Question

For a given reaction, energy of activation for forward reaction $E _{ af }$ is $80 kJ mol ^{-1}$ and $\Delta H =-40 kJ mol ^{-1}$. A catalyst lowers $E _{ af }$ to $20 kJ mol ^{-1}$. The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

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

Answer 1

Answer: (D)

$\Delta H = E _{ f }- E _{ b } $
$-40=80- E _{ b }$
$E _{ b }=120 kJ / mol $,
Catalyst lower the $E _{ f }$ to $20 kJ / mol$ for forward reaction then $E _{ f }^{\prime}=20 kJ / mol$
We know catalyst decreases the activation energy equal amount in both direction.
$E _{ b }^{\prime}=120-60=60 kJ / mol $
$\frac{ E _{ b }}{ E _{ b }^{\prime}}=\frac{120}{60}=2$