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Calculate the enthalpy change for the following reaction: `CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l)` given, enthalpies of formation of `CH_(4),CO_(2) and H_(2)O` are `-74.8kJ` `mol^(-1),-393.5kJ" "mol^(-1) and -286.2kJ" "ml^(-1)` respectively. |
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Answer» `DeltaH^(@)=DeltaH_(f("products"))^(@)-DeltaH_(f("reactants"))^(@)` `=[DeltaH_(f(CO_(2)))^(@)+2DeltaH_(f(H_(2)O))^(@)]-[DeltaH_(f(CH_(4)))^(@)+2DeltaH_(f(O_(2)))^(@)]` `=[-393.5+2xx(-286.2)]=[-74.8+2xx0]` `=-393.5-572.4+74.8` `=891.1kJ` |
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