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NCERT Solutions for Class 12 Chemistry chapter 4-Chemical Kinetics

NCERT Solutions for Class 12 Chemistry

NCERT Solutions for class-12 Chemistry Chapter 4 Chemical Kinetics is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-12 in NCERT textbook, also do read theory of this Chapter 4 Chemical Kinetics while going before solving the NCERT questions. Our Physics Wallah team Prepared Other Subjects NCERT Solutions for class 12.

Chapter 4 Chemical Kinetics

Answer the following Questions.

Question1. For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Solution :
Average rate of reaction NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 6.67 × 10−6 M s−1

Question2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L−1 to 0.4 mol L−1 in 10 minutes. Calculate the rate during this interval?

Solution :
 

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 0.005 mol L−1 min−1

= 5 × 10−3 M min−1

Question3. For a reaction, A + B → Product; the rate law is given by, r  = k [A]½ [B]2. What is the order of the reaction?

Solution :
The order of the reaction r  = k [A]½ [B]2

= 1 / 2 + 2

= 2.5

Question4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Solution :
The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]2 (1)

Let [X] = a mol L−1, then equation (1) can be written as:

Rate1 = k .(a)2

= ka2

If the concentration of X is increased to three times, then [X] = 3a mol L−1

Now, the rate equation will be:

Rate = k (3a)2

= 9(ka2)

Hence, the rate of formation will increase by 9 times.

Question5. A first order reaction has a rate constant 1.15 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g?

Solution :
From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10−3 s−1

We know that for a 1st order reaction,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 444.38 s

= 444 s (approx)

Question6. Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Solution :
We know that for a 1st order reaction,

t½  = 0.693 / k

It is given that t1/2 = 60 min

k = 0.693 / t½

  = 0.693 / 60

  = 0.01155 min-1

   = 1.155 min-1

Or

k = 1.925 x 10-2 s-1

Question7. What will be the effect of temperature on rate constant?

Solution :
The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

k = Ae - Ea / RT

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

Ea is the activation energy

Question8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.

Solution :
It is given that T1 = 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k1 = k and that of k2 = 2k

Also, R = 8.314 J K−1 mol−1

Now, substituting these values in the equation:

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 52897.78 J mol−1

= 52.9 kJ mol−1

Question9. The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Solution :
In the given case:

Ea = 209.5 kJ mol−1 = 209500 J mol−1

T = 581 K

R = 8.314 JK−1 mol−1

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

x = e−Ea / RT

⇒Inx= −Ea / RT

⇒logx=−Ea / 2.303RT

⇒logx= −209500Jmol−1 / 2.303 × 8.314JK−1mol−1×581

=−18.8323

Now,x= Antilog (−18.8323)

=1.471×10−19

Question10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N2O (g) Rate = k[NO]2

(ii) H2O2 (aq) + 3 I (aq) + 2 H+ → 2 H2O (l) + I3- Rate = k[H2O2][I]

(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2

(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl]

Solution :
(i) Given rate = k [NO]2

Therefore, order of the reaction = 2

Dimension of  k = Rate / [NO]2

= mol L-1 s-1 / (mol L-1)2

= mol L-1 s
-1 / mol2 L-2

= L mol-1s-1

(ii) Given rate = k [H2O2] [I]

Therefore, order of the reaction = 2

Dimension of

k = Rate  / [H2O2][I - ]

= mol L-1 s-1  / (mol L-1) (mol L-1)

= L mol-1 s-1

 

(iii) Given rate = k [CH3CHO]3/2

Therefore, order of reaction = 3 / 2

Dimension of  k = Rate / [CH3CHO]3/2

=  mol L-1 s-1   / (mol L-1)3/2

= mol L
-1   s-1     / mol3/2  L-3/2

= L½ mol-½  s-1

(iv) Given rate = k [C2H5Cl]

Therefore, order of the reaction = 1

Dimension of k = Rate /  [C2H5Cl]

= mol L-1 s-1   / mol L-1

= s-1

 

Question11. For the reaction:

2A + B → A2B

the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.

Solution :
The initial rate of the reaction is

Rate = k [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2

= 8.0 × 10−9 mol−2 L2 s−1

When [A] is reduced from 0.1 mol L−1 to 0.06 mol−1, the concentration of A reacted = (0.1 − 0.06) mol L−1 = 0.04 mol L−1

Therefore, concentration of B reacted 1/2 x 0.04 mol L-1 = 0.02 mol L−1

Then, concentration of B available, [B] = (0.2 − 0.02) mol L−1

= 0.18 mol L−1

After [A] is reduced to 0.06 mol L−1, the rate of the reaction is given by,

Rate = k [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

= 3.89 mol L−1 s−1

Question12. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1?

Solution :
The decomposition of NH3 on platinum surface is represented by the following equation.

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

 

= 7.5 × 10−4 mol L−1 s−1

Question13. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by

Rate = k [CH3OCH3]3/2

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

rate = k (PCH3OCH3)3/2

If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?

