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Question

A gas mixture of 3.67 litre of ethylene and methane on complete combustion at 25oC produces 6.11 litre of CO2. Find out the heat evolved on burning 1 litre of the gas mixture. The heats of combustion of ethylene and methane are 1423 and 891 kJ mol1 at 25oC.

A
60kJ
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B
-50.90kJ
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C
30kJ
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D
100kJ
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Solution

The correct option is B -50.90kJ
Solution:- (B) 50.90KJ
Combustion of CH4-
CH41 vol.+2O2CO21 vol.+2H2O
Combustion of C2H4-
C2H41 vol.+3O22CO22 vol.+2H2O
Let V litre be the volume of CH4 in the mixture.
Total volume of mixture =3.67L(Given)
Volume of C2H4 in the mixture =(3.67V)L
Volume of CO2 formed during combustion of C2H4=V+2(3.67V)
But given that the volume of CO2 formed during combustion of C2H4 is 6.11L
V+2(3.67V)=6.11
V=1.23L
Therefore,
Volume of CH4 in the mixture =1.23L
Volume of C2H4 in the mixture =3.671.23=2.44L
Now,
Volume of CH4 in 1L of mixture =1.233.67
Volume of C2H4 in 1L of mixture =2.443.67
As we know that volue of 1 mole of any gas at 25 and 1atm=22.4273×298=24.45L[By using charle's law]
Given that:-
heat of combustion of methane =891KJ/mol
Heat of combustion of ethylene =1423KJ/mol
Heat evolved due to the combustion of CH4=1.23×(891)3.67×24.45=12.213KJ
Heat evolved due to the combustion of C2H4=2.44×(1423)3.67×24.45=38.694KJ
Total heat evolved =(12.213)+(38.694)=50.907KJ
Hence the correct answer is 50.90KJ.

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