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

Let, volume of `C_(2)H_(4)` and `CH_(4)` be .a. and .b. L respectively at `25^(@)` C.
`therefore a+b=3.67` litre………..(1)
Also, `C_(2)H_(4) + 3O_(2) to 2CO_(2) + 2H_(2)O`
`CH_(4) + 2O_(2) to CO_(2) + 2H_(2)O`
`therefore` Volume of `CO_(2)` produced by .a. L `C_(2)H_(4) = 2a`
Volume of `CO_(2)` produced by .b. L `CH-(4)=b`
`therefore 2a + b=6.11`...............(2)
On solving Eqs. (1) and (2), we get
a=2.44 L
b=1.23 L
`therefore` Volume of `C_(2)H_(4)` in 1 L mixture at `25^(@) C = 2.44/3.67 xx 1`
Volume of `CH_(4)` in 1 L mixture at `25^(@) C = 1.23/3.67 xx 1`
`=0.3352` L
Now, volume of 1 mole `C_(2)H_(4)` at `0^(@) C = 22.4` L
`therefore` Volume of 1 mole `C_(2)H_(4)` at `0^(@) C = (22.4 xx 298)/273 = 24.45` L
Similarly, volume of 1 mole `CH_(4)` at `25^(@) C = (22.4 xx 298)/273 = 24.45` L
`therefore 24.45` litre `C_(2)H_(4)` produces `=+1423 kJ` at `25^(@)` C
`therefore 0.6648` litre `C_(2)H_(4)` produces `=(+1423 xx 0.6648)/24.45 = +38.69` kJ
Similarly, 24.45 litre `CH_(4)` produces `=+891` kJ at `25^(@)` C
`therefore 0.3352` litre `CH_(4)` produces `=(+891 xx 0.03352)/24.45 = +12.22` kJ
Total heat produced = 38.89 + 12.22
=50.91 kJ