NCERT Class 12 Chemistry Chapter 4 Chemical Kinetics

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NCERT Class 12 Chemistry Chapter 4 Chemical Kinetics

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 12 Chemistry Chapter 4 Chemical Kinetics Solutions are part of All Subject Solutions. Here we have given NCERT Class 12 Chemistry Part: I, Part: II Notes. NCERT Class 12 Chemistry Chapter 4 Chemical Kinetics Notes, NCERT Class 12 Chemistry Textbook Solutions for All Chapters, You can practice these here.

Chemical Kinetics

Chapter: 4

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

(i) 3NO(g) → N₂O (g) Rate = k[NO]²

Ans: Rate = k[NO]²

Order of Reaction:

Order with respect to NOis 2

Overall order of the reaction is 2.

The dimensions of the rate constants [k] is mol⁻¹Ls⁻¹

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

Ans: Order of Reaction:

Rate = k[H₂​O₂​][I⁻].

Order with respect to H₂O₂ is 1

Order with respect to I⁻ is 1.

Overall order of the reaction is 2.

The dimensions of the rate constants [k] is mol⁻¹Ls⁻¹

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

Ans: Order of Reaction:

[Rate] = k[CH₃​CHO]^3/2

The reaction order with respect to CH₃CHO is 3/2​.

The overall order of the reaction is 3/2.

The dimensions of the rate constants [k] = mol^−1/2L^1/2s⁻¹

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

Ans: Order of Reaction:

[Rate] = k[C₂​H_5​Cl]

The reaction order with respect to C₂H₅Cl is 1.

The overall order of the reaction is 1.

The dimensions of the rate constants [k] = s⁻¹

2. 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. 

Ans: 2A + B → A2​B

Rate expression: Rate = k[A][B]² 

k= (2.0 × 10⁻⁶ mol) × (0.2 mol L^-1)²

= 8.0 × 10^-9 mol L^-1 s^-1

 Rate After Reduction of [A]

New concentration: [A] = 0.06 mol L⁻¹[A] = 0.06, 

[B] = 0.2 mol L⁻¹

New concentration of B: 0.2−0.02

= 0.18 mol L⁻¹

R = (2.0 × 10−6 mol−2L2s⁻¹) × (0.06mol L⁻¹) × (0.18mol L⁻¹)²

= 0.0324 mol ² L⁻²

Ratenew​ = 3.89 × 10−9mol L⁻1s⁻¹

3. 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? 

Ans: 

For the zero order reaction rate: K

= 2.5 × x10⁻⁴ mol/L/s

N2 = d[N₂] / dt 

= 2.5  × 10^–4 mol/L/s

Rate of production of 

d[H2​]/dt = 3 × k

= 3 × 2.5 ×1 0⁻⁴ 

= 7.5 × x10⁻⁴ molL/s

4. 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 (P_CH3O_CH3 )^3/2

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

Ans: Unit of rate of reaction = bar min⁻¹

Units of Rate (K) = Rate​/ (PCH₃​OCH₃​​)^3/2

= Units of Rate Constant (K): bar min⁻¹/ (bar)^3/2

= bar^-1/2 min⁻¹

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

Ans: The various factors which affecting the rate of a chemical reaction are:

(i) Concentration of reactants.

(ii) Temperature of reaction.

(iii) Nature of reactants.

(iv) Surface area.

(v) Exposure to radiation.

(vi) Presence of catalyst.

6. 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.

Ans: The initial rate of the reaction is:

Rate = k[A]² = ka²

When construction of A is double 

I.e. [A] = 2a 

Rate of reaction becomes 4 times.

(ii) reduced to half? 

Ans: If the concentration of the reactant is reduced to half, the rate of reaction becomes one fourth.

I.e  [A] = ½(a)

Rate = k(a/2)² 

= 1/4ka²

7. What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? 

Ans: The rate constant of a reaction increases with increase of temperature and becomes nearly double for every 10° rise of temperature. This is also called temperature coefficient. It is the ratio of rate constants of the reaction at two temperatures differing by ten degrees. According to Arrhenius equation,

k = Ae ^-EART

where Ea represents the activation energy of the reaction and A represents the frequency factor.

8. In a pseudo first order reaction in water, the following results were obtained:

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

Ans: (b) The average rate of reaction between the time interval 30 to 60 is

Average rate =

= 0.14/30 

= – 0.00467 

= -4.67 ×10^-3Ms^-1

Minus sign shows that rate of reaction is decreasing with time as conc. Of ester is decreasing with time. 

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

(i) Write the differential rate equation. 

Ans: Rate = k[A] [B]²

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

Ans: Rate = k[A][3B]² = 9k [A] [B]²

Thus, the rate of the reaction will increase by a factor of 9 when the concentration of B is increased three times

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

Ans: Rate = k[2A] [2B]² = 8k [A] [B]

Thus, the rate of the reaction will increase by a factor of 8 when the concentrations of both A and B are doubled.

