Final answer:
The enthalpy change of the reaction is -484 kJ/mol.
Step-by-step explanation:
To calculate the enthalpy change (ΔH) of the reaction, we need to use bond energies. Bond energy is the energy required to break a bond in one mole of a gaseous substance. The enthalpy change is then calculated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.
In this case, we have 2 moles of H-H bonds and 1 mole of O=O bonds in the reactants. The bond energy for H-H is 432 kJ/mol and the bond energy for O=O is 498 kJ/mol. On the product side, we have 4 moles of O-H bonds. The bond energy for O-H is 464 kJ/mol. Plugging these values into the equation, we get:
ΔH = (2 * 432 kJ/mol + 1 * 498 kJ/mol) - (4 * 464 kJ/mol)
ΔH = -484 kJ/mol