Solution :
If pressure is measured in bar and time in minutes, then

Unit of rate = bar min−1

Rate = k [CH3OCH3]3/2

⇒ k =Rate / [CH3OCH3]3/2

Therefore, unit of rate constants(k) = bar min−1 / bar3/2

= bar-½ min -1

Question14. Mention the factors that affect the rate of a chemical reaction.

Solution :
The factors that affect the rate of a reaction are as follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst

15. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i) doubled (ii) reduced to half?

Solution :
Letthe concentration of the reactant be [A] = a

Rate of reaction, R = k [A]2

= ka2

(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R = k(2a)2

= 4ka2

= 4 R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a , then the rate of the reaction would be

R = k(1/2a)2

= 1/4 Ka2

= 1/4 R

Therefore, the rate of the reaction would be reduced to

16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?

Solution :
When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

k=Ae−Ea/RT

Where,

k = rate constant,

A = Frequency factor / Arrhenius factor,

R = gas constant

T = temperature

Ea = activation energy for the reaction.

Question17. In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s

0

30

60

90

[Ester]mol L−1

0.55

0.31

0.17

0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Solution :
(i) Average rate of reaction between the time interval, 30 to 60 seconds, d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10−3 mol L−1 s−1

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]º / [R]

For t = 30 s, k1

= 1.911 × 10−2 s−1

For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31

= 1.957 × 10−2 s−1

For t = 90 s,  k3 = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 - 2s - 1

= 2.075 × 10−2 s−1

Then, average rate constant, k = k1 + k2+ k3  / 3

= 1.911 × 10 - 2  + 1.957 × 10 - 2 + 2.075 × 10 - 2 / 3

= 1.981 x 10-2 s - 1

 

Question18. A reaction is first order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

Solution :
(i) The differential rate equation will be

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

Question19. In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

A/ mol L−1

0.20

0.20

0.40

B/ mol L−1

0.30

0.10

0.05

r0/ mol L−1 s−1

5.07 × 10−5

5.07 × 10−5

1.43 × 10−4

What is the order of the reaction with respect to A and B?

Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

Dividing equation (iii) by (ii), we obtain

= 1.496

= 1.5 (approximately)

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

Question20. The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Experiment

A/ mol L−1

B/ mol L−1

Initial rate of formation of D/mol L−1 min−1

I

0.1

0.1

6.0 × 10−3

II

0.3

0.2

7.2 × 10−2

III

0.3

0.4

2.88 × 10−1

IV

0.4

0.1

2.40 × 10−2

Determine the rate law and the rate constant for the reaction.

Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

 

 

Dividing equation (iv) by (i), we obtain

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

Dividing equation (iii) by (ii), we obtain

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

 

Question21. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment

A/ mol L−1

B/ mol L−1

Initial rate/mol L−1 min−1

I

0.1

0.1

2.0 × 10−2

II

--

0.2

4.0 × 10−2

III

0.4

0.4

--

IV

--

0.2

2.0 × 10−2

Solution :
The given reaction is of the first order with respect to A and of zero order with respect to B.

Therefore, the rate of the reaction is given by,

Rate = k [A]1 [B]0

⇒ Rate = k [A]

From experiment I, we obtain

2.0 × 10−2 mol L−1 min−1 = k (0.1 mol L−1)

⇒ k = 0.2 min−1

From experiment II, we obtain

4.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.2 mol L−1

From experiment III, we obtain

Rate = 0.2 min−1 × 0.4 mol L−1

= 0.08 mol L−1 min−1

From experiment IV, we obtain

2.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.1 mol L−1

Question22. Calculate the half-life of a first order reaction from their rate constants given below:

(i) 200 s−1 (ii) 2 min−1 (iii) 4 years−1

Solution :
(i) Half life, t 1/2 = 0.693 / k

= 0.693 / 200 s-1

= 3.47×10 -3 s (approximately)

(ii) Half life, t 1/2 = 0.693 / k

 = 0.693 / 2 min-1

= 0.35 min (approximately)

(iii) Half life, t 1/2 = 0.693 / k

= 0.693 / 4 years-1

= 0.173 years (approximately)

Question23. The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Solution :
Here,  k = 0.693 / t1/2

= 0.693 / 5730 years-1

It is known that,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Question24. The experimental data for decomposition of N2O5

[2N2O5 → 4NO2 + O2]  in gas phase at 318K are given below:

t(s)

0

400

800

1200

1600

2000

2400

2800

3200

102 × [N2O5]  mol L-1

1.63

1.36

1.14

0.93

0.78

0.64

0.53

0.43

0.35

(i) Plot [N2O5] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log [N2O5] and t.

(iv) What is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii).

