Test Index

NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

© examsnet.com
Question : 27
Total: 39
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log‌k=14.34 –1.25×104K∕T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:  
According to Arrhenius equation, k=Ae−Ea∕RT
or, ln‌k=ln‌A−
Ea
RT
or, log‌k=log‌A−
Ea
2.303RT
...(i)
Given equation is
‌‌‌‌‌log‌k=14.34−1.25×104
K
T
...(ii)
Comparing (i) with (ii),
Ea
2.303RT
=
1.25×104 K
T
or, ‌‌Ea=2.303R×1.25×104K
=2.303×(8.314)×1.25×104
=239.34 kJ mol−1
When t1∕2=256‌min,
k=
0.693
256×60
=4.51×10−5s−1
Substituting this value in the given equation,
log‌ (4.51×10−5 )=14.34−
1.25×104 K
T
i.e., (−5+0.6542)=14.34−
1.25×104 K
T
or,
1.25×104 K
T
=18.6858 or, T=669K
© examsnet.com
Go to Question:
Prev Question
Next Question