Chemical Kinetics Question 310
Question: A reaction rate constant is given by $ k=1.2\times 10^{14}{e^{-25000/RT}}se{c^{-1}} $ . It means
Options:
A) log k versus log T will give a straight line with a slope as $ -25000 $
B) log k versus T will give a straight line with slope as 25000
C) log k versus 1/T will give a straight line with slope as $ -25000/R $
D) log k versus 1/T will give a straight line
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ k=1.2\times 10^{14}{e^{-25000/RT}}{{\sec }^{-1}} $ or $ \log k=\log 1.2\times 10^{14}-\frac{25000}{R}.\frac{1}{T} $ Equation of straight line $ slope=-\frac{25000}{R} $
