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The energies of activation forward and reverse reaction for $A_{2} + B_{2} ⟶ 2 AB$ are $180 kJ {mol}^{- 1}$ and $200 {kJmol}^{- 1}$ respectively. The pressure of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 {kJmol}^{- 1}$ The magnitude of enthalpy change of the reaction $(A_{2} + B_{2} ⟶ 2 AB)$ in the presence of catalyst will be (in $kJ {mol}^{- 1}$ ) :

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280

$-$ 20

300

answer is B.

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Detailed Solution

Initially

$Δ H = E_{a f} - E_{a b}$

$= 180 - 200 = - 20 kJ / mol$

Now we no that on addition of $∆ H$ catalyst Romains same

$\Rightarrow New △ H = - 20$

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