Chemistry, asked by Parvani8896, 11 months ago

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

Answers

Answered by kumarshivamsingh516
0

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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