$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ H = - 22 k c a l$
Activation energy, $E_{a}$ for the given reaction is $70 k c a l$ . Find the activation energy for $2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$ in $k c a l$ .

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Solution

The given reaction is an exothermic reaction since $Δ H = - v e$
$E_{a_{F}}$ =Activation energy of the forward reaction
$E_{a_{B}}$ = Activation energy of the backward reaction

Enthalpy change can be expressed also in terms of activation energy,

$Δ H = E_{a_{F}} - E_{a_{B}}$
$\Rightarrow - 22 = 70 - E_{a_{B}}$
$\Rightarrow E_{a_{B}} = 92 kcal$

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