tranque102791 tranque102791
  • 16-06-2017
  • Chemistry
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What is the change in enthalpy for the following reaction? 2H2O2(aq) 2H2O(l) + O2(g) Given: H2O: ∆H= -242 kJ H2O2: ∆H= -609 kJ

-286 kJ-572 kJ572 kJ286 kJ

Respuesta :

dec77cat dec77cat
  • 16-06-2017
ΔH = {2ΔH(H₂O) + ΔH(O₂)} - 2Δ(H₂O₂)

ΔH(H₂O) = -242 kJ
ΔH(O₂,g) = 0
Δ(H₂O₂) = -609 kJ

ΔH = 2*(-242) + 0 - 2*(-609) = 734 kJ
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