**Chemical Kinetics**

**Hints to Subjective Problems**

**LEVEL – I**

- Relate the effect of change of conc. of reactant on rate of reaction
- AsH3 (g) ⎯→ As (s) + H2 (g)
- See the time taken by reactants at different time intervals.
- CaCO3 (s) + 2 HCl (l) ⎯→ CaCl2 (s) + H2O (l) + CO2 (g)
- Change the equation in straight line equation and then compare
- First, find out the value of rate constant.
- Get the value of rate.
- Rate constant in both cases will be same. Equate the equations.
- Relate both rates with the help of graph
- Effect of temperature on rate constant.
- Relate the effect of change of conc. of reactant on rate of reaction.
- Radioactive decay is 1st order reaction
- Get the value of time interval.
- Originally present U238 = U238 + Pb206

(Present) (Present)

**LEVEL – II**

- (c) See how change in conc. of product is related to change in conc. of reactant.
- Find out the value of K2/K1
- First find out the value of rate constant at 450 oC
- Compare the equation with Arrhenius equation.
- Find out the values of a and a – x in terms of P0 and Pt
- Time taken to sour is inversely proportional to rate and hence rate constant at that temperature.
- Get the value of t.

**LEVEL – III**

- Percentage of isopropyl bromide by 2nd order mechanism will increase with increase in conc. of [OH–]

% =

- Relate Pt and P0 to a and a- x
- AB is not stable, find out the rate of formation of C in terms of conc. of A & B
- Rate constant includes conc. of H+ which remains constant.

**Solutions to Subjective Problems **

**LEVEL – I**

- i) Above data shows that at constant concentration of B, when concentration of A is doubled, (in first and second run) the rate of reaction is also doubled i.e. Rate α [A]1. Similarly when concentration of A is kept constant and concentration of B is doubled (in first and third run) the rate remains unchanged i.e. Rate α [B]0.

∴ Order w.r.t. A is 1

Order w.r.t. B is o

Rate law – Rate = K[A]1[B]0

- ii) 002 = K × 0.01

∴ K = 0.2 min–1

iii) t1/2 = for a first order reaction

∴ t1/2 = = 3.465 minutes

- iv) Rate = k × (A)1 (B)0

∴ Rate = 0.2 × 0.31 × 0.40 = 0.06 mol/L min.

- On heating arsine produces hydrogen gas as,

AsH3(g) → As(s) +

According to this equation one mole of arsine produces 3/2 moles of hydrogen or x moles of arsine produces 3π/2 moles of hydrogen. It means the increased pressure is due to x/2 moles. Since Pressure is directly proportional to number of moles, hence

P0α a and Pt – P0 α x/2

or P0 – 2(Pt – P0) α a – x

Therefore the first order equation

k =

may be written as

k =

=

Putting the values at different times, we get

k =

=

= **3.987 ****×**** 10****–2**** hr****–1**

k =

=

= 4.023 × 10–2 hr–1

k =

= **4.023 ****×**** 10****–2**** hr****–1**

The average value of specific reaction rate

**k = 4.011 ****×**** 10****–2**** hr****–1**

For half life period

t1/2 =

= = **17.28 hr**

- t1/2 α …(1)

We see that t1/2 is constant i.e. 2 hours whether the initial concentration is A0 A0/2 or A0/4. In other words it takes as long to go from [A0]/2 to [A0]/4 as it takes to go from [A0] to [A0]/2. In fact it would take the same length of time to halve any value of [A] i.e.

t1/2 α …(2)

By comparing equation (1) and (2)

n = 1 = 0

or n = 1 order is one

- K =

when t = 1

K = = 2.303 log 1.277

= 2.303 × 0.10619

= 0.2446 min–1

When t = 2

K =

=

= 0.2625 min–1

When t = 3

K =

= = = 0.279 min–1

Since we get reasonably constant value of K, which shows reaction is 1st order

Average rate constant = 0.262 min-1

- 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 =

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.

- a) For a first order reaction

K =

=

=

= × 0.187

= 1.3256 × 10–3 sec–1

- b) 325 × 10–3 =

∴ t =

= × 0.523

= 0.908 × 103 sec

- c) 3256 × 10–3 =

t = log10

=

= 1.737 × 103sec

- rate = 0.2 × 10-3 = 2 × 10-4 mole/litre/min

10-3 =

= 1.22

∴x = or 18 % of a

- Let t1 and t2 be the time required for 50% and 99.9% completion, respectively of the reaction.