10. 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:

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

Ans: r₀​ = k[A]m[B]n

Given,

[A] and [B] are the concentrations of reactants A and B,

m and n are the orders of the reaction with respect to A and B, respectively,

k is the rate constant.

r₀​ is the initial rate of reaction,

Dividing equation (i) by equation (ii)

= 1

[3]_y = [3]⁰ 

= 1 

y = 0

Now dividing equation (ii) by Equation (iii) 

[½]^x = ½.82

2x = 2.82

X log 2 = log 2.82

X = 1.4957 ≃1.5 

Order with respect to A: 1.5

Order with respect to B: 0.

11. The following results have been obtained during the kinetic studies of the reaction: 2A + B ➡ C + D

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

Ans: The rate law expression: k[A]^x[B]^y

Here:

k is the rate constant,

m and n are the reaction orders with respect to A and B,

[A] and [B] are the concentrations of the reactants.

(Rate₁​) = 6.0 × 10⁻³ = k(0.1)^x (0.1)^y……………(i)

(Rate₂​) = 7.2 × 10⁻² = k(0.3)^x (0.2)^y……………(ii)

(Rate₃​) = 2.88 × 10⁻¹ = k(0.3)^x (0.4)^y………….(iii)

(Rate₄​) = 2.40 × 10⁻² = k(0.4)^x (0.1)^y…………(iv)

Divide equation (ii) by equation (i)

Or

¼ = (0.1)^x/(0.4)^y

= (¼)^x

∴ x = 1

From experiments II and III, we get

Or 

¼ = (0.2)^y / (0.4)^y 

= (½)y

∴ y = 2

The rate law expression is, k[A][B]²

Substitute values in equation (i)

= 6.0 mol^-2 min^-1

K = 6.0mol^-2 L² min^-1.

12. 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:

Ans: Rate law for the reaction = k[A]¹[B]⁰ = k[A]

From the table for Experiment I, we know

Rate = 2.0 × 10⁻² mol L⁻¹ min⁻¹

[A] = 0.1mol L⁻¹

[B] = 0.1mol L⁻¹

For I nd Experiment

Rate = k[A]

2.0 × 10⁻² = k × 0.1

k = (2.0 × 10⁻²​)/0.1

= 0.2Min^-1

For II nd Experiment 

Rate = 4.0 × 10⁻² mol L⁻¹ min⁻¹

[B] = 0.2mol L⁻¹

4.0 × 10⁻² = 0.2 × [A] 

[A] = 4.0 ×10⁻²​/0.2 

= 0.2mol L⁻¹ 

For III nd Experiment 

[A] = 0.4mol L⁻¹

[B] = 0.4mol L⁻¹

Rate = k[A]

= 0.2 × 0.4 mol L⁻¹ min ⁻¹

= 8.0 × 10⁻² mol L⁻¹ min⁻¹

For IV nd Experiment 

Rate = k[A]

Rate = 2.0 × 10⁻² mol L⁻¹ min⁻¹

[B] = 0.2mol L⁻¹

= 2.0 × 10⁻² mol L⁻¹ min⁻¹

= 0.2 × [A]

[A] = (2.0 × 10⁻²)/0.2

= 0.1mol L⁻¹

13. 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 

Ans: The half-life t_1/2 for a first-order reaction is given by the formula:

t_1/2​ = 0.693/k

Where:

t_1/2​ is the half-life,

k is the rate constant.

(a) k = 200s⁻¹

t_1/2​ = 0.693/200 

= 3.465 × 10^-3 second 

(b) k = 2min⁻¹

t_1/2​ = 0.693/2min⁻¹

= 0.3465 min.

(c) k = 4 years⁻¹

t_1/2​ = 0.693/4years⁻¹

= 0.17325 years

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

Ans: N_t ​= N₀​⋅e ^−λt

k = 1.209 × 10⁻⁴years⁻¹,

NO = 100 (original amount),

N = 80(remaining amount).

λ is the decay constant,

t is the time elapsed.

λ = 0.693 /t_1/2

= 0.693/5730

= 1.21 × 10^-4 year^-1

All radioactive nuclear are first order process,

Therefore, decay constant,

= 1845 year

So, the estimated age of the sample is approximately 1844 years

15. The experimental data for decomposition of 

[N₂O₅ [2N₂O₅  → 4NO₂ + O₂]

in gas phase at 318K are given below:

(i) Plot [N₂O₅ ] against t. 

Ans: 

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

Ans: Initial concentration ([N₂O₅]₀) = 1.63 × 10² mol⁻¹

Half of this value = 1.63/2 = 0.815 × 10² mol⁻¹

Time for half of initial concentration (i.e half life period) from plot is 1440 s

Hence, T_1/2 =1440s

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

Ans: 

(iv) What is the rate law? 

