CBSE Class 12 Chemistry 2018 Solved Paper

Question : 13

Total: 26

A first order reaction is 50% completed in 40 minutes at 300K and in 20 minutes at 320K. Calculate the activation energy of the reaction.
‌‌ (Given : ‌log‌2=0.3010,log‌4=0.6021, R=8.314‌JK−1mol−1‌ ) ‌

Solution:

‌k2=0.693∕20‌, ‌
‌k1=0.693∕40
‌log‌

=‌

2.303R

[‌

−‌

]
‌k2∕k1=2
‌log‌2=‌

2.303×8.314

[‌

320−300

320×300

]
‌Ea=27663.8J∕mol‌ or ‌27.66‌kJ∕mol
‌

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