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A gas mixture of `3.67L` of ethylene and methane on complete combustion at `25^(@)C` produces `6.11 L` of `CO_(2)`. Find out the heat evolved on buring `1L` of the gas mixture. The heats of combustion of ethylene and methane are `-1423` and `-891kJ mol^(-1)`, respectively, at `25^(@)C`. |
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Answer» `underset("a litre")(C_(2)H_(4))+3O_(2)tounderset("2a litre")(2CO_(2))+2H_(2)O` `underset((3.67-a)litre)(CH_(4))+2O_(2)tounderet((3.67-a)litre)(CO_(2))+2H_(2)O` Given, `2a+3.67-a=6.11` a=2.44 litre Volume of ethylene in mixture =2.44 litre Volume of methane in mixture=1.23 litre Volume of ethylene in 1 litre mixture `=(2.44)/(3.67)=0.6649` litre Volume of methane in 1 litre mixture `=(1.23)/(3.67)=0.3351` litre 24.45 litre of a gas at `25^(@)C` correspond to 1 mole Thus, heat evolved by burning 0.6649 litre of ethylene thus, heat evolved by burning 0.6649 litre of ethylene `=-(1424)/(24.5)xx0.6649=-38.69kJ` and heat evolved by burning 0.3351 litre of methane `=-(891)/(24.45)xx0.3351=-12.21kJ` ltBrgt so, total heat evolved by buring 1 litre of mixture. `=-38.69-12.21` `=-50.90kJ` |
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