JEE Main Previous Year Questions (2016- 2024): Chemical Kinetics & Nuclear Chemistry PDF Download

Q.1. The number of correct statement/s from the following is       (JEE Main 2023)
(a) Larger the activation energy, smaller is the value of the rate constant.
(b) The higher is the activation energy, higher is the value of the temperature coefficient.
(c) At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature
(d) A plot of ln  is a straight line with slope equal to

Ans. d
K = Ae^–Ea/RT
Here, Ea ­↑ K↓
lnK = lnA –Ea/RT
slope of lnK vs 1/T = -Ea/R
The higher is the activation energy, higher is the value of the temperature coefficient.

Q.2. A student has studied the decomposition of a gas AB₃ at 25^∘C. He obtained the following data.     (JEE Main 2023)
The order of the reaction is
(a) 0 (zero)
(b) 0.5
(c) 1
(d) 2

Ans. d

Q.3. Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. X_s and k are constants having appropriate units.          (JEE Advanced 2022)
(a) I → P; II → Q; III → S; IV →T
(b) I → R; II → S; III → S; IV →T
(c) I → P; II → Q; III → Q; IV → R
(d) I → R; II → S; III → Q; IV → R

Ans. a

Q.4. Assuming 1μg of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ × 10⁻¹μg.
[Given: ln 10 = 2.303; log 2 = 0.30]      (JEE Main 2022)

Ans. 1

Q.5. The reaction between X and Y is first order with respect to X and zero order with respect to Y.
Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)      (JEE Main 2022)

Ans. 40
r = k[X][Y]⁰ = k[X]
Using I & II 

Using I & III 

Q.6. For a reaction, given below is the graph of. The activation energy for the reaction is equal to ____________ calmol⁻¹. (nearest integer)
(Given: R = 2calK⁻¹ mol⁻¹)      (JEE Main 2022)

Ans. 8
Slope K = Ae^−Ea/RT


E_a = 4R = 8Cal/mol

Q.7. For the given first order reaction
A → B
the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to ___________. (Nearest integer)      (JEE Main 2022)

Ans. 100


K = 2.303
 
A₀ → initial concentration of reactant
A_t → concentration of reactant at time t
 

Q.8. Ka for butyric acid (C₃H₇COOH) is 2 × 10⁻⁵. The pH of 0.2M solution of butyric acid is __________ × 10⁻¹. (Nearest integer)
[Given log⁡2 = 0.30]      (JEE Main 2022)

Ans. 27
K_a of Butyric acid ⇒ 2 × 10⁻⁵PKa = 4.7
pH of 0.2M solution
 

= 2.35 + 0.35 = 2.7
pH = 27 × 10⁻¹

Q.9.
If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x × 10⁻⁶ s⁻¹. The value of x is ______________. (Nearest Integer)      (JEE Main 2022)

Ans. 165

 
= 165 × 10⁻⁶ s⁻¹

Q.10. 2NO + 2H₂ → N₂ + 2H₂O
The above reaction has been studied at 800^∘C. The related data are given in the table below:
The order of the reaction with respect to NO is ___________.       (JEE Main 2022)

Ans. 2
Let the rate of reaction (r) is as
r = K[NO]ⁿ[H₂]^m
From 1^st  data
0.135 = K[40]ⁿ ⋅ (65.6)^m...(1)
From 2^nd data
0.033 = K(20.1)ⁿ ⋅ (65.6)^m...(2)
On dividing equation (1) by equation (2)

4 = (2)ⁿ
∴ n = 2
∴ Order of reaction w.r.t. NO is 2.

Q.11. For a reaction A → 2 B + C the half lives are 100 s and 50 s when the concentration of reactant A is 0.5 and 1.0 mol L⁻¹ respectively. The order of the reaction is ______________ . (Nearest Integer)       (JEE Main 2022)

Ans. 2

t_1/2 = 100sec  a₀ = 0.5
t_1/2 = 50sec  a₀ = 1

(2) = (2)ⁿ⁻¹
n − 1 = 1
n = 2

Q.12. For the decomposition of azomethane.
CH₃N₂CH₃(g) → CH₃CH₃(g) + N₂(g), a first order reaction, the variation in partial pressure with time at 600 K is given as
The half life of the reaction is __________ × 10⁻⁵ s. [Nearest integer]        (JEE Main 2022)

Ans. 2
For first order reaction,
ln ⁡A = ln⁡ A₀ − kt 
Hence Slope = −k
−k = −3.465 × 10⁴


t_1/2 = 2 × 10⁻⁵ s

Q.13. The half life for the decomposition of gaseous compound A is 240 s when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be 4.0 min. The order of the reaction is ______________. (Nearest integer)        (JEE Main 2022)

