The activation energy for the reaction
Question:

The activation energy for the reaction

2HI(g)  H2 + I2(g)

is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Solution:

In the given case:

Ea = 209.5 kJ mol−1 = 209500 J mol−1

T = 581 K

R = 8.314 JK−1 mol−1

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

$x=e^{-E_{a} / \mathrm{R} T}$

$\Rightarrow \operatorname{In} x=-\frac{E_{a}}{\mathrm{R} T}$

$\Rightarrow \log x=-\frac{E_{a}}{2.303 \mathrm{R} T}$

$\Rightarrow \log x=-\frac{209500 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 581}=-18.8323$

Now, $x=$ Antilog $(-18.8323)$

$=1.471 \times 10^{-19}$