JEE Main Previous Year Questions (2025): Chemical Kinetics

 Page 1


JEE Main Previous Year Questions 
(2025): Chemical Kinetics 
Q1: ?? ? ?? 
The molecule A changes into its isomeric form B by following a first order kinetics at a 
temperature of 1000 K . If the energy barrier with respect to reactant energy for such 
isomeric transformation is ?????? . ???? ???? ?? ?? ?? - ?? and the frequency factor is ????
????
, the 
time required for ???? % molecules of ?? to become ?? is _ _ _ _ picoseconds (nearest 
integer). [ ?? = ?? . ?????? ?? ?? - ?? ?? ?? ?? - ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 69 
Solution: 
To determine the time required for 50% of molecule A to change into its isomeric form B , 
follow these steps: 
Half-life Formula for First Order Kinetics: 
The half-life ( ?? 1 / 2
 ) for a first-order reaction is given by: 
?? 1 / 2
=
0 . 693
?? 
Calculate the Rate Constant (K): 
The rate constant ?? can be calculated using the Arrhenius equation: 
?? = ?? · ?? -
?? ?? ????
 
Given: 
?? ( frequency factor ) = 10
20
 
?? ?? ( activation energy ) = 191 . 48 kJ / m o l = 191 .48 × 10
3
 J / m ol 
?? ( universal gas constant ) = 8 . 314 J / m ol · K 
?? = 1000 K 
Substitute the values into the Arrhenius equation: 
?? = 10
20
× ?? -
1 91 . 48 ×10
3
8 . 3 1 4 × 1 0 0 0
 
Simplify this calculation: 
?? = 10
20
× ?? - 23 . 0 3 1
 
Simplifying further by recognizing that ?? - 23 . 0 3 1
 is a very small number, gives: 
?? ˜
10
20
10
10
= 10
10
s ec
- 1
 
Calculate the Half-life: 
Using the calculated value of ?? : 
?? 1 / 2
=
0 . 6 9 3
10
10
= 6 . 93 × 10
- 11
 seconds 
Page 2


JEE Main Previous Year Questions 
(2025): Chemical Kinetics 
Q1: ?? ? ?? 
The molecule A changes into its isomeric form B by following a first order kinetics at a 
temperature of 1000 K . If the energy barrier with respect to reactant energy for such 
isomeric transformation is ?????? . ???? ???? ?? ?? ?? - ?? and the frequency factor is ????
????
, the 
time required for ???? % molecules of ?? to become ?? is _ _ _ _ picoseconds (nearest 
integer). [ ?? = ?? . ?????? ?? ?? - ?? ?? ?? ?? - ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 69 
Solution: 
To determine the time required for 50% of molecule A to change into its isomeric form B , 
follow these steps: 
Half-life Formula for First Order Kinetics: 
The half-life ( ?? 1 / 2
 ) for a first-order reaction is given by: 
?? 1 / 2
=
0 . 693
?? 
Calculate the Rate Constant (K): 
The rate constant ?? can be calculated using the Arrhenius equation: 
?? = ?? · ?? -
?? ?? ????
 
Given: 
?? ( frequency factor ) = 10
20
 
?? ?? ( activation energy ) = 191 . 48 kJ / m o l = 191 .48 × 10
3
 J / m ol 
?? ( universal gas constant ) = 8 . 314 J / m ol · K 
?? = 1000 K 
Substitute the values into the Arrhenius equation: 
?? = 10
20
× ?? -
1 91 . 48 ×10
3
8 . 3 1 4 × 1 0 0 0
 
Simplify this calculation: 
?? = 10
20
× ?? - 23 . 0 3 1
 
Simplifying further by recognizing that ?? - 23 . 0 3 1
 is a very small number, gives: 
?? ˜
10
20
10
10
= 10
10
s ec
- 1
 