Ans. 1
(t_1/2)_A = 240 s when P = 500 torr 
(t_1/2)_A = 4min = 4 × 60 = 240 sec when P = 250 torr 
If means half-life is independent of concentration of reactant present. 
∴ Order of reaction = 1

Q.14. For the reaction P → B, the values of frequency factor A and activation energy E_A are 4 × 10¹³ s⁻¹ and 8.3 kJ mol⁻¹ respectively. If the reaction is of first order, the temperature at which the rate constant is 2 × 10⁻⁶ s⁻¹ is _____________ × 10⁻¹ K.
(Given: ln 10 = 2.3, R = 8.3 J K⁻¹ mol⁻¹, log2 = 0.30)        (JEE Main 2022)

Ans. 225

Q.15. The equation
k = (6.5 × 10¹²s⁻¹)e^−26000K/T
is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol⁻¹. [nearest integer]
(Given: R = 8.314 J K⁻¹ mol⁻¹)        (JEE Main 2022)

Ans. 216

Q.16. The activation energy of one of the reactions in a biochemical process is 532611 J mol⁻¹. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k₃₀₀ = x × 10⁻³ k₃₁₀. The value of x is ____________.
[Given: ln⁡10 = 2.3, R = 8.3 J K⁻¹ mol⁻¹]        (JEE Main 2022)

Ans. 1

Q.17. A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)
(Given: antilog 0.125 = 1.333, antilog 0.693 = 4.93)        (JEE Main 2022)

Ans. 75

Q.18. For a first order reaction A → B, the rate constant, k = 5.5 × 10⁻¹⁴ s⁻¹. The time required for 67% completion of reaction is x × 10⁻¹ times the half life of reaction. The value of x is _____________ (Nearest integer)
(Given: log 3 = 0.4771)        (JEE Main 2022)

Ans. 16

Q.19. The solubility product of a sparingly soluble salt A₂X₃ is 1.1 × 10⁻²³. If specific conductance of the solution is 3 × 10⁻⁵ S m⁻¹, the limiting molar conductivity of the solution is x× 10⁻³ S m² mol⁻¹. The value of x is ___________.        (JEE Main 2022)

Ans. 3

Q.20. It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol⁻¹. [nearest integer]
(Given ln10 = 2.3, R = 8.3 J K⁻¹ mol⁻¹, log 2 = 0.30)       (JEE Main 2022)

Ans. 59

Q.21. pH value of 0.001 M NaOH solution is ____________.        (JEE Main 2022)

Ans. 11

Q.22. The rate constant for a first order reaction is given by the following equation: 

The activation energy for the reaction is given by ____________ kJ mol⁻¹. (In nearest integer) (Given: R = 8.3 J K⁻¹ mol⁻¹)        (JEE Main 2022)

Ans. 166

Q.23. Catalyst A reduces the activation energy for a reaction by 10 kJ mol⁻¹ at 300 K. The ratio of rate constants, is e^x. The value of x is ___________. [nearest integer]
[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K⁻¹ mol⁻¹]        (JEE Main 2022)

Ans. 4

Q.24. A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ___________ s.
(Given: ln 2 = 0.693)        (JEE Main 2022) 

Ans. 200

Q.25. At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was found to be 170 s. The order of the reaction is ____________. [integer answer]         (JEE Main 2022)

Ans. 0

Q.26. For a given chemical reaction
γ₁A + γ₂B → γ₃C + γ₄D
Concentration of C changes from 10 mmol dm⁻³ to 20 mmol dm⁻³ in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm⁻³ s⁻¹. Therefore, the rate of reaction is _____________ mmol dm⁻³ s⁻¹. (Nearest Integer)         (JEE Main 2022) 

Ans. 1





r₂ = 2r₁
r₄ = 1.5r₂ = 3r₁



r₄ = 9r₃ = 3r₁
⇒ r₁ = 3r₃
3r₃ A + 6r₃ B → r₃C + 9r₃D

= 1 m⋅mol dm⁻³sec⁻¹

Q.27. The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 - 1000 K. The data has been analysed by plotting ln k vs 10³ / T graph. The value of activation energy for the reaction is ___________ kJ mol⁻¹. (Nearest integer)
(Given: R = 8.31 J K⁻¹ mol⁻¹)         (JEE Main 2022)

Ans. 154



∴ E_a = 18.5 × 8.31 × 1000 ≃ 154 kJ mol⁻¹

Q.28. At 30^∘C, the half life for the decomposition of AB₂ is 200 s and is independent of the initial concentration of AB₂. The time required for 80% of the AB₂ to decompose is

Given: log⁡ 2 = 0.30    log 3 = 0.48         (JEE Main 2022)
(a) 200 s
(b) 323 s
(c) 467 s
(d) 532 s

Ans. c
Since, half-life is independent of the initial concentration of AB₂. Hence, the reaction is "First Order". 



t = 467 s

Q.29. K_a1, K_a2 and K_a3 are the respective ionization constants for the following reactions (a), (b) and (c).

