Chemistry Enthalpy Question, explain pls Calculate the heat of reaction for 2 H2 + O2 = 2 H2O. Bond energies H2 436 kJ/mol, O2 498, HO 463?

Refer to the equation below. In the reaction, 2 mole of H-H bonds and 1 mole of O=O are broken, and 4 mole of H-O bonds are formed.

Heat change for bond breaking = +(436×2 + 498) kJ = +1370 kJ
Heat change for bond formation = -463 × 4 kJ = -1852 kJ

Heat of reaction = (+1370 - 1852) kJ = -482 kJ

In using bond energies to calculate enthalpy changes, the enthalpy change for the reaction is calculated as:
Delta H = Sum of bond energies of bonds broken - sum of bond energies of bonds formed

In this reaction, you break 2 H-H bonds and 1 O=O bond, and you form 4 H-O bonds. so:

Delta H = 2(436) + 498 - 4(463) = -482 kJ/mol

Bonds breaking release energy, bond formation takes in energy.
Two moles of h2 breaking puts out 2(436)kj
O2 breaking let's out 498kj
Two HO bonds are broken in H2O, so two moles takes up 4(463)kj, rather than releasing it, so this number will be negative.
This gives the equation
2(436)+498-4(463)=-482
So, the total energy is -482kj making it and endothermic reaction