Calculate the Half-life: 
Using the calculated value of ?? : 
?? 1 / 2
=
0 . 6 9 3
10
10
= 6 . 93 × 10
- 11
 seconds 
Convert to Picoseconds: 
Since 1 second = 10
12
 picoseconds: 
?? 1 / 2
= 6 . 93 × 10
- 11
× 10
12
 picoseconds = 69 . 3 picoseconds 
Therefore, the time required for 50% of the molecules of A to become B is approximately 69 
picoseconds (nearest integer). 
Q2: For the thermal decomposition of ?? ?? ?? ?? ( ?? ) at constant volume, the following 
table can be formed, for the reaction mentioned below. 
?? ?? ?? ?? ?? ( ?? ) ? ?? ?? ?? ?? ?? ( ?? ) + ?? ?? ( ?? ) 
Sr. No. Time/s Total pressure/(atm) 
1 0 0.6 
2 100 ' x ' 
?? = _ _ _ _ × ????
- ?? ?????? [nearest integer] 
Given : Rate constant for the reaction is ?? . ?????? × ????
- ?? ?? - ?? . 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 897 
Solution: 
K
N
2
O
5
= 2 × 4 . 606 × 10
- 2
 S
- 1
 
2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g ) 
?? ?? 0 . 6 0 0 
?? ?? 0 . 6 - ?? ?? ?? 2
 
2 × 4 . 606 × 10
- 2
=
2 . 303
100
log ?
0 . 6
0 . 6 - P
 
4 log
10
?
0 . 6
0 . 6 - P
 
10
4
=
0 . 6
0 . 6 - P
 
? 0 . 6 × 10
4
- 10
4
P = 0 . 6 
? 10
4
P = 0 . 6 ( 10
4
- 1 ) 
P = ( 6000 - 0 . 6 ) × 10
- 4
 
= 5999 . × 10
- 4
 
= 0 . 59994 
P
Total 
= 0 . 6 +
P
2
 
= 0 . 6 + 0 . 29997 
= 0 . 89997 
= 899 . 97 × 10
- 3
 
Ans. 900 
Page 3


JEE Main Previous Year Questions 
(2025): Chemical Kinetics 
Q1: ?? ? ?? 
The molecule A changes into its isomeric form B by following a first order kinetics at a 
temperature of 1000 K . If the energy barrier with respect to reactant energy for such 
isomeric transformation is ?????? . ???? ???? ?? ?? ?? - ?? and the frequency factor is ????
????
, the 
time required for ???? % molecules of ?? to become ?? is _ _ _ _ picoseconds (nearest 
integer). [ ?? = ?? . ?????? ?? ?? - ?? ?? ?? ?? - ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 69 
Solution: 
To determine the time required for 50% of molecule A to change into its isomeric form B , 
follow these steps: 
Half-life Formula for First Order Kinetics: 
The half-life ( ?? 1 / 2
 ) for a first-order reaction is given by: 
?? 1 / 2
=
0 . 693
?? 
Calculate the Rate Constant (K): 
The rate constant ?? can be calculated using the Arrhenius equation: 
?? = ?? · ?? -
?? ?? ????
 
Given: 
?? ( frequency factor ) = 10
20
 
?? ?? ( activation energy ) = 191 . 48 kJ / m o l = 191 .48 × 10
3
 J / m ol 
?? ( universal gas constant ) = 8 . 314 J / m ol · K 
?? = 1000 K 
Substitute the values into the Arrhenius equation: 
?? = 10
20
× ?? -
1 91 . 48 ×10
3
8 . 3 1 4 × 1 0 0 0
 
Simplify this calculation: 
?? = 10
20
× ?? - 23 . 0 3 1
 
Simplifying further by recognizing that ?? - 23 . 0 3 1
 is a very small number, gives: 
?? ˜
10
20
10
10
= 10
10
s ec
- 1
 