The relationship between K_a1, K_a2 and K_a3 is given as         (JEE Main 2022)
(a) K_a3 = K_a1 + K_a2
(b) K_a3 = K_a1 − K_a2 
(c) K_a3 = K_a1 / K_a2 
(d) K_a3 = K_a1 × K_a2

Ans. d





K_a3 = K_a1 × K_a2

Q.30. The equilibrium constant for the reversible reaction

2A(g) ⇌ 2B(g) + C(g) is K₁

K₁ and K₂ are related as:         (JEE Main 2022) 
(a)
(b)
(c)
(d)

Ans. c

Q.31. The solubility of AgCl will be maximum in which of the following?         (JEE Main 2022)
(a) 0.01 M KCl
(b) 0.01 M HCl
(c) 0.01 M AgNO₃
(d) Deionised water

Ans. d
In deionized water no common ion effect will take place so maximum solubility. 

Q.32. For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is
(Given: ln 10 = 2.303 and log 2 = 0.3010)         (JEE Main 2022)
(a) 1.12
(b) 2.43
(c) 3.32
(d) 33.31

Ans. c
A → Products
For a first order reaction, 

Time for 90% conversion, 

Q.33. For the following reaction,the rate of reaction isTwo moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are)
(Use: ln 2 = 0.693)      (JEE Advanced 2021)
(a) The rate constant, k, of the reaction is 13.86 × 10⁻⁴ s⁻¹.
(b) Half-life of X is 50 s.
(c)
(d)

Ans. b, c, d

As the concentration of reactant becomes half at t = 50 s. So, half-time of reaction is 50 s.
Given, 

So, options (b), (c) and (d) are correct.

Q.34. Which one of the following given graphs represents the variation of rate constant (k) with temperature (T) for an endothermic reaction?     (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. a
According to Arrhenius equation

The graph will varies as

Q.35. For a reaction of order n, the unit of the rate constant is:      (JEE Main 2021)
(a) A mol¹⁻ⁿ L¹⁻ⁿ s
(b) mol¹⁻ⁿ L²ⁿ s⁻¹ 
(c) mol¹⁻ⁿ Lⁿ⁻¹ s⁻¹ 
(d) mol¹⁻ⁿ L¹⁻ⁿ s⁻¹

Ans. c
Rate = k[A]ⁿ
comparing units 

⇒ k = mol¹⁻ⁿ L¹⁻ⁿ s⁻¹

Q.36. For the following graphs, 
Choose from the options given below, the correct one regarding order of reaction is:      (JEE Main 2021)
(a) (b) Zero order (c) and (e) First order
(b) (a) and (b) Zero order (e) First order 
(c) (b) and (d) Zero order (e) First order 
(d) (a) and (b) Zero order (c) and (e) First order

Ans. b
For Zero order reactions
rate = K [Reactant]⁰
⇒ r = k 



For First order reaction →
r = K [Concentration] 

Reactant concentration after time t →

Q.37. Isotope(s) of hydrogen which emits low energy β⁻ particles with t_1/2 value > 12 years is/are       (JEE Main 2021)
(a) Protium 
(b) Tritium 
(c) Deuterium 
(d) Deuterium and Tritium

Ans. b

Q.38. Given below are two statements:
Statement I: The pH of rainwater is normally − 5.6.
Statement II: If the pH of rainwater drops below 5.6, it is called acid rain.
In the light of the above statements, choose the correct answer from the options given below:        (JEE Main 2021)
(a) Both Statement I and Statement II are true. 
(b) Both Statement I and Statement II are false. 
(c) Statement I is true but Statement II is false. 
(d) Statement I is false but Statement II is true.

Ans. a
The pH of rain water is normally 5.6. If it pH drop below 5.6, it is called acid rain.
∴ Both statements are correct. 

Q.39. For the reaction A → B, the rate constant k(in s⁻¹) is given by 

The energy of activation in kJ mol⁻¹ is ____________. (Nearest integer) [Given: R = 8.314 J K⁻¹ mol⁻¹]        (JEE Main 2021)

Ans. 47
Given,

We know 

= 47.29 = 47 (Nearest integer)

Q.40. According to the following figure, the magnitude of the enthalpy change of the reaction
A + B → M + N in kJ mol⁻ is equal to ___________. (Integer answer)         (JEE Main 2021) 

Ans. 45
ΔH = E_Product − E_Reactant
= 15 − (15 + 45)
= −45 KJ/mol
|ΔH| = 45 KJ/mol 

Q.41. For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ____________. (Integer answer)         (JEE Main 2021) 

