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The heats of combustion of `C_(2)H_(4(g)), C_(2)H_(6(g))` and `H_(2(g))` are `-1405,-1558.3` and `-285.6 kJ` respectively. Calculate heat of hydrogenation of ethylene. |
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Answer» `C_(2)H_(4(g))+H_(2(g)) rarrC_(2)H_(6(g)),DeltaH=?` Given, `C_(2)H_(4(g))+3O_(2(g))rarr 2CO_(2(g))+2H_(2)O_((g)), DeltaH=-1409.5kJ ...(1)` `C_(2)H_(6)+(7)/(2)O_(2(g))rarr 2CO_(2(g))+3H_(2)O_((g)), DeltaH=-1558.3kJ ..(2)` `H_(2)+(1)/(2)O_(2(g)) rarrH_(2)O_((g)),DeltaH=-285kJ ......(3)` Adding `eqs. (1)` and `(3)` `C_(2)H_(4(g))+(7)/(2)O_(2(g))+H_(2(g)) rarr 2CO_(2(g))+3H_(2)O_((g)), DeltaH=-1695.1kJ ....(4)` Subtracting `eq. (2)` from `(4)` `ul( {:(C_(2)H_(6(g)),+(7)/(2)O_(2(g)),rarr,2CO_(2(g)),+3H_(2)O_((g)),,DeltaH=-1558.3kJ .....(5)),(-,-,,-,-," +"):} ) ` `C_(2)H_(4(g))+H_(2(g))rarr C_(2)H_(6(g)), DeltaH=-136.8kJ` `:.` Heat of hydrogenation of `C_(2)H_(4)=-136.8kJ` |
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