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`.

Answer

5557 kJ

Explanation

Cobustion of `C_(2)H_(4)` and `CH_(4)` takes places as `underset(1 vol)(C_(2)H_(4)) + underset(2 vol)(3 O_(2)) rarr 2CO_(2) + 2 H_(2)O`
`underset(1 vol)(CH_(4)) + underset(1 vol)(2 O_(2)) rarr CO_(2) + 2H_(2)O`
Let the volume of `CH_(4)` in the mixture be x litre. then the volume of `C_(2)H_(2)` in mixture `= (3.67 - x)` litre
Volume of `CO_(2)` produced by x L of `CH_(4) = xL`
Volume of `CO_(2)` produced by `(3.67 - x) L` of `C_(2)H_(4) = 2 (3.67 - x)L`
`:.` Total volume of `CO_(2)` produced `x + 2 (3.67 - x) L` or `6.11 = x + 2 (3.67 - x) L rArr x = 12.3 L`
Thus volume of `C_(2)H_(4)` is mixture `= (3.67 - 1.23) L + 2.44 L`
Volume of `CH_(4)`/litre of mixture `= (1.23)/(3.67) = 0.335 L`
Volume of `C_(2)H_(4)`/litre of mixture `= (2.44)/(3.67) = 0.665 L`
Since volume of 1 mole of a gas at NTP `= 22.4 L`
So heat evolved due to combustion of 0.665 L of `C_(2)H_(4) = - (0.665 xx 1423)/(22.40) = - 42.32 kJ`
Hence total heat evolved `= 13.32 + (- 42.25) = 55.57 kJ`, Total heat evolved = 5557 kJ
Negative sign indicates evolution of heat