Ans. 2

Q.42. The first order rate constant for the decomposition of CaCO₃ at 700 K is 6.36 × 10⁻³s⁻¹ and activation energy is 209 kJ mol⁻¹. Its rate constant (in s⁻¹) at 600 K is x × 10⁻⁶. The value of x is ___________. (Nearest integer)
[Given R = 8.31 J K⁻ mol⁻¹; log 6.36 × 10⁻³ = −2.19, 10^−4.79 = 1.62 × 10⁻⁵]         (JEE Main 2021)

Ans. 16
K₇₀₀ = 6.36 × 10⁻³s⁻¹;
K₆₀₀ = x × 10⁻⁶s⁻¹
E_a = 209 kJ/mol
Applying;



log(6.36 × 10⁻³) − logK₆₀₀ = 2.6
⇒ logK₆₀₀ = −2.19 − 2.6 = −4.79
⇒ K₆₀₀ = 10^−4.79 = 1.62 × 10⁻⁵
= 1.62 × 10⁻⁶
= x × 10⁻⁶
⇒ x = 16

Q.43. The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is
2K₂Cr₂O₇ + 8H₂SO₄ + 3C₂H₆O → 2Cr₂(SO₄)₃ + 3C₂H₄O₂ + 2K₂SO₄ + 11H₂O
If the rate of appearance of Cr₂(SO₄)₃ is 2.67 mol min⁻¹ at a particular time, the rate of disappearance of C₂H₆O at the same time is _____________ mol min⁻¹. (Nearest integer)         (JEE Main 2021)

Ans. 4

 

⇒ Rate of disappearance of C₂H₆O = 4.005 mol/min. 

Q.44. The reaction rate for the reaction
[PtCl₄]²⁻ + H₂O ⇌ [Pt(H₂O)Cl₃]⁻ + Cl⁻
was measured as a function of concentrations of different species. It was observed that

where square brackets are used to denote molar concentrations. The equilibrium constant K_c = ____________ . (Nearest integer)         (JEE Main 2021) 

Ans. 50
[PtCl₄]²⁻ + H₂O ⇌ [Pt(H₂O)Cl₃]⁻ + Cl⁻

Q.45. The overall stability constant of the complex ion [Cu(NH₃)₄]²⁺ is 2.1 × 10¹³. The overall dissociations constant is y × 10⁻¹⁴. Then y is __________. (Nearest integer)         (JEE Main 2021)  

Ans. 5

Given k_s = 2.1 × 10¹³
K_d = 1/k_s = 4.7 × 10⁻¹⁴
∴ y = 4.7 ≈ 5 

Q.46. The following data was obtained for chemical reaction given below at 975 K.
2NO_(g) + 2H_2(g) → N_2(g) + 2H₂O_(g)
The order of the reaction with respect to NO is ___________. [Integer answer]         (JEE Main 2021)  

Ans. 1
7 × 10⁻⁹ = K × (8 × 10⁻⁵)^x (8 × 10⁻⁵)^y ......... (1)
2.1 × 10⁻⁸ = K × (24 × 10⁻⁵)^x (8 × 10⁻⁵)^y ....... (2)  

Q.47. For the first order reaction A → 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is __________ min. (Round off to the nearest integer).
[Use: ln 2 = 0.69, ln 10 = 2.3]
Properties of logarithms: ln x^y = y ln x;
ln⁡(x/y) = n⁡x − ln⁡y
(Round off to the nearest integer)         (JEE Main 2021)  

Ans. Between 600 and 700



Ans. 600 to 700

Q.48. For a chemical reaction A → B, it was found that concentration of B is increased by 0.2 mol L⁻ in 30 min. The average rate of the reaction is ____________ × 10⁻¹ mol L⁻¹ h⁻¹. (in nearest integer)         (JEE Main 2021)  

Ans. 4
A → B


= 0.4 = 4 × 10⁻¹ mol / L × hr  

Q.49.
In the above first order reaction the initial concentration of N₂O₅ is 2.40 × 10⁻² mol L⁻¹ at 318 K. The concentration of N₂O₅ after 1 hour was 1.60 × 10⁻² mol L⁻¹. The rate constant of the reaction at 318 K is ______________ × 10⁻³ min⁻¹. (Nearest integer)
[Given: log 3 = 0.477, log 5 = 0.699]          (JEE Main 2021)

Ans. 7

Q.50. PCl₅(g) → PC_l3(g) + Cl₂(g)
In the above first order reaction the concentration of PCl₅ reduces from initial concentration 50 mol L⁻¹ to 10 mol L⁻¹ in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x × 10⁻² min⁻¹. The value of x is __________. [Given log5 = 0.6989]          (JEE Main 2021)

Ans. 1

t = 0
50M
t = 120min
10 M



= 1.3413 × 10⁻² min⁻¹
1.34 ⇒ Nearest integer = 1 

Q.51. The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ___________ × 10⁻³ min⁻¹. (Nearest integer)
[Use: ln 10 = 2.303; log₁₀ 3 = 0.477; property of logarithm : log x^y = y log x]          (JEE Main 2021)

Ans. 106
Unit of rate constant is min⁻¹, so it must be a first order reaction. For first order reaction,

k is the rate constant
t is the time
A₀ is the initial conc.
A_t is the conc. at time, t
Using formula,
A₀ = 100, A_t = 90 min 1 min

= 2.303 × (log 10 − 2 log 3)
= 2.303 × (1 − 2 × 0.477)
= 0.10593
= 105.93 × 10⁻³
≈ 106
Hence, the rate constant for viral inactivation is 106.

