In: Other
A gaseous fuel mixture stored at 745 mmHg and 298 K contains only methane(CH4) and propane (C3H8). When 12.6 L of this fuel mixture burns, it produces 761 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?
From the ideal gas equation
Total number of moles of mixture
n = PV/RT
n = [(745/760 atm) x (12.6 L)] / [(0.08206 L-atm/mol-K) x (298K)]
n = 0.5051 moles
Combustion of methane
CH4 (g) + 2 O2 (g) ? CO2(g) + 2 H2O (g)
Enthalpy of combustion = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants
H1 = 2*Hf(H2O) + Hf(CO2) - 2*Hf(O2) - Hf(CH4)
Combustion of propane
C3H8 (g) + 5 O2 (g) ? 3 CO2(g) + 4 H2O (g)
Enthalpy of combustion = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants
H2 = 4*Hf(H2O) + 3*Hf(CO2) - 5*Hf(O2) - Hf(C3H8)
Let the moles of methane combusted = X mol
Moles of propane combusted = 0.5051 - X
Mixture generates = 761 kJ
H1 * X + H2 * (0.5051 - X) = 761
(802.5 kJ/mol) (X) + (2043.9 kJ/mol) (0.5051 - X) = 761 kJ
802.5X + 1032.37 - 2043.9X = 761
-1241.4 X = - 271.37
X = 0.2186
Moles of CH4 = 0.2186 mol
Moles of C3H8 = 0.5051 - 0.2186 = 0.2865 mol
Mol fraction of CH4 = moles of CH4/total moles
= 0.2186/0.5051
= 0.4328