Calculate the Half-life: 
Using the calculated value of ?? : 
?? 1 / 2
=
0 . 6 9 3
10
10
= 6 . 93 × 10
- 11
 seconds 
Convert to Picoseconds: 
Since 1 second = 10
12
 picoseconds: 
?? 1 / 2
= 6 . 93 × 10
- 11
× 10
12
 picoseconds = 69 . 3 picoseconds 
Therefore, the time required for 50% of the molecules of A to become B is approximately 69 
picoseconds (nearest integer). 
Q2: For the thermal decomposition of ?? ?? ?? ?? ( ?? ) at constant volume, the following 
table can be formed, for the reaction mentioned below. 
?? ?? ?? ?? ?? ( ?? ) ? ?? ?? ?? ?? ?? ( ?? ) + ?? ?? ( ?? ) 
Sr. No. Time/s Total pressure/(atm) 
1 0 0.6 
2 100 ' x ' 
?? = _ _ _ _ × ????
- ?? ?????? [nearest integer] 
Given : Rate constant for the reaction is ?? . ?????? × ????
- ?? ?? - ?? . 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 897 
Solution: 
K
N
2
O
5
= 2 × 4 . 606 × 10
- 2
 S
- 1
 
2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g ) 
?? ?? 0 . 6 0 0 
?? ?? 0 . 6 - ?? ?? ?? 2
 
2 × 4 . 606 × 10
- 2
=
2 . 303
100
log ?
0 . 6
0 . 6 - P
 
4 log
10
?
0 . 6
0 . 6 - P
 
10
4
=
0 . 6
0 . 6 - P
 
? 0 . 6 × 10
4
- 10
4
P = 0 . 6 
? 10
4
P = 0 . 6 ( 10
4
- 1 ) 
P = ( 6000 - 0 . 6 ) × 10
- 4
 
= 5999 . × 10
- 4
 
= 0 . 59994 
P
Total 
= 0 . 6 +
P
2
 
= 0 . 6 + 0 . 29997 
= 0 . 89997 
= 899 . 97 × 10
- 3
 
Ans. 900 
 Given : 2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g )
t = 0 0 . 6 0 0
t = 100 s 0 . 6 - x x x / 2
 
P
Total 
= 0 . 6 +
x
2
 
As given in equation 
K
r
= 4 . 606 × 10
- 2
s e c
- 1
 
(Here language conflict in Q) 
( K
r
=
KA
2
 not considered) 
K
r
t ? = ln ?
0 . 6
0 . 6 - x
4 . 606 ? × 10
- 2
× 100 = 2 . 303lo g ?
0 . 6
0 . 6 - x
P
Total 
? = 0 . 6 +
0 . 594
2
= 0 . 89 7 atm
? = 897 × 10
- 3
 a tm
 
Q3: Consider a complex reaction taking place in three steps with rate constants ?? ?? , ?? ?? 
and ?? ?? respectively. The overall rate constant ?? is given by the expression ?? = v
?? ?? ?? ?? ?? ?? . 
If the activation energies of the three steps are 60,30 and ???? ???? ?? ?? ?? - ?? respectively, 
then the overall energy of activation in ???? ?? ?? ?? - ?? is _ _ _ _ . (Nearest integer) 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 20 
Solution: 
To determine the overall energy of activation for the given complex reaction with rate constants 
?? 1
, ?? 2
, and ?? 3
, we start with the expression for the overall rate constant ?? : 
?? = v
?? 1
?? 3
?? 2
 
The rate constant can also be expressed in terms of the Arrhenius equation: 
?? · ?? - ?? ?? / ????
= v
?? 1
?? - ?? ?? 1
/ ????
· ?? 3
?? - ?? ?? 3
/ ????
?? 2
?? - ?? ?? 2
/ ????
 
