NCERT Solution Class 12th Chemistry Chapter – 4 Chemical Kinetics Question & Answer

NCERT Solution Class 12th Chemistry Chapter – 4 Chemical Kinetics

NCERT Solution Class 12th Chemistry Chapter – 4 Chemical Kinetics

?Chapter – 4?

✍Chemical Kinetics✍

?Question & Answer?

1.(i) 2NO(g)→N₂​O(g);    Rate=k[NO]²
(ii) H₂​O₂​(aq)+3I−(aq)+2H+→2H2O(I)+3I;    Rate=k[H₂​O₂​][I−]
(iii) CH₃​CHO(g)→CH₄​(g)+CO(g);    Rate=k[CH₃​CHO]3/2
(iv) C_2​H₅​Cl(g)→C₂​H₄​(g)+HCl(g);    Rate = k[C₂​H_5​Cl]

Answer:  (i) 3NO(g)→N₂​O(g)Rate=k[NO]₂
Order w.r.t NO is 2 and overall order is 2.
Dimensions of the rate constant
molL⁻¹/s​ = k(molL⁻¹)²   k = mol⁻¹L¹s⁻¹

(ii) H₂​O₂​(aq)+3I−(aq)+2H+→2H2O(I)+I₃−​Rate=k[H₂​O_2​][I−]
Order w.r.t H₂​O₂​ is 1, order w.r.t I− is 1 and overall order is 2.
Dimensions of the rate constant
molL⁻¹/s​ = k(molL⁻¹)²   k = mol⁻¹L¹s⁻¹

(iii) CH₃​CHO(g)→CH₄​(g)+CO(g)Rate=k[CH_3​CHO]^3/2
Order w.r.t CH3​CHO is 3/2 and overall order is 3/2.
Dimensions of the rate constant
molL⁻¹/s​ = k(molL⁻¹)^3/2   k = mol^−1/2L^1/2s⁻¹

(iv) C₂​H_5​Cl(g)→C₂​H_4​(g)+HCl(g)Rate=k[C₂​H_5​Cl]
Order w.r.t C₂​H_5​Cl is 1, and overall order is 1.
Dimensions of the rate constant
molL⁻¹/s​ = k(molL⁻¹)¹   k = s⁻¹​

Q 2. For the reaction: 2A+B→A₂​B the rate = k[A][B]² with k=2.0×10⁻⁶mol⁻²L²s⁻¹. Calculate the initial rate of the reaction when [A]=0.1molL−1,[B]=0.2molL⁻¹. Calculate the rate of reaction after $[A]$ is reduced to 0.06molL⁻¹.

Answer:  The initial rate of the reaction is :
Rate=k[A][B]²

⟹(2.0×10⁻⁶mol⁻²L²s⁻¹)(0.1molL−1)(0.2molL⁻¹)²
⟹8.0×10−9molL⁻¹s⁻¹

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

Therefore, concentration of B reacted=21​×0.04molL⁻¹=0.02molL⁻¹Then, concentration of B available, [B]=(0.2−0.02) molL⁻¹=0.18molL⁻¹

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

Rate=k[A][B]² = (2.0×10⁻⁶ mol⁻²L²s⁻¹)(0.06molL⁻¹)(0.18molL⁻¹)2=3.89×10⁻⁹ molL⁻¹s⁻¹

Q 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?

Answer:  2NH₃​→N₂​+3H_2​
Rate of reaction = −1/2​ d/dt[NH₃​] = d/td​[N₂​] = 1/3 d/dt​[H₂​] = k

For a zero order reaction,
Rate = k = 2.5×10⁻⁴M/s

Rate of production of N₂​ = d/dt[N_2​] = 2.5×10₋₄M/s
Rate of production of H_2​ = d/dt​[H₂​] = 3×2.5×10⁻⁴M/s = 7.5×10⁻⁴M/s

Q 4. The decomposition of dimethyl ether leads to the formation of CH₄, H₂and 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_CH3OCH3)^3/2

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

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

Answer: The factors which are responsible for the effect in chemical reaction’s rate are:

(a) Reaction temperature

(b) Presence of a catalyst

(c) The concentration of reactants (pressure in case of gases)

(d) Nature of the products and reactants

(e) Radiation exposure

(f) Surface area

Q 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 (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

Q 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?

Answer: The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.