Solution :
NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

(ii) Time corresponding to the concentration, 1630x102 / 2 mol L-1 = 81.5 mol L-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t(s)

102 × [N2O5]  mol L-1

Log[N2O5]

0

1.63

− 1.79

400

1.36

− 1.87

800

1.14

− 1.94

1200

0.93

− 2.03

1600

0.78

− 2.11

2000

0.64

− 2.19

2400

0.53

− 2.28

2800

0.43

− 2.37

3200

0.35

− 2.46

Class 12 NCERT Solutions for Chemical Kinetics

(iv) The given reaction is of the first order as the plot, Log[N2O5v/s t, is a straight line. Therefore, the rate law of the reaction is

Rate = k [N2O5]

(v) From the plot, Log[N2O5] v/s t, we obtain

- k /2.303

Again, slope of the line of the plot  Log[N2O5v/s t is given by

- k / 2.303. = -0.67 / 3200

Therefore, we obtain,

- k / 2.303  = - 0.67 / 3200

⇒ k = 4.82 x 10-4 s-1

 

(vi) Half-life is given by,

t½ = 0.693 / k

= 0.639 / 4.82x10-4 s

=1.438 x 103

This value, 1438 s, is very close to the value that was obtained from the graph.

25. The rate constant for a first order reaction is 60 s−1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Solution :
It is known that,

Class 12 NCERT Solutions for Chemical Kinetics

 

Question26. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Solution :
,

Class 12 NCERT Solutions for Chemical Kinetics

 

Therefore, 0.7814 μg of 90Sr will remain after 10 years.

Again,

Class 12 NCERT Solutions for Chemical Kinetics

 

Therefore, 0.2278 μg of 90Sr will remain after 60 years.

Question27. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Solution :
For a first order reaction, the time required for 99% completion is

t1 = 2.303/k Log 100/100-99

= 2.303/k Log 100

= 2x 2.303/k

For a first order reaction, the time required for 90% completion is

t2 = 2.303/k Log 100 / 100-90

= 2.303/k Log 10

= 2.303/k

Therefore, t1 = 2t2

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

Question28. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Solution :
For a first order reaction,

t = 2.303/k Log [R] º / [R]

k = 2.303/40min  Log 100 / 100-30

= 2.303/40min  Log 10 / 7

= 8.918 x 10-3 min-1

 

Therefore, t1/2 of the decomposition reaction is

t1/2 = 0.693/k

=  0.693 / 8.918 x 10-3  min

= 77.7 min (approximately)

= 77.7 min (approximately)

Question29. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t (sec)

P(mm of Hg)

0

35.0

360

54.0

720

63.0

Calculate the rate constant.

Solution :
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

Class 12 NCERT Solutions for Chemical Kinetics

After time, t, total pressure, Pt = (Pº - p) + p + p

⇒ Pt = (Pº + p)

⇒ p =
Pt  - P0

therefore, Pº - p = P0  - Pt  - P0

= 2P0 − Pt

For a first order reaction,

k = 2.303/t  Log P0 /P0  - p

=   2.303/t  Log
P0 / 2 P0  -  Pt

When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0

= 2.175 × 10−3 s−1

When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0

= 2.235 × 10−3 s−1

Hence, the average value of rate constant is

k = (2.175 × 10 - 3  + 2.235 × 10 - 3 ) / 2   s - 1

= 2.21 × 10−3 s−1

Question30. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

SO2Cl2(g)  →  SO2(g) + Cl2(g)

Experiment

Time/s−1

Total pressure/atm

1

0

0.5

2

100

0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.

Solution :
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.

Class 12 NCERT Solutions for Chemical Kinetics

After time, t, total pressure, Pt = (Pº - p) + p + p

⇒ Pt = (Pº + p)

⇒ p = Pt  - Pº

therefore, Pº - p = Pº  - Pt  - Pº

= 2 Pº -  Pt

For a first order reaction,

k = 2.303/t  Log  Pº / Pº  - p

=   2.303/t  Log  Pº / 2 Pº  -  Pt

 

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6

= 2.231 × 10 - 3s - 1

 

When Pt= 0.65 atm,

P0+ p= 0.65

⇒ p= 0.65 - P0

= 0.65 - 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is

PSOCL2 = P0 - p

= 0.5 - 0.15

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate = k(pSOCL2)

= (2.23 × 10 - 3s - 1) (0.35 atm)

= 7.8 × 10 - 4 atm s - 1

 

Question31. The rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C

0

20

40

60

80

105 X K /S-1

0.0787

1.70

25.7

178

2140

Draw a graph between ln k and 1/T and calculate the values of A and Ea.

Predict the rate constant at 30º and 50ºC.

Solution :
From the given data, we obtain

T/°C

0

20

40

60

80

T/K

273

293

313

333

353

1/T / k-1

3.66×10−3

3.41×10−3

3.19×10−3

3.0×10−3

2.83 ×10−3

105 X K /S-1

0.0787

1.70

25.7

178

2140

ln k

−7.147

− 4.075

−1.359

−0.577

3.063

Class 12 NCERT Solutions for Chemical Kinetics

Slope of the line,

NCERT Solutions  class 12 chemistry

In k= - 2.8

Therefore, k = 6.08x10-2s-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10-3 K

In k = - 0.5

Therefore, k = 0.607 s-1

Question32. The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Solution :
k = 2.418 × 10−5 s−1

T = 546 K

Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1

According to the Arrhenius equation,

NCERT Solutions  class 12 chemistry

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 1012 s−1 (approximately)

Question33. Consider a certain reaction A → Products with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.

