Question

The activation energy for the reaction-
$H_2{O}_2\to H_{2}O+\frac{1}{2}{O}_2$
is $18Kcal/mol$ at $300K$. calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Anti log $(-13.02)=9.36\times {10}^{-14}$

• A. $9.36\times {10}^{-14}$
• B. $1.2\times {10}^{-12}$
• C. $4.2\times {10}^{-30}$
• D. $5.2\times {10}^{-15}$

Answer 1

The correct option is A $9.36\times {10}^{-14}$

We know that Arrhenius equation for calculation of energy of activation of a reaction is given as:

$K=A$ $e^{-Ea/RT}$ $-\left(i\right)$

where, $K$= Rate constant

$A$= pre exponential factor or frequency factor

$Ea$= Activation energy

$R$= Ideal gas constant

$T$= Temperature in K (Kelvin)

Now, the fraction of moleucle having energy equal to or greater than activation energy is given by:-

$f=e^{-Ea/RT}$ $-\left(ii\right)$

Given, $Ea=18KCal/mol$ , $T=300K$

$R=2Cal{K}^{-1}mo{l}^{-1}$

Using these values in equation $\left(ii\right)$

So, $p=e^{-18\times 1000/2\times 300}$ $[\because 1KCal=1000Cal]$

Taking $\log$ on both sides:-

$\log f=\frac{18\times 1000}{2\times 300}=-30$

$\Rightarrow f=$ $Antilog$ $(-30)={10}^{-30}$