Test Index

NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

© examsnet.com
Question : 29
Total: 39
The time required for 10% completion of the 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.
Solution:  
t=
2.303
k1
‌log‌ ‌
[R]0
90
100
[R]0
,
t=
2.303
k2
‌log‌ ‌
[R]0
75
100
[R]0
t=
2.303
k1
‌log‌ ‌
10
9
,
t=
2.303
k2
‌log‌
4
3
‌=
2.303
k1
‌log‌ ‌
10
9
‌=
2.303
k2
‌log‌ ‌
4
3
⇒
k2
k1
=
log‌
4
3
log‌ ‌
10
9
=
log‌1.333
log‌1.111
=
0.1249
0.0457
=2.733
log‌
k2
k1
=
Ea
2.303R
 (
T2−T1
T1T2
 )
⇒log‌2.733=
Ea
2.303×8.314
 (
308−298
298×308
)
Ea=
2.303×8.314×308×298
10
×0.4367

‌‌=
19.147×308×298
10
×0.4367
‌‌‌=76.75 kJ mol−1
ln‌k=ln‌A−
Ea
RT
log‌k=log‌A−
Ea
2.303RT
=log‌ (4×1010)−
76.75×1000
2.303×8.314×318
=10.6021−
76750
6088.746
=10.6021−12.6051=−2.003
k= Antilog (−2.003)=9.93×10−3
© examsnet.com
Go to Question:
Prev Question
Next Question