Question

The rate constant for two parallel reactions were found to be $1.0\times {10}^{-2}{\text{dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$ and $3.0\times {10}^{-2}{\text{dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$. If the corresponding energies of activation of the parallel reactions are $60.0{\text{kJ mol}}^{-1}$ and $70.0{\text{kJ mol}}^{-1}$ respectively, what is the apparent overall energy of activation?

• A. $130.0{\text{kJ mol}}^{-1}$
• B. $67.5{\text{kJ mol}}^{-1}$
• C. $100.0{\text{kJ mol}}^{-1}$
• D. $65.0{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $67.5{\text{kJ mol}}^{-1}$

We know that, the overall energy of activation for the parallel reaction can be calculated as:

$E_a=\frac{E{a}_1\times k_1+E{a}_2\times k_2}{(k_1+k_2)}$

Given,
$E_{a1}=60{\text{kJ mol}}^{-1}$
$E_{a2}=70{\text{kJ mol}}^{-1}$
$k_1=1.0\times {10}^{-2}{\text{dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$
$k_2=3.0\times {10}^{-2}{\text{dm}}^3{\text{mol}}^{-1}{\text{s}}^{-1}$

$\therefore E_a=\frac{1.0\times {10}^{-2}\times 60+3\times {10}^{-2}\times 70}{[1.0\times {10}^{-2}+3.0\times {10}^{-2}]}$

$Ea=67.5{\text{kJ mol}}^{-1}$