Q.52. A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is _________ min. (Round off to the Nearest Integer). [Use: ln 2 = 0.69; ln 10 = 2.3]           (JEE Main 2021)

Ans. 10
Given t_1/2 = 1 min 

From formula we know, 




⇒ t = 10 min 

Q.53. 2NO(g) + Cl₂(g) ⇌ 2NOCl(s)
This reaction was studied at −10^∘ and the following data was obtained
[NO]₀ and [Cl₂]₀ are the initial concentrations and r₀ is the initial reaction rate.
The overall order of the reaction is __________. (Round off to the Nearest Integer).           (JEE Main 2021)

Ans.  3
We know,
Rate = K[A]^x[B]^y
Here from,
Exp 1 : 0.18 = K[0.1]^x[0.1]^y ..... (1)
Exp. 2 : 0.35 = K[0.1]^x[0.2]^y .... (2)
Exp. 3 : 1.40 = K[0.2]^x[0.2]^y .... (3)
(2) ÷ (3)


⇒ x = 2
(1) ÷ (2)

⇒ y = 1
∴ x + y = 3 

Q.54. The reaction 2A + B₂ → 2AB is an elementary reaction.
For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ____________. (Round off to the Nearest Integer).          (JEE Main 2021)

Ans. 27
Let initial concentration of A & B is a & b 

Q.55. For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ × 10⁻³ s⁻¹. (Round off to the Nearest Integer). [Given: log₁₀2 = 0.301, ln10 = 2.303]          (JEE Main 2021)

Ans. 2




k = 2 × 10⁻³s⁻¹

Q.56. A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer).           (JEE Main 2021)

Ans. 108
Initially : [A₀] = [B₀] = a
After time 't' min : [A] = 16[B]

Q.57. The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 × 10⁻³ s⁻¹ and the activation energy E_a = 11.488 kJ mol⁻¹, the rate constant at 200 K is ____________ × 10⁻⁵ s⁻¹. (Round off to the Nearest Integer). (Given : R = 8.314 J mol⁻¹ K⁻¹)            (JEE Main 2021)

Ans. 10

k₁ (at 200 K) = ?
k₂ (at 300 K) = 1 × 10⁻³s⁻¹



⇒  k1 = 10 × 10⁻⁵s⁻¹

Q.58. If the activation energy of a reaction is 80.9 kJ mol⁻¹, the fraction of molecules at 700 K, having enough energy to react to form products is e^−x. The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K⁻¹ mol⁻¹]             (JEE Main 2021)

Ans. 14
E_a = 80.9 kJ/mol
Fraction of molecules able to cross energy barrier = e^−E_a/RT = e^−x

⇒ x ≃ 14 

Q.59. The rate constant of a reaction increases by five times on increase in temperature from 27^∘C to 52^∘C. The value of activation energy in kJ mol⁻¹ is _________. (Rounded off to the nearest integer)
[R = 8.314 J K⁻¹mol⁻¹]             (JEE Main 2021)

Ans. 52
T₁ = (273 + 27) = 300 K, T₂ = (273 + 52) = 325 K
Given, temperature coefficient of the reaction,




E_a = 52194.78 J mol⁻
= 52.194 kJ mol⁻¹
≃ 52 kJ mol⁻¹

Q.60. For the reaction, aA + bB → cC + dD, the plot of log k vs 1/T is given below:

The temperature at which the rate constant of the reaction is 10^-4 s^-1 is _________ K. (Rounded off to the nearest integer)
[Given: The rate constant of the reaction is 10^-5 s^-1 at 500 K.]             (JEE Main 2021)

Ans. 526
 

Q.61. Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25^∘C. After 9 h, the fraction of sucrose remaining is f. The value of log₁₀(1/f) is ________ × 10⁻². (Rounded off to the nearest integer)
[Assume: ln 10 = 2.303, ln 2 = 0.693]             (JEE Main 2021)

Ans. 81


At t = 0, a = [A]₀      (initial conc.)
t = 9h, a − x = [A]_t         [conc. at time t]
For using 1st order equation, 



x = 81.24 or x ≈ 81 

Q.62. Gaseous cyclobutene isomerises to butadiene in a first order process which has a 'k' value of 3.3 × 10^-4 s^-1 at 153^°C. The time in minutes it takes for the isomerization to proceed 40% to completion at this temperature is ______.
(Rounded off to the nearest integer)             (JEE Main 2021)

