CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For a reaction, consider the plot of ln k versus 1/T given in the figure where k is rate constant and T is temperature. If the rate constant of this reaction at 500 K is 104 s1, then the rate constant at 600 K is:

Given :
lne(10)2.303

A
10 s1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
106 s1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
102 s1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
103 s1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 103 s1

According to Arrhenius equation,
k=AeEa/RT
Taking ln on both sides,

ln k=lnAEaRT
This equation is in the form of y=mx+c
The plot of ln k vs 1T gives a strraight line with negative slope.
Slope =EaR
yintercept=ln A
From the graph we know,
Slope =m=EaR=6909 K
Let k1 and k2 are the rate constant of the reaction at temperature T1 and T2

k1k2=AeEa/RT1AeEa/RT2

104k2=eEaR(1T21T1)

104k2=eEaR(16001500)

104k2=e(6909)(1/3000)

104k20.1
k2=103 s1


flag
Suggest Corrections
thumbs-up
6
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Simple and Compound Interest
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon