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Calculate the standard enthalpy of combustion of `CH_(4)`, it standard enthalpies of formation of `CH_(4(g)),H_(2)O_((l)),` and `CO_(2(g))` are `-74.81kJ mol^(-1), -285.83kJ mol^(-1)` and `-393.51kJ mol^(-1)` respectively. |
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Answer» The desired equation is `:` `CH_(4)+2O_(2)rarr CO_(2)+2H_(2)O, DeltaH_(1)^(@)=? …(1)` Given that `:` `C_((s))+2H_(2) rarr CH_(4(g)),` `Delta_(2)^(@)=-74.81kJ …(2)` `H_(2(g))+(1)/(2)O_(2(g))rarr H_(2)O_((l)),` `DeltaH_(3)^(@)=-285.83kJ ...(3)` `C_((s))+O_(2(g)) rarr CO_(2(g)),` `DeltaH_(4)^(@)=-393.51kJ ....(4)` The obtain `eq. (1)` following mathematical operation must be performed. `2xx(3)+(4)-(1)` `2H_(2(g))+O_(2(g))+C_((s))+O_(2(g))-C_((s))-2H_(2) rarr 2H_(2)O_((l))+CO_(2(g))-CH_(4(g))` or `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l)) ...(i)` Same mathematical treatment is done with `DeltaH` values `DeltaH_(1)^(@)=[2xxDeltaH_(3)^(@)+DeltaH_(4)^(@)]-[DeltaH_(2)^(@)]` `=[2xx(-285.83)+(-393.5)]-[-74.81]` Thus, enthalpy of combustion of `CH_(4)` `(i.e., DeltaH_(1)^(@)=-890.36kJ)` |
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