Question

A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25ºC produces 6.11 litres of CO₂. Find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are –1423 and –891 kJ mol⁻¹ at 25ºC.

Answer 1

Combustion of C2H4 and Ch4 takes place as follows :

C2H4 + 3O2 → 2CO2 + 2H2O

1 vol 2 vol.

CH4 + 2O2 → CO2 + 2H2O

1 vol. 1 vol

Let the vol. of CH4 in mixture = x l

∴ Vol. of C2H4 in the mixture = (3.67 - x) l

Vol. of CO2 produced by x l of CH4 = x l and

Vol. of CO2 produced by (3.67 - x) l of C2H4 = 2(3.67 - x) l

∴ Total vol. of CO2 produced = x + 2 (3.67 - x)

Or 6.11 = x + 2(3.67 - x) or x = 1.23 l

∴ Vol. of CH4 in the mixture = 1.23 l

and Vol. of C2H4 in the mixture = 3.67 – 1.23 = 2.44 l

Vol. of CH4 per litre of the mixture = 1.23/3.67 = 0.335 l

Vol. of C2H4 per litre of the mixture = 2.44/3.67 = 0.665 l

Now we know that volume of 1 mol. Of any gas at 25°C (298 K) = 22.4 * 298/273 = 24.45 l

[∵ Volume at NTP = 22.4L]

Heat evolved due to combustion of 0.335 l of CH4 = - 0.335 * 891/24.45 = - 12.20 kJ [given, heat evolved by combustion of 1 l = 891 kJ]

Similarly, heat evolved due to combustion due to combustion of 0.665 l of C2H4

= - 0.665 * 1423/24.45 = - 38.70 kJ

∴ Total heat evolved = 12.20 + 338.70 = 50.90 kJ

Hope it would help................