Change in enthalpy for reaction
2H2O2(ℓ)→2H2O(ℓ)+O2(g)
If heat of formation of H2O2(ℓ) and H2O(ℓ) are -188 & -286 KJ/mol respectively :-
Answers
Answered by
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If you dump some sodium chloride into some water, the solution is conductive. That’s because the salt dissociates into sodium and chlorine ions after it hits the water. Ions are mobile and have a charge, and thus the solution can easily support electrical current.
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Answered by
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Answer:
Explanation:
2
H₂O₂(l)
→ 2
H₂
O₂ (l)+
O
2
(g)
ΔH=?
2H₂O₂(l)→2H2O2(l)+O2(g)
ΔH=?
ΔH=[2×Δ
H
f of H₂
O(l)+(Δ
H
f of O₂ )]
ΔH=[2×ΔHfofH2O(l)+(ΔHfofO2)]
−(2×Δ
H
f of H₂
O₂ (l))]
−(2×ΔHf of H₂O₂ (l))]
=[(2×−286)+(0)−(2×−188)]
=[(2×−286)+(0)−(2×−188)]
=[−572+376]
=[−572+376]
=−196kJ/mol
=−196kJ/mol
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