Question

Determine the higher heating value of methyl alcohol (CH_3 OH) given that its enthalpy of formation is -238 000 kj/kg-mol. Assume that the gasoline at the pumps is a mixture of 90% octane and 10% methyl alcohol, determine the mixtures heating value.

Answer 1

CH3OH+ 1.5O2--->CO2(g)+ 2H2O(l)

since heat of formation of O2=0

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

Heat of +combustion = 1 heat of formation of CO2 +2* heat of formation of H2O- 1* heat of formation of CH3OH

= -393.5*1000+2*(-285.8*1000) +238*1000 KJ =727100Kj/Kgmole =727100/32 = 22722 Kj/Kg

this is the higher heating value since water is in liquid state,

2. Heat of combustion of octane C818+25/2O2---> 8CO2 + 9H2O(l) deltaH= -5460 Kj/mole=

heat of combustion of CH3=727.1 Kj/mole

heat of combustion of 90% octane and 10% CH3OH= 0.9*5460+0.1*727.1 Kj/mole=4987 Kj/mole

molar mass of 90% octane and 10% methyl alcohol = 0.9*114+0.1* 32 = 105.8gm/mole

Heating value= 4987*1000/ 105.8 Kj/Kg=47136 Kj/Kg