QJEEM19Chc7f9: JEE Main 2019 Chemistry - Class 12 JEEM19Chc7f9 Solution: If a reaction follows the Arrhenius equation, the plot ln k vs…

According to Arrhenius equation,
k = A$$e^{-\frac{E_a}{RT}}$$

or ln k = ln A - $$\frac{E_a}{RT}$$

Comparing the above equation with straight line equation,

y = mx + c,

we get, slope or gradient (m) = –E_a
and Intercept (c) = ln A

Also given that slope or gradient (m) = -y

$$\therefore$$ -y = –E_a

$$\Rightarrow$$ E_a = y

So the activation energy of the reactant, E_a = y unit