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
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Explanation:
2H
2
O(l)⟶2H
2
O(l)+O
2
(g)
△H
f
[H
2
O
2
(l)]=−188 kJ/mol
For 2 moles H
2
O
2
=−2×188 kJ/mol
△H
f
[H
2
O(l)]=−286 kJ/mol
For 2 moles H
2
O=−2×286 kJ/mol
△H
r
=(2×−286)−(2×−188)
=−196 kJ/mol.
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