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What is the enthalpy change for 2H₂O₂(l) → 2H₂O(l) + O₂(g) if heat of formation of H₂O₂(l) and H₂O(l) are -188 and -286 kJ/mol respectively?
- -196 kJ/mol
- +948 kJ/mol
- +196 kJ/mol
- -948 kJ/mol
Correct answer: -196 kJ/mol
Solution
To find the enthalpy change for 2H₂O₂(l) → 2H₂O(l) + O₂(g), we need to use the heat of formation of H₂O₂(l) and H₂O(l). The heat of formation of H₂O(l) is -286 kJ/mol, and the heat of formation of H₂O₂(l) is -188 kJ/mol. Therefore, the enthalpy change for 2H₂O₂(l) → 2H₂O(l) + O₂(g) is 2 * (-286) - 2 * (-188) = -196 kJ/mol.
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