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the enthalpy of formation of Methane carbon dioxide and water are - 74.8 393.5 , - 286.2 kJ respectively calculate the enthalpy of combustion of Methane at ordinary temperature?
Dear Student CH4 + 2O2 → CO2 +2 H2O this is the reaction of combustion of methane.C + O2 → CO2 ∆H1 = -393 Kj/molH2 + 1/2O2→ H2O; ∆ H2 = 286.2 Kj/mol C + 2H2 → CH4 ; ∆H3 = -74.8 kj/mol CH4 → C + 2H2 , ∆H3' = 74.8 C + O2 → CO2 ∆ H1 = -393 2( H2 + 1/2O2 → H2O ; ∆H2 = 2× 286.2 =============================add this reactions, CH4 + 2O2 → CO2 +2H2O ∆H =( 74.6 -393 +2×286.2 ) Kj/mol =254 Kj/mol RegardsArun (askIItians forum expert)
CH4 + 2O2 → CO2 +2 H2O this is the reaction of combustion of methane.C + O2 → CO2 ∆H1 = -393 Kj/molH2 + 1/2O2→ H2O; ∆ H2 = 286.2 Kj/mol C + 2H2 → CH4 ; ∆H3 = -74.8 kj/mol CH4 → C + 2H2 , ∆H3' = 74.8 C + O2 → CO2 ∆ H1 = -393 2( H2 + 1/2O2 → H2O ; ∆H2 = 2× 286.2 =============================add this reactions, CH4 + 2O2 → CO2 +2H2O ∆H =( 74.6 -393 +2×286.2 ) Kj/mol =254 Kj/mol
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