The given reaction is the formation of 2 moles of $H_2$ molecules from 4 moles of H atoms: $4H(g) \rightarrow 2H_2(g); \Delta H = -869.6 \text{ kJ}$

The enthalpy change for the formation of 1 mole of $H_2$ (which involves forming 1 mole of H-H bonds) is: $2H(g) \rightarrow H_2(g); \Delta H = \frac{-869.6}{2} = -434.8 \text{ kJ}$

The bond dissociation energy is defined as the energy required to break one mole of covalent bonds to form products in the gas phase . This process is the exact reverse of the bond formation reaction: $H_2(g) \rightarrow 2H(g)$

Therefore, the enthalpy change for bond dissociation is the positive value of the bond formation enthalpy: $\Delta_{bond}H = +434.8 \text{ kJ}$.