A gas mixture of 3.67 lit of ethylene and methane on complete combusti

1.	A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at 1 atmpressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in k], during thiscombustion. (ΔΗ-(CH4)--890 kl mol-1 and (AHc(CH)--1423 kJ mol-11.
Answer» follows : C2H4+ 3O2→ 2CO2+ 2H2O 1 vol 2 vol. CH4+ 2O2→ CO2+ 2H2O 1 vol. 1 vol Let the vol. of CH4in mixture = xl ∴ Vol. of C2H4in the mixture = (3.67 - x)l Vol. of CO2produced by xlof CH4= xland Vol. of CO2produced by (3.67 - x)lof C2H4= 2(3.67 - x)l ∴ Total vol. of CO2produced = x + 2 (3.67 - x) Or 6.11 = x + 2(3.67 - x) or x = 1.23l ∴ Vol. of CH4in the mixture = 1.23l and Vol. of C2H4in the mixture = 3.67 – 1.23 = 2.44l Vol. of CH4per litre of the mixture = 1.23/3.67 = 0.335l Vol. of C2H4per 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 = 24.45l [∵ Volume at NTP = 22.4L] Heat evolved due to combustion of 0.335lof CH4= - 0.335 * 891/24.45 = - 12.20 kJ [given, heat evolved by combustion of 1l= 891 kJ] Similarly, heat evolved due to combustion due to combustion of 0.665lof C2H4 = - 0.665 * 1423/24.45 = - 38.70 kJ ∴ Total heat evolved = 12.20 + 338.70 =50.90 kJ

A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at 1 atmpressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in k], during thiscombustion. (ΔΗ-(CH4)--890 kl mol-1 and (AHc(CH)--1423 kJ mol-11.

Answer»

follows :

C2H4+ 3O2→ 2CO2+ 2H2O

1 vol 2 vol.

CH4+ 2O2→ CO2+ 2H2O

1 vol. 1 vol

Let the vol. of CH4in mixture = xl

∴ Vol. of C2H4in the mixture = (3.67 - x)l

Vol. of CO2produced by xlof CH4= xland

Vol. of CO2produced by (3.67 - x)lof C2H4= 2(3.67 - x)l

∴ Total vol. of CO2produced = x + 2 (3.67 - x)

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

∴ Vol. of CH4in the mixture = 1.23l

and Vol. of C2H4in the mixture = 3.67 – 1.23 = 2.44l

Vol. of CH4per litre of the mixture = 1.23/3.67 = 0.335l

Vol. of C2H4per 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 = 24.45l

[∵ Volume at NTP = 22.4L]

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

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

= - 0.665 * 1423/24.45 = - 38.70 kJ

∴ Total heat evolved = 12.20 + 338.70 =50.90 kJ