Q. The energies of activation for forward and reverse reactions for $A_{2} + B_{2} \rightleftharpoons 2AB$ are 180 kJ $mol^{- 1}$ and 200 kJ $mol^{- 1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ $mol^{- 1}$. The enthalpy change of the reaction ($A_{2} + B_{2} \rightarrow 2AB$) in the presence of catalyst will be (in kJ $mol^{- 1}$)
A) 300
B) 120
C) 280
D) 20
Explanation
(d)
$A_{2} + B_{2} \rightleftharpoons 2AB$
$E_{a}$(forward)= 180 kJ $mol^{- 1}$
$E_{a}$(backward)= 200 kJ $mol^{- 1}$
In the presence of catalyst
$E_{a}$(forward)=180-100=80 kJ $mol^{- 1}$
$E_{a}$(backward)=200-100=100 kJ $mol^{- 1}$
$\mathrm{\Delta}H = E_{a}(forward) - E_{a}(backward)$
$= 80 - 100$
$= - 20\ kJ\ mol^{-}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.