Final answer:
The enthalpy changes for the reaction H₂(g) + ½O₂(g) → H₂O(l) is -286 kJ for forming 1 mole of liquid water. When considering the reversible reaction for 2 moles of water, the change would be -572 kJ. The provided equation in the question seems to contain a typo, and we use the correct value of -286 kJ per mole of water formation for calculations.
Step-by-step explanation:
To calculate the enthalpy change for the formation of water from hydrogen and oxygen, we use the given reaction 2H₂O(l) = 2H₂(g) + O₂(g) ΔH= +576 kJ and the formation enthalpy for liquid water.
For the reaction H₂(g) + ½O₂(g) → H₂O(l), the enthalpy change is -286 kJ for 1 mole of liquid water formed from its elements. Since the provided reaction equation involves the decomposition of 2 moles of liquid water, we must reverse the reaction and also consider that the enthalpy change would be for 2 moles when forming water.
Therefore, the enthalpy change for the reaction H₂(g) + ½O₂(g) → H₂O(l) would be -2 × 286 kJ = -572 kJ.
The initial equation given in the question has different values for enthalpy change; hence, it appears the question might involve a typographical error. To align with the known thermodynamic properties of water formation, we will utilize the correct enthalpy value of -286 kJ per mole of water as the basis for our enthalpy change calculation in this context.