [A]3

=

is k = lit2 mol–2 min–1

- t½ = 140 days

Given 1120 days ∴ x = = 8

∴ After 1120 day, = of it will remain

- 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

- k = k[2]–1/2 [2]3/2 = k[2]3/2–1/2 = k = [2]1 = 2k
- Average energy of reactants + activation energy = threshold energy
- 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

- 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.

- 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

6