Ans. 26

It is a first order isomerisation reaction. Integrated rate law for 1st order reaction is

Here,
k = rate constant
[A]₀ = initial concentration
[A]_t = concentration at time 't'
Given, k = 3.3 × 10⁻⁴ s⁻¹
[A]₀ = 100 ⇒ [A]_t = 100 − 40 = 60
Put values in Eq. (i), we get

t = 1547.95 s = 25.79 min (1 min = 60 s or 1s = 1/60 min) = 26 minutes 

Q.63. During the nuclear explosion, one of the products is ⁹⁰Sr with half-life of 6.93 years. If 1 µg of ⁹⁰Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically?    (2020)

Ans. 23.03

Given, half-life of ⁹⁰Sr = 6.93 year; amount = 1 × 10^–6g

From the first order kinetic
 
For 90% decay of ⁹⁰Sr,

⇒ t = 23.03 log 10
⇒ t = 23.03 years

Q.64. For the reaction
2H₂(g) + 2NO(g) → N₂(g) + 2H₂O(g),
the observed rate expression is, rate =. The rate expression for the reverse reaction is    (2020)
(a) k_b[N₂][H₂O]²
(b)
(c) k_b [N₂][H₂O]
(d)

Ans. d

Given, rate expression for reaction,
2H₂ (g) + 2NO(g) → N₂(g) + 2H₂O(g)

r_f=k_f[NO]²[H₂]

Now,
⇒

Rearranging the above equation,

=> r_f = r_b

Q.65. The rate of a certain biochemical reaction at physiological temperature (T) occurs 10⁶ times faster with enzyme than without it. The change in the activation energy upon adding enzyme is    (2020)
(a) −6(2.303)RT
(b) −6RT
(c) +6(2.303)RT
(d) +6RT

Ans. a

From Arrhenius equation without enzyme
    (1)
With enzyme
    (2)
On Dividing Eq. (1) by (2), we get

⇒     (2)
Taking log of Eq. (3), we get
E'_a - E_a = - 6(2.303)RT

Q.66.Consider the following plots of rate constant versus 1/T for four different reactions. Which of the following orders is correct for the activation energies of these reactions?    (2020)

(a) E_b > E_a > E_d > E_c
(b) E_a > E_c > E_d > E_b
(c) E_c > E_a > E_d > E_b
(d) E_b > E_d > E_c > E_a

Ans. c

From Arrhenius equation without enzyme

Taking log of Eq. (1), we get

By plotting log k against 1/T,  the slope = and intercept = log A
From the graph of four different reaction,

The plot having high value of slop will have high value of activation energy. Thus, the correct order for activation energy of the reactions is: E_c > E_a > E_d > E_b.

Q.67. For the following reactions

it was found that the E_a is decreased by 30 kJ mol^–1 in the presence of catalyst. If the rate remains unchanged, the activation energy for catalyzed reaction is (Assume pre exponential factor is same)    (2020)
(a) 75 kJ mol^–1
(b) 105 kJ mol^–1
(c) 135 kJ mol^–1
(d) 198 kJ mol^–1

Ans. a

 Given, E_al = E_a and E_a2 = E_a - 30
T₁ = 700 K and T₂ = 500 K
    (1) [Given, Rate remains unchanged]
Substituting the values in Eq. (1), we get

⇒ 500E_a = 700E_a - 2100
⇒ 2100 = 700E_a - 500E_a
⇒ 2100 = 200E_a
⇒
⇒ E_a = 105 KJ mol^-1
Thus, activation energy in presence of catalyst is Ea₂ = E_a - 30
⇒ E_a2 = 105 - 30 = 75 kJ mol^-1

Q.68. A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activation energy (in kJ mol⁻¹) for this process is closest to ________.    (2020)

Ans. 3.98

Using Arrhenius equation at two different temperature,
    (1)
Since, milk splits after 60 min. at 300 K and after 40 min. at 400 K.
Substituting the values in Eq. (1), we get

   
⇒ E_a = 3.984 kJ mol⁻¹

Q.69. The following results were obtained during kinetic studies of the reaction;
2A + B → Products

The time (in minutes) required to consume half of A is:    (2019)
(a) 5
(b) 10
(c) 1
(d) 100

Ans. a

Q.70. For the reaction, 2A + B → products, when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L^-1s^-1 to 2.4 mol L^-1s^-1. When the concentration of A alone is doubled, the rate increased from 0.3 L^-1s^-1 to 0.6 mol L^-1s^-1. Which one of the following statements is correct?    (2019)
(a) Total order of the reaction is 4
(b) Order of the  reaction with respect to B is 2
(c) Order of the reaction with respect to B is 1
(d) Order of the reaction with respect to A is 2

