Question

The Activation Energy For The Reaction $2\mathrm{HI}\left(\mathrm{g}\right)\to {\mathrm{H}}_2+{\mathrm{I}}_2\left(\mathrm{g}\right)\mathrm{Is}209.5{\mathrm{Mol}}^{-1}\mathrm{at}581\mathrm{K}.$Calculate The Fraction Of Molecules Of Reactants Having Energy Equal To Or Greater Than The Activation Energy.

Answer 1

Solution :

Step 1: Given data:

The Arrhenius Equation is given by,

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

Here in this equation, Ea is the activation energy, and the term, $k=e^{\frac{-E_a}{RT}}$ is the fraction of molecules of reactants having energy equal to or greater than the activation energy.

Step 2: Formula applied:

Therefore,

$$\begin{gathered} e^{\frac{-E_a}{RT}}=e^{\frac{-209500}{8.314\times 581}} \\ =e^{-43.4} \end{gathered}$$

Step 3: Calculating the Fraction Of Molecules :

Now, calculating the values of the Fraction Of Molecules

= $\frac{1}{anti\log (43.4)}$

$=1.47\times {10}^{-19}$

Hence The Fraction Of Molecules Of Reactants Having Energy Equal To Or Greater Than The Activation Energy are $1.47\times {10}^{-19}$.