Question

If a reaction follows the Arrhenius equation, the plot $lnK$ vs $\frac{1}{\left(RT\right)}$ gives a straight line with a gradient $(-y)$ unit. The energy required to activate the reactant is :

• A. y unit
• B. -y unit
• C. yR unit
• D. y/R unit

Answer 1

The correct option is A y unit

We have,

$K=A{e}^{\frac{-E_a}{RT}}$

$\therefore lnK=ln\left(A{e}^{\frac{-E_a}{RT}}\right)$

$\therefore lnK=lnA-Ea\frac{1}{RT}$

Compare it with $y=mx+c$ we get slope, $m=-E_a=-y\left(Given\right)$

$\therefore E_a=y$