Chemical Kinetics

 

Hints to Subjective Problems

LEVEL – I

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

                (Present) (Present)

 

LEVEL – II

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

LEVEL – III

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

% =

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

 

Solutions to Subjective Problems 

 

LEVEL – I

1. 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

1. ii) 002 = K × 0.01

∴ K = 0.2 min–1

iii) t1/2 = for a first order reaction

∴ t1/2 = = 3.465 minutes

1. iv) Rate = k × (A)1 (B)0

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

 

2. 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

 

3. 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

 

4. 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

5. 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.

 

6. a) For a first order reaction

K =

=

=

= × 0.187

= 1.3256 × 10–3 sec–1

1. b) 325 × 10–3 =

∴ t =

= × 0.523

= 0.908 × 103 sec

1. c) 3256 × 10–3 =

t = log10

=

= 1.737 × 103sec

 

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

10-3 =

= 1.22 

∴x = or 18 % of a

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

Then t1 = , and

t2 =

∴

 

9. 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

 

10. 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, k2 = 0.35 mol−1 dm3 s−1

 

11. 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

 

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

t =   =

N  = 6.1 x 10-7g 

 

13. 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

 

14. 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.

 

15. Pb present  = U present =

No = U present + U decayed

= 0.1 + 0.1= 0.2

and N = 0.1

∴ t = =

t = 4.5 × 109  yrs.

LEVEL – II

 

1. 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

 

2. a) K =

= = 1.386 × 10–3 sec–1

1. b) K =

=

1. 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

 

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 EB = EA

Therefore, log

From equation  (i) 

log = log2 = log

or =

 

or k1 = = 0.327 min–1

 

4. 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

 

6. 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

 

7. (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

 

8. 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 

 

9. 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

∴

10. (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

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

We get, 

k = = 0.0197 min–1

1. b) The half-life period is

t1/2 = = 31.18 min

1. 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

 

11. 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

 

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

 

12. 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

 

13. 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 108 yrs.

 

14. 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

 

15. =

∴ t = = 41 months

=

∴ t1/2 = = 79.7 months

LEVEL – III

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

=

1. a) When [OH–] = 0.001 M % = × 100 = 1.9%
2. b) When [OH–] = 0.01 M % = × 100 = 16.4%
3. c) When [OH–] = 0.1 M, 66.2%
4. d) When [OH–] = 1.0 M, 95.1%
5. e) When [OH–] = 5.0M % = 99.0% 

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

 

2. 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

 

3. 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

 

4. 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 %

 

5. a) log

log

Ea = 22.013 KJ mol-1

Now, log k = log A –

∴log A = 10.735

A = 5.4 × 1010 sec-1

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

 

8. 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

 

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

Therefore,

= K

∴

∴

 

10. 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

 

11. 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%

 

12. % 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

 

13. 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

 

14. 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.

 

15. 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 × 109 yrs. 

 

Solutions to Objective Problems 

 

LEVEL – I

 

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

 

3. Rate = k [COS]1 [H2O]0

 

5.

∴

∴ K’ = 2K

 

6. All factors are important.

 

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

 

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

 

9. Order = + 3/2 – 1 = 1/2

 

10. Overall order = 1.5 – 1 + 0 = 0.5

 

11. 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)

 

12. For a 3rd order reaction 

Rate = k [A]3

=

is k = lit2 mol–2 min–1

∴ (B)

 

13. t½ = 4 days 

Given 16 days ∴ x = = 4

∴ After 16 day, = of it will remain 

∴ (B) 

 

14. 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)

 

15. R = k[2]–1/2 [2]3/2 = k[2]3/2–1/2 = k = [2]1 = 2k

∴ (B)

 

16. Average energy of reactants + activation energy = threshold energy 

 

17. = temperature coefficient = 2 to 3

 

18. 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

 

19. 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)

 

20. 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

 

1. 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

∴(C)

 

2. D is wrong because has positive value as it is a product 

 

3.

when t = t1/2    A = 0.5 A0

then t1/2 =

 

4. (D) is wrong as Arhenius equation does not include ΔGo
5. fraction remaining =

n = no. of t1/2

= = =

 

7. log

∴ Ea = = 14.7 KJ

 

8. 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

 

10. 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)

 

11. 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 %

 

13. a A + b B ⎯→ Product

 

14. 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)

 

15. R1 = k [RCl]1 [NaOH]0

R2 = k [0.5 RCl]1 [NaOH]

∴ R2 = 0.5 R1

 

16.

∴or

∴ t = 20 minutes

 

17. Conc. time–1 = K(conc.)1.5

∴ K = (conc.)–0.5 time–1

 

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

 

19. fraction remaining =

=

20. on integration

t =

∴ t =

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