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.

Chapter: 4

Part – I

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

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(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−1/2L1/2s−1

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

Ans: Order of Reaction:

[Rate] = k[C2​H5​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  → A2

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 L1s−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  × 104 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 (PCH3OCH3 )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:

t/s0306090
[A]/ mol L–10.550.310.170.085

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:

A/ mol L–10.200.200.40
B/ mol L–10.300.100.05
r 0 /mol L–1s –15.07 × 10–55.07 × 10–51.43 × 10–4

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,

A (mol L⁻¹)0.200.200.40
B (mol L⁻¹)0.300.100.05
r0​ (mol L⁻¹ s⁻¹)5.07×10−55.07×10−51.43×10−4

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

Experiment[A]/mol L–1 [B]/mol L–1Initial rate of formation of D/mol L–1 min–1
I0.10.16.0 × 10–3
II0.30.27.2 × 10–2
III0.30.42.88 × 10–1
IV0.40.12.40 × 0–2

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:

Experiment[A]/ mol L–1[B]/ mol L–1Initial rate/ mol L–1 min–1
I0.10.12.0 × 10–2
II0.24.0 × 10–2
III0.40.4
IV0.22.0 × 10–2

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

Experiment[A]/ mol L–1[B]/ mol L–1Initial rate/ mol L–1 min–1
I0.10.12.0 × 10–2
II0.20.24.0 × 10–2
III0.40.48.0 × 10–2
IV0.10.22.0 × 10–2

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 t1/2 for a first-order reaction is given by the formula:

t1/2​ = 0.693/k

Where:

t1/2​ is the half-life,

k is the rate constant.

(a) k = 200s−1

t1/2​ = 0.693/200 

= 3.465 × 10-3 second 

(b) k = 2min−1

t1/2​ = 0.693/2min−1

= 0.3465 min.

(c) k = 4 years−1

t1/2​ = 0.693/4years−1

= 0.17325 years

14. The half-life for radioactive decay of 14C 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: Nt ​= 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 /t1/2

= 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:

t/s0400800120016002400280032003200
102×[N2O5 ]/mol-1 1.631.361.140.930.780.640.530.430.35

(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, T1/2 =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: t1/2​ = 0.693/k

Given ​

k = 4.84 × 10−4s−1

 t1/2​ = (0.693)/(4.84 × 10−4)

t1/2​ = (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.

t15/16 = ?

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 (No/N)

t =10 years

No = 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,

t99% = 4.602/k…………..(i)

When reaction is 90% complete,

(2.303/k)log10 = 2.303/k ………….(ii)

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

t99%/ t90% = (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 t1/2. 

Ans: 

Where:

t = 40 min,

x = 30% decomposition

[A0​−x] = 100−30 = 70%.

A = 100.

k = 8.19 × 10−3 min−1

t1/2​ = 0.693/(8.919 × 10−3

= 77.6 minutes.

Therefore the half-life t1/2 is indeed 77.6 minutes.

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

t (sec)P(mm of Hg) 
035.0
36054.0
72063.0

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)

ExperimentTime/s–1Total pressure/atm
100.5
21000.6

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

Ans: SO2Cl2(g) → SO2(g)Cl2(g)

Att = 0P000
Att = tPo-PPP

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

= Pt = (Po+P)

= p= Pt-Po

∴ Po – p = Po – Pt -Po

= 2Po – Pt

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) = 2Pi -Pt

= 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:

T/°C020406080
105 × k/s-10.07871.7025.71782140

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)

Temperature (°C)Temperature (K)1/T​ (K−1)k (×103 s−1)lnk
0273.150.003660.0787-2.553
20293.150.003411.700.526
40313.150.0031925.73.440
60333.150.003001785.182
80353.150.0028321407.329

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

t1/2 = 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 t1/2​ 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 Ea.

Ans: According to Arrhenius equation

k = AeRT−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 t1/2 = 256 min 

For I st order reaction: 

t1/2​ = 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 = 10C = 283

Ea​ = 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)

= 24.19C

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:

k1​ 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.

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