Chemistry - Chemical Kinetics Question with Solution | TestHub

According to Arrhenius concept,

$\mathrm{K}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}$

Given that,

${\mathrm{K}}_{}=\left({\mathrm{K}}_1.{\mathrm{K}}_2\right)/{\mathrm{K}}_3$

Using Arrhenius equation we can expand above equation as

${\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}\left(\mathrm{net}\right)}/\mathrm{RT}}=\frac{{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}1}/\mathrm{RT}}\times {\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}2}/\mathrm{RT}}}{{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}3}/\mathrm{RT}}}$

Hence,

 $$\begin{gathered} {\mathrm{E}}_{\mathrm{a}\left(\mathrm{net}\right)}={\mathrm{E}}_{\mathrm{a}1}+{\mathrm{E}}_{\mathrm{a}2}-{\mathrm{E}}_{\mathrm{a}3} \\ \Rightarrow {\mathrm{E}}_{\mathrm{a}\left(\mathrm{net}\right)}=40+50-60 \\ =30\mathrm{kJ}/\mathrm{mol} \end{gathered}$$