Question

Arrhenius proposed a quantitative relationship between rate constant and temperature as K = Ae \(\cfrac{-Ea}{RT}\) or ln k = In A - E_a / RT.

What will be the slope of graph when ln k versus 1/T is plotted?

(a) \(\cfrac{E_a}{R}\)

(b) - \(\cfrac{E_a}{R}\)

(c) \(\cfrac{1}{T}\)

(d) ln A

Answer 1

Correct option:  (b) - \(\cfrac{E_a}{R}\)

According to Arrhenius equation, we

know that k = Ae - \(\cfrac{E_a}{TR}\)

Taking log on both sides,

In k = In A - \(\cfrac{E_a}{TR}\)

On compairing with straight line equation, y = mx + c ,

Slope = - \(\cfrac{E_a}{R}\)