If activation energy| Chemical Kinetics

JEE MAIN 2021

26.02.2021_S2

Question

If the activation energy of a reaction is $80.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the fraction of molecules at 700 K , having enough energy to react to form products is $\mathrm{e}^{-x}$. The value of $x$ is $\_\_\_\_$ (Rounded off to the nearest integer) $\left[\right.$ Use $\left.\mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

Write Your Answer

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Fraction of molecules to have enough energy to react $=e^{-\mathrm{Ea}^2 / R T}$ $$ \begin{aligned} & \text { So, } x=\frac{E_a}{R T} \\ & =\frac{80.9 \times 10^3}{8.31 \times 700} \\ & =13.9 \end{aligned} $$

Question Tags

JEE Main

Chemistry

Easy

Start Preparing for JEE with Competishun

Report Issue