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A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25^oC 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 25^oC.

Shane Macguire

12 Years agoGrade upto college level

2 Answers

Deepak Patra

12 Years ago

Hello Student,

Please find the answer to your question

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

Thanks
jitender
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