Ans. b

∴ y = 2

Q.71. Consider the given plots for a reaction obeying Arrhenius equation (0°C < T < 300°C): (K and E_a are rate constant and activation energy, respectively)

Choose the correct option:    (2019)
(a) I is right but II is wrong
(b) Both I and II are correct
(c) I is wrong but II is right
(d) Both I and II are wrong

Ans. b
From Arrhenius equation,
K= Ae^-Ea/RT
So, as E_a increases, e^-Ea/RT decreases, K decreases and as T increases, E_a/RT decreases, e^-Ea/RT increases,

Q.72. For an elementary chemical reaction,  the expression for d[A]/dt is:    (2019)
(a) k₁[A₂] - k_-1[A]²
(b) 2k₁[A₂] - k_-1[A]²
(c) k₁ [A₂] + k_-1[A]²
(d) 2k₁[A₂] - 2k_-1[A]²

Ans. d

⇒

Q.73. If a reaction follows the Arrhenius equation, the plot lnk vs 1/(RT) gives straight line with a gradient (-y) unit. The energy required to activate the reactant is:    (2019)
(a) y/R unit
(b) y unit
(c) yR unit
(d) - y unit

Ans. b
From Arrhenius equation,


slope = -y (given)
-y = -Ea
⇒  E_a = y

Q.74. The reaction 2X → B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be:    (2019)
(a) 9.0 h
(b) 12.0 h
(c) 18.0 h
(d) 7.2 h

Ans. c
For zero order reaction

t = 18 hrs

Q.75. Decomposition of X exhibits a rate constant of 0.05  μg/year. How many years are required for the decomposition of 5 μg of X into 2.5  μg?    (2019)
(a) 50
(b) 25
(c) 20
(d) 40

Ans. a
Rate constant of decomposition of X= 0.05 mg/year. Unit of rate constant confirms that the decomposition of X is a zero order reaction.
For zero order kinetics,

Q.76. For a reaction, consider the plot of In k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10^-5 s^-1, then the rate constant at 500 K is:    (2019)
(a) 10^-6 s^-1
(b) 2 x 10^-4 s^-1
(c) 10^-4 s^-1
(d) 4 x 10^-4 s^-2

Ans. c
From Arrhenius equation,

Q.77. For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below.
The rate law for the reaction is:    (2019)
(a) Rate = k[A][B]²
(b) Rate = k[A]²[B]²
(c) Rate = k[A][B]
(d) Rate = k[A]²[B]

Ans. a
2A + B → C
Rate = k[A]^x [B]^y
Exp-1, 0.045 = k[0.05]^x [0.05]^y    ...(i)
Exp-2, 0.090 = k[0.1]^x [0.05]^y    ...(ii)
Exp-3, 0.72 = k[0.2]^x [0.1]^y    ...(iii)
Divide equation (i) by equation (ii)

Divide equation (i) by equation (iii)

Rate law = k[A]¹ [B]².

Q.78. For a reaction scheme  if the rate of formation of B is set to be zero then the concentration of B is given by:    (2019)
(a) (k₁ - k₂) [A]
(b) k₁k₂ [A]
(c) (k₁ + k₂) [A]
(d) (k₁/k₂) [A]

Ans. d

Q.79. The given plots represents the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are:    (2019)

(a) 1, 0
(b) 1, 1
(c) 0, 1
(d) 0, 2

Ans. a

For First order reaction

Q.80. A bacterial infection in an internal wound grows as N²(t) = N₀ exp(t), where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as dN/dt = -5N². What will be the plot of N₀/N vs. t after 1 hour?    (2019)
(a)
(b)
(c)
(d)

Ans. c
When drug is administered bacterial growth is given by dN/dt = -5N²
On integrating the above equation,

The above equation is similar to straight line equation with positive slope.
Thus N₀/N_t increases linearly with t.

Q.81. For the reaction of H₂ with I₂, the rate constant is 2.5 x 10^-4 dm³ mol^-1 s^-1 at 327°C and 1.0 dm³ mol^-1 s^-1 at 527°C. The activation energy for the reaction, in kJ mol^-1 is: (R = 8.314J K^-1 mol^-1)    (2019)
(a) 166
(b) 150
(c) 72
(d) 59

Ans. a

Q.82. In the following reaction: xA → yB

‘A' and ‘B’ respectively can be:    (2019)
(a) n-Butane and Iso-butane
(b) C₂H₂ and C₆H₆
(c) C₂H₄ and C₄H₈
(d) N₂O₄ and NO₂

Ans. c
xA → yB

Comparing this equation with the equation given in question. We get,

∴ x/y = 2
∴ The reaction is of type 2A → B.
Hence, option (c) is correct.

