Question

For a reaction, given below is the graph of ln k vs 1/T. The activation energy for the reaction is equal to____ $cal mol^{–1}$. (Nearest integer)

(Given: $R = 2 cal K^{-1} mol^{-1}$)

Answer 1

$Slope$ =$−\frac{20}{5}$

$ln \space k=ln \space A −\frac{E_a}{RT}$

∴$\frac{E_a}{R}=\frac{20}{5}$

$⇒$ $\frac{20R}{5}=8 \space cal mol^{−1}$