Question

A reaction is given, which energy of activation for forward reaction is $\left(E_{af}\right)$ is $80kJmo{l}^{-1}$.Enthalpy change is $\Delta H=-40kJmo{l}^{-1}$ for the reaction. A catalyst lowers $E_{af}$ to $20kJ$$mo{l}^{-1}$.What will be the ratio of energy of activation for reverse reaction before and after addition of catalyst:

• A. $2.0$
• B. $0.5$
• C. $1.2$
• D. $1.0$

Answer 1

The correct option is A $2.0$

Enthalpy change, $\Delta H=E_f-E_b\Rightarrow -40=80-E_b$

$E_b=120kJ/mol$
The catalyst lowers the $E_{af}to20kJ/mol$ for forward reaction then $E_{af}^{{}^{\prime }}=20kJ/mol$
As we know that a catalyst decreases the activation energy by equal amount in both direction, the decrease in both the directions is by $60kJ/mol$

$E_b^{{}^{\prime }}=(120-60)=60kJ/mol$

$\frac{E_b}{E_b^{{}^{\prime }}}=\frac{120}{60}=2$

Hence, (a) is correct.