By comparing the exponential terms from both sides, we have: 
?? ?? ????
=
1
2
(
?? ?? 1
????
+
?? ?? 3
????
-
?? ?? 2
????
) 
This simplifies to: 
?? ?? =
?? ?? 1
+ ?? ?? 3
- ?? ?? 2
2
 
Substituting the given activation energies - ?? ?? 1
= 60 kJ m ol
- 1
, ?? ?? 2
= 30 kJ m ol
- 1
, and ?? ?? 3
=
10 kJ m ol
- 1
 : 
Page 4


JEE Main Previous Year Questions 
(2025): Chemical Kinetics 
Q1: ?? ? ?? 
The molecule A changes into its isomeric form B by following a first order kinetics at a 
temperature of 1000 K . If the energy barrier with respect to reactant energy for such 
isomeric transformation is ?????? . ???? ???? ?? ?? ?? - ?? and the frequency factor is ????
????
, the 
time required for ???? % molecules of ?? to become ?? is _ _ _ _ picoseconds (nearest 
integer). [ ?? = ?? . ?????? ?? ?? - ?? ?? ?? ?? - ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 69 
Solution: 
To determine the time required for 50% of molecule A to change into its isomeric form B , 
follow these steps: 
Half-life Formula for First Order Kinetics: 
The half-life ( ?? 1 / 2
 ) for a first-order reaction is given by: 
?? 1 / 2
=
0 . 693
?? 
Calculate the Rate Constant (K): 
The rate constant ?? can be calculated using the Arrhenius equation: 
?? = ?? · ?? -
?? ?? ????
 
Given: 
?? ( frequency factor ) = 10
20
 
?? ?? ( activation energy ) = 191 . 48 kJ / m o l = 191 .48 × 10
3
 J / m ol 
?? ( universal gas constant ) = 8 . 314 J / m ol · K 
?? = 1000 K 
Substitute the values into the Arrhenius equation: 
?? = 10
20
× ?? -
1 91 . 48 ×10
3
8 . 3 1 4 × 1 0 0 0
 
Simplify this calculation: 
?? = 10
20
× ?? - 23 . 0 3 1
 
Simplifying further by recognizing that ?? - 23 . 0 3 1
 is a very small number, gives: 
?? ˜
10
20
10
10
= 10
10
s ec
- 1
 
Calculate the Half-life: 
Using the calculated value of ?? : 
?? 1 / 2
=
0 . 6 9 3
10
10
= 6 . 93 × 10
- 11
 seconds 
Convert to Picoseconds: 
Since 1 second = 10
12
 picoseconds: 
?? 1 / 2
= 6 . 93 × 10
- 11
× 10
12
 picoseconds = 69 . 3 picoseconds 
Therefore, the time required for 50% of the molecules of A to become B is approximately 69 
picoseconds (nearest integer). 
Q2: For the thermal decomposition of ?? ?? ?? ?? ( ?? ) at constant volume, the following 
table can be formed, for the reaction mentioned below. 
?? ?? ?? ?? ?? ( ?? ) ? ?? ?? ?? ?? ?? ( ?? ) + ?? ?? ( ?? ) 
Sr. No. Time/s Total pressure/(atm) 
1 0 0.6 
2 100 ' x ' 
?? = _ _ _ _ × ????
- ?? ?????? [nearest integer] 
Given : Rate constant for the reaction is ?? . ?????? × ????
- ?? ?? - ?? . 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 897 
Solution: 
K
N
2
O
5
= 2 × 4 . 606 × 10
- 2
 S
- 1
 
2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g ) 
?? ?? 0 . 6 0 0 
?? ?? 0 . 6 - ?? ?? ?? 2
 
2 × 4 . 606 × 10
- 2
=
2 . 303
100
log ?
0 . 6
0 . 6 - P
 
4 log
10
?
0 . 6
0 . 6 - P
 
10
4
=
0 . 6
0 . 6 - P
 
? 0 . 6 × 10
4
- 10
4
P = 0 . 6 
? 10
4
P = 0 . 6 ( 10
4
- 1 ) 
P = ( 6000 - 0 . 6 ) × 10
- 4
 
= 5999 . × 10
- 4
 
= 0 . 59994 
P
Total 
= 0 . 6 +
P
2
 
= 0 . 6 + 0 . 29997 
= 0 . 89997 
= 899 . 97 × 10
- 3
 
Ans. 900 
 Given : 2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g )
t = 0 0 . 6 0 0
t = 100 s 0 . 6 - x x x / 2
 
