The energies of activation for $A_{2} + B_{2} ⇌ 2 A B$ are $180 k J$ $m o l^{- 1}$ forward and $200 k J$ $m o l^{- 1}$ reverse reactions respectively. The presence of catalyst lowers the activation energies of both (forward and reverse) reactions $(A_{2} + B_{2} ⇌ 2 A B)$ by $100 k J$ $m o l^{- 1} .$ The magnitude of enthalpy change of the reaction in the presence of catalyst will be (in kJ $m o l^{- 1})$ :

300

No worries! We‘ve got your back. Try BYJU‘S free classes today!

120

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 20

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $20$
Presence of catalyst lowers the activation energy of both forward & backward reaction by equal amounts :
Hence, it does not cause any change in enthalpy of a reaction.
$Δ H = E_{f} - E_{b}$
$= 200 - 180 = 20 k J / m o l$
This is the value of enthalpy change with and without the catalyst.
Option C is correct.

Suggest Corrections

Similar questions

Q. The energies of activation for forward and reverse reactions for

(A_{2} + B_{2} ⇌ 2 A B)

are

180 k J

mol

^{- 1}

and

200 k J

mol

^{- 1}

respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by

100 k J

mol

^{- 1}

. The enthalpy change of the reaction

(A_{2} + B_{2} \to 2 A B)

in the presence of catalyst will be (in kJ mol

^{- 1}

)

The energies of activation for forward and reverse reactions for $A_{2} + B_{2} ⇌ 2 A B$ are $180 k J m o l^{- 1}$ and $200 k J m o l^{- 1}$ respectively. The presence of a catalyst lowers activation energy of both reactions by $100 k J m o l^{- 1}$ . The enthalpy change of the reaction $(A_{2} + B_{2} \to 2 A B)$ in the presence of a catalyst will be (in kJ $m o l^{- 1}$ )