Then t1 = , and

t2 =

∴

- 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**

- k1 = 5.03 × 10−2 mol−1 dm3 s−1 at T1 = 289 K

k2 = 6.71 mol−1 dm3 s−1 at T2 = 333 K

log =

On solving we get, Ea = 88.914 kJ

The rate constsnt at 305 K may be determined from the relation:

log

log

On solving we get, **k****2**** = 0.35 mol****−****1**** dm****3**** s****−****1**

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

i.e. R = k [NO]x [Cl2]y y = Prder 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**

- 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**

- Rate constant for a first order reaction is given by

K =

Here a [C]α – [C]0

and a – x = [C]α – [C]t

∴ K =

When t = 1

K = = 2.303 log = 2.303 log 1.124

= 2.303 × 0.05 = 0.1169 min–1

When t = 2

K = = = = 0.10757 min–1

Average rate constant = 0.1122 min-1

Thus we see that putting value of conc. of HCl at various time gives reasonably constant value of K for 1st order indicating reaction is of 1st order.

- 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**** yrs.**

**LEVEL – II**

- 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**

- a) K =

= = 1.386 × 10–3 sec–1

- b) K =

=

- c) K =

Now 2M → L + N

Initial M0 0 0

At time t M0 – x

Here = 0.02

∴ x = 2 × 0.02 = 0.04

K =

=

= = 0.016097 s–1

- 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 EB = EA

Therefore, log

From equation (i)

log = log2 = log

or =

or k1 = = **0.327 min****–1**

- 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

5.

= 1.64

or 2.303 log = 1.64

∴ = 100.712 = 5

- For first order reaction,

K =

K653 = = 1.925 × 10–3 min–1

K723 = ?

log K723 – log K653 =

log K723 – log 1.925 × 10–3 min–1

= ×

log K723 – (0.2844 – 3) = 1.55

log K723 = 2.8344

K273 = 6.82 × 10–2 min–1

For first order reaction,

t = = = 33.768 min log 4

= 33.768 × 0.602 = **20.33 min**

- (i) According to Arrhenius equation

K =

or loge K = logeA + loge

log K = log A – —– (i)

Given that

Log K(s–1) = 14.34 – —— (ii)

On comparing equation (i) and (ii)we can say that,

or Ea = 1.25 × 104 K × 2.303 × R

= 1.25 × 104 K × 2.303 × 8.314 JK–1 mol–1 = 23.934 × 104 J mol–1

= **239.34 kJ mol****–1**

- If reactions 1st and 2nd have A1 = 5A2 and K1 = 100 K2 at room temperature, find out Ea1 – Ea2.

Given A1 5A2 and K1 = 100 K2

∴

∴ Ea2 – Ea1 = 2.303 log20 × 1.987 × 10–3 cal mol–1lit–1 × 298 K

= 1.7745 K cal mol–1

- K =

Let K = Rate constant for catalysed reaction

K’ = Rate constant for non catalysed reaction

EA = Activation energy for catalysed reaction

E*A = Activation energy for non-catalysed reaction

=

= 14.57

∴

- (CH3)2COOC(CH3)3 ⎯→ 2CH3COCH3 + C2H6

P0 0 0 t = 0

P0 – P 2P P t = t

Total pressure at time t, Pt = P0 + 2P

So, 256.0 = 169.3 + 2P, ∴ P = 43.35 torr, and

P0 – P = 169.3 – 43.35 = 125.95 torr

- a) Now using the integrated rate expression for the first order reaction

We get,

k = = 0.0197 min–1

- b) The half-life period is

t1/2 = = 31.18 min

- c) Substituting the known parameters is the integrated rate expression, we get,

0.0197 =

Upong solving: P = 27.50 torr

∴ Pt = 169.23 + 2 × 27.50 = 224.3 torr

- a) K =

A → 3B

Initial A0 0

After time t A0 – x 3x

Given [A0] = 0.04 mol L–1 and 3x = 0.03 mol L-1

∴ x = = 0.01

Now K =

= = 0.0959 min–1

- b) 3x = 0.06

∴0.0959 =

∴ x = 0.02

0.0959 = log2 = 7.228 minute

Δt = 7.228 – 3 = 4.228 minutes more are required

- CH3–OCH3(g) ⎯→ CH4(g) + H2(g) + CO(g)

