Slope of the straight line obtained by plotting `log_(10)` k against 1/T represents what term ?

Options

A. `-E_(a)`
B. `-2303E_(a)//R`
C. `-E_(a)//2303R`
D. `-E_(a)//R`

Answer

C

Explanation

According to Arrhenius equation
`k=Ae^(E_(a)//RT)`
By taking log on both sides
`logk=log_(e)A-(E_(a))/(RT)`
`log_(10)k=log_(10)A-(E_(a))/(2303RT)`
Comparing it with `y=mx+c`
A plot between `log_(10)` k and `((1)/(T))` is a straight line and slope is given by

Slope `=-(E_(a))/(2.303R)`
(slope has negative value)