JEE Advanced Previous Year Questions (2021-2023): Chemical Equilibrium | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Advanced Previous Year Questions 
(2021-2023): Chemical Equilibrium 
2023 
Numerical Type 
JEE Advanced 2023 Paper 1 Online 
Q1. The plot of ???? ?? ? ?? ?? versus ?? / ?? for a reversible reaction ?? ( ?? ) ? ?? ( ?? ) is shown. 
 
Pre-exponential factors for the forward and backward reactions are ????
????
 ?? - ?? and 
????
????
 ?? - ?? , respectively. If the value of ???? ?? ? ?? for the reaction at ?????? ?? is 6 , the value of 
| ???? ?? ? ?? ?? | at ?????? ?? is 
[ ?? = equilibrium constant of the reaction 
?? ?? = rate constant of forward reaction 
?? ?? = rate constant of backward reaction ] 
Ans: 5 
 
For reaction A ( g ) ? P ( g ) 
log ? k
f
=
- E
f
2 . 3 0 3 R T
+ log ? A
f
 [Arrhenius equation for forward reaction] 
From plot when, 
1
 T
= 0 . 002 , log ? k
f
= 9 
? 9 =
- ?? ?? 2 . 303 ?? ( 0 . 002 ) + log ? ( ?? ?? ) 
Given : ?? ?? = 10
15
 s
- 1
 
Page 2


JEE Advanced Previous Year Questions 
(2021-2023): Chemical Equilibrium 
2023 
Numerical Type 
JEE Advanced 2023 Paper 1 Online 
Q1. The plot of ???? ?? ? ?? ?? versus ?? / ?? for a reversible reaction ?? ( ?? ) ? ?? ( ?? ) is shown. 
 
Pre-exponential factors for the forward and backward reactions are ????
????
 ?? - ?? and 
????
????
 ?? - ?? , respectively. If the value of ???? ?? ? ?? for the reaction at ?????? ?? is 6 , the value of 
| ???? ?? ? ?? ?? | at ?????? ?? is 
[ ?? = equilibrium constant of the reaction 
?? ?? = rate constant of forward reaction 
?? ?? = rate constant of backward reaction ] 
Ans: 5 
 
For reaction A ( g ) ? P ( g ) 
log ? k
f
=
- E
f
2 . 3 0 3 R T
+ log ? A
f
 [Arrhenius equation for forward reaction] 
From plot when, 
1
 T
= 0 . 002 , log ? k
f
= 9 
? 9 =
- ?? ?? 2 . 303 ?? ( 0 . 002 ) + log ? ( ?? ?? ) 
Given : ?? ?? = 10
15
 s
- 1
 
? ? 9 =
- ?? ?? 2 . 303 ?? ( 0 . 002 ) + 15
? ?
?? ?? 2 . 303 ?? =
6
0 . 002
= 3000
 
Now, K =
k
f
k
b
=
A
f
A
b
e
- ( E
f
- E
b
) / RT
 
log ? K = -
1
2 . 303
( E
f
- E
b
)
????
+ log ? (
10
15
10
11
) 
At 500 K 
? ? 6 =
- ( E
f
- E
b
)
500R ( 2 . 303 )
+ 4
? ? ( 1000R ) ( 2 . 3 0 3 ) = E
b
- E
f
? ? ( 1000R ) ( 2 . 3 0 3 ) = E
b
- 3000 ( 2 . 3 0 3 R )
 
? E
b
= 4000R ( 2 . 303 ) … ( ?? ) 
Now k
b
= A
b
e
- E
b
/ RT
 
? log ? k
b
=
- E
b
2 . 3 0 3 RT
+ log ? A
b
 
At 250 K, 
?? log ? k
b
= -
4000
250
+ log ? ( 10
11
) [From equation (1)] 
? = - 16 + 11 = - 5
 
| log ? k
b
| = 5