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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?

1. -196 kJ/mol
2. +948 kJ/mol
3. +196 kJ/mol
4. -948 kJ/mol

Correct answer: -196 kJ/mol

Solution

The enthalpy change is calculated using the formula ΔH = ΣΔHf(products) - ΣΔHf(reactants). For the reaction, ΔH = [2(-286) + 0] - [2(-188)] = -572 + 376 = -196 kJ/mol.

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