Question

A gaseous fuel mixture stored at 745 mmHg and 298 Kcontains only methane(CH4) and propane (C3H8). When 12.3 L of this fuel mixture burns, it produces 780. kJ of heat. The enthalpy of combustion for CH4(g) is −802.5kJ. The enthalpy of combustion for C3H8(g) is −2043.9kJ.

What is the mole fraction of methane in the mixture? (Assume that the water produced by the combustion is in the gaseous state.)

Answer 1

let n= no of moles of mixture= PV/RT, P= Pressure in atm = 745/760 atm =0.98 atm, V= 12.3L, T= 298K

R=0.0821 L.atm/mole.K, n= 0.98*12.3/(0.0821*298)= 0.493 moles

heat of formation (KJ/mole)
Heat of formation of methane CH4(g) -74.6
Heat of formation of propane C3H8 -103.85
Heat of formation of CO2(g) -393.5
Heat of formation of H2O(g) -241.8 and O2=0

heat of combustion of CH4 is CH4(g)+ O2 (g)----->CO2(g) +2H2O(g)

= sum of heat of formation of products- sum of heat of formation of reactants

=1* heat of formation of CO2+2* heat of formation of H2O(g)- { 1*heat of formation of CH4+1*heat of formation of O2} where 1,2,1 and 1 are coefficients of CO2, H2O, CH4 and O2 repsectively.

Hence = 1*(-393.5)+2*(-241.8)- (-74.6)= -802.5, so heat liberated / mole of CH4=-802.5 Kj

for combustion of C3H8, C3H8+ O2------->3CO2+ 4H2O

heat of combustion = 3*(-393.5)+4*(-241.8)- (-103.85)=-2043.9Kj

this is the heat of combustion/ mole

let x= mole of CH4, 0.493-x= mole of C3H8

hence 780 = x*802.5+(0.493-x)*2043.9

780 = x*802.5 +1007.6- 2043.9x

x*(2043.9-802.5)= 1007.6-780

x= 0.183

mole fraction of CH4= 0.183/0.493 =0.37