Ans: As plot of log (N₂O₅) vs. time is a straight line, hence it is a reaction of first order i.e., rate law is 

Thus, the rate law is: Rate = k[N₂O₅]

Here k is the rate constant.

(v) Calculate the rate constant.

Ans: = -(k/2303)

Slope = (-2.46 -(1.79))/3200-0 

= -0.67 / 3200

I.e., k / 2303 = 0.67 / 3200

K = (0.67/3200) × 2.303

= 4.82 × 10^-4 mol L^-1 s^-1

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

Ans: t_1/2​ = 0.693/k

Given ​

k = 4.84 × 10⁻⁴s⁻¹

 t_1/2​ = (0.693)/(4.84 × 10⁻⁴)

t_1/2​ = (0.693)/0.000484s

= 1432 seconds.

16. 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? 

Ans: For the first-order reaction we have:

Here:

The rate constant k = 60 s⁻¹.

t_15/16 = ?

Rate constant of first order reaction is given by

Where 15/16th the reaction is over then

= 0.046 S 

= 4.6 × 10^-2sec.

17. During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1mg 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. 

Ans: λ = 0.693 / t1/2 

= 0.693/28.1

= 0.0247 year-1

(i) After 10 years, let xμ be the concentration of ⁹⁰Sr.

λ = (2.303/t) log (N_o/N)

t =10 years

N_o = 1 microgram 

= 1 × 10-6g,

N = ?

(ii) After 60 years, let yμ be the concentration of ⁹⁰Sr.

N = (1 × 10^-6)/4.400 

= 0.227ug.

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

Ans: 

When reaction is 99% complete,

t_99% = 4.602/k…………..(i)

When reaction is 90% complete,

(2.303/k)log10 = 2.303/k ………….(ii)

Divide equation (1) by (2), we get

t_99%/ t_90% = (4.606/h) × (k/2.303)

= 2 

The time required for 99% completion of a first-order reaction is indeed twice the time required for the completion of 90% of the reaction.

19. A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2. 

Ans: 

Where:

t = 40 min,

x = 30% decomposition

[A₀​−x] = 100−30 = 70%.

A = 100.

k = 8.19 × 10⁻³ min⁻¹

t_1/2​ = 0.693/(8.919 × 10⁻³) 

= 77.6 minutes.

Therefore the half-life t_1/2 is indeed 77.6 minutes.

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

Calculate the rate constant. 

Ans: (CH₃​)2​CHN = N(CH₃​)2​ → N₂ ​+ C₆​H1₄​

Given:

Initial pressure P0 = 35.0 mm of Hg

Pressure at t = 360 sect = 3,

Pt​ = 54.0mm of Hg

Pressure at t = 720 sect

Pt​ = 63.0mm of Hg

Where:

P0​ is the initial pressure, 35.0mm of Hg,

Pt​ is the pressure at time t, 54.0mm of Hg.

= 2.2 × 10^-3s^-1

K = 2.303 / 720s

= 2.2 ×10^-3s^-1.

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

SO₂Cl₂(g) → SO₂(g)Cl₂(g)

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

Ans: SO₂Cl₂(g) → SO₂(g)Cl₂(g)

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

= Pt = (P_o+P)

= p= P_t-P_o

∴ P_o – p = P_o – P_t -P_o

= 2P_o – P_t

Initial pressure, P₀ = 0.6 atm

Total pressure at t = 100 sect is 0.6 atm

Total pressure we need to evaluate is 0.65 atm

t = 100 sect 

= 2.23 × 10-3s-1

For total pressure of 0.65atm

Rate of reaction = kp(SO₂CI₂)

From the reaction stoichiometry 

p(SO₂​CI₂) = 2P_i -P_t

= 1 atm -0.65atm.

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

= 2.23 × 10^-3s^-1 × 0.35atm

= 7.8 × 10^-4 s^-1 atm.

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

Draw a graph between ln k and 1/T and calculate the values of A and Ea . Predict the rate constant at 30° and 50°C. 

Ans: Convert the Celsius temperatures to Kelvin by adding 273.15: 

T0 ​= 0°C = 273.15 K

T20​ = 20°C = 293.15 K

T40​ = 40°C = 313.15 K

T60​ = 60°C = 333.15 K

T80​ = 80°C = 353.15 K

Calculation of 1/T and Ink:

1​/T (reciprocal of temperature in Kelvin)

In k (natural logarithm of the rate constant)

Slope =  -2.4 / 0.00047 

= Ea/ 2.303R

Therefor Activation energy 

(Ea) = (2.4 × 2.303 × 8.314J mol^-1) / (0.00047)

= 97875J mol 

= 97.875kJmol^-1

Ea = 97.875kJmol^-1

Now, 

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

Ans: According to Arrhenius equation

Log k = log A-(Ea/2.303Rt)

K = 2.418 × 10-5s^-1

Ea = 179.98 kJ mol^-1

= 179900 J mol^-1

R = 8.314 JK-1 mol^-1

T = 546K

= 12.5916

A = Antilog 12.5916

= 3.9 × 10¹²s^-1

A = 3.9 × 10¹²s^-1.

