Final answer:
When 2.00 mol of NaOH is formed in the reaction 2Na₂O₂ (s) + 2H₂O (l) → 4NaOH (s) + O₂ (g) with a standard enthalpy change of -X kJ, then -X/2 kJ of heat are released. This is because the given ΔH° corresponds to 4 moles of NaOH, so for 2 moles, it would be halved.
Step-by-step explanation:
The question is asking for the amount of heat released when 2.00 mol of NaOH is formed in the given reaction, with the standard enthalpy change (ΔH°) provided as -X kJ.
According to stoichiometry, 4 moles of NaOH are produced for every 2 moles of Na₂O₂, as per the balanced chemical equation: 2Na₂O₂ (s) + 2H₂O (l) → 4NaOH (s) + O₂ (g). Therefore, the given enthalpy change (ΔH°) is for the formation of 4 moles of NaOH.
Since the enthalpy change is extensive, it depends on the amount of reactants and products. If ΔH° is -X kJ for 4 moles of NaOH, then for 2 moles of NaOH, half of the heat (energy) will be released, making the answer -X/2 kJ. Therefore, when 2.00 mol of NaOH is formed, -X/2 kJ are released.