Initial 0.4 0 0 0

at time t = 0.4–p p p p

Suppose initial moles = a

and at any time number of moles = a – x

As ideal gas behaviour so moles α pressure

∴a α 0.40 & (a-x) α (0.40 – p)

Now calculate P which is the pressure corresponding to the moles of reactant decomposed. Total pressure is the summation of the pressure of entire system. Total pressure = 0.4 – p + p + p + p

Pt = 0.4 + 2p

∴p = 0.175 atm

Pt = 0.4 + 2 × 0.175 = 0.75 atm

- Uranium present = = 2.1 x 10-3g atom

Pb present =

Pb from Uranium decay = = 0.109 x 10-3g atom

∴N = 2.1 x 10-3g atom

N0 = (2.1 x 10-3 + 0.109 x 10-3)g atom = 2.2 x 10-3g atom

Now t = =

**∴**** t ** **=** **3.3 x 10****8**** yrs.**

- At a given temperature greater is the time taken by reactants to react, slower is the rate. Since rate constant is rate of a reaction when concentrations of reactants are unity, it is also inversely related to time taken.

∴

∴log

or log 4.667 =

Ea = = 43.16 KJ/mol

- =

∴ t = = 41 months

=

∴ t1/2 = = 79.7 months

**LEVEL – III**

- % of isopropyl bromide which reacts by second order mechanism is given by

=

- a) When [OH–] = 0.001 M % = × 100 = 1.9%
- b) When [OH–] = 0.01 M % = × 100 = 16.4%
- c) When [OH–] = 0.1 M, 66.2%
- d) When [OH–] = 1.0 M, 95.1%
- e) When [OH–] = 5.0M % = 99.0%

Preferentially i.e. at high concentration of [OH–] reaction follows second order mechanism.

- 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] =

1.3 ×10-5 + 9 × 1.3 × 10-5 =

0.2 =

or 100.2 = 1 +

or = 0.525

- 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**

- EA for the reaction = 73.04 kJ/mole

K =

When 20% decomposition then

x= or 0.2a

∴ K =

=

=

= 0.011159 min–1

or K313 = 4.07 × 0.01159 = 0.0454 min–1

K =

∴ = 100.4 = 2.5

or or % decomposition is 60 %

- a) log

log

Ea = 22.013 KJ mol-1

Now, log k = log A –

∴log A = 10.735

A = 5.4 × 1010 sec-1

- b) N2O5 (g) ⎯→ 2 NO2 (g) + ½ O2 (g)

initial P0 0 0

at time t P0 – x 2x 0.5x

Pt = P0 + 1.5 x

X = = 240 mm Hg

∴ mole fraction of N2O5 decomposed = = = 0.4

6.

= 0.15

∴ = 1.41

or x = 71 % of a

but in thyroid gland % radioactivity is 67.77 % of a

∴ fraction of radioactive iodide migrated to thyroid gland = = 0.953

since rate of diffusion of stable as well as radioactive iodide is same

∴0.1 × 0.953 = 0.0953 mg of iodide has migrated to thyroid gland.

7.

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**

- t1/2 = 14.3 days

A = λ N.

Now = q–A Where q = rate of formation

= q–λN A = rate of decay

= t

= 0.925 × 107 sec = 10.7 days

- Since K–1 >> K2, AB remains in equilibrium with A and B

Therefore,

= K

∴

∴

- N0 = 1g atom t1/2 = 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

**∴** **239****Pu = 38.96% and ****240****Pu = 61.04%**

- % Zinc recovered = %62 Zn recovered

= = 82.3

Total zinc = = 0.243 g

The mass of the added 62Zn+2 is negligible, hence 0.243 g is the mass of the Zn+2 in the original sample = **0.0744M Zn****+2**

- 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.**

- 90Th232 ⎯→ 82Pb208 + 6 2He4 + 4–1e0

6 × 22400 ml of He is formed by 232 g Th decay

∴ 8 × 10–5 ml of He is formed by g Th decay

= 1.38 × 10–7 g Th decay

at t = 5, sample has Th = 5 × 10–7g

at t = 0, sample had Th = 5 × 10–7 g + 1.38 × 10–7 = 6.38 × 10–7 g

∴ t = =**4.89 ****×**** 10****9**** yrs.**

**Solutions to Objective Problems **

**LEVEL – I**

- At least two molecules ( one of C12H22O11 and one of H2O) are required for reaction to occur
- For 1st order reaction value of k depends only on unit of time

- Rate = k [COS]1 [H2O]0

5.