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

K = Ae ^{-E_a / RT}

where, kis the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

E_a is the energy of activation for the reaction

Q 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.

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 – 3mol L ^{– 1}s ^{– 1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]_º / [R]

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

= 1.911 × 10 ^{– 2}s ^{– 1}

For t = 60 s, k₂ = 2.303/ 60 log 0.55 / 0.17

= 1.957 × 10 ^{– 2}s ^{– 1}

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

= 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}

Q 9. 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: (a) The differential rate equation will be

(b) If the concentration of B is increased three times, then

Therefore, the reaction rate will be increased by 9 times.

(c) When the concentrations of both A and B are doubled,

Therefore, the rate of reaction will increase 8 times.

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:

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

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

Then,

Dividing equation (i) by (ii), we get

y = 0

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

=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.

Q 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.

Answer: 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,

Rate = k\left [ A \right ]^{x} \left [ B \right ]^{y}k[A]x[B]y

According to the question,

Dividing equation (4) by (1), we get

Dividing equation (3) by (2), we get

From experiment 1, we get

​

= 6.0 L² mol^-2 min^-1

From experiment 2, we get

​

= 6.0 L² mol^-2 min^-1

From experiment 1, we get

​

= 6.0 L² mol^-2 min^-1

From experiment 1, we get

​

= 6.0 L² mol^-2 min^-1

Thus, rate constant, k = 6.0 L² mol^-2 min^-1

Q 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:

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 x 10^-2mol L^-1min^-1= k (0.1 mol L^-1)

⇒ k = 0.2 min^-1

From experiment II, we obtain

4.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.2 mol L^-1

From experiment III, we obtain

Rate = 0.2 min^-1 x 0.4 mol L^-1

= 0.08 mol L^-1min^-1

From experiment IV, we obtain

2.0 x 10^-2mol L^-1min^-1= 0.2 min^-1[A]

⇒ [A] = 0.1 mol L^-1

Q 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

Answer: (i) Half life, t_½ = 0.693 / k

= 0.693 / 200 s^-1

= 3.47 ×10 ^-3 s (approximately)

(ii) Half life, t_½ = 0.693 / k

= 0.693 / 2 min^-1

= 0.35 min (approximately)

(iii) Half life,t_½ = 0.693 / k

= 0.693 / 4 years^-1

= 0.173 years (approximately)

Q 14. 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.

Answer: Here,

K = 0.693 / t_½

= 0.693 / 5730 years^-1

It is known that,

= 1845 years (approximately)

Hence, the age of the sample is 1845 years.

Q 15. The experimental data for decomposition of N₂O₅

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

(a) Plot [N₂O₅] against t.

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

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

(d) What is the rate law ?

(e) Calculate the rate constant.

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

Answer:  

(a)

(b) 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.

(c)

(d) 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]

(e) From the plot,  log[N₂O₅]

v/s t, we obtain

Slope = -2.46 -(1.79) / 3200-0

= -0.67 / 3200

Again, slope of the line of the plot log[N₂O₅] v/s t is given by

– k / 2.303

.Therefore, we obtain,

– k / 2.303  = – 0.67 / 3200

⇒ k = 4.82 x 10^-4 s^-1

(f) half life is given by,

t_½ = 0.693 / k

= 0.639 / 4.82×10^-4 s

=1.438 x 10³ s

Or we can say

1438 S

Which is very near to what we obtain from graph.

Q 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 1/16^th value?

Answer:

Hence, the required time is 4.6 x 10⁻² s 

Q 17. During the nuclear explosion, one of the products is ⁹⁰Sr with a 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

Q 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.

Answer: For a first order reaction, the time required for 99% completionis

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.

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

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, t_1/2 of the decomposition reaction is

t_1/2 = 0.693/k

=  0.693 / 8.918 x 10^-3  min

= 77.7 min (approximately)

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

Calculate the rate constant.

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

After time, t, total pressure, P_t = (P_º – p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t  – P_º

therefore, P_º – p = P_º  – P_t  – P_º

= 2 P_º –  P_t

For a first order reaction,

k = 2.303/t  Log  P_º / P_º  – p

=   2.303/t  Log  P_º / 2 P_º  –  P_t

When t = 360 s, k = 2.303 / 360s log 35.0 / 2×35.0 – 54.0

= 2.175 × 10 ^{– 3} s ^{– 1}

When t = 720 s, k = 2.303 / 720s log 35.0 / 2×35.0 – 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}

Q 21. The following data were obtained during the first order thermal decomposition of SO₂Cl₂at a constant volume.