P
Total 
= 0 . 6 +
x
2
 
As given in equation 
K
r
= 4 . 606 × 10
- 2
s e c
- 1
 
(Here language conflict in Q) 
( K
r
=
KA
2
 not considered) 
K
r
t ? = ln ?
0 . 6
0 . 6 - x
4 . 606 ? × 10
- 2
× 100 = 2 . 303lo g ?
0 . 6
0 . 6 - x
P
Total 
? = 0 . 6 +
0 . 594
2
= 0 . 89 7 atm
? = 897 × 10
- 3
 a tm
 
Q3: Consider a complex reaction taking place in three steps with rate constants ?? ?? , ?? ?? 
and ?? ?? respectively. The overall rate constant ?? is given by the expression ?? = v
?? ?? ?? ?? ?? ?? . 
If the activation energies of the three steps are 60,30 and ???? ???? ?? ?? ?? - ?? respectively, 
then the overall energy of activation in ???? ?? ?? ?? - ?? is _ _ _ _ . (Nearest integer) 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 20 
Solution: 
To determine the overall energy of activation for the given complex reaction with rate constants 
?? 1
, ?? 2
, and ?? 3
, we start with the expression for the overall rate constant ?? : 
?? = v
?? 1
?? 3
?? 2
 
The rate constant can also be expressed in terms of the Arrhenius equation: 
?? · ?? - ?? ?? / ????
= v
?? 1
?? - ?? ?? 1
/ ????
· ?? 3
?? - ?? ?? 3
/ ????
?? 2
?? - ?? ?? 2
/ ????
 
By comparing the exponential terms from both sides, we have: 
?? ?? ????
=
1
2
(
?? ?? 1
????
+
?? ?? 3
????
-
?? ?? 2
????
) 
This simplifies to: 
?? ?? =
?? ?? 1
+ ?? ?? 3
- ?? ?? 2
2
 
Substituting the given activation energies - ?? ?? 1
= 60 kJ m ol
- 1
, ?? ?? 2
= 30 kJ m ol
- 1
, and ?? ?? 3
=
10 kJ m ol
- 1
 : 
?? ?? =
60 + 10 - 30
2
=
40
2
= 20 kJ mo l
- 1
 
Therefore, the overall energy of activation is 20 kJ / m ol . 
Q4: For the reaction ?? ? products. 
 
 
The concentration of ?? at 10 minutes is _ _ _ _ × ????
- ?? ?? ?? ?? ?? - ?? (nearest integer). The 
reaction was started with ?? . ?? ?? ?? ?? ?? - ?? of ?? . 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 2435 
Solution: 
Order of Reaction: 
Since ?? 1 / 2
? [ ?? ]
0
, the reaction follows a zero-order kinetics. 
Half-life Equation for Zero-order Reaction: 
The half-life ( ?? 1 / 2
 ) is calculated as: 
?? 1 / 2
=
[ ?? ]
0
2 ?? 
Given the slope from the graph is 76.92 , which equals 
1
2 ?? . This implies: 
?? =
1
2 × 76 . 92
 
Concentration of ?? at ???? Minutes: 
Apply the zero-order kinetics equation: 
[ ?? ]
10
= - ???? + [ ?? ]
0
 
Substituting the values: 
[ ?? ]
10
= - (
1
2 × 76 . 92
) × 10 + 2 . 5 = 2 . 435 m ol L
- 1
 
Page 5


JEE Main Previous Year Questions 
(2025): Chemical Kinetics 
Q1: ?? ? ?? 
The molecule A changes into its isomeric form B by following a first order kinetics at a 
temperature of 1000 K . If the energy barrier with respect to reactant energy for such 
isomeric transformation is ?????? . ???? ???? ?? ?? ?? - ?? and the frequency factor is ????
????
, the 
time required for ???? % molecules of ?? to become ?? is _ _ _ _ picoseconds (nearest 
integer). [ ?? = ?? . ?????? ?? ?? - ?? ?? ?? ?? - ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Ans: 69 
Solution: 
To determine the time required for 50% of molecule A to change into its isomeric form B , 
follow these steps: 
Half-life Formula for First Order Kinetics: 
The half-life ( ?? 1 / 2
 ) for a first-order reaction is given by: 
?? 1 / 2
=
0 . 693
?? 
Calculate the Rate Constant (K): 
The rate constant ?? can be calculated using the Arrhenius equation: 
?? = ?? · ?? -
?? ?? ????
 
