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`.

Explanation

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 above reaction `:`
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`
`:. ` 0.34 litre of` CH_(4)` will give heat `= (891)/(24.45) xx0.34 kJ = 12.39 kJ`
`:. `Total heat produced `= 38.41 +12.39 =50.38 kJ`