Chemistry - Chemical Kinetics Question with Solution | TestHub

$\Delta \text{H}={\text{E}}_{\text{f}}-{\text{E}}_{\text{b}}$

$-40=80-{\text{E}}_{\text{b}}$

${\text{E}}_{\text{b}}=120\text{kJ/mol,}$

Catalyst lower the${\text{E}}_{\text{f}}$to$20\text{kJ/mol}$for forward reaction then${\text{E}}_{\text{f}}^{\prime }=\text{20 kJ/mol}$

We know catalyst decreases the activation energy equal amount in both direction.

${\text{E}}_{\text{b}}^{\prime }=\left(120-60\right)=\text{60 kJ/mol}$

$\frac{{\text{E}}_{\text{b}}}{{\text{E}}_{\text{b}}^{\prime }}=\frac{120}{60}=2$