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CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

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Question : 28
Total: 35
A first order reaction is 50% complete in 30 minutes at 300K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction.
[R=8.314 J K−1mol−1 ]
[Given :log‌2=0.3010,log‌3=0.4771, log‌4=0.6021]
Solution:     👈: Video Solution
K1 at 27°C or 300 K=
0.693
30min
=0.0231min−1
K2 at 47°C or 320K1=
0.693
10 min
=0.0693 min−1
Using Arrhenius equation:
log(
K2
K1
)=
Ea
2.303R
 (
T2−T1
T1T2
)
log(
0.0693
0.0231
)=
Ea
2.303×8.314×10−3 kj mol−1 K−1
(
20
300×320
)

Ea=43.85 kj∕ mol.
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