    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 10−4 s−1, 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 10−3 s−1​According to Arrhenius equation, k=Ae−Ea/RT Taking ln on both sides, ​ ln k=lnA−EaRT 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 y−intercept=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=Ae−Ea/RT1Ae−Ea/RT2​ 10−4k2=eEaR(1T2−1T1)​ 10−4k2=eEaR(1600−1500) 10−4k2=e(6909)(−1/3000)​ 10−4k2≈0.1​ k2=10−3 s−1​  Suggest Corrections  5      Similar questions
Join BYJU'S Learning Program
Select...  Related Videos   Simple and Compound Interest
MATHEMATICS
Watch in App  Explore more
Join BYJU'S Learning Program
Select...