Solution :
k = 2.0 × 10−2 s−1

T = 100 s

[A]o = 1.0 moL−1

Since the unit of k is s−1, the given reaction is a first order reaction.

Therefore, k = 2.303/t  Log  [A]º / [A]

⇒2.0 × 110-2 s-1  = 2.303/100s  Log  1.0 / [A]

⇒2.0 × 110-2 s-1  = 2.303/100s  ( - Log [A] )

⇒ - Log [A] = -  (2.0 x 10-2 x 100) /   2.303

⇒ [A] = antilog [-  (2.0 x 10-2 x 100)  /  2.303]

= 0.135 mol L−1 (approximately)

Hence, the remaining concentration of A is 0.135 mol L−1.

Question34. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Solution :For a first order reaction,

k = 2.303/t  Log  [R]º / [R]

It is given that, t1/2 = 3.00 hours

Therefore, k = 0.693 / t1/2

= 0.693 / 3  h-1

= 0.231 h - 1

Then, 0.231 h - 1 = 2.303 / 8h  Log  [R]º / [R]

NCERT Solutions  class 12 chemistry

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Question35. The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s−1) e−28000 K/T Calculate Ea.

Solution :
The given equation is

k = (4.5 × 1011 s−1) e−28000 K/T (i)

Arrhenius equation is given by,

k= Ae -Ea/RT  (ii)

From equation (i) and (ii), we obtain

Ea  / RT  =  28000K / T

⇒ Ea  = R x 28000K

= 8.314 J K−1 mol−1 × 28000 K

= 232792 J mol−1

= 232.792 kJ mol−1

Question36. The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k = 14.34 − 1.25 × 104 K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Solution :
Arrhenius equation is given by,

k= Ae -Ea/RT  

⇒In k = In A - Ea/RT

⇒In k = Log A - Ea/RT

⇒ Log k = Log A - Ea/2.303RT         (i)

 

The given equation is

Log k = 14.34 - 1.25 104 K/T             (ii)

From equation (i) and (ii), we obtain

Ea/2.303RT  = 1.25 104 K/T

= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1

= 239339.3 J mol−1 (approximately)

= 239.34 kJ mol−1

Also, when t1/2 = 256 minutes,

k = 0.693 / t1/2

= 0.693 / 256

= 2.707 × 10 - 3 min - 1

= 4.51 × 10 - 5s - 1 

= 2.707 × 10−3 min−1

= 4.51 × 10−5 s−1

It is also given that, log k = 14.34 − 1.25 × 104 K/T

NCERT Solutions  class 12 chemistry

= 668.95 K

= 669 K (approximately)

Question37. The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?

Solution :
From Arrhenius equation, we obtain

log k2/k1 = Ea / 2.303 R (T2 - T1) / T1T2

Also, k1 = 4.5 × 103 s−1

T1 = 273 + 10 = 283 K

k2 = 1.5 × 104 s−1

Ea = 60 kJ mol−1 = 6.0 × 104 J mol−1

Then,

NCERT Solutions  class 12 chemistry

= 297 K

= 24°C

Hence, k would be 1.5 × 104 s−1 at 24°C.

Question38. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s−1. Calculate k at 318 K and Ea.

Solution :
For a first order reaction,

t = 2.303 / k log a  / a - x

At 298 K, t = 2.303 / k log 100  / 90

= 0.1054 / k

At 308 K, t' = 2.303 / k' log 100  / 75

= 2.2877 / k'

According to the question,

t = t'

0.1054 / k  =  2.2877 / k'

⇒ k' k  = 2.7296

 

From Arrhenius equation, we obtain

NCERT Solutions  class 12 chemistry

To calculate k at 318 K,

It is given that, A = 4 x 1010 s-1, T = 318K

Again, from Arrhenius equation, we obtain

NCERT Solutions  class 12 chemistry

Therefore, k = Antilog (-1.9855)

= 1.034 x 10-2 s -1

Question39. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Solution :
From Arrhenius equation, we obtain

NCERT Solutions  class 12 chemistry

Hence, the required energy of activation is 52.86 kJmol−1.


 

 

NCERT Solutions For Class 12 Chemistry Chapter Wise.

 


  1. Chapter-1 The Solid States

  2. Chapter-2 Solutions

  3. Chapter-3 Electrochemistry

  4. Chapter-4 Chemical Kinetics

  5. Chapter-5 Surface Chemistry

  6. Chapter-6 General Priciples and Processes of Isolation of Elements

  7. Chapter-7 The P Block Elements

  8. Chapter-8 The D and F Elements

  9. Chapter-9 Coordination Compounds

  10. Chapter-10 Haloalkanes and Haloarenes

  11. Chapter-11 Alcohols, Phenols and Ethers

  12. Chapter-12 Aldehydes, ketones and Carboxylics Acids

  13. Chapter-13 Amines

  14. Chapter-14 Biomolecules

  15. Chapter-15 Polymers

  16. Chapter-16 Chemistry In Everyday Life



 

Question4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Solution :
The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]2 (1)

Let [X] = a mol L−1, then equation (1) can be written as:

Rate1 = k .(a)2

= ka2

If the concentration of X is increased to three times, then [X] = 3a mol L−1

Now, the rate equation will be:

Rate = k (3a)2

= 9(ka2)

Hence, the rate of formation will increase by 9 times.

Question5. A first order reaction has a rate constant 1.15 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g?

Solution :
From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10−3 s−1

We know that for a 1st order reaction,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 444.38 s

= 444 s (approx)

Question6. Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Solution :
We know that for a 1st order reaction,

t½  = 0.693 / k

It is given that t1/2 = 60 min

k = 0.693 / t½

  = 0.693 / 60

  = 0.01155 min-1

   = 1.155 min-1

Or

k = 1.925 x 10-2 s-1

Question7. What will be the effect of temperature on rate constant?

Solution :
The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

k = Ae - Ea / RT

Where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

Ea is the activation energy

Question8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.

Solution :
It is given that T1 = 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k1 = k and that of k2 = 2k

Also, R = 8.314 J K−1 mol−1

Now, substituting these values in the equation:

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 52897.78 J mol−1

= 52.9 kJ mol−1

Question9. The activation energy for the reaction 2HI(g) → H2 + I2(g) is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Solution :
In the given case:

Ea = 209.5 kJ mol−1 = 209500 J mol−1

T = 581 K

R = 8.314 JK−1 mol−1

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

x = e−Ea / RT

⇒Inx= −Ea / RT

⇒logx=−Ea / 2.303RT

⇒logx= −209500Jmol−1 / 2.303 × 8.314JK−1mol−1×581

=−18.8323

Now,x= Antilog (−18.8323)

=1.471×10−19

Question10. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N2O (g) Rate = k[NO]2

(ii) H2O2 (aq) + 3 I (aq) + 2 H+ → 2 H2O (l) + I3- Rate = k[H2O2][I]

(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2

(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl]

Solution :
(i) Given rate = k [NO]2

Therefore, order of the reaction = 2

Dimension of  k = Rate / [NO]2

= mol L-1 s-1 / (mol L-1)2

= mol L-1 s
-1 / mol2 L-2

= L mol-1s-1

(ii) Given rate = k [H2O2] [I]

Therefore, order of the reaction = 2

Dimension of

k = Rate  / [H2O2][I - ]

= mol L-1 s-1  / (mol L-1) (mol L-1)

= L mol-1 s-1

 

(iii) Given rate = k [CH3CHO]3/2

Therefore, order of reaction = 3 / 2

Dimension of  k = Rate / [CH3CHO]3/2

=  mol L-1 s-1   / (mol L-1)3/2

= mol L
-1   s-1     / mol3/2  L-3/2

= L½ mol-½  s-1

(iv) Given rate = k [C2H5Cl]

Therefore, order of the reaction = 1

Dimension of k = Rate /  [C2H5Cl]

= mol L-1 s-1   / mol L-1

= s-1

 

Question11. For the reaction:

2A + B → A2B

the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.

Solution :
The initial rate of the reaction is

Rate = k [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2

= 8.0 × 10−9 mol−2 L2 s−1

When [A] is reduced from 0.1 mol L−1 to 0.06 mol−1, the concentration of A reacted = (0.1 − 0.06) mol L−1 = 0.04 mol L−1

Therefore, concentration of B reacted 1/2 x 0.04 mol L-1 = 0.02 mol L−1

Then, concentration of B available, [B] = (0.2 − 0.02) mol L−1

= 0.18 mol L−1

After [A] is reduced to 0.06 mol L−1, the rate of the reaction is given by,

Rate = k [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

= 3.89 mol L−1 s−1

Question12. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1?

Solution :
The decomposition of NH3 on platinum surface is represented by the following equation.

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

 

= 7.5 × 10−4 mol L−1 s−1

Question13. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by

Rate = k [CH3OCH3]3/2

The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,

rate = k (PCH3OCH3)3/2

If the pressure is measured in bar andtime in minutes, then what are the units of rate and rate constants?

Solution :
If pressure is measured in bar and time in minutes, then

Unit of rate = bar min−1

Rate = k [CH3OCH3]3/2

⇒ k =Rate / [CH3OCH3]3/2

Therefore, unit of rate constants(k) = bar min−1 / bar3/2

= bar-½ min -1

Question14. Mention the factors that affect the rate of a chemical reaction.

Solution :
The factors that affect the rate of a reaction are as follows.

(i) Concentration of reactants (pressure in case of gases)

(ii) Temperature

(iii) Presence of a catalyst

15. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i) doubled (ii) reduced to half?

Solution :
Letthe concentration of the reactant be [A] = a

Rate of reaction, R = k [A]2

= ka2

(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R = k(2a)2

= 4ka2

= 4 R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a , then the rate of the reaction would be

R = k(1/2a)2

= 1/4 Ka2

= 1/4 R

Therefore, the rate of the reaction would be reduced to

16. What change would happen in the rate constant of a reaction when there is a change in its temperature? How can this temperature effect on rate constant be represented quantitatively?

Solution :
When a temperature of 10∘ rises for a chemical reaction then the rate constant increases and becomes near to double of its original value.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

k=Ae−Ea/RT

Where,

k = rate constant,

A = Frequency factor / Arrhenius factor,

R = gas constant

T = temperature

Ea = activation energy for the reaction.

Question17. In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s

0

30

60

90

[Ester]mol L−1

0.55

0.31

0.17

0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Solution :
(i) Average rate of reaction between the time interval, 30 to 60 seconds, d[ester] / dt

= (0.31-0.17) / (60-30)

= 0.14 / 30

= 4.67 × 10−3 mol L−1 s−1

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]º / [R]

For t = 30 s, k1

= 1.911 × 10−2 s−1

For t = 60 s, k1 = 2.303/ 30 log 0.55 / 0.31

= 1.957 × 10−2 s−1

For t = 90 s,  k3 = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 - 2s - 1

= 2.075 × 10−2 s−1

Then, average rate constant, k = k1 + k2+ k3  / 3

= 1.911 × 10 - 2  + 1.957 × 10 - 2 + 2.075 × 10 - 2 / 3

= 1.981 x 10-2 s - 1

 

Question18. A reaction is first order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

Solution :
(i) The differential rate equation will be

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

Question19. In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

A/ mol L−1

0.20

0.20

0.40

B/ mol L−1

0.30

0.10

0.05

r0/ mol L−1 s−1

5.07 × 10−5

5.07 × 10−5

1.43 × 10−4

What is the order of the reaction with respect to A and B?

Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

Dividing equation (iii) by (ii), we obtain

= 1.496

= 1.5 (approximately)

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

Question20. The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Experiment

A/ mol L−1

B/ mol L−1

Initial rate of formation of D/mol L−1 min−1

I

0.1

0.1

6.0 × 10−3

II

0.3

0.2

7.2 × 10−2

III

0.3

0.4

2.88 × 10−1

IV

0.4

0.1

2.40 × 10−2

Determine the rate law and the rate constant for the reaction.

Solution :
Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

 

 

Dividing equation (iv) by (i), we obtain

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

Dividing equation (iii) by (ii), we obtain

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

 

Question21. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment

A/ mol L−1

B/ mol L−1

Initial rate/mol L−1 min−1

I

0.1

0.1

2.0 × 10−2

II

--

0.2

4.0 × 10−2

III

0.4

0.4

--

IV

--

0.2

2.0 × 10−2

Solution :
The given reaction is of the first order with respect to A and of zero order with respect to B.

Therefore, the rate of the reaction is given by,

Rate = k [A]1 [B]0

⇒ Rate = k [A]

From experiment I, we obtain

2.0 × 10−2 mol L−1 min−1 = k (0.1 mol L−1)

⇒ k = 0.2 min−1

From experiment II, we obtain

4.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.2 mol L−1

From experiment III, we obtain

Rate = 0.2 min−1 × 0.4 mol L−1

= 0.08 mol L−1 min−1

From experiment IV, we obtain

2.0 × 10−2 mol L−1 min−1 = 0.2 min−1 [A]

⇒ [A] = 0.1 mol L−1

Question22. Calculate the half-life of a first order reaction from their rate constants given below:

(i) 200 s−1 (ii) 2 min−1 (iii) 4 years−1

Solution :
(i) Half life, t 1/2 = 0.693 / k

= 0.693 / 200 s-1

= 3.47×10 -3 s (approximately)

(ii) Half life, t 1/2 = 0.693 / k

 = 0.693 / 2 min-1

= 0.35 min (approximately)

(iii) Half life, t 1/2 = 0.693 / k

= 0.693 / 4 years-1

= 0.173 years (approximately)

Question23. The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Solution :
Here,  k = 0.693 / t1/2

= 0.693 / 5730 years-1

It is known that,

NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Question24. The experimental data for decomposition of N2O5

[2N2O5 → 4NO2 + O2]  in gas phase at 318K are given below:

t(s)

0

400

800

1200

1600

2000

2400

2800

3200

102 × [N2O5]  mol L-1

1.63

1.36

1.14

0.93

0.78

0.64

0.53

0.43

0.35

(i) Plot [N2O5] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log [N2O5] and t.

(iv) What is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii).

Solution :
NCERT Solutions for Class 12  Chemistry Chapter 4 - Chemical Kinetics

(ii) Time corresponding to the concentration, 1630x102 / 2 mol L-1 = 81.5 mol L-1 is the half life. From the graph, the half life is obtained as 1450 s.

(iii)

t(s)

102 × [N2O5]  mol L-1

Log[N2O5]

0

1.63

− 1.79

400

1.36

− 1.87

800

1.14

− 1.94

1200

0.93

− 2.03

1600

0.78

− 2.11

2000

0.64

− 2.19

2400

0.53

− 2.28

2800

0.43

− 2.37

3200

0.35

− 2.46

Class 12 NCERT Solutions for Chemical Kinetics

(iv) The given reaction is of the first order as the plot, Log[N2O5v/s t, is a straight line. Therefore, the rate law of the reaction is

Rate = k [N2O5]

(v) From the plot, Log[N2O5] v/s t, we obtain

- k /2.303

Again, slope of the line of the plot  Log[N2O5v/s t is given by

- k / 2.303. = -0.67 / 3200

Therefore, we obtain,

- k / 2.303  = - 0.67 / 3200

⇒ k = 4.82 x 10-4 s-1

 

(vi) Half-life is given by,

t½ = 0.693 / k

= 0.639 / 4.82x10-4 s

=1.438 x 103

This value, 1438 s, is very close to the value that was obtained from the graph.

