InterviewSolution
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InterviewSolution
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Calculate the standard enthalpy of formation of acetylene from the following data `:` `C_((g))+O_(2(g))rarr CO_(2(g)),DeltaH^(@)=-393kJ mol^(-1)` `H_(2(g))+(1)/(2)O_(2(g)) rarr H_(2)O_((l)),DeltaH^(@)=-285.8kJ mol^(-1)` `2C_(2)H_(2(g))+5O_(2(g))rarr 4CO_(2(g))+2H_(2)O_((l)),DeltaH^(@)=-2598.8kJ mol^(-1)`. |
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Answer» `{:(,Delta_(r )H^(@)),(4xx(a)4C("graphite")+4O_(2)(g)rarr4CO_(2)(g),bar(4(-393.5 kJ mol^(-1)))),(2xx(b)2H_(2)(g)+O_(2)rarr2H_(2)O(l),2(-285.8 kJ mol^(-1))),((-c)4CO_(2)(g)+2H_(2)O(l)rarr2C_(2)H_(2)(g)+5O_(2)(g)+,2598.8 kJ mol^(-1)):}/{:(4C("graphite")+2H_(2)(g)rarr2C_(2)H_(2)(g),Delta_(r )H^(@)=+453.2 kJ mol^(-1)),(or 2C("graphite")+H_(2)(g)rarrC_(2)H_(2)(g),Delta_(r )H^(@)=+226.6 kJ mol^(-1)):}` Since the above thermochemical equation represents the synthesis of `C_(2)H_(2)` from its elements, we have `Delta_(r )H^(@)(C_(2)H_(2))=+226.6 kJ mol^(-1)`. |
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