Question -
A gas mixture 3.67L in volume contains C2H4 and CH4 is proportion of 2:1 by moles and is at 25∘C and 1atm. If the ΔHC (C2H4) and ΔHC(CH4) are −1400 and −900kJ/mol find heat evolved on burning this mixture
Options:
A .  20.91 kJ
B .  50.88 kJ
C .  185 kJ
D .  160 kJ
Answer: Option C : C Volume of C2H4 = 3.67×23=2.45Lit Volume of CH4 = 1.22Lit Mole of C2H5 = PV = nRTn = PVRT = 1×2.450.082×298=n=0.1mole Mole of CH4 = nCH4 = 1×1.220.082×298=0.05mole ∵ Energy released due to combusting 1 mole C2H4=1400KJ ∴ Energy released due to combusting 1 mole C2H4 = 1400×0.1=140kJ Similar energy released in combusting 0.05 mole CH4=−900×0.05=45 So that heat realized = 140+45 = 185kJ/mole
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