25. The rate constant for a first order reaction is 60 s−1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Solution :
It is known that,

Class 12 NCERT Solutions for Chemical Kinetics

 

Question26. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Solution :
,

Class 12 NCERT Solutions for Chemical Kinetics

 

Therefore, 0.7814 μg of 90Sr will remain after 10 years.

Again,

Class 12 NCERT Solutions for Chemical Kinetics

 

Therefore, 0.2278 μg of 90Sr will remain after 60 years.

Question27. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Solution :
For a first order reaction, the time required for 99% completion is

t1 = 2.303/k Log 100/100-99

= 2.303/k Log 100

= 2x 2.303/k

For a first order reaction, the time required for 90% completion is

t2 = 2.303/k Log 100 / 100-90

= 2.303/k Log 10

= 2.303/k

Therefore, t1 = 2t2

Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

Question28. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Solution :
For a first order reaction,

t = 2.303/k Log [R] º / [R]

k = 2.303/40min  Log 100 / 100-30

= 2.303/40min  Log 10 / 7

= 8.918 x 10-3 min-1

 

Therefore, t1/2 of the decomposition reaction is

t1/2 = 0.693/k

=  0.693 / 8.918 x 10-3  min

= 77.7 min (approximately)

= 77.7 min (approximately)

Question29. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t (sec)

P(mm of Hg)

0

35.0

360

54.0

720

63.0

Calculate the rate constant.

Solution :
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

Class 12 NCERT Solutions for Chemical Kinetics

After time, t, total pressure, Pt = (Pº - p) + p + p

⇒ Pt = (Pº + p)

⇒ p =
Pt  - P0

therefore, Pº - p = P0  - Pt  - P0

= 2P0 − Pt

For a first order reaction,

k = 2.303/t  Log P0 /P0  - p

=   2.303/t  Log
P0 / 2 P0  -  Pt

When t = 360 s, k = 2.303 / 360s log 35.0 / 2x35.0 - 54.0

= 2.175 × 10−3 s−1

When t = 720 s, k = 2.303 / 720s log 35.0 / 2x35.0 - 63.0

= 2.235 × 10−3 s−1

Hence, the average value of rate constant is

k = (2.175 × 10 - 3  + 2.235 × 10 - 3 ) / 2   s - 1

= 2.21 × 10−3 s−1

Question30. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

SO2Cl2(g)  →  SO2(g) + Cl2(g)

Experiment

Time/s−1

Total pressure/atm

1

0

0.5

2

100

0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.

Solution :
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.

Class 12 NCERT Solutions for Chemical Kinetics

After time, t, total pressure, Pt = (Pº - p) + p + p

⇒ Pt = (Pº + p)

⇒ p = Pt  - Pº

therefore, Pº - p = Pº  - Pt  - Pº

= 2 Pº -  Pt

For a first order reaction,

k = 2.303/t  Log  Pº / Pº  - p

=   2.303/t  Log  Pº / 2 Pº  -  Pt

 

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2x0.5 - 0.6

= 2.231 × 10 - 3s - 1

 

When Pt= 0.65 atm,

P0+ p= 0.65

⇒ p= 0.65 - P0

= 0.65 - 0.5

= 0.15 atm

Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is

PSOCL2 = P0 - p

= 0.5 - 0.15

= 0.35 atm

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate = k(pSOCL2)

= (2.23 × 10 - 3s - 1) (0.35 atm)

= 7.8 × 10 - 4 atm s - 1

 

Question31. The rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C

0

20

40

60

80

105 X K /S-1

0.0787

1.70

25.7

178

2140

Draw a graph between ln k and 1/T and calculate the values of A and Ea.

Predict the rate constant at 30º and 50ºC.

Solution :
From the given data, we obtain

T/°C

0

20

40

60

80

T/K

273

293

313

333

353

1/T / k-1

3.66×10−3

3.41×10−3

3.19×10−3

3.0×10−3

2.83 ×10−3

105 X K /S-1

0.0787

1.70

25.7

178

2140

ln k

−7.147

− 4.075

−1.359

−0.577

3.063

Class 12 NCERT Solutions for Chemical Kinetics

Slope of the line,

NCERT Solutions  class 12 chemistry

In k= - 2.8

Therefore, k = 6.08x10-2s-1

Again when T = 50 + 273K = 323K,

1/T = 3.1 x 10-3 K

In k = - 0.5

Therefore, k = 0.607 s-1

Question32. The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Solution :
k = 2.418 × 10−5 s−1

T = 546 K

Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1

According to the Arrhenius equation,

NCERT Solutions  class 12 chemistry

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 1012 s−1 (approximately)

Question33. Consider a certain reaction A → Products with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.

