Question

In a reaction carried out at $400K,0.0001$% of the total number of collisions are effective. The energy of activation of the reaction is:

• A. zero
• B. $7.37kcal/mol$
• C. $9.212kcal/mol$
• D. $11.05kcal/mol$

Answer 1

Answer: (D)

$11.05kcal/mol$

The fraction of molecules having energy equal to or greater than $E_a$ is :
$x=\frac{n}{N}=e^{-E_a/RT}(x=\frac{n}{N}=0.0001\%=\frac{0.0001}{100}={10}^{-6})$

$\therefore \log x=\frac{-E_a}{2.3\times RT}$
$⟹log{10}^{-6}=\frac{-E_a}{2.3\times 2\times 400}$
$⟹E_a=6\times 2.3\times 800$
$⟹E_a=11.05kcal/mol$