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Given: Fe(s) + H2O(g) à FeO(s) + H2(g)∆H = -24.7 kJ
2 FeO(s) + 1/2O2(g) à Fe2O3(s)∆H = -289.2 kJ
H2(g) + 1/2O2(g) à H2O(g)∆H = -241.8 kJ
Calculate the enthalpy change, ∆H for the following reaction:
2Fe(s) + 3H2O(g) à Fe2O3(s) + 3H2(g)
isnt it products- reactants?
im getting [-289.2 + 3(241.8)] - [2(24.7)+3(-241.8)]
the correct answer is -96.8kJ
i dont get thier explanation
2 FeO(s) + 1/2O2(g) à Fe2O3(s)∆H = -289.2 kJ
H2(g) + 1/2O2(g) à H2O(g)∆H = -241.8 kJ
Calculate the enthalpy change, ∆H for the following reaction:
2Fe(s) + 3H2O(g) à Fe2O3(s) + 3H2(g)
isnt it products- reactants?
im getting [-289.2 + 3(241.8)] - [2(24.7)+3(-241.8)]
the correct answer is -96.8kJ
i dont get thier explanation
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