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

Ans: According to Arrhenius equation

k = 2.1418 × 10^-5s^-1

Ea = 179.9kJmol^-1

= 179900 J^-1 mol^-1

R = 8.314 JK^-1

T = 546K

= -4.6184 + 17.21

= 12.5916

A = Antilog 12.5916

= 3.9 × 10¹²s^-1

A = 3.9 × 10¹²s^-1.

25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t _1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Ans: K = 2.0 × 10^-2s^-1

t_1/2 = 100 sec, a = 1.0 molL^-1

Or 0.8684 = log (1/ (1-x))

Antilog 0.8684 = 1/(1-x)

Or 7.386 (1-x) =1

Or 1-x = 1/7.386 

= 0.135

∴ Concentration of A remaining after 100S = 0.135M.

The relationship between the rate constant k and the half-life t_1/2​ for a first-order reaction is

K = 0.693 / t _1/2

Given 

t _1/2 = 3.00 hours

t = 8 hours.

For the first order reaction, 

K = 0.693. 3.00 

= 0.231hours^-1

Now

Calculation of  first-order decay equation to find the fraction remaining after 8 hours:

a/(a-x) = 6.345

Or Fraction left = (a-x)/a 

= 1/6.345 = 0.157M.

26. The decomposition of hydrocarbon follows the equation 

k = (4.5 × 1011s –1) e ^-28000K/T 

Calculate E_a.

Ans: According to Arrhenius equation

k = A^{eRT−Ea​​}

Here:

k is the rate constant,

A is the pre-exponential factor (here, 4.5 × 1011s⁻¹),

Ea​ is the activation energy,

R is the universal gas constant (8.314 J/mol\cdotpK 8.314),

T is the temperature in Kelvin.

On comparing both equations,

-(Ea/RT) = -28000K/T

Ea = (28000k) × R

= 28000 × 8.314 J/mol^-1

= 232792J/mol^-1

Thus, the activation energy (Ea​) is approximately 232.8 kJ/mol.

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

log k = 14.34 – 1.25 × 10⁴K/T 

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

Ans: The Arrhenius equation helps in determining the activation energy for a reaction.

K = Ae^{– Ea/Rt}

Calculation of activation energy Ea According to Arrhrnius equation,

On comparing both equation,

Ea/2.303Rt = 1.25 × 10⁴ K × 2.303 × 8.314 jK^-1 Mol^-1

= 23.93 × 10⁴ J mol^-1

= 239.3 kJ mol^-1.

Calculation of required temperature if t_1/2 = 256 min 

For I st order reaction: 

t_1/2​ = 0.693/k

= 0.693/ (256 min)

= 4.51 × 10^-5 s^-1

According to Arrhenius theory

Log4.51 × 10⁻⁵ = 14.341.25 × 104K/T

-4.35 = 14.341.25 × 104K/T

T = 669 K

Therefore, the temperature at which the half-life period is 256 minutes is 669 K.

28. 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 × 10⁴s^–1?

Ans: 

k₁​ = 4.5 × 10³s⁻¹ 

k₂​ = 1.5 × 10⁴s⁻¹

T₁ = 10^∘C = 283

E_a​ = 60kJ/mol = 60000J/mol.

1-(283/T2) = 0.04776

Or

T₂ = 283/1-0.04776 

= 283/0.95224

T₂ = 297.19 K

= (297.19 -273.0)

= 24.19^∘C

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

Ans: Calculation of Activation energy (Ea) 

Here:

k_1​ is the rate constant at T₁ = 298 

k₂ is the rate constant at T₂ = 308 

A is the pre-exponential factor

Ea​ is the activation energy

R is the gas constant (8.314 J/mol·K)

For 10% completion of the reaction, we have (i)

For 25% completion of the reaction, we have(ii)

Dividing Rq (ii) by (i)

According to Arrhenius theory: 

Calculation of rate constant(K)

Logk = 10.6021-12.5870

= 1.9849

K = Antilog (-1.9849)

= Antilog (2^–.0151)

= 1.035 × 10^-2s^-1

Ea = 76.640kJ mpl^-1

k = 1.035 V10^-2s¹.

30. 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.

Ans: According to Arrhenius equation:

= 5.2863 × 10⁴ J mol^-1

= 52.863 kJ mol^-1.