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 $m o l^{- 1}$ )

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280

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Solution

The correct option is A
20

$A_{2} + B_{2} ⇌ 2 A B$
$E a_{1} = 180 k J m o l^{- 1} E a_{2} = 200 k J m o l^{- 1}$
Enthalpy of $π x^{n} = E a_{2} - E a_{1} = 200 - 180 = 20 k J m o l^{- 1}$

By having a catalyst, both activation energies lower by the same amount, so the difference doesn't change
Ans $\to 20 k J m o l^{- 1}$

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Similar questions

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

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180 k J

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(A_{2} + B_{2} ⇌ 2 A B)

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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 $m o l^{- 1}$ )