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Multiple Choice
Barium dioxide decomposes into barium oxide and oxygen gas according to the following reaction: 2 BaO2 → 2 BaO + O2. If 37.6 grams of BaO2 decompose and 182 kJ of heat is absorbed in this process, what is the enthalpy change (ΔH) of this reaction in kJ/mol of BaO2?
A
-182 kJ/mol
B
+91 kJ/mol
C
+182 kJ/mol
D
-91 kJ/mol
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검증된 단계별 안내
1
First, determine the molar mass of BaO2. Barium (Ba) has an atomic mass of approximately 137.33 g/mol, and oxygen (O) has an atomic mass of approximately 16.00 g/mol. Therefore, the molar mass of BaO2 is calculated as: M(BaO2) = 137.33 g/mol + 2 * 16.00 g/mol.
Next, calculate the number of moles of BaO2 that decompose. Use the formula: moles = mass / molar mass. Here, the mass of BaO2 is given as 37.6 grams.
The problem states that 182 kJ of heat is absorbed during the decomposition of 37.6 grams of BaO2. This means the reaction is endothermic, and the enthalpy change (ΔH) is positive.
To find the enthalpy change per mole of BaO2, divide the total heat absorbed by the number of moles of BaO2 calculated in step 2. Use the formula: ΔH (kJ/mol) = total heat absorbed (kJ) / moles of BaO2.
Finally, interpret the sign of ΔH. Since the reaction absorbs heat, ΔH should be positive, indicating an endothermic reaction. Compare your calculated ΔH with the given options to find the correct answer.