CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
7
You visited us 7 times! Enjoying our articles? Unlock Full Access!
Question

It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be _______kJ mol1. [nearest integer]
(Given ln10=2.3,R=8.3 J K1 mol1,log 2=0.30)

A
59.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
59.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
59
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

T1=300 K
T2=309 K
Since the rate constant doubles with every 9K rise in temperature:
K2=2K1
Activation energy, Ea=?
logK2K1=Ea2.3R[T2T1T2.T1]
log2=Ea2.3×8.3[9309×300]
Ea=0.3×309×300×2.3×8.39
=58988.1 J/mol
=59 kJ/mol

flag
Suggest Corrections
thumbs-up
15
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Arrhenius Equation
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon