Question

A reaction rate constant is given by : $k=1.2\times 10^{14}{e}^{\frac{-25000}{RT}}se{c}^{-1}$. It means:

• A. log $k$ versus log $T$ will give a straight line with a slope as 25000
• B. log k versus $\log T$ will give a straight line with a slope as $-25000$
• C. log $k$ versus $T$ will give a straight line with a slope as $-25000$
• D. log $k$ versus $1/T$ will give a straight line

Answer 1

Answer: (D)

Correct answer is (d) log $k$ versus $1/T$ will give a straight line