For the reaction A_2 + B_2 rightleftarrows 2A | vedclass

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(D) The enthalpy change $(\Delta H)$ of a reaction is given by the difference between the activation energy of the forward reaction $(E_{af})$ and the

Vedclass · Accepted Answer

$-20 \ kJ \ mol^{-1}$

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(D) The enthalpy change $(\Delta H)$ of a reaction is given by the difference between the activation energy of the forward reaction $(E_{af})$ and the activation energy of the backward reaction $(E_{ab})$.
$\Delta H = E_{af} - E_{ab}$
Initially,$\Delta H = 180 \ kJ \ mol^{-1} - 200 \ kJ \ mol^{-1} = -20 \ kJ \ mol^{-1}$.
$A$ catalyst lowers the activation energy of both the forward and backward reactions by the same amount $(100 \ kJ \ mol^{-1})$.
New $E_{af}' = 180 - 100 = 80 \ kJ \ mol^{-1}$.
New $E_{ab}' = 200 - 100 = 100 \ kJ \ mol^{-1}$.
New $\Delta H' = E_{af}' - E_{ab}' = 80 - 100 = -20 \ kJ \ mol^{-1}$.
Since a catalyst does not change the enthalpy of the reaction,the value remains $-20 \ kJ \ mol^{-1}$.

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