Barium dioxide decomposes into barium oxide and oxygen gas according to the following reaction: 2 BaO2 → 2 BaO + O2. If 27.3 grams of BaO2 decompose and 168 kJ of heat is absorbed in this process, what is the enthalpy change (ΔH) of this reaction in kJ/mol rxn?

• A. -84 kJ/mol
• B. +168 kJ/mol
• C. +84 kJ/mol
• D. -168 kJ/mol

Answer: (C)

Verified step by step guidance

1. First, determine the molar mass of barium dioxide (BaO2). Calculate it by adding the atomic masses of barium (Ba) and two oxygen (O) atoms. Use the periodic table to find these values: Ba = 137.33 g/mol, O = 16.00 g/mol.
2. Calculate the number of moles of BaO2 that decompose. Use the formula: \( \text{moles of BaO2} = \frac{\text{mass of BaO2}}{\text{molar mass of BaO2}} \). Substitute the given mass of BaO2 (27.3 grams) and the molar mass calculated in the previous step.
3. The reaction given is: 2 BaO2 → 2 BaO + O2. Notice that the stoichiometry of the reaction shows that 2 moles of BaO2 decompose to produce 1 mole of O2. Therefore, the enthalpy change (ΔH) given for the reaction is for 2 moles of BaO2.
4. Since 168 kJ of heat is absorbed for the decomposition of the given amount of BaO2, calculate the enthalpy change per mole of reaction. Use the formula: \( \Delta H = \frac{\text{heat absorbed}}{\text{moles of BaO2}} \).
5. Finally, adjust the calculated ΔH to reflect the enthalpy change per mole of reaction as per the stoichiometry of the balanced equation. Since the reaction involves 2 moles of BaO2, divide the calculated ΔH by 2 to find the enthalpy change per mole of reaction.