Question

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

• A. zero
• B. 7.37 k cal/mol
• C. 9.212 k cal/mol
• D. 11.05 k cal/mol

Answer 1

Answer: (D)

11.05 k cal/mol

We know that, Arrhenius equation for calculation of energy of activation of reaction with rate constant $K$ and temeperature $T$ is

$K=A$ $e^{-Ea/RT}$

where, $Ea$= Arrhenius activation energy

$A$= pre exponential factor (frequency factor)

Now, $e^{-Ea/K_{B}T}$= Fraction of collision having more than activation energy

where, $K_B=$ Boltzmann constant

Given, $T=400$ $K$ and effective collision= $0.0001$%

$\Rightarrow$ Effective Collision= $e^{-Ea/K_{B}T}$

$\Rightarrow$ $0.0001$%= $e^{-Ea/1.3\times {10}^{-23}\times 400}$

$\Rightarrow$ ${10}^{-6}$= $e^{-Ea/1.3\times {10}^{-23}\times 400}$

$\Rightarrow$ $2.303\times \log {10}^{-6}$= $\frac{-Ea}{1.3\times {10}^{-23}\times 400}$

$\Rightarrow$ $2.303\times (-6)$= $\frac{-Ea}{1.3\times {10}^{-23}\times 400}$

$\Rightarrow$ $Ea$= $1.3\times {10}^{-23}\times 400\times 6\times 2.303=7.19\times {10}^{-20}$ $J/mol$