∴

∴ K’ = 2K

- All factors are important.

- Only change in temperature or activation energy changes value of rate constant.

- According to law of mass action rate α product of active masses of reactants

- Order = + 3/2 – 1 = 1/2

- Overall order = 1.5 – 1 + 0 = 0.5

- 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

**∴**** (D)**

- For a 3rd order reaction

Rate = k [A]3

=

is k = lit2 mol–2 min–1

**∴**** (B)**

- t½ = 4 days

Given 16 days ∴ x = = 4

∴ After 16 day, = of it will remain

**∴**** (B) **

- 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

**∴**** (D)**

- R = k[2]–1/2 [2]3/2 = k[2]3/2–1/2 = k = [2]1 = 2k

**∴**** (B)**

- Average energy of reactants + activation energy = threshold energy

- = temperature coefficient = 2 to 3

- When [B] doubles keeping [A] constant rate becomes 8 times

∴ rate α [B]3

When [A] doubles, keeping [B] constant rate remains same

∴ rate α [A]o

∴Rate = k [A]o [B]3

- 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**

**∴****(A)**

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

∴time needed = 20 × 3 = 60 min.

**LEVEL – II**

- 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

**∴****t****1/2**** = 16 minutes**

**∴****(C)**

- D is wrong because has positive value as it is a product

3.

when t = t1/2 A = 0.5 A0

then t1/2 =

- (D) is wrong as Arhenius equation does not include ΔGo
- fraction remaining =

n = no. of t1/2

= = =

- log

∴ Ea = = 14.7 KJ

- Here, half life, t1/2 = 5 years

Decomposed amount = 99.9%

Let the original amount, No = 100

Left amount after time t, = 100 – 99.9 = 0.1

N = 0.1

kt = 2.303 log

or,

or,

or, t =

**= 49.85 years**

**∴****(D)**

9 b) is wrong because molecules cannot react in fractional form to form activated complex

- Let original amount No = 100

Since, disappeared amount = 75%

∴left amount, N = 100 – 75 =25

∴Since, where n = no. of half lives

or,

or, 22 = 2n, here n = 2

Total time = n × t1/2

1.386 hours = 2 × t1/2,

t1/2 = 0.693 hours

= 2494.8 sec.

k = = **2.7 ****×**** 10****–4**** sec****–1**

**∴**** (C)**

- Let the initial amount of the reactant = No =100

Let the rest amount of the reactant = N

As we know,

k × t = 2.303 log

or,

or,

or, 1.414 =

or, N= 70.72%

decomposed amount = 1100 – 70.72 = **29.28%**

**∴**** (B)**

12.

∴ = 1.41

∴ x = = 0.29 a or 29 %

- a A + b B ⎯→ Product

- The reaction is A(g) → 2B(g) + C(g)

at t =0 pressure PA 0 0

Pressure after P4 – x 2x x

10 minutes

Total pressure after 10 minutes, PT = PA + PB + PC

or, PT = (PA – x) + 2x + x

= PA + 2x = 180

PA = 90 mm

90 + 2x = 180 or, X = 45mm

∴PA – x = 90-45 = 45 mm

Since, kt = 2.303 log

or, k × 10 = 2.303 log

k × 10 = 0.693

or, k = = 0.0693 min-1

= **= 1.155 x 10****-3**** sec****-1**

**∴**** (A)**

- R1 = k [RCl]1 [NaOH]0

R2 = k [0.5 RCl]1 [NaOH]

∴ R2 = 0.5 R1

16.

∴or

∴ t = 20 minutes

- Conc. time–1 = K(conc.)1.5

∴ K = (conc.)–0.5 time–1

- Catalyst lower EA for both forward and backward reaction by equal amount

- fraction remaining =

=

- on integration

t =

∴ t =

this is an equation for straight line with negative intercept and positive slope