Given: 
?? ( frequency factor ) = 10
20
 
?? ?? ( activation energy ) = 191 . 48 kJ / m o l = 191 .48 × 10
3
 J / m ol 
?? ( universal gas constant ) = 8 . 314 J / m ol · K 
?? = 1000 K 
Substitute the values into the Arrhenius equation: 
?? = 10
20
× ?? -
1 91 . 48 ×10
3
8 . 3 1 4 × 1 0 0 0
 
Simplify this calculation: 
?? = 10
20
× ?? - 23 . 0 3 1
 
Simplifying further by recognizing that ?? - 23 . 0 3 1
 is a very small number, gives: 
?? ˜
10
20
10
10
= 10
10
s ec
- 1
 
Calculate the Half-life: 
Using the calculated value of ?? : 
?? 1 / 2
=
0 . 6 9 3
10
10
= 6 . 93 × 10
- 11
 seconds 
Convert to Picoseconds: 
Since 1 second = 10
12
 picoseconds: 
?? 1 / 2
= 6 . 93 × 10
- 11
× 10
12
 picoseconds = 69 . 3 picoseconds 
Therefore, the time required for 50% of the molecules of A to become B is approximately 69 
picoseconds (nearest integer). 
Q2: For the thermal decomposition of ?? ?? ?? ?? ( ?? ) at constant volume, the following 
table can be formed, for the reaction mentioned below. 
?? ?? ?? ?? ?? ( ?? ) ? ?? ?? ?? ?? ?? ( ?? ) + ?? ?? ( ?? ) 
Sr. No. Time/s Total pressure/(atm) 
1 0 0.6 
2 100 ' x ' 
?? = _ _ _ _ × ????
- ?? ?????? [nearest integer] 
Given : Rate constant for the reaction is ?? . ?????? × ????
- ?? ?? - ?? . 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 897 
Solution: 
K
N
2
O
5
= 2 × 4 . 606 × 10
- 2
 S
- 1
 
2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g ) 
?? ?? 0 . 6 0 0 
?? ?? 0 . 6 - ?? ?? ?? 2
 
2 × 4 . 606 × 10
- 2
=
2 . 303
100
log ?
0 . 6
0 . 6 - P
 
4 log
10
?
0 . 6
0 . 6 - P
 
10
4
=
0 . 6
0 . 6 - P
 
? 0 . 6 × 10
4
- 10
4
P = 0 . 6 
? 10
4
P = 0 . 6 ( 10
4
- 1 ) 
P = ( 6000 - 0 . 6 ) × 10
- 4
 
= 5999 . × 10
- 4
 
= 0 . 59994 
P
Total 
= 0 . 6 +
P
2
 
= 0 . 6 + 0 . 29997 
= 0 . 89997 
= 899 . 97 × 10
- 3
 
Ans. 900 
 Given : 2 N
2
O
5
( g ) ? 2 N
2
O
4
( g ) + O
2
( g )
t = 0 0 . 6 0 0
t = 100 s 0 . 6 - x x x / 2
 
P
Total 
= 0 . 6 +
x
2
 
As given in equation 
K
r
= 4 . 606 × 10
- 2
s e c
- 1
 
(Here language conflict in Q) 
( K
r
=
KA
2
 not considered) 
K
r
t ? = ln ?
0 . 6
0 . 6 - x
4 . 606 ? × 10
- 2
× 100 = 2 . 303lo g ?
0 . 6
0 . 6 - x
P
Total 
? = 0 . 6 +
0 . 594
2
= 0 . 89 7 atm
? = 897 × 10
- 3
 a tm
 