Solution :
k = 2.0 × 10−2 s−1

T = 100 s

[A]o = 1.0 moL−1

Since the unit of k is s−1, the given reaction is a first order reaction.

Therefore, k = 2.303/t  Log  [A]º / [A]

⇒2.0 × 110-2 s-1  = 2.303/100s  Log  1.0 / [A]

⇒2.0 × 110-2 s-1  = 2.303/100s  ( - Log [A] )

⇒ - Log [A] = -  (2.0 x 10-2 x 100) /   2.303

⇒ [A] = antilog [-  (2.0 x 10-2 x 100)  /  2.303]

= 0.135 mol L−1 (approximately)

Hence, the remaining concentration of A is 0.135 mol L−1.

Question34. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Solution :For a first order reaction,

k = 2.303/t  Log  [R]º / [R]

It is given that, t1/2 = 3.00 hours

Therefore, k = 0.693 / t1/2

= 0.693 / 3  h-1

= 0.231 h - 1

Then, 0.231 h - 1 = 2.303 / 8h  Log  [R]º / [R]

NCERT Solutions  class 12 chemistry

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Question35. The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s−1) e−28000 K/T Calculate Ea.

Solution :
The given equation is

k = (4.5 × 1011 s−1) e−28000 K/T (i)

Arrhenius equation is given by,

k= Ae -Ea/RT  (ii)

From equation (i) and (ii), we obtain

Ea  / RT  =  28000K / T

⇒ Ea  = R x 28000K

= 8.314 J K−1 mol−1 × 28000 K

= 232792 J mol−1

= 232.792 kJ mol−1

Question36. The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k = 14.34 − 1.25 × 104 K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Solution :
Arrhenius equation is given by,

k= Ae -Ea/RT  

⇒In k = In A - Ea/RT

⇒In k = Log A - Ea/RT

⇒ Log k = Log A - Ea/2.303RT         (i)

 

The given equation is

Log k = 14.34 - 1.25 104 K/T             (ii)

From equation (i) and (ii), we obtain

Ea/2.303RT  = 1.25 104 K/T

= 1.25 × 104 K × 2.303 × 8.314 J K−1 mol−1

= 239339.3 J mol−1 (approximately)

= 239.34 kJ mol−1

Also, when t1/2 = 256 minutes,

k = 0.693 / t1/2

= 0.693 / 256

= 2.707 × 10 - 3 min - 1

= 4.51 × 10 - 5s - 1 

= 2.707 × 10−3 min−1

= 4.51 × 10−5 s−1

It is also given that, log k = 14.34 − 1.25 × 104 K/T

NCERT Solutions  class 12 chemistry

= 668.95 K

= 669 K (approximately)

Question37. The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?

Solution :
From Arrhenius equation, we obtain

log k2/k1 = Ea / 2.303 R (T2 - T1) / T1T2

Also, k1 = 4.5 × 103 s−1

T1 = 273 + 10 = 283 K

k2 = 1.5 × 104 s−1

Ea = 60 kJ mol−1 = 6.0 × 104 J mol−1

Then,

NCERT Solutions  class 12 chemistry

= 297 K

= 24°C

Hence, k would be 1.5 × 104 s−1 at 24°C.

Question38. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s−1. Calculate k at 318 K and Ea.

Solution :
For a first order reaction,

t = 2.303 / k log a  / a - x

At 298 K, t = 2.303 / k log 100  / 90

= 0.1054 / k

At 308 K, t' = 2.303 / k' log 100  / 75

= 2.2877 / k'

According to the question,

t = t'

0.1054 / k  =  2.2877 / k'

⇒ k' k  = 2.7296

 

From Arrhenius equation, we obtain

NCERT Solutions  class 12 chemistry

To calculate k at 318 K,

It is given that, A = 4 x 1010 s-1, T = 318K

Again, from Arrhenius equation, we obtain

NCERT Solutions  class 12 chemistry

Therefore, k = Antilog (-1.9855)

= 1.034 x 10-2 s -1

Question39. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Solution :
From Arrhenius equation, we obtain

NCERT Solutions  class 12 chemistry

Hence, the required energy of activation is 52.86 kJmol−1.

NCERT Solutions For Class 12 Chemistry Chapter Wise.

  1. Chapter 1 The Solid States

  2. Chapter 2 Solutions

  3. Chapter 3 Electrochemistry

  4. Chapter 4 Chemical Kinetics

  5. Chapter 5 Surface Chemistry

  6. Chapter 6 General Priciples and Processes of Isolation of Elements

  7. Chapter 7 The P Block Elements

  8. Chapter 8 The D and F Elements

  9. Chapter 9 Coordination Compounds

  10. Chapter 10 Haloalkanes and Haloarenes

  11. Chapter 11 Alcohols, Phenols and Ethers

  12. Chapter 12 Aldehydes, ketones and Carboxylics Acids

  13. Chapter 13 Amines

  14. Chapter 14 Biomolecules

  15. Chapter 15 Polymers

  16. Chapter 16 Chemistry In Everyday Life

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