Question #160788

Find the Enthalpies of Combustion of gasoline (C8H18), methane (CH4), methnol (CH3OH), ethanol (C2H5OH), propane (C3H8) and kerosene (C15H32). Then determine the Energy-to-Pollution Ratios (EPRs) of each fuel and compare them. Discuss which fuels are most rational to use...if we insist upon burnign hydrocarbons to obtain energy.


1
Expert's answer
2021-02-03T07:16:08-0500

Enthalpy of combustion is calculated using the following formula,

ΔHrxn0=ΔHf0(products)ΔHf0(reactants)\Delta H^0_{rxn}=\sum\Delta H^0_f(products)-\sum\Delta H^0_f(reactants)

Gasoline

2C8H18(l)+25O2(g)16CO2(g)+18H2O(l)ΔHc0=4730kJ/g2C_8H_{18(l)}+25O_{2(g)}\to16CO_{2(g)}+18H_2O_{(l)}\newline \Delta H^0_c=4730kJ/g

Methane

CH4g+O2gCO2g+2H2Ol+ΔHrxn0=[(393.52286)(74.85)=890.7kJ/gCH_{4g}+O_{2g}\to CO_{2g}+2H_2O_l+\Delta H^0_{rxn}\newline =[(-393.5-2*286)-(-74.85)\newline= 890.7kJ/g

Methanol

CH3OHg+32O2gCO2g+2H2Ol=726kJ/gCH_3OH_g+\dfrac{3}{2}O_{2g}\to CO_{2g}+2H_2O_l\newline \newline =726kJ/g

Ethanol

C2H5OHg+3O2g2CO2g+3H2Ol=1368kJ/gC_2H_5OH_g+3O_{2g}\to 2CO_{2g}+3H_2O_l\newline =1368kJ/g

Propane

C3H8g+5O2g3CO2g+4H2Ol=2220kJ/gC_3H_{8g}+5O_{2g} \to 3CO_{2g} +4H_2O_l\newline =2220kJ/g





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