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A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25^(@)C produces 6.11 litres of CO_(2). Find out the amount of heat evolvedon burning one litre of the mixture. The heatsof combustion of ethylene and methane are - 1423 and - 891kJ mol^(-1) at 25^(@)C |
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Answer» Solution :Combustion REACTION of ETHYLENE and methane are `C_(2)H_(4) + 3O_(2)rarr 2CO_(2) + 2H_(2)O , Delta H = - 1423kJ` `CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2)O , Delta H = - 891 kJ` Suppose volume of `C_(2)H_(4)` in the mixture of `C_(2)H_(4)` in the mixture`=X` litres. Then volume of `CH_(4)` in the mixture `= ( 3.67 -x) ` litres From the abovereaction `:` 1 litres of `C_(2) H_(4)` gives `CO_(2)= 2 `litres `:. ` x litre of `C_(2)H_(4)` gives ` CO_(2) = 2 x `litres 1 litre of `CH_(4)` gives `CO_(2) = 1 ` litres `:.( 3.67 - x)` litres of `CH_(4)` gives `CO_(2) = ( 3.67 - x)` litres Total `CO_(2)` produced `= 2 x + ( 3.67 - x ) = ( 3.67 + xx) ` litres `:. 3.67+ x = 6.11` or ` x= 2.44` `:. ` 1 litre of the mixture will contain `C_(2)H_(4) = ( 2.44)/( 3.67 )= 0.66 `litre and ` CH_(4) = 1 - 0.66 = 0.34 ` litre Volume of 1 mole `C_(2)H_(4)` or `CH_(4)` at `0^(@)C= 22.4L` Volume of 1 mole `C_(2)H_(4)` or `CH_(4)` at `25^(@)C = ( 22.4 xx 298 ) /( 273)= 24.45 L` 24.45 litres of `C_(2)H_(4)`at `25^(@)C` give heat `= 1423kJ` `:. 0.66` litre of `C_(2) H_(4)` will give heat`= ( 1423)/( 24.45) xx 0.66 kJ = 3841kJ` `24.45` litre of`CH_(4)` give heat`= 891 kJ` `:. ` 0.34litre of` CH_(4)` will give heat`= (891)/(24.45) xx0.34 kJ = 12.39 kJ` `:. `Total heatproduced `= 38.41 +12.39 =50.38 kJ` |
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