Test Index

NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

© examsnet.com
Question : 23
Total: 39
The rate constant for the decomposition of a hydrocarbon is 2.418×10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Solution:  
Here, k=2.418×10−5 s−1,
Ea=179.9 kJ mol−1,
T=546 K,A=?
According to Arrhenius equation,
log‌A=log‌k+
Ea
2.303RT

=log‌ (2.418×10−5 )+
179.9
2.303×8.314×10−3×546
=(−5+0.3834)+17.2081
=12.5924 s−1
or, A= Antilog (12.5924) s−1
=3.912×1012 s−1
© examsnet.com
Go to Question:
Prev Question
Next Question