question_answer

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

A) log k versus log T will give a straight line with a slope as \[-25000\]

B) log k versus 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

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 \[slope=-\frac{25000}{R}\]