Question

A gaseous fuel mixture contains 20.7% methane (CH4), 44.0% ethane (C2H6) and the rest propane (C3H8) by volume.

Part A When the fuel mixture contained in a 1.57 L tank, stored at 756 mmHg and 298 K, undergoes complete combustion, how much heat is emitted? (Assume that the water produced by the combustion is in the gaseous state.) Express your answer with the appropriate units.

Answer 1

from the given data, using gas law, total moles of gas mixture can be calculated using gas law equation

n= PV/RT, R= gas constant =0.0821 L.atm/mole.K,

P= pressure in atm   , 760 mm Hg= 1atm, 756 mm Hg= 756/760 atm=0.995 atm, T= 298K, V= 1.57 L

n= 0.995*1.57/(0.0821*288)=0.064 moles

given the fuel contains ( since the mixture is gas mixture, the % is specified in mole%)

CH4 :20.7%, moles of methane= 0.064*0.207=0.013248

Ethane is 44%, moles of ethane= 0.064*0.44 = 0.02816

% of propane =100-% of ethane-% of ethane= 100-20.7-44=35.3%

moles of propane = 0.064*0.353 =0.022592

the standard heat of formation data (KJ/mole), CO2: = -393.5 Kj/mole, H2O(g)= -241.8, CH4= -74.8, C2H6= -84.7 and C3H8= -103.8, O2=0

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

the combustion of all these fuels is represented as

CH4+ 2O2---->CO2 + 2H2O, deltaH= heat of combustion 1* standard heat of formation of CO2+2* standard heat of formation of H2O- standard heat of formation of CH4= 1*(-393.5)+2*(-241.8)-{(1*-74.80)+2*0}= -802.3 Kj

1,2,1 and 2 are coefficients of CO2, H2O, CH4 and O2 respectively.

C2H6+ 3.5O2------>2CO2+ 3H2O, deltaH= 2*(-393.5)+3*(-241.8)- {1*(-84.7)+3.5*0}=-1427.7 Kj

C3H8+ 5O2-------->3CO2+4H2O, deltaH= 3*(-393.5+4*(-241.8)- { 1*(-103.8)+5*0} =-2043.9 Kj

when 1 mole of above fuels are burnnt

the heat of combustion (KJ) : CH4= -802.3, C2H6= -1427.7 and C3H8= -2043.9

hence overall heat of combstion when thee mixture containing 0.013248, 0.02816 and 0.022592 moles of CH4, C2H6 and C3H8 are combusred

=0.013248*(-802.3)+0.02816*(-1427.7)+ 0.022592*(-2043.9)= -97 Kj