Q3: Consider a complex reaction taking place in three steps with rate constants ?? ?? , ?? ?? 
and ?? ?? respectively. The overall rate constant ?? is given by the expression ?? = v
?? ?? ?? ?? ?? ?? . 
If the activation energies of the three steps are 60,30 and ???? ???? ?? ?? ?? - ?? respectively, 
then the overall energy of activation in ???? ?? ?? ?? - ?? is _ _ _ _ . (Nearest integer) 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 20 
Solution: 
To determine the overall energy of activation for the given complex reaction with rate constants 
?? 1
, ?? 2
, and ?? 3
, we start with the expression for the overall rate constant ?? : 
?? = v
?? 1
?? 3
?? 2
 
The rate constant can also be expressed in terms of the Arrhenius equation: 
?? · ?? - ?? ?? / ????
= v
?? 1
?? - ?? ?? 1
/ ????
· ?? 3
?? - ?? ?? 3
/ ????
?? 2
?? - ?? ?? 2
/ ????
 
By comparing the exponential terms from both sides, we have: 
?? ?? ????
=
1
2
(
?? ?? 1
????
+
?? ?? 3
????
-
?? ?? 2
????
) 
This simplifies to: 
?? ?? =
?? ?? 1
+ ?? ?? 3
- ?? ?? 2
2
 
Substituting the given activation energies - ?? ?? 1
= 60 kJ m ol
- 1
, ?? ?? 2
= 30 kJ m ol
- 1
, and ?? ?? 3
=
10 kJ m ol
- 1
 : 
?? ?? =
60 + 10 - 30
2
=
40
2
= 20 kJ mo l
- 1
 
Therefore, the overall energy of activation is 20 kJ / m ol . 
Q4: For the reaction ?? ? products. 
 
 
The concentration of ?? at 10 minutes is _ _ _ _ × ????
- ?? ?? ?? ?? ?? - ?? (nearest integer). The 
reaction was started with ?? . ?? ?? ?? ?? ?? - ?? of ?? . 
JEE Main 2025 (Online) 2nd April Morning Shift 
Ans: 2435 
Solution: 
Order of Reaction: 
Since ?? 1 / 2
? [ ?? ]
0
, the reaction follows a zero-order kinetics. 
Half-life Equation for Zero-order Reaction: 
The half-life ( ?? 1 / 2
 ) is calculated as: 
?? 1 / 2
=
[ ?? ]
0
2 ?? 
Given the slope from the graph is 76.92 , which equals 
1
2 ?? . This implies: 
?? =
1
2 × 76 . 92
 
Concentration of ?? at ???? Minutes: 
Apply the zero-order kinetics equation: 
[ ?? ]
10
= - ???? + [ ?? ]
0
 
Substituting the values: 
[ ?? ]
10
= - (
1
2 × 76 . 92
) × 10 + 2 . 5 = 2 . 435 m ol L
- 1
 
Final Concentration: 
Convert to scientific notation: 
[ ?? ]
10
= 2435 × 10
- 3
 m ol L
- 1
 
Thus, the concentration of A at 10 minutes is approximately 2435 × 10
- 3
 m ol L
- 1
. 
Q5: For the reaction ?? ? ?? the following graph was obtained. The time required (in 
seconds) for the concentration of ?? to reduce to ?? . ?? ?? ?? - ?? (if the initial concentration 
of ?? was ???? ?? ?? - ?? ) is (Nearest integer) 
Given : ???? ?? ? ?? = ?? . ???????? 
 
 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 43 
Solution: 
To determine the time required for the concentration of A to decrease from an initial value of 
50 g L
- 1
 to 2 . 5 g L
- 1
 in the reaction A ? B, we assume first-order kinetics. Although the graph 
does not provide a clear indication of the reaction order over the intervals 0 - 5 , 5 - 10, and 
10 - 15 seconds, where the order appears to be zero, we'll proceed with the assumption of 
first-order kinetics, since the graph is not a straight line.