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Enthalpy change for the reaction, $4H(g) \rightarrow 2H_2(g)$ is −869.6 kJ. The dissociation energy of H–H bond is
A
+217.4 kJ
B
−434.8 kJ
C
−869.6 kJ
D
+434.8 kJ
Detailed Solution
$4H(g) \rightarrow 2H_2(g)$; $\Delta H = -869.6$ kJ. In this reaction 2 moles of H–H bonds are formed.
Energy released in forming 1 mole of H–H bonds = $\frac{869.6}{2} = 434.8$ kJ
Bond dissociation is the reverse process: $H_2(g) \rightarrow 2H(g)$, which absorbs the same amount of energy.
Bond dissociation energy is therefore positive (endothermic).
Dissociation energy of the H–H bond = +434.8 kJ
Energy released in forming 1 mole of H–H bonds = $\frac{869.6}{2} = 434.8$ kJ
Bond dissociation is the reverse process: $H_2(g) \rightarrow 2H(g)$, which absorbs the same amount of energy.
Bond dissociation energy is therefore positive (endothermic).
Dissociation energy of the H–H bond = +434.8 kJ