Q.83. NO, required for a reaction is produced by the decomposition of N₂O₅ in CCl₄ as per the equation,
2N₂O₅(g) → 4NO₂(g) + O₂(g).
The initial concentration of N₂O₅ is 3.00 mol L^-1 and it is 2.75 mol L^-1 after 30 minutes. The rate of formation of NO₂ is:    (2019)
(a) 4.167 x 10^-3 mol L^-1 min^-1
(b) 1.667 x 10^-2 mol L^-1 min^-1
(c) 8.333 x 10^-3 mol L^-1 min^-1
(d) 2.083 x 10^-3 mol L^-1 min^-1

Ans. b

According to the question
 

Q.84. At 518⁰C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s^-1 when 5% had reacted and 0.5 Torr s^-1 when 33% had reacted. The order of the reaction is:    (2018)
(a) 2
(b) 3
(c) 1
(d) 0

Ans. a

Q.85. N₂O₅ decomposes to NO₂ and O₂ and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 minute at constant temperature will be:    (2018)
(a) 106.25 mm Hg 
(b) 116.25 mm Hg 
(c) 136.25 mm Hg 
(d) 175.0 mm Hg 

Ans. a
2N₂O₅ → 4NO₂ + O₂
p − 2x         4x         x
p_t = p − 2x + 4x + x
p_t = p + 3x 

at  t = 0,   pt = p = 50 mm Hg
at  t = 50 mm,   p_t = 87.5 mm Hg
p + 3x = 87.5
p = 87.5 − 3x
50 = 87.5 − 3x
12.5 = x 

p − 2x = 50 − 2(12.5) = 25 

Since K will remain same  


50 = 50 × 4 − 8y
50 = 200 − 8y
8y = 150
y =  18.75
pt = p + 3y
= 50 + 3 (18.73) = 106.25 mm Hg 

Q.86. If 50 % of a reaction occurs in 100 second and 75 % of the reaction occurs in 200 second, the order of this reaction is:    (2018)
(a) 2
(b) 3
(c) Zero
(d) 1

Ans. d

First order reaction as half life is constant. 

Q.87. Two reactions R₁ and R₂ have identical pre - exponential factors. Activation energy of R₁ exceeds that of R₂ by 10 kJ mol^–1. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K, then ln (k₁/ k₂) is equal to
(R = 8.314 J mole^–1 K^–1)    (2017)
(a) 8
(b) 12
(c) 6
(d) 4

Ans. d

Q.88. The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.    (2017)
(a) 4.92 K
(b) 9.84 K
(c) 19.67 K
(d) 2.45 K

Ans. a
 ...(i)
 ...(ii)

= 304.92

Q.89. The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:
(Assume activation energy and pre-exponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol^-1 K^-1)    (2017)
(a) 53.6 kJ mol^-1 
(b) 214.4 kJ mol^-1 
(c) 107.2 kJ mol^-1 
(d) 26.8 kJ mol^-1

Ans. c



= 107165.79 J = 107.165 KJ

Q.90. Decomposition of H₂O₂ follows a first order reaction. In fifty minutes the concentration of H₂O₂ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H₂O₂ reaches 0.05 M, the rate of formation of O₂ will be:    (2016)
(a) 6.93 x 10^-2 mol min^-1 
(b) 6.93 x 10^-4 mol min^-1 
(c) 2.66 L min^-1 at STP 
(d) 1.34 x 10^-2 mol min^-1 

Ans. b

For a first order reaction
K = 2.303/t log a/(a-x)
Given a = 0.5, (a-x) = 0.125, t = 50 min
∴ k = 2.303/50 log 0.5/0.125
= 2.78 x 10^-2 min^-1
r = d[H₂O₂] = 2.78 x 10^-2 x 0.05
= 1.386 x 10^-3 mol min^-1
Now

Q.91. The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process show below:
O₃(g) + Cl*(g) → O₂(g) + ClO*(g) ...(i)
k_i = 5.2 × 10⁹ L mol^-1 s^-1 
ClO*(g) + O*(g) → O₂(g) + Cl*(g) ...(ii)
[K_ii = 2.6 x 10¹⁰ L mol^-1s^-1]
The closest rate constant for the overall reaction O₃(g) + O*(g) → 2O₂(g) is:    (2016)
(a) 1.4 × 10²⁰ L mol^-1 s^-1
(b) 5.2 × 10⁹ L mol^-1 s^-1
(c) 3.1 × 10¹⁰ L mol^-1 s^-1
(d) 2.6 × 10¹⁰ L mol^-1 s^-1

Ans. b
The rate constant of overall reaction depends slowest step. Hence equation(i) is slowest step. Option(b) is correct.

Q.92. The rate law for the reaction below is given by the expression k [A][B]
A + B → Product
If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be:    (2016)
(a) 9 k
(b) 3 k
(c) k/3
(d) k

Ans. d
Rate constant is independent of concentration.