Question

In the first-order reaction a→b+c carried out in 27c if 3.8 *(10)^-16 % of the reactant molecules exist in the activated state, the activation energy of the reaction is

Answer 1

Answer : The activation energy of the reaction is, 88.56 KJ

Explanation :

Let us consider the limit as  $Ea\rightarrow  0$

The formula will be:

$\underset{Ea\rightarrow 0}\lim=A\times e^{\frac{-Ea}{RT}}=k$

If the activation energy is zero then the rate constant is equal to the frequency factor. That means 100%  of the molecules are activated.

Thus,  $\frac{k}{A}\times 100\%$  is the percent of molecules that can do the reaction.  

$\frac{k}{A}\times 100\%=3.8\times 10^{-16}\%$

$\frac{k}{A}=3.8\times 10^{-16}=e^{\frac{-Ea}{RT}}$

This means that,

$e^{\frac{-Ea}{RT}}=3.8\times 10^{-16}$

${\frac{-Ea}{RT}}=\ln(3.8\times 10^{-16})$

$Ea=88.56KJ$

Therefore, the activation energy of the reaction is, 88.56 KJ