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| Chapter | 4. Chemical Kinetics |
| Subject | Chemistry |
| Textbook | Chemistry I |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 4: Chemical Kinetics.
NCERT Solutions for Class 12 Chemistry Chapter 4
Chemical Kinetics Solutions
Q1) 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/
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl]
Answer)
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
Q2) 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.
Answer) 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
Q3) 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?
Answer)
Q4)
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?
Answer)
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
Q5) Mention the factors that affect the rate of a chemical reaction.
Answer) The factors that affect the rate of a reaction are as follows:
- Concentration of reactants (pressure in case of gases)
- Temperature
- Presence of a catalyst
Q6) 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?
Answer) Let the 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 1/4th.
Q7) 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?
Answer) 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.
Q8) In a pseudo first order hydrolysis of ester in water, the following results were obtained:
| t/s | 0 | 30 | 60 | 90 |
| [Ester]mol / L | 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.
Answer)
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
Q9) 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?
Answer)
Q10) 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?
Answer)
Q11) 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.
Answer)
Q12) 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 |
Answer) 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
Q13) 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
Answer 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)
Q14) 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.
Here, k = 0.693 / t1/2
= 0.693 / 5730 years-1
It is known that,
t = 2.303/k log[R]0/[R]
= 2.303/0.693/5730 log 100/80
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.
Q15) 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).
Answer)
i)
ii) Time corresponding to the concentration, 1630×102 / 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 |
iv) The given reaction is of the first order as the plot, Log[N2O5] v/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[N2O5] v/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 / 4.82×10-4 s
=1.438 x 103
This value, 1438 s, is very close to the value that was obtained from the graph.
Q16) 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?
Answer)
Q17) 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.
Answer)
Therefore, 0.7814 μg of 90Sr will remain after 10 years.
Again,
Therefore, 0.2278 μg of 90Sr will remain after 60 years.
Q18) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer)
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.
Q19) A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Answer) 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)
Q20) 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.
Answer)
Q21) 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.
Answer) The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.
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 / 2×0.5 – 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
Q22) 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.
Answer)
Q23) 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.
Answer) 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.
Q24) 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?
Answer) for the first order
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
Q25) The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s−1) e−28000 K/T. Calculate Ea.
Answer) 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
Q26) 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?
Answer) 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
Q27) 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?
Answer)
Q28) 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.
Answer)
Q29) 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.
Answer)
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