SO₂Cl₂₍g)  →  SO₂₍g) + Cl₂(g)

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

Answer: The thermal decomposition of SO₂Cl₂at a constant volume is represented by the following equation.

SO₂Cl₂ (g) → SO₂ (g) + Cl₂ (g)

After time, t, total pressure,P_t = (P_º – p) + p + p

⇒ P_t = (P_º + p)

⇒ p = P_t  – P_º

therefore, P_º – p = P_º  – P_t  – P_º

= 2 P_º –  P_t

For a first order reaction,

k = 2.303/t  Log  P_º / P_º  – p

=   2.303/t  Log  P_º / 2 P_º  –  P_t

When t= 100 s,

k = 2.303 / 100s log 0.5 / 2×0.5 – 0.6

= 2.231 × 10 ^{– 3}s ^{– 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

p_SOCL2 = 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(p_SOCL2)

= (2.23 × 10 ^{– 3}s ^{– 1}) (0.35 atm)

= 7.8 × 10 ^{– 4}atm s ^{– 1}

Q 22. 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 the pre-exponential factor?

Answer: k= 2.418 × 10^-5 s^-1 

T= 546 K

E_a= 179.9 kJ mol ^{– 1} = 179.9 × 10³J mol ^{– 1}

According to the Arrhenius equation,

= (0.3835 – 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s ^{– 1}(approximately)

Q 23. 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.

Answer: k= 2.0 × 110^-2 s^-1

T= 100 s

[A]_o= 1.0 moL ^{– 1}

Since the unit of kis 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}.

Q 24. 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 a sample of sucrose remains after 8 hours?

Answer: For a first order reaction,

k = 2.303/t  Log  [R]_º / [R]

It is given that, t_1/2 = 3.00 hours

Therefore, k = 0.693 / t_1/2

= 0.693 / 3  h^-1

= 0.231 h ^{– 1}

Then, 0.231 h ^{– 1} = 2.303 / 8h  Log  [R]_º / [R]

= 0.158

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

Q 25. The decomposition of hydrocarbon follows the equation

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T} Calculate E_a.

Answer: The given equation is

k = (4.5 x 10¹¹ s^-1) e^{-28000 K/T}           (i)

Arrhenius equation is given by,

k= Ae ^-E_a/RT                                                (ii)

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

E_a  / RT  =  28000K / T

⇒ E_a  = R x 28000K 

= 8.314 J K ^{– 1}mol ^{– 1}× 28000 K

= 232792 J mol ^{– 1}

= 232.792 kJ mol ^{– 1}

Q 26. The rate constant for the first order decomposition of H₂O₂ is given by the following equation:

log k = 14.34 – 1.25 x 10⁴ K/T

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

Answer: Arrhenius equation is given by,

k= Ae ^-E_a/RT  

⇒In k = In A – E_a/RT

⇒In k = Log A – E_a/RT

⇒ Log k = Log A – E_a/2.303RT         (i)

The given equation is

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

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

E_a/2.303RT  = 1.25 10⁴ K/T  

⇒ E_a  =1.25 × 10⁴K × 2.303 × R

= 1.25 × 10⁴K × 2.303 × 8.314 J K ^{– 1}mol ^{– 1}

= 239339.3 J mol ^{– 1} (approximately)

= 239.34 kJ mol – 1

Also, when t_1/2= 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10 ^{– 3} min ^{– 1}

= 4.51 × 10 ^{– 5}s ^{– 1}

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

= 668.95 K

= 669 K (approximately)

Q 27. 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: From Arrhenius equation, we obtain

log k₂/k₁ = E_a / 2.303 R (T₂ – T₁) / T₁T₂

Also, k₁ = 4.5 × 10³ s ^{– 1}

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s ^{– 1}

E_a = 60 kJ mol ^{– 1} = 6.0 × 10⁴ J mol ^{– 1}

Then,

= 297 K = 24°C

Q 28. 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 E_a.

Answer: 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

 To calculate k at 318 K,

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

Again, from Arrhenius equation, we obtain

Therefore, k = Antilog (-1.9855)

= 1.034 x 10^-2 s ^-1

Q 29. 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: From Arrhenius equation, we obtain