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06/10/2020 5:10 pm
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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?
(a) -196 kJ/mol
(b) + 948 kJ/mol
(c) + 196 kJ/mol
(d) -948 kJ/mol
1 Answer
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06/10/2020 5:14 pm
(a) -196 kJ/mol
Explanation:
2H2O2(l) → 2H2O(l) + O2(g) DH = ?
DH = [2' DHf of H2O(l) + (DHf of O2(g)) - (2' DHf of H2O2(l))]
= [(2 x -286) + (0) - (2 x -188)]
